My parents (now in their eighties) were taught the square root by hand method in grade 4. I learned it in my 40s when reading a book on abacus calculations. I wasn't quite sure why the algorithm worked and meant to derive a proof eventually. What you presented here is essentially the same algorithm. So now I don't have to figure it out. THANKS!
Your comment tempted me to ask, did you mean you were intending on deriving a proof on your own, and if so, what math-background did you have involving proofs? Or, were you going to just research the proof derivation.
@@AhirZamanSairi I imagine he means a proof on his own. If you have had three semesters of calc, this is a thing you can do all on your own if you wish.
@@AhirZamanSairi I was going to figure out - on my own - why the square root by hand method works. I have a bachelor maths degree from ages ago, where I studied real and complex analysis along with lots of measure theory. Back then it was theorem, proof, theorem, proof, ... and it was truly loads of fun.
Lol I'm 51 but my school also taught long-hand square roots. Our entire county was stuck in the 1950s though....rather still is....1950s plus the internet, now.
for those wondering,not only this is an application of newton's method,this method is also called linearlization(1st degree taylor polynomial).just like binomial approximation(also another form of this formula only with a = 1),or generally any approximation that "brings down" the exponent is used in many areas of physics and math.
Especially non European ancient civilizations, which often have their accomplishment understated to such a degree it feels like no one did science and philosophy other than the Greeks
@ I would argue Indians Chinese and Arabs get plenty recognition though it’s probably an understatement considering their massive achievements. African Oceanic Pre Columbian and Middle Eastern civilizations though are massively underrated I hope they gain more traction with the expansion of history
@@luisfilipe2023 Middle Easterners also get a great deal of credit wdym. However yeah you never hear about African or Pacific islander discoveries often. Hell even south Americans get a great deal of recognition for agriculture and astronomy. The idea that no one but western advancements get recognition was maybe valid in the 70s. However since the 90s at least everything from college to media definitely has been giving credit where it's due for the most part. I don't even know who people claim discovered this stuff in Europe. However I constantly hear about Middle Eastern and indian mathematicians. You can't even talk about this stuff without people like this guy making sure we all know how awesome everyone but Europe is.
@@luisfilipe2023 I mean I think the middle east gets credit if for nothing else then the ancient egytpians which the europeans where obsessed with for centuries(hence why the US has so many places named after egypt(Cairo illinois or Memphis Tenessee come to mind). I think the reason why africa doesn't get any credit is because we don't have any written accounts so its hard to know what they knew, and many of their buildings where not built from materials that actually hold up over time like other groups did(I mean they had straw huts which don't last long, though if you've seen how they are built you would absolutely consider them at the very least done with great artistry(I mean they weave reliefs and meanders into the walls its actually really cool). The other is most groups where not particurally keen on preserving their nations history either because of more pressing concerns, corruption or religious reasons(many areas are more concerned with acquiring resources hence why I believe it was sudan who just flooded an entire region filled with pyramids so they could build a dam, the ottoman turks where not particuarlly fond of anything not related to their religion(and to be fair neither where christians) etc.). As for south america I do believe they get some credit due to Manchu Pechu and terra petra, as do the aztecs I think. I do believe that more credit should be given to polynesians as we have evidence they navigated the pacific and reached south america before anyone and an anthropologist used their navigation system to get their by the same style of boat they would have used. But again, I think the major issue is going to be lack of records and a natural bias to place one's own group above others which is why if you talk to greeks they would tell you they invented everything and asians would do much the same thing as do most every group.
Here is another way of looking at the process. Think of the graph of y = x^2. It has a point on the curve (x, 17) and x is of course the square root of 17. We know that (4, 16) is pretty close. Mentally draw a tangent line at that point. The derivative of Y= x^2 is 2x, so the slope at our initial guess is 2*4=8. Delta x = delta Y / slope = 1/8. So the 2nd approximation is 4 + (17-16)/8.
Isnt it also related to the wuadrstoc formula..that's a thought/observation I had. The b/2a since the number you are looking for is between x and x plus 1or in this case greater than a but less than a plus 1..That's a neat observation and may be relevant right?
@@major__kongisn't newton's method iterating this process, though? this isn't newton's method, this is just linear approximation (or, the taylor polynomial p_n where n=1)
This works, but accuracy increases in single iterations of newton's method for larger numbers, so you would get a more accurate calculation if you use larger numbers, and nearby perfect squares. For example: √999 ≈ 32 + (999 - 1024) / 64 To simplify, I would say do something similar to what you suggest. (999-1024) / 64 = 25/64 Multiply by 10. 250 / 64 = 256 / 64 - 3/32 The last term is pretty close to 1/10, so: 250 / 64 ≈ 3.9 Then divide by 10. 25 / 64 ≈ 0.39 So we're looking for 32 - 0.39 = (3200 - 39)/100, which yields 31.61. This squares to 999.19. The method you describe is great if you don't know a nearby perfect square, though.
@@MegaMinerd Right. This is what I do when I don't have a calculator. This is how I calculate square roots when I'm teaching and don't want to go grab one. I was just sharing further refinements.
@@bhatkrishnakishor The general formula is: √x ≈ k + (x - k^2) / (2k) If you're trying to find √9.999, then the closest perfect square is 9, so: √9.999 ≈ 3 + (9.999 - 9) / (2 * 3) = 3 + 0.999 / 6
I coded in Intel assembly a version of the Newton Raphson method. For the square root of a, take an estimate x. Then take a/x and average it with your original estimate. That is, a' = ( a + (x/a)) / 2 . Repeat until you are close enough. This method converges rapidly. For the initial estimate, the Intel processor is very helpful. Floating point numbers are stored in quasi-scientific notation according to the IEEE standard. All you do is divide the exponent by two (right shift 1) and remove the 1 from the first bit of the mantissa. This is close enough for the method to converge rapidly. This method is about as fast as the supplied C library sqrt call, which uses a completely different algorithm that doesnt require a hardware floating point divide function.
if x is the sqrt(a) then a/x == sqrt(a) if x is approx_sqrt(a) then a/x == sqrt(a) + error average(x,a/x) == average(x,sqrt(a) + error) = average(sqrt(a)+error_x , sqrt(a)+error_aoverx) == sqrt(a) + average(error_x,error_aoverx)
@@starpawsyThat’s not assembly code, it’s pseudocode. Assembly code is much more low level, the algorithm would probably take several hundred lines of code, since you need to make everything from scratch.
@@KindledAsh Thanx for the mansplaining. My implementation did not take "several hundred" lines. And it certainly was in Intel assembler. My description was, of necessaity, pseudocode.
Thanks for showing the table at 0:33. It shows clearly that the numbers from 1 to 59 were represented as two digits, where the leftmost digit shows the number of tens and the rightmost digit shows the number of ones. I.e., these numbers are represented very similar to the way we do, except that they didn't use a symbol for zero, they used different sets of digits for the tens and for the ones, and their maximum number of tens was 5.
Each of those is a single digit. The tens and units within a digit are not separate digits, they're subparts of the same digit. It's a common misconception about cuneiform numbers.
This is the direct application of Newton's method to find the zero of a function: Let f(x)=x²-a, so that we want to find x such that f(x)=0 x²=a we then have x_n+1 = x_n - f(x_n)/f'(x_n) It becomes x_n+1 = x_n - (x_n²-a)/(2*x_n), which corresponds to the algorithm described in the video
@@voicutudor7331 me too. A simpler summary of this formula is that your next guess is the mean of x and a/x. Hence, x_n+1=(x_n+a/x_n)/2. If for sqrt(2) you know a good start is 7/5, then next one is 7/10+5/7=99/70 which is good for 4 decimal places.
Not good because you are introducing calculus instead of focusing on the fundamentals such as multiplication, division, addition and subtraction. Why adding calculus here is bad? Because computers can't do calculus. You need to think in a practical way sometimes.
The method I learned in grade school is to take the average of a and s/a. This gives the same result as Presh's "a + (s - a^2)/(2a)", but I find "(a + s/a) / 2" to be more intuitive. If "a" is a bit low, then s/a will be a bit high (and vice versa), so halfway between them will be closer than either of them.
I just came up with this method from Presh's method with a bit of simplification. Then I extrapolated it to the nth root for fun. I agree that it's more intuitive, but I needed the geometric intuition to expand to the nth root because I was otherwise not weighting the summands correctly. (a + s / a^(n - 1) ) / 2 doesn't work, but averaging (n - 1) a's with s / a^(n - 1) does. The result is nth√s ≈ ( (n - 1)a + s / a^(n - 1) ) / n
This is probably the best way to calculate a square root when you're programming a computer in assembly language and have no library function to do it for you. It's fast and the convergence is quadratic - the number of correct digits doubles with each iteration. Adaptation to n-th roots is fairly simple, although the convergence is not as rapid. Even if you have a library function available this will normally be a bit faster, and can be tailored to the application for extra speed.
I was taught a different method, which is a lot like long division. You draw a radical sign over the number you want to sqrt. The number is then split up by putting commas after every two digits, spreading out from the decimal point. During the process, you always bring down the next two digits then repeat the process. Along the way you'll get increasingly longer trial divisors. The beauty of this method is that it uses a single, simple layout, & you never have to average anything, or sometimes subtract instead of adding anything.
Isn't 111 closer to 11² = 121 than 10²=100? at ~5:54; So sqrt(111) ~~11 - 10/22 = 11 - 5 / 11 = 10.545454... which is obviously closer to 10.536... (sqrt(111)) than 10.55 (the approximation given in the video by starting with 10²)
The closer you are to the squared value, the more accurate the adjustment value will be it seems. Since 111 is closer to 11² then using 11 + adjustment will be more accurate than 10 + adjustment indeed
When performing a mental calculation, the choice tends to be driven by ease of calculation. 10 in Roman numerals is a decimal position, x 10 is simply a numeral shift.
This is pretty cool. But also: you can start with any non-zero number and it will converge, usually within 8 or ten steps. Obviously, the closer your guess, the quicker it converges. But for the given solution, you need to know the nearest squares, so you could just use sqrt(x) ≈ [root of lower square] + (x-[lower square])/([upper square] - [lower square]) and it will be good to a few decimal places, and less than 1% error for numbers larger than 12. For example: sqrt(70) = 8.3666 ≈ 8 + (70-64)/(81-64) = 8+6/17 = 8.3529, the error is less that 0.2%. That should be good enough for anything you do, unless you are going to the moon or something.
I was thinking while watching, that you should derive the formula for the square root approximation, and you did later. The table of squares was good to show the values to consider. I had to derive a square root procedure starting with the algebraic form ( a + b ) then squaring it then factoring out ( b ) from the form (( a )( a ) + ( b )( 2a + b )). It gets more complicated after that. Most instructors teach an arithmetic procedure that takes the ( 2a + b ) to compute the next iteration of the approximation. I spent some time putting the procedure into a spreadsheet to do the calculations to six figures. There is more to be said. Sorry, don't have the time to discuss it here. Great Video. Thanks for doing this. History helps. ;D
If you have all the two-digit squares memorized, you can add two zeros to the radical and make a better estimate. I would change 69 to 6900, and since I know that 83^2 is 6889, I know that my approximation will be 83. plus 11/83+84, or 11/167. So I have 83.0. I can continue calculating by trial and error, doubling the number of significant digits each time. Once I calculated a square root to twenty-four significant digits, all in my head, which took me about three days.
Thanks for sharing this finding. I am a big fan of square root and higher roots. I first learned extracting a square root of a number via a long division from my high school chemistry teacher. I was totally hooked afterward. I was so excited that I couldn't help doing square root of any number every day. Then, I wanted to know how to extract a cube root of a number. But, I couldn't find anyone who can show me. I told my father about it. One day he surprised me with a math book he bought from a book store, and there it was, an instruction on how to extract a cube root of a number via a long division. I was just totally overwhelm. The book also showed how to extract any root of a number using binomial expansion (the holy grail). The binomial expansion technique is what I use today to extract a square root. Even though I know how to extract a cube root and higher, I use calculator to save time. I love this stuff.😊
Actually the 2 can be better explained as the distance between squares. (x+1)²-x²=2x+1. Thus is a discrete problem so calculus doesn't even make sense. Talk about regurgitating and not thinking.
In other words , newton was “inspired “ by Babylon method for this , just like he was inspired from Sanskrit text for 3 laws of motion and concept of gravity
I remember using this method in college. In those days (early 70s) calculators were just coming out and had limited functionality. Mine was a Comodore something which I paid over $90 for. It had no square root function, just add, subtract, multiply and divide. Plus if you moved a switch, you could store the display in memory. I was the only one in my class who had one. I would run through the formula the first time, store the result in memory and use that value for the guess for the second run. Amazing results. My classmates were using their slide rules! Actually I would use my own slide rule for the first approximation. I had a piece of paper taped to the back of the calculator which gave me the formulas for the trig functions. Just an old man taking a trip down memory lane. Thought I would share. Always enjoy your videos.
This kinda reminds me of when I worked out how to square any 2 digit number in my head back in high school: For example if you want to square 32, start by squaring 30 to get 900. Then rough tune by adding the double of 30 twice to get 1020. Then square 2 to get 4 and add to get 1024. If you want to square 44, multiply 4 by 5 and slap 2 0s on the end to get 2000. Then subtract the double of 40 once to get 1920. Then add the square of the 1s digit, 4 x 4, to get 1936. If you want to square 79, start by squaring 80 to get 6400. Then subtract the double of 80 once to get 6240. Then add the square of the 1s place, in this case -1, to get 6241. The process of ballparking by squaring the nearest multiple of 10, rough tuning by adding or subtracting multiples of twice the nearest multiple of 10, and fine tuning by adding the square of the 1s place is very effective. Putting in the midway shortcut of multiplying the 10s digit by 1 greater than itself and putting 2 0s to get the midway rough tuning means you will never have to rough tune up or down more than twice. As a bonus, if you have to multiply 2 numbers that are both odd or both even, take the square of their midpoint and then subtract the square of the offset. 87 x 93 for example. Since their midpoint is 90, we square 90 to get 8100. Since 87 and 93 are both offset from 90 by 3, we subtract the square of 3, 9, from 8100 to get 8091.
Many have pointed out that this is what you get if you use Newton's method (a.k.a. Taylor approximation of order one). It is also the result of the following: Let x be a first approximation of √s. Then we would very much like to calculate the geometric mean of x and s/x, as that is exactly √s. But we can't. What we CAN do is to calculate the arithmetic mean. And that yields exactly this method. As an added bonus, both Taylor approximation and the mean method have natural ways to bound their respective errors. Taylor has Taylor's theorem. And we know the geometric mean lies between the arithmetic and harmonic means.
@@gregorymorse8423 I don't know which Newton's method you are taking about. But the one I'm talking about here takes your approximation to a zero of a function, finds the tangent of the function at that x-value, and uses the zero of that tangent as the next approximation. Repeat until you're close enough. If you say this is a continuous method, presumably you're referring to the finding of the tangent?
Its called approximation in calculas. Not only square root, we can calculate approx value of any function with any argument. The expression goes like this : f(x + ∆x) = f(x) + ∆x*f'(x) Where f'(x) is first derivative of f(x) w.r.t x The more further apart x and ∆x are , the more accurate the answer is.
I've implemented this algorithm a couple times before, but I never thought of it as something that can be calculated in your head. You have changed that today 🙂
Nice video! I teach this method in my numerical analysis class. A few comments: 1) Note that the formula sqrt(s) ~= a + (s-a^2)/2a can be simplified as sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a). So in other words, given your approximation a, you take the average of a and s/a. The idea being that if a is bigger than sqrt(s), s/a will be proportionally smaller (and vice-versa if a is smaller than sqrt(s) ), so the average should get you closer to the true value 2) As others have noted, this is a special case of newton's method for finding the root of f(x) where f(x) = x^2 - s. One of the main selling points of Newton's method is that it is very fast -- if you feed your updated estimate to sqrt(s) back into the formula, you get about double the total number of accurate digits every time (provided you are close enough to the answer). So if you were to continue the estimate to sqrt(2) shown in the example, you would get the sequence: 1 3/2 = 1.5 (1 digit correct) 17/12 = 1.41666... (3 digits correct) 577/408 = 1.414215... (5 digits correct -- this is the last estimate shown in the video) 665857/470832 = 1.414213562... (10 digits correct!) This is called quadratic convergence, and you can prove why this happens for Newton's method in general (not just for this example, but most functions f(x) also) 3) It generalizes to higher roots as well. For example, to approximate the cube root of s, pick an approximation a and then compute: cuberoot(s) ~= 1/3(2a + s/a^2) (This is what you get from applying Newton's method to f(x) = x^3 - s instead of x^2 - s). So for example to approximate cuberoot(2) = 1.259921050 you would get the sequence 1 1/3(2 + 2) = 4/3 = 1.3333... (1 digit correct) 1/3(8/3 + 2/(16/9) = 91/72 = 1.263888... (3 digits correct) 1126819/894348 = 1.259933493 (5 digits correct) 2146097524939083451/1703358734191174242 = 1.259921050 (10 digits)
This has nothing to do with Newton Raphson approximation which is a continuous not discrete method. Its astounding how many math flunkies are here and don't watch or understand the video or the Babylonian method.
Nice explanation, but we need calculator for calculate the "width fraction" 1126819/894348 .... For grade 4 students need some efforts, if calculator is not allowed.....
It completely destroyed my perception of the history of mathematics. Babylonian mathematicians almost 4000 years ago knew mathematics thousands of times better than many people today.
Tbf, the people who did this were probably part of some elite merchant group or around the equivalent of our upper middle class nowadays. Not necessarily nobles, but certainly not your average person whose daily life only really needs the four basic arithmetic operations (tho, of course, that's when we compute stuff ourselves; our phones and computers nowadays rely on more complex stuff afaik lol)
@johnbjorkman4144 I mean, if you're taking the world population and not just the wealthy countries (or heck, even if you _are_ just taking the wealthy countries), that's always been the case. It's not like people are suddenly dumber on average today, people with no access to education or people who are difficult to teach (whether due to genetics, family/social issues, mental health issues, etc.) have always existed XD
@@moondust2365 yea, elite no kidding right? all of their other needs must have been attended to for them to afford the time and effort to devise the method then have it recorded.
2:00 the Babylonians also developed the Saros cycles, based on their correct geocentric cosmology (earth being level and stationary). The Saros cycles are still being used today.
I created a spreadsheet with columns for both methods: rounded down to a whole number (FLOOR) + the positive fraction, and rounded up to a whole number (CEILING) + the negative fraction. The FLOOR method gives a closer approximation for numbers closer to the lower perfect square (like 27) and the CEILING method gives a closer approximation for numbers closer to the upper perfect square. Since all square numbers have a difference of an odd number, the number exactly between two perfect squares gives the same approximation with both methods. I created one column for each method to calculate the difference between the approximation and the actual square root. The approximation is always greater than the actual root. I copied the formulas in rows from 1 to 1024, then averaged both difference columns. They weren’t the same. The average difference of the FLOOR method is 0.008761… and the average difference of the CEILING method is 0.002423… I can’t explain why the CEILING method ends up being more accurate on average. Edit: When I copied the formulas into 10,000 rows, the CEILING method is only about 7% more accurate than the FLOOR method. Approaching infinity, would the two methods achieve the same accuracy?
It's even more fun in copilot AI. Give it this command: Generate a table with two columns, A and B. In column A, start with number 1 and for each row add 0.01 until you reach number 10. Column B is the square of A. It should generate a column with 1000 row and enough precision for most practical purposes.
I came up with a pretty unique and simple method to compute square roots myself. Eg: square root of 17. find the closest square number=4. do 17/4 . Then find the average of 4 and 17/4 assume it to be a. Divide 17 by a and add a and 17/a. repeat continuously until you get the answer. Currently in add math, when i forgot calculator, i used this method to approximate my answers lol.
This is actually the same method as he shows in the video. His formula sqrt(s) ~= a + (s-a^2)/2a simplifies to sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a), which is taking the average of the two quantities as you said.
as many pointed out, this can be seen as an average between the estimate (x) and S/x; an easiest way to visualize it is to consider these values as sides of a rectangle, having the same area S of the original square, and after every iteration the two sides converge to a common value, becoming a square.
For sqrt(2) one could instead compute sqrt(200) using e.g. the proposed method, and divide the result by 10. Sqrt(200) is approximately equal to sqrt(196) which is 14. Then add 4/28, that'll give approximately 14.142, and divide by 10 to get our sqrt(2).
a general form of this equation can actually not only be introduced(newton's method),if you know programming,this method can actually be implemented in programming languages like c,python,java,....And this method applies for all real univariate polynomials(any kind of function such that the initial guess is closer than a zero of the derivative in the open set resembling a circle with a radius r and center the zero we are talking about)
To make it more accurate without doing multiple calculations, just multiply it first. So instead of calculating sqrt(2), calculate sqrt(200), because sqrt(200) is equal to 10 times sqrt(2). Using the method shown, sqrt(200) is approximately 14 + 4/28 = 14.142857... Divide that by 10 and you get that sqrt(2) is approximately 1.143, which is close enough
Another way to get a better value to approximate root(2) is to observe that root(98) = root(49)*root(2) = 7*root(2). Using the method, root(98) = approx 10 - 2/20 = 99/10. Dividing by 7 gives root(2) = approx 99/70 = 1.4142857... which is already a lot better than 1.5. Edit: If you use root(200) = 10*root(2) and 14*2 = 196 we get root(200) approx = 14 + 4/28 = 14.142857... so dividing by 10 for root(2) gives the same approximation as the root(98) example.
A simple trick to get better approximations is to consider not only squares of integers in the initial table, but also squares of integers plus 0.5. These can be calculated easily: (a+0.5)² = a(a+1) + 0.25, e.g. 3.5² = 3*4 + 0.25 = 12.25
Newtons method to find x = √y. find integer x, so that x^2 is close to y then repeatedly evaluate the expression: x ← ½(x + y/x) For y=17 we get: x = 4; x = 4.125; x = 4.12310606060606; x = 4.12310562561768; x = 4.12310562561766... Wolfram says √17 = 4.12310562561766054982140985597407798... (same to 15 digits) Note: the typical Newton's method expression: x ← x - (x^2 - y)/2x simplifies to: x ←½(x + y/x)
I divised this method in highschool and was able to calculate a bunch of digits too which I then crosschecked with my calculator. It's so cool to know now that this method has such a long history!! Yea learn something every day.
This confused the heck out of me because, at 12:53, he said we had another figure whose total area was EQUAL to s (emphasis mine). But we don't. We have a side that can be used to form a square, but that square's area is not equal to s, it's bigger, which is why it's an approximation.
I found this method always gives a slight overestimate of the square root, so one must always subtract after the first step. I simplified to the following equivalent formulas which I prefer: √s ≈ (a² + s) ÷ (2a) = (a + s / a) ÷ 2. Intuitively, √s is between a and s / a so it seems right that averaging these values would approach √s. My formulas have fewer operations and avoid user error since subtraction isn't commutative. And you don't need to consider whether your estimate is too high.
And now, using the geometric approach shown in this video, I extended it to cube roots with ³√s ≈ a + (s - a³) / (3a²). Seeing a pattern, I've extrapolated the nth root of s ≈ a + (s - a^n) / (n • a^(n - 1)), or alternatively, the nth root of s ≈ ((n - 1)a + x^n / a^(n - 1))/ n. I'm sure this is an old result, but it was fun to figure out.
The trick uses the first and second binomial formulas: (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b². The difference to the real value is the missing b² part. In this case: (4 + 1/8)² = 4² + 2 · 4 · 1/8 + 1/8² = 16 + 1 + 1/64. And the 1/16 is tiny in comparison to the 17.
I take exception that you are choosing to round some values and not round other values to make it seem like they were more accurate than they actually were. For example, at 5:09, you show 8.312 and 8.307. You did not round the first number up, but you rounded the second number up. It should be 8.3125 and 8.3066, and if you rounded them both, would be 8.313 and 8.307...
@@azrobbins01 You can't round everything in the same direction, that's a different operation (floor and ceiling). He just opted to round xx25 up to xx30 but a xx24 would be rounded to xx20. 5 is the exact middle and you can choose how to round it, as long as you're consistent.
@@jimmyh2137 How often do you see people rounding 6 up, but 5 down? Is this a common practice? If you did this at work, would people think it is normal?
@@azrobbins01 6 is always rounded up. 6-7-8-9 up 1-2-3-4 down 5 is the "problem", could be argued for either side. In this case it's exactly 5, makes sense to round down. It's 100% normal to round 5 down.
Amazing capabilities so long ago. Makes one wonder how they arrived at such a system and why... Seems math and numbers in general were more advanced then I would have ever thought. Thumbs Up!
This is a special case of the equation of the tangent line y=f(a)+f'(a)*(x-a) Insert f(a)=sqrt(a) and f'(a)=1/(2*sqrt(a)) to get y=sqrt(a)+(x-a)/(2*sqrt(a)) Insert a=4 and sqrt(a)=2 to get y=2+(1-4)/(2*2)=1,25
I know more interesting way, allowing to calculate square roots with arbitrary precision. It's iterative and can be performed the best way on a mechanical adding machine or on a paper. This method is found in a paper by S. V. Savich and another but quite similar was is used on actual old Friden calculators.. You perform division by sequential subtraction (like it's usually done on paper), but: 1. You track not only the remainder but also count operations. 2. You change the subtrahend with each turn. 3. You subtract sequential odd numbers starting from 1 to 17, until intermediate remainder becomes higher that subsequent subtrahend, so you got a remainder and move to less significant positions. Almost like with a general division giving a long fraction. Example: you need to calculate a square root of 2. You start with 1. 2-1=1. There was 1 operation. Then you could try to subtract 3, but remainder is 1 and you would cross zero, so instead you write down current count of performed subtractions (1) to the first digit of the root, then decrease the current digit of the subsequent subtrahend by 1 (3 becomes 2), and append 1 at the right (getting 21). Then you move this intermediate subtrahend one position to the right respective to the remainder. So you start subtracting this two-digit number summarizing 20 and sequence of odd numbers (1, 3, 5...), from three digit number 100 (the first digit of the remainder and two trailing zeroes to the right). You subtract 21, 23, 25, 27. This way you perform 4 subtractions until you got 004, from which you can't subtract 29. OK. you decrease this to 28, write down your 4 operations as the next digit of the root, append 1 at the right side of the subtrahend: 281 and shift the whole subtrahend one position to the right again. This time you subtract 281 from 400, you get 119 in the reminder, so you have performed subtraction 1 time. 119 is less than 283, so you write down 1 to the root, decrease 283 to 282, append new one (2821), shift right and start subtracting that from 11900 starting with 2821, then 2823, etc up to 2827, getting just 604 in the remainder... So this way you can calculate a square root with any precision, performing as many subtractions as you get as the sum of all digits of the root, and each time you deal with 1 digit longer intermediate subtrahend and by the way it's twice the result. If the rightmost digit of this subtrahend reaches 9, the next time you should add 2 to this rightmost digit with carry to the left, so nnn89 will becomes nnn91. The explanation looks complicated and random like numerology but actually it's pretty intuitive and straightforward once you get the principle. So that is: you subtract odd numbers, you accumulate the subtrahend by appending new digits starting with 1 at the right side, you count your operations and write them to the root, and you track your remainder, each time moving to the right once it becomes less that the subtrahend. I'm very curious if there is a good description for expanding this algorithm to binary.
As a math geek I was excited to show this to my fellow students after I learned it in 7th grade. Teacher let me do a whole presentation. Fellow students were unimpressed. 🙄
Order of operations is often taught poorly. My experience with math before taking responsibility for educating myself was a series of terrible teachers and wasted time. I think this is a common experience.
Nice! What's the name of the method where you make an estimate, divide the original number (dividend) by the estimate (divisor), then average the divisor and quotient (so the two factors when one is your estimate), make the average your new estimate, and repeat?
Lots of people mentioned that this is essentially Newton's method / a Taylor approximation. But strangely, apparently no one mentioned that this method is very similar to a far older method, which has been known as "Heron's method" for ages. And I _was_ taught that at school.
I'm amazed that anyone was even thinking about square roots 3600 years ago. You could go into some high schools in the US today where the students don't know what a square root is.
That is interesting. I now know at least 5 ways to do sqrt(X) by hand. 1) Newtons method as done in school way back when (Estimate divide average) 2) The variation on Newton that gives you 1/sqrt(X) which you then multiply X by to get sqrt(X) 3) The "classical" one that involves a lot of dividing by estimated squares 4) A microcontroller one that needs only shifts and adds and subtracts to do. 5) This new one Computers that can multiply but can't divide tend to use method #2 I haven't used #3 in so long I would take some time to remember it. A variation of the microcontroller one is the way I might do it by hand if I had to.
Thanks!!!!!!!!!!!!!!! I thought this was a simple method and then I started reading all of the comments. It seems that if there is an easy method for non-mathematicians to use there are an equal number of complicated methods put forth by highly skilled mathematicians.
When you write all this stuff out it's almost two pages for one square root. It makes you appreciate calculators and admire how people did this before calculators.
There is a geometric construction the ancients could use to obtain the square root of a length, using semicircle geometry. It requires defining a unit-length (which pins numerics onto a physical length). The method is good for obtaining accurate square-roots for lengths between 0.1X and 10X the unit length.
It took a moment, but my brain suddenly went 'click' as I realised why this method felt familiar; it corresponds to the Newton-Raphson method applied to finding the root(s) of X² - a
quick summary : let x be the number and y be the square closest to it sqrt (x) = [x+y]/(2 sqrt(y) ) -----(1) to derive this, simply use [f(y+h) - f(y)]/h = f'(y) ---------(2) and let y+h=x , f(y) = sqrt (y) => as x and y get closer the equation (2) becomes close to the derivative of f(y) at a given y also, this approximation holds very good for larger numbers
Hi Presh, It's a bit late but still. It's a really interesting approach. I tried to write an algorithm by myself some time ago but I used kinda "dichotomy": I searched for the interval of two consecutive integers whose squares are upper and lower bound of the given integer and took their arithmetic mean. If the square of this AM is less then the given integer, I assigned AM to the lower bound, otherwise - to the upper bound. Did as many iteration as needed to get a desired precision level. 🤔
Calculating square root by hand is derived from this method. Of further interest is applying this scheme to binary numbers. It dramatically simplifies to appending -01 to the right of the existing answer to test against the remainder. If it "fits", append 1 to the right of the answer. If not, append 0 & discard the result of the subtraction.
5:20 Do you mean the closest number without going over? Here you say 111 is closest to 100, which is false. 111-100 = 11 while 121-111=10 which means 111 is closer to 121 than 100. But then at 6:11 you say that 23 is closest to 25 (which it is), so now I am confused. Did we say 100 in the first example because our gable only goes up to 100?
I thought at first this might be a video on how to calculate square roots directly though a process similar to long division. Let's take for example the number 579 and calculate it's square root. We write it down like we're doing long division: ．．． √５７９ The first step is to pair up digits together from the decimal point outward, so for instance a number like 1234567 would be paired into 1(23)(45)(67). ．．． √５（７９） Now the first digit of the square root will be whatever digit's square is closest to the first digit or pair of digits in the original without going over. 5 is the first digit in the original above, so the first digit in the square root is 2 and we write it as shown here ．．．．２ ．．２ √５（７９） Now similar to long division, we multiply the current solution by the placeholder on the left, subtract, and bring down the next pair of numbers: ．．．．２ ．．２ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） Next multiply the placeholder by two and add in "x" like so: ．．．．２ｘ ．４ｘ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） "x" will be the largest digit that x times "4x" is still less than 1(79). In this case, x=4 since 44*4 = 176 is the closest we can get without going over: ．．．．２４ ．４４ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） And we continue this process. Multiply the new digit by the current placeholder and subtract: ．．．．２４ ．４４ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） ．．．．１（７６） ．．．．_ ．．．．．．．３（００） Multiply the newest digit by two and add in an "x" ．．．．２４．ｘ ．４８ｘ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） ．．．．１（７６） ．．．．_ ．．．．．．．３（００） We need the largest x such that "48x" times x is still less than 300. In this case x=0 (since 480 is already more than 300). ．．．．２４．０ ．４８０ √５（７９） ．．．．４ ．．．．_ ．．．．１（７９） ．．．．１（７６） ．．．．_ ．．．．．．．３（００） ．．．．．．．．．．０ ．．．．．．．．．．_ ．．．．．．．３（００）（００） One more time for this example. Multiply the newest digit by 2 (0*2 = 0), then add in x ．．．．２４．０ｘ ．４８０ ｘ√５（７９） ．．．．４ ．．．．_ ．．．．１（７９） ．．．．１（７６） ．．．．_ ．．．．．．．３（００） ．．．．．．．．．．０ ．．．．．．．．．．_ ．．．．．．．３（００）（００） We want the largest x such that "480x" times x is still less than 30000, which turns out to be x=6. ．．．．２４．０６ ．４８０ ６√５（７９） ．．．．４ ．．．．_ ．．．．１（７９） ．．．．１（７６） ．．．．_ ．．．．．．．３（００） ．．．．．．．．．．０ ．．．．．．．．．．_ ．．．．．．．３（００）（００） ．．．．．．．２（８８）（３６） ．．．．．．．_ ．．．．．．．．（１１）（６４） And so on, continuing the process as long as desired. The nice thing about this method is, while it's not necessarily the fastest algorithm to getting an approximation, it does literally gives you the exact rounded off decimal version of the square root at every digit and it works for any number you can write down.
I made a method to find square roots when I was in 3rd grade… it was iffy but had only slight inaccuracies I took the same table of squares and used that to do the following: If we are trying to find the square root of 17: 17 is between 16 and 25 So we will have 4. 25 - 16 = 9 And 17 - 16 = 1 So the square root will be 4 1/9 I know it’s not the best but it’s 3rd grade me so…
As many have said, this is just repeatedly using Newton's method. But it is also a consequence of the generalized binomial theorem: (1+x)^r = 1 + rx + r(r-1)/2 x^2 + ... for all real numbers r, given that the sum converges. Using r = 1/2 gives the square root technique described in the video, and we can in fact use r = 1/3 to approximate cube roots well!
I was expecting the same technique seen in another recent video: use the difference from the lower square as the numerator, and the difference between the two squares as the denominator. Thus: √17 = 4 + 1/(25-16) = 4.11 ... √69 = 8 + 5/(81-64) = 8.29 ... √111 = 10 + 11/(121-100) = 10.52 Base ten has the advantage of more easily multiplying both sides of an equation. If I were approximating √2, I would instead multiple by 100 under the radical (10^2), then divide the result by 10. √200 ~= 14, so: √200 = 14 + 4/(225-196) = 14.1379 >>> 1.41379 (by the other video's method) √200 = 14 + 4/28 = 14.142857 >>> 1.4142857 (by this video's method)
I think it would have been helpful to mention, that this directly corresponds to the binomial formula: √b = a+x => b = a² + 2ax + x² hence x ≈ (b-a²)/2a neglecting the x² term as it's supposedly much smaller than the rest. So it's not black math magic but something you're probably quite familiar. But maybe I'm weird and this would not have helped at all... ¯\_(ツ)_/¯
I tested a cubic root with no success out of curiosity before seeing the geometrical explanation in the end of the video and now I both love this method and also got curious if there is a universal method to calculate "x root of y" and I hope this not too much of a nightmare to find out.
Remember the dark middle ages? Back when nearly nobody could write or read. We barely have memories and records from the dark ages. Look at humanity now and you realize we are heading in the same direction. Digitalization leads to no recording and humanity is slowly getting stupider. And yet ancient pottery is still found, even completely intact at times.
So in simpler terms the formula (a+b)^2= a^2 + 2ab + b^2 is used and since we take 'a' as the whole number that when squared results in a number closest to the number for which we seek to find the square root of and as for 'b' we take that as the fractional term whose square is negligible hence 'b^2' can be omitted in the expansion of (a+b)^2 = r with r being the number for which we wish to find the square root of. Resulting in a^2 + 2ab ~ r
1:54 I am sorry but what marks that this is a decimal number, did they have any decimal seperator? like it can't be coincidence that the tablet has the aproximation but without a seperator asystem seems to complicated to use.
Hmmm. The only way I ever extracted a square root was guess-and-check. For something like 97, you start with the nearest square to it which is 81, so 9 is the first digit. 2nd digit, guess pretty big, because 97's close to 100, so 9.7^2 = etc. Higher? Lower? Slowly, methodically, build each digit. Basically, at each step, find the least upper bound on the digit that keeps it less than 100 in the product. This method seems more efficient, but I don't quite understand it.
My parents (now in their eighties) were taught the square root by hand method in grade 4. I learned it in my 40s when reading a book on abacus calculations. I wasn't quite sure why the algorithm worked and meant to derive a proof eventually. What you presented here is essentially the same algorithm. So now I don't have to figure it out. THANKS!
Your comment tempted me to ask, did you mean you were intending on deriving a proof on your own, and if so, what math-background did you have involving proofs? Or, were you going to just research the proof derivation.
Send thank you notes to Isaac Newton ;)
@@AhirZamanSairi I imagine he means a proof on his own. If you have had three semesters of calc, this is a thing you can do all on your own if you wish.
@@AhirZamanSairi I was going to figure out - on my own - why the square root by hand method works. I have a bachelor maths degree from ages ago, where I studied real and complex analysis along with lots of measure theory. Back then it was theorem, proof, theorem, proof, ... and it was truly loads of fun.
Lol I'm 51 but my school also taught long-hand square roots. Our entire county was stuck in the 1950s though....rather still is....1950s plus the internet, now.
for those wondering,not only this is an application of newton's method,this method is also called linearlization(1st degree taylor polynomial).just like binomial approximation(also another form of this formula only with a = 1),or generally any approximation that "brings down" the exponent is used in many areas of physics and math.
What’s funny is I was thinking this is so much like Newton’s method, but it’s not. What are the odds?
We don’t give ancient people nearly as much credit as they deserve
Especially non European ancient civilizations, which often have their accomplishment understated to such a degree it feels like no one did science and philosophy other than the Greeks
@ I would argue Indians Chinese and Arabs get plenty recognition though it’s probably an understatement considering their massive achievements. African Oceanic Pre Columbian and Middle Eastern civilizations though are massively underrated I hope they gain more traction with the expansion of history
@@luisfilipe2023 Middle Easterners also get a great deal of credit wdym. However yeah you never hear about African or Pacific islander discoveries often. Hell even south Americans get a great deal of recognition for agriculture and astronomy. The idea that no one but western advancements get recognition was maybe valid in the 70s. However since the 90s at least everything from college to media definitely has been giving credit where it's due for the most part. I don't even know who people claim discovered this stuff in Europe. However I constantly hear about Middle Eastern and indian mathematicians. You can't even talk about this stuff without people like this guy making sure we all know how awesome everyone but Europe is.
@@luisfilipe2023 I mean I think the middle east gets credit if for nothing else then the ancient egytpians which the europeans where obsessed with for centuries(hence why the US has so many places named after egypt(Cairo illinois or Memphis Tenessee come to mind). I think the reason why africa doesn't get any credit is because we don't have any written accounts so its hard to know what they knew, and many of their buildings where not built from materials that actually hold up over time like other groups did(I mean they had straw huts which don't last long, though if you've seen how they are built you would absolutely consider them at the very least done with great artistry(I mean they weave reliefs and meanders into the walls its actually really cool). The other is most groups where not particurally keen on preserving their nations history either because of more pressing concerns, corruption or religious reasons(many areas are more concerned with acquiring resources hence why I believe it was sudan who just flooded an entire region filled with pyramids so they could build a dam, the ottoman turks where not particuarlly fond of anything not related to their religion(and to be fair neither where christians) etc.).
As for south america I do believe they get some credit due to Manchu Pechu and terra petra, as do the aztecs I think. I do believe that more credit should be given to polynesians as we have evidence they navigated the pacific and reached south america before anyone and an anthropologist used their navigation system to get their by the same style of boat they would have used. But again, I think the major issue is going to be lack of records and a natural bias to place one's own group above others which is why if you talk to greeks they would tell you they invented everything and asians would do much the same thing as do most every group.
@@luisfilipe2023 Middle easterns get a lot more credit than any other in my experience
Here is another way of looking at the process. Think of the graph of y = x^2. It has a point on the curve (x, 17) and x is of course the square root of 17. We know that (4, 16) is pretty close. Mentally draw a tangent line at that point. The derivative of Y= x^2 is 2x, so the slope at our initial guess is 2*4=8. Delta x = delta Y / slope = 1/8. So the 2nd approximation is 4 + (17-16)/8.
So Newton's method
Isnt it also related to the wuadrstoc formula..that's a thought/observation I had. The b/2a since the number you are looking for is between x and x plus 1or in this case greater than a but less than a plus 1..That's a neat observation and may be relevant right?
@@major__kongisn't newton's method iterating this process, though? this isn't newton's method, this is just linear approximation (or, the taylor polynomial p_n where n=1)
Well articulated.
So easy 😂😂😂😢😢😢😢😢😢😮😮😮😮😮😮😮😮😮😮😮😮😮😮
Walk like an Egyptian, and root like a Babylonian. 💀💀💀
is that a jojo reference?!
lmao!
@@cr0n0us_4it is a Song
Okay Ali
Words to live by
For large numbers, divide by 100 before and multiply 10 after.
√999?
√9.99 ≈ 3 + .99/6
√999 ≈ 30 + 9.9/6 = 31 + 3.9/6 = 31.65
Actual value 31.6069
This works, but accuracy increases in single iterations of newton's method for larger numbers, so you would get a more accurate calculation if you use larger numbers, and nearby perfect squares.
For example:
√999 ≈ 32 + (999 - 1024) / 64
To simplify, I would say do something similar to what you suggest.
(999-1024) / 64 = 25/64
Multiply by 10.
250 / 64 = 256 / 64 - 3/32
The last term is pretty close to 1/10, so:
250 / 64 ≈ 3.9
Then divide by 10.
25 / 64 ≈ 0.39
So we're looking for 32 - 0.39 = (3200 - 39)/100, which yields 31.61. This squares to 999.19.
The method you describe is great if you don't know a nearby perfect square, though.
I figure the point of such methods is when you have no calculator, which I'm helping further with.
@@MegaMinerd Right. This is what I do when I don't have a calculator. This is how I calculate square roots when I'm teaching and don't want to go grab one. I was just sharing further refinements.
Why divide by 6? Please clarify.
@@bhatkrishnakishor The general formula is:
√x ≈ k + (x - k^2) / (2k)
If you're trying to find √9.999, then the closest perfect square is 9, so:
√9.999 ≈ 3 + (9.999 - 9) / (2 * 3) = 3 + 0.999 / 6
I coded in Intel assembly a version of the Newton Raphson method. For the square root of a, take an estimate x. Then take a/x and average it with your original estimate. That is, a' = ( a + (x/a)) / 2 . Repeat until you are close enough. This method converges rapidly.
For the initial estimate, the Intel processor is very helpful. Floating point numbers are stored in quasi-scientific notation according to the IEEE standard. All you do is divide the exponent by two (right shift 1) and remove the 1 from the first bit of the mantissa. This is close enough for the method to converge rapidly. This method is about as fast as the supplied C library sqrt call, which uses a completely different algorithm that doesnt require a hardware floating point divide function.
if x is the sqrt(a) then a/x == sqrt(a)
if x is approx_sqrt(a) then a/x == sqrt(a) + error
average(x,a/x) == average(x,sqrt(a) + error) = average(sqrt(a)+error_x , sqrt(a)+error_aoverx) == sqrt(a) + average(error_x,error_aoverx)
@@StevenSiew2 That's more assembly coding than my attention span is capable of LOL LOL LOL.
@@starpawsyThat’s not assembly code, it’s pseudocode. Assembly code is much more low level, the algorithm would probably take several hundred lines of code, since you need to make everything from scratch.
@@KindledAsh Thanx for the mansplaining. My implementation did not take "several hundred" lines. And it certainly was in Intel assembler. My description was, of necessaity, pseudocode.
@@starpawsy Yeah, I was thinking of x86 assembly, and I meant it would be more complicated, not that long. I missed that, sorry.
Thanks for showing the table at 0:33. It shows clearly that the numbers from 1 to 59 were represented as two digits, where the leftmost digit shows the number of tens and the rightmost digit shows the number of ones. I.e., these numbers are represented very similar to the way we do, except that they didn't use a symbol for zero, they used different sets of digits for the tens and for the ones, and their maximum number of tens was 5.
Not exactly as they did have a symbol for 10 which could not be expressed only with 1..9.
Each of those is a single digit. The tens and units within a digit are not separate digits, they're subparts of the same digit.
It's a common misconception about cuneiform numbers.
This is the direct application of Newton's method to find the zero of a function:
Let f(x)=x²-a, so that we want to find x such that f(x)=0 x²=a
we then have x_n+1 = x_n - f(x_n)/f'(x_n)
It becomes x_n+1 = x_n - (x_n²-a)/(2*x_n), which corresponds to the algorithm described in the video
Fascinating
i was searching for a comment that realised this 😂
@@voicutudor7331 me too. A simpler summary of this formula is that your next guess is the mean of x and a/x. Hence, x_n+1=(x_n+a/x_n)/2. If for sqrt(2) you know a good start is 7/5, then next one is 7/10+5/7=99/70 which is good for 4 decimal places.
Not good because you are introducing calculus instead of focusing on the fundamentals such as multiplication, division, addition and subtraction. Why adding calculus here is bad? Because computers can't do calculus. You need to think in a practical way sometimes.
@@TemplarX2 its newton iteration, not complex integration. this is how the formula is derived.
The method I learned in grade school is to take the average of a and s/a. This gives the same result as Presh's "a + (s - a^2)/(2a)", but I find "(a + s/a) / 2" to be more intuitive. If "a" is a bit low, then s/a will be a bit high (and vice versa), so halfway between them will be closer than either of them.
Known as Heron's method.
I just came up with this method from Presh's method with a bit of simplification. Then I extrapolated it to the nth root for fun. I agree that it's more intuitive, but I needed the geometric intuition to expand to the nth root because I was otherwise not weighting the summands correctly. (a + s / a^(n - 1) ) / 2 doesn't work, but averaging (n - 1) a's with s / a^(n - 1) does. The result is nth√s ≈ ( (n - 1)a + s / a^(n - 1) ) / n
This is probably the best way to calculate a square root when you're programming a computer in assembly language and have no library function to do it for you. It's fast and the convergence is quadratic - the number of correct digits doubles with each iteration. Adaptation to n-th roots is fairly simple, although the convergence is not as rapid. Even if you have a library function available this will normally be a bit faster, and can be tailored to the application for extra speed.
Very nice! This reminds me of high school days! Enjoyed very much haha
5:30 111 is closer to 121 than to 100
It doesn't make a difference which way you go, you just have to add a negative if you go the other direction
I was taught a different method, which is a lot like long division. You draw a radical sign over the number you want to sqrt. The number is then split up by putting commas after every two digits, spreading out from the decimal point. During the process, you always bring down the next two digits then repeat the process. Along the way you'll get increasingly longer trial divisors.
The beauty of this method is that it uses a single, simple layout, & you never have to average anything, or sometimes subtract instead of adding anything.
There is a similar algorithm for cube roots. We learned this in school.
Isn't 111 closer to 11² = 121 than 10²=100? at ~5:54; So sqrt(111) ~~11 - 10/22 = 11 - 5 / 11 = 10.545454... which is obviously closer to 10.536... (sqrt(111)) than 10.55 (the approximation given in the video by starting with 10²)
Yes, but he didn't have 11 on his table
The closer you are to the squared value, the more accurate the adjustment value will be it seems. Since 111 is closer to 11² then using 11 + adjustment will be more accurate than 10 + adjustment indeed
I noticed that right away also.
When performing a mental calculation, the choice tends to be driven by ease of calculation. 10 in Roman numerals is a decimal position, x 10 is simply a numeral shift.
same I was about to type this
This is pretty cool. But also: you can start with any non-zero number and it will converge, usually within 8 or ten steps. Obviously, the closer your guess, the quicker it converges. But for the given solution, you need to know the nearest squares, so you could just use sqrt(x) ≈ [root of lower square] + (x-[lower square])/([upper square] - [lower square]) and it will be good to a few decimal places, and less than 1% error for numbers larger than 12. For example: sqrt(70) = 8.3666 ≈ 8 + (70-64)/(81-64) = 8+6/17 = 8.3529, the error is less that 0.2%. That should be good enough for anything you do, unless you are going to the moon or something.
I was thinking while watching, that you should derive the formula for the square root approximation, and you did later. The table of squares was good to show the values to consider. I had to derive a square root procedure starting with the algebraic form ( a + b ) then squaring it then factoring out ( b ) from the form (( a )( a ) + ( b )( 2a + b )). It gets more complicated after that. Most instructors teach an arithmetic procedure that takes the ( 2a + b ) to compute the next iteration of the approximation. I spent some time putting the procedure into a spreadsheet to do the calculations to six figures. There is more to be said. Sorry, don't have the time to discuss it here. Great Video. Thanks for doing this. History helps. ;D
If you have all the two-digit squares memorized, you can add two zeros to the radical and make a better estimate. I would change 69 to 6900, and since I know that 83^2 is 6889, I know that my approximation will be 83. plus 11/83+84, or 11/167. So I have 83.0. I can continue calculating by trial and error, doubling the number of significant digits each time.
Once I calculated a square root to twenty-four significant digits, all in my head, which took me about three days.
Wow!
Yeah memorizing all the squares to 99 seems easier 😀
Impressive rooting, but I guess not what she said 😂
Thanks for sharing this finding. I am a big fan of square root and higher roots. I first learned extracting a square root of a number via a long division from my high school chemistry teacher. I was totally hooked afterward. I was so excited that I couldn't help doing square root of any number every day. Then, I wanted to know how to extract a cube root of a number. But, I couldn't find anyone who can show me. I told my father about it. One day he surprised me with a math book he bought from a book store, and there it was, an instruction on how to extract a cube root of a number via a long division. I was just totally overwhelm. The book also showed how to extract any root of a number using binomial expansion (the holy grail). The binomial expansion technique is what I use today to extract a square root. Even though I know how to extract a cube root and higher, I use calculator to save time. I love this stuff.😊
This is also essentially the same as the Newton's method. The 2 in the bottom can be explained by the fact that the derivative of x² is 2x.
Actually the 2 can be better explained as the distance between squares. (x+1)²-x²=2x+1. Thus is a discrete problem so calculus doesn't even make sense. Talk about regurgitating and not thinking.
In other words , newton was “inspired “ by Babylon method for this , just like he was inspired from Sanskrit text for 3 laws of motion and concept of gravity
I remember using this method in college. In those days (early 70s) calculators were just coming out and had limited functionality. Mine was a Comodore something which I paid over $90 for. It had no square root function, just add, subtract, multiply and divide. Plus if you moved a switch, you could store the display in memory. I was the only one in my class who had one. I would run through the formula the first time, store the result in memory and use that value for the guess for the second run. Amazing results. My classmates were using their slide rules! Actually I would use my own slide rule for the first approximation. I had a piece of paper taped to the back of the calculator which gave me the formulas for the trig functions.
Just an old man taking a trip down memory lane. Thought I would share. Always enjoy your videos.
This kinda reminds me of when I worked out how to square any 2 digit number in my head back in high school:
For example if you want to square 32, start by squaring 30 to get 900. Then rough tune by adding the double of 30 twice to get 1020. Then square 2 to get 4 and add to get 1024.
If you want to square 44, multiply 4 by 5 and slap 2 0s on the end to get 2000. Then subtract the double of 40 once to get 1920. Then add the square of the 1s digit, 4 x 4, to get 1936.
If you want to square 79, start by squaring 80 to get 6400. Then subtract the double of 80 once to get 6240. Then add the square of the 1s place, in this case -1, to get 6241.
The process of ballparking by squaring the nearest multiple of 10, rough tuning by adding or subtracting multiples of twice the nearest multiple of 10, and fine tuning by adding the square of the 1s place is very effective. Putting in the midway shortcut of multiplying the 10s digit by 1 greater than itself and putting 2 0s to get the midway rough tuning means you will never have to rough tune up or down more than twice.
As a bonus, if you have to multiply 2 numbers that are both odd or both even, take the square of their midpoint and then subtract the square of the offset. 87 x 93 for example. Since their midpoint is 90, we square 90 to get 8100. Since 87 and 93 are both offset from 90 by 3, we subtract the square of 3, 9, from 8100 to get 8091.
It is (a+b)(a-b)=a^2-b^2=90^2-3^2=8100-9=8091.
Many have pointed out that this is what you get if you use Newton's method (a.k.a. Taylor approximation of order one). It is also the result of the following: Let x be a first approximation of √s. Then we would very much like to calculate the geometric mean of x and s/x, as that is exactly √s. But we can't. What we CAN do is to calculate the arithmetic mean. And that yields exactly this method.
As an added bonus, both Taylor approximation and the mean method have natural ways to bound their respective errors. Taylor has Taylor's theorem. And we know the geometric mean lies between the arithmetic and harmonic means.
I'm sure the Ancient Babylonians knew this Taylor guy intimately
Wrong. Newton's method uses continuous approximation. This is a discrete method.
@@gregorymorse8423 I don't know which Newton's method you are taking about. But the one I'm talking about here takes your approximation to a zero of a function, finds the tangent of the function at that x-value, and uses the zero of that tangent as the next approximation. Repeat until you're close enough.
If you say this is a continuous method, presumably you're referring to the finding of the tangent?
Its called approximation in calculas. Not only square root, we can calculate approx value of any function with any argument. The expression goes like this :
f(x + ∆x) = f(x) + ∆x*f'(x)
Where f'(x) is first derivative of f(x) w.r.t x
The more further apart x and ∆x are , the more accurate the answer is.
That is great, thanks. Made my day
Slight correction, the bigger del(x)/x is the less accurate the answer.
I've implemented this algorithm a couple times before, but I never thought of it as something that can be calculated in your head. You have changed that today 🙂
You should mention that is exactly what you get if you apply Newtons' method.
We should focus on the Babylonians, Newton already has enough
@@ZanderzMcCluerya but that’s the rigorous reason why the method works.
it's lovely how the geometric proof shows the logical step towards "completing the square" wonderful
Nice video! I teach this method in my numerical analysis class. A few comments:
1) Note that the formula sqrt(s) ~= a + (s-a^2)/2a can be simplified as sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a). So in other words, given your approximation a, you take the average of a and s/a. The idea being that if a is bigger than sqrt(s), s/a will be proportionally smaller (and vice-versa if a is smaller than sqrt(s) ), so the average should get you closer to the true value
2) As others have noted, this is a special case of newton's method for finding the root of f(x) where f(x) = x^2 - s. One of the main selling points of Newton's method is that it is very fast -- if you feed your updated estimate to sqrt(s) back into the formula, you get about double the total number of accurate digits every time (provided you are close enough to the answer). So if you were to continue the estimate to sqrt(2) shown in the example, you would get the sequence:
1
3/2 = 1.5 (1 digit correct)
17/12 = 1.41666... (3 digits correct)
577/408 = 1.414215... (5 digits correct -- this is the last estimate shown in the video)
665857/470832 = 1.414213562... (10 digits correct!)
This is called quadratic convergence, and you can prove why this happens for Newton's method in general (not just for this example, but most functions f(x) also)
3) It generalizes to higher roots as well. For example, to approximate the cube root of s, pick an approximation a and then compute:
cuberoot(s) ~= 1/3(2a + s/a^2)
(This is what you get from applying Newton's method to f(x) = x^3 - s instead of x^2 - s). So for example to approximate cuberoot(2) = 1.259921050 you would get the sequence
1
1/3(2 + 2) = 4/3 = 1.3333... (1 digit correct)
1/3(8/3 + 2/(16/9) = 91/72 = 1.263888... (3 digits correct)
1126819/894348 = 1.259933493 (5 digits correct)
2146097524939083451/1703358734191174242 = 1.259921050 (10 digits)
Nice! I also spotted the (surprising to me) equivalence of the well known (and more intuitive) "average s with s/a" method.
This has nothing to do with Newton Raphson approximation which is a continuous not discrete method. Its astounding how many math flunkies are here and don't watch or understand the video or the Babylonian method.
Nice explanation, but we need calculator for calculate the "width fraction" 1126819/894348 .... For grade 4 students need some efforts, if calculator is not allowed.....
It completely destroyed my perception of the history of mathematics. Babylonian mathematicians almost 4000 years ago knew mathematics thousands of times better than many people today.
Tbf, the people who did this were probably part of some elite merchant group or around the equivalent of our upper middle class nowadays. Not necessarily nobles, but certainly not your average person whose daily life only really needs the four basic arithmetic operations (tho, of course, that's when we compute stuff ourselves; our phones and computers nowadays rely on more complex stuff afaik lol)
@@moondust2365 and, of course, saying 'better than many people today' is a pretty low bar. 😉
@johnbjorkman4144 I mean, if you're taking the world population and not just the wealthy countries (or heck, even if you _are_ just taking the wealthy countries), that's always been the case. It's not like people are suddenly dumber on average today, people with no access to education or people who are difficult to teach (whether due to genetics, family/social issues, mental health issues, etc.) have always existed XD
I think it's impressive that even the "elite mathematicians" 4000 years ago knew more than the average person today.
@@moondust2365 yea, elite no kidding right? all of their other needs must have been attended to for them to afford the time and effort to devise the method then have it recorded.
2:00 the Babylonians also developed the Saros cycles, based on their correct geocentric cosmology (earth being level and stationary). The Saros cycles are still being used today.
I created a spreadsheet with columns for both methods: rounded down to a whole number (FLOOR) + the positive fraction, and rounded up to a whole number (CEILING) + the negative fraction.
The FLOOR method gives a closer approximation for numbers closer to the lower perfect square (like 27) and the CEILING method gives a closer approximation for numbers closer to the upper perfect square. Since all square numbers have a difference of an odd number, the number exactly between two perfect squares gives the same approximation with both methods.
I created one column for each method to calculate the difference between the approximation and the actual square root. The approximation is always greater than the actual root.
I copied the formulas in rows from 1 to 1024, then averaged both difference columns. They weren’t the same.
The average difference of the FLOOR method is 0.008761… and the average difference of the CEILING method is 0.002423…
I can’t explain why the CEILING method ends up being more accurate on average.
Edit: When I copied the formulas into 10,000 rows, the CEILING method is only about 7% more accurate than the FLOOR method. Approaching infinity, would the two methods achieve the same accuracy?
It's even more fun in copilot AI. Give it this command: Generate a table with two columns, A and B. In column A, start with number 1 and for each row add 0.01 until you reach number 10. Column B is the square of A. It should generate a column with 1000 row and enough precision for most practical purposes.
I am going to go over this until I get it. Im in my 70s why weren't we taught his school. thank you.
Ben! This is a great tutorial with awesome trial and error with detailed explanations. You are a great maker!
I love that oxymoron : “result is an accurate approximation”. Love that video too. 71 going to 72, still learning and enjoying it.
Hats off to those Babylonians.
They really had it going on when it came to math and ASTRONOMY...!!!
I came up with a pretty unique and simple method to compute square roots myself. Eg: square root of 17. find the closest square number=4. do 17/4 . Then find the average of 4 and 17/4 assume it to be a. Divide 17 by a and add a and 17/a. repeat continuously until you get the answer. Currently in add math, when i forgot calculator, i used this method to approximate my answers lol.
This is actually the same method as he shows in the video. His formula sqrt(s) ~= a + (s-a^2)/2a simplifies to sqrt(s) ~= a + s/2a - a/2 = 1/2(a + s/a), which is taking the average of the two quantities as you said.
@@tdhumphr I see
Enjoyed the video. Was hoping to see how we arrive at the extra terms since the first example had 4 terms
as many pointed out, this can be seen as an average between the estimate (x) and S/x; an easiest way to visualize it is to consider these values as sides of a rectangle, having the same area S of the original square, and after every iteration the two sides converge to a common value, becoming a square.
I have enjoyed this video thoroughly. Thank you very much for sharing.
For sqrt(2) one could instead compute sqrt(200) using e.g. the proposed method, and divide the result by 10. Sqrt(200) is approximately equal to sqrt(196) which is 14. Then add 4/28, that'll give approximately 14.142, and divide by 10 to get our sqrt(2).
a general form of this equation can actually not only be introduced(newton's method),if you know programming,this method can actually be implemented in programming languages like c,python,java,....And this method applies for all real univariate polynomials(any kind of function such that the initial guess is closer than a zero of the derivative in the open set resembling a circle with a radius r and center the zero we are talking about)
I was taught how to calculate the square root by hand in middle school. I still remember it and use it to these days.
To make it more accurate without doing multiple calculations, just multiply it first. So instead of calculating sqrt(2), calculate sqrt(200), because sqrt(200) is equal to 10 times sqrt(2). Using the method shown, sqrt(200) is approximately 14 + 4/28 = 14.142857... Divide that by 10 and you get that sqrt(2) is approximately 1.143, which is close enough
I subbed because of this vid. Thank you!
I wish that had been taught in the 70’s and 80’s when I went to school. Way easier than the now long forgotten hit and miss method I was taught.
Another way to get a better value to approximate root(2) is to observe that root(98) = root(49)*root(2) = 7*root(2).
Using the method, root(98) = approx 10 - 2/20 = 99/10. Dividing by 7 gives root(2) = approx 99/70 = 1.4142857... which is already a lot better than 1.5.
Edit: If you use root(200) = 10*root(2) and 14*2 = 196 we get root(200) approx = 14 + 4/28 = 14.142857... so dividing by 10 for root(2) gives the same approximation as the root(98) example.
Every nerd knows that the square root of 69 is ate something.
What
❤
Lol... here we go
The use of BCE and CE is a sign of historical dishonesty.
A simple trick to get better approximations is to consider not only squares of integers in the initial table, but also squares of integers plus 0.5. These can be calculated easily: (a+0.5)² = a(a+1) + 0.25, e.g. 3.5² = 3*4 + 0.25 = 12.25
Newtons method to find x = √y. find integer x, so that x^2 is close to y then repeatedly evaluate the expression:
x ← ½(x + y/x) For y=17 we get: x = 4; x = 4.125; x = 4.12310606060606; x = 4.12310562561768; x = 4.12310562561766... Wolfram says √17 = 4.12310562561766054982140985597407798... (same to 15 digits)
Note: the typical Newton's method expression: x ← x - (x^2 - y)/2x simplifies to: x ←½(x + y/x)
I divised this method in highschool and was able to calculate a bunch of digits too which I then crosschecked with my calculator. It's so cool to know now that this method has such a long history!! Yea learn something every day.
This confused the heck out of me because, at 12:53, he said we had another figure whose total area was EQUAL to s (emphasis mine). But we don't. We have a side that can be used to form a square, but that square's area is not equal to s, it's bigger, which is why it's an approximation.
I always used the Babylonian method as a good introduction to teaching Euler approximation methods.
I found this method always gives a slight overestimate of the square root, so one must always subtract after the first step.
I simplified to the following equivalent formulas which I prefer: √s ≈ (a² + s) ÷ (2a) = (a + s / a) ÷ 2. Intuitively, √s is between a and s / a so it seems right that averaging these values would approach √s. My formulas have fewer operations and avoid user error since subtraction isn't commutative. And you don't need to consider whether your estimate is too high.
And now, using the geometric approach shown in this video, I extended it to cube roots with ³√s ≈ a + (s - a³) / (3a²). Seeing a pattern, I've extrapolated the nth root of s ≈ a + (s - a^n) / (n • a^(n - 1)), or alternatively, the nth root of s ≈ ((n - 1)a + x^n / a^(n - 1))/ n. I'm sure this is an old result, but it was fun to figure out.
The trick uses the first and second binomial formulas:
(a + b)² = a² + 2ab + b² and
(a - b)² = a² - 2ab + b².
The difference to the real value is the missing b² part.
In this case: (4 + 1/8)² = 4² + 2 · 4 · 1/8 + 1/8² = 16 + 1 + 1/64.
And the 1/16 is tiny in comparison to the 17.
I’m going to teach my kid this. What a great way to use square roots and such a simple method.
I take exception that you are choosing to round some values and not round other values to make it seem like they were more accurate than they actually were.
For example, at 5:09, you show 8.312 and 8.307. You did not round the first number up, but you rounded the second number up. It should be 8.3125 and 8.3066, and if you rounded them both, would be 8.313 and 8.307...
he just rounded 5 down to 0 instead of up
@@jimmyh2137 lol. Same argument. Round them both the same direction.
@@azrobbins01 You can't round everything in the same direction, that's a different operation (floor and ceiling).
He just opted to round xx25 up to xx30 but a xx24 would be rounded to xx20.
5 is the exact middle and you can choose how to round it, as long as you're consistent.
@@jimmyh2137 How often do you see people rounding 6 up, but 5 down? Is this a common practice? If you did this at work, would people think it is normal?
@@azrobbins01 6 is always rounded up.
6-7-8-9 up
1-2-3-4 down
5 is the "problem", could be argued for either side. In this case it's exactly 5, makes sense to round down.
It's 100% normal to round 5 down.
Amazing capabilities so long ago. Makes one wonder how they arrived at such a system and why... Seems math and numbers in general were more advanced then I would have ever thought. Thumbs Up!
This is a special case of the equation of the tangent line
y=f(a)+f'(a)*(x-a)
Insert f(a)=sqrt(a) and f'(a)=1/(2*sqrt(a)) to get
y=sqrt(a)+(x-a)/(2*sqrt(a))
Insert a=4 and sqrt(a)=2 to get
y=2+(1-4)/(2*2)=1,25
I know more interesting way, allowing to calculate square roots with arbitrary precision. It's iterative and can be performed the best way on a mechanical adding machine or on a paper. This method is found in a paper by S. V. Savich and another but quite similar was is used on actual old Friden calculators.. You perform division by sequential subtraction (like it's usually done on paper), but: 1. You track not only the remainder but also count operations. 2. You change the subtrahend with each turn. 3. You subtract sequential odd numbers starting from 1 to 17, until intermediate remainder becomes higher that subsequent subtrahend, so you got a remainder and move to less significant positions. Almost like with a general division giving a long fraction.
Example: you need to calculate a square root of 2. You start with 1. 2-1=1. There was 1 operation. Then you could try to subtract 3, but remainder is 1 and you would cross zero, so instead you write down current count of performed subtractions (1) to the first digit of the root, then decrease the current digit of the subsequent subtrahend by 1 (3 becomes 2), and append 1 at the right (getting 21). Then you move this intermediate subtrahend one position to the right respective to the remainder. So you start subtracting this two-digit number summarizing 20 and sequence of odd numbers (1, 3, 5...), from three digit number 100 (the first digit of the remainder and two trailing zeroes to the right). You subtract 21, 23, 25, 27. This way you perform 4 subtractions until you got 004, from which you can't subtract 29. OK. you decrease this to 28, write down your 4 operations as the next digit of the root, append 1 at the right side of the subtrahend: 281 and shift the whole subtrahend one position to the right again. This time you subtract 281 from 400, you get 119 in the reminder, so you have performed subtraction 1 time. 119 is less than 283, so you write down 1 to the root, decrease 283 to 282, append new one (2821), shift right and start subtracting that from 11900 starting with 2821, then 2823, etc up to 2827, getting just 604 in the remainder... So this way you can calculate a square root with any precision, performing as many subtractions as you get as the sum of all digits of the root, and each time you deal with 1 digit longer intermediate subtrahend and by the way it's twice the result. If the rightmost digit of this subtrahend reaches 9, the next time you should add 2 to this rightmost digit with carry to the left, so nnn89 will becomes nnn91. The explanation looks complicated and random like numerology but actually it's pretty intuitive and straightforward once you get the principle. So that is: you subtract odd numbers, you accumulate the subtrahend by appending new digits starting with 1 at the right side, you count your operations and write them to the root, and you track your remainder, each time moving to the right once it becomes less that the subtrahend.
I'm very curious if there is a good description for expanding this algorithm to binary.
I'm amazed at how someone so far back as Babylon could figure this out. Something most of us modern individuals could not do.
Anyone today can do this. Babylon happens to be an early example of surviving durable records
As a math geek I was excited to show this to my fellow students after I learned it in 7th grade. Teacher let me do a whole presentation.
Fellow students were unimpressed. 🙄
3 thousand years later... Videos appear on RUclips "90% of people will not be able to answer correctly 2+2x2"
WTH went wrong???
Order of operations is often taught poorly. My experience with math before taking responsibility for educating myself was a series of terrible teachers and wasted time. I think this is a common experience.
Nice! What's the name of the method where you make an estimate, divide the original number (dividend) by the estimate (divisor), then average the divisor and quotient (so the two factors when one is your estimate), make the average your new estimate, and repeat?
Heron's method.
Extremely awesome and helpful! Thank you!
Lots of people mentioned that this is essentially Newton's method / a Taylor approximation. But strangely, apparently no one mentioned that this method is very similar to a far older method, which has been known as "Heron's method" for ages. And I _was_ taught that at school.
I'm amazed that anyone was even thinking about square roots 3600 years ago. You could go into some high schools in the US today where the students don't know what a square root is.
My dad taught me the fast way to calculate the diagonal by multiplying aside by square root of two that was a very fun lesson.
That is interesting. I now know at least 5 ways to do sqrt(X) by hand.
1) Newtons method as done in school way back when (Estimate divide average)
2) The variation on Newton that gives you 1/sqrt(X) which you then multiply X by to get sqrt(X)
3) The "classical" one that involves a lot of dividing by estimated squares
4) A microcontroller one that needs only shifts and adds and subtracts to do.
5) This new one
Computers that can multiply but can't divide tend to use method #2
I haven't used #3 in so long I would take some time to remember it.
A variation of the microcontroller one is the way I might do it by hand if I had to.
Thanks!!!!!!!!!!!!!!! I thought this was a simple method and then I started reading all of the comments. It seems that if there is an easy method for non-mathematicians to use there are an equal number of complicated methods put forth by highly skilled mathematicians.
When you write all this stuff out it's almost two pages for one square root. It makes you appreciate calculators and admire how people did this before calculators.
it is amazing to see these tricks ancient cultures used to do math.
There is a geometric construction the ancients could use to obtain the square root of a length, using semicircle geometry. It requires defining a unit-length (which pins numerics onto a physical length). The method is good for obtaining accurate square-roots for lengths between 0.1X and 10X the unit length.
It took a moment, but my brain suddenly went 'click' as I realised why this method felt familiar; it corresponds to the Newton-Raphson method applied to finding the root(s) of X² - a
quick summary :
let x be the number and y be the square closest to it
sqrt (x) = [x+y]/(2 sqrt(y) ) -----(1)
to derive this, simply use
[f(y+h) - f(y)]/h = f'(y) ---------(2)
and let y+h=x , f(y) = sqrt (y)
=> as x and y get closer the equation (2) becomes close to the derivative of f(y) at a given y
also, this approximation holds very good for larger numbers
This can also be related to the pell's equation, the ratio of its solutions will give you the root of the coefficient of y🎉
Hi Presh,
It's a bit late but still. It's a really interesting approach. I tried to write an algorithm by myself some time ago but I used kinda "dichotomy": I searched for the interval of two consecutive integers whose squares are upper and lower bound of the given integer and took their arithmetic mean. If the square of this AM is less then the given integer, I assigned AM to the lower bound, otherwise - to the upper bound. Did as many iteration as needed to get a desired precision level. 🤔
Calculating square root by hand is derived from this method. Of further interest is applying this scheme to binary numbers. It dramatically simplifies to appending -01 to the right of the existing answer to test against the remainder. If it "fits", append 1 to the right of the answer. If not, append 0 & discard the result of the subtraction.
5:20
Do you mean the closest number without going over?
Here you say 111 is closest to 100, which is false. 111-100 = 11 while 121-111=10 which means 111 is closer to 121 than 100.
But then at 6:11 you say that 23 is closest to 25 (which it is), so now I am confused.
Did we say 100 in the first example because our gable only goes up to 100?
Thank you .. And this is the R code
sqrt_iterative
I learned this as "Newton's method of approximation" you can derive it with a little differential calculus.
5:20 No, it's closest to 11^2 = 121. 111 is 10 less than 121, and 11 more than 100. So 111 is closer to 121 than to 100.
I thought at first this might be a video on how to calculate square roots directly though a process similar to long division.
Let's take for example the number 579 and calculate it's square root. We write it down like we're doing long division:
．．． √５７９
The first step is to pair up digits together from the decimal point outward, so for instance a number like 1234567 would be paired into 1(23)(45)(67).
．．． √５（７９）
Now the first digit of the square root will be whatever digit's square is closest to the first digit or pair of digits in the original without going over. 5 is the first digit in the original above, so the first digit in the square root is 2 and we write it as shown here
．．．．２
．．２ √５（７９）
Now similar to long division, we multiply the current solution by the placeholder on the left, subtract, and bring down the next pair of numbers:
．．．．２
．．２ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
Next multiply the placeholder by two and add in "x" like so:
．．．．２ｘ
．４ｘ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
"x" will be the largest digit that x times "4x" is still less than 1(79). In this case, x=4 since 44*4 = 176 is the closest we can get without going over:
．．．．２４
．４４ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
And we continue this process. Multiply the new digit by the current placeholder and subtract:
．．．．２４
．４４ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
．．．．１（７６）
．．．．_
．．．．．．．３（００）
Multiply the newest digit by two and add in an "x"
．．．．２４．ｘ
．４８ｘ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
．．．．１（７６）
．．．．_
．．．．．．．３（００）
We need the largest x such that "48x" times x is still less than 300. In this case x=0 (since 480 is already more than 300).
．．．．２４．０
．４８０ √５（７９）
．．．．４
．．．．_
．．．．１（７９）
．．．．１（７６）
．．．．_
．．．．．．．３（００）
．．．．．．．．．．０
．．．．．．．．．．_
．．．．．．．３（００）（００）
One more time for this example. Multiply the newest digit by 2 (0*2 = 0), then add in x
．．．．２４．０ｘ
．４８０ ｘ√５（７９）
．．．．４
．．．．_
．．．．１（７９）
．．．．１（７６）
．．．．_
．．．．．．．３（００）
．．．．．．．．．．０
．．．．．．．．．．_
．．．．．．．３（００）（００）
We want the largest x such that "480x" times x is still less than 30000, which turns out to be x=6.
．．．．２４．０６
．４８０ ６√５（７９）
．．．．４
．．．．_
．．．．１（７９）
．．．．１（７６）
．．．．_
．．．．．．．３（００）
．．．．．．．．．．０
．．．．．．．．．．_
．．．．．．．３（００）（００）
．．．．．．．２（８８）（３６）
．．．．．．．_
．．．．．．．．（１１）（６４）
And so on, continuing the process as long as desired. The nice thing about this method is, while it's not necessarily the fastest algorithm to getting an approximation, it does literally gives you the exact rounded off decimal version of the square root at every digit and it works for any number you can write down.
The automatic captions call you "Press Tow Walker"
I made a method to find square roots when I was in 3rd grade… it was iffy but had only slight inaccuracies
I took the same table of squares and used that to do the following:
If we are trying to find the square root of 17:
17 is between 16 and 25
So we will have 4.
25 - 16 = 9
And 17 - 16 = 1
So the square root will be 4 1/9
I know it’s not the best but it’s 3rd grade me so…
finally, this is what i need in my life, thank you
As many have said, this is just repeatedly using Newton's method. But it is also a consequence of the generalized binomial theorem: (1+x)^r = 1 + rx + r(r-1)/2 x^2 + ... for all real numbers r, given that the sum converges. Using r = 1/2 gives the square root technique described in the video, and we can in fact use r = 1/3 to approximate cube roots well!
I was expecting the same technique seen in another recent video: use the difference from the lower square as the numerator, and the difference between the two squares as the denominator. Thus:
√17 = 4 + 1/(25-16) = 4.11 ... √69 = 8 + 5/(81-64) = 8.29 ... √111 = 10 + 11/(121-100) = 10.52
Base ten has the advantage of more easily multiplying both sides of an equation. If I were approximating √2, I would instead multiple by 100 under the radical (10^2), then divide the result by 10. √200 ~= 14, so:
√200 = 14 + 4/(225-196) = 14.1379 >>> 1.41379 (by the other video's method)
√200 = 14 + 4/28 = 14.142857 >>> 1.4142857 (by this video's method)
So, they had the "numerical methods" approach that we still use in computers today. That is so cool.
I think it would have been helpful to mention, that this directly corresponds to the binomial formula:
√b = a+x => b = a² + 2ax + x² hence
x ≈ (b-a²)/2a neglecting the x² term as it's supposedly much smaller than the rest.
So it's not black math magic but something you're probably quite familiar. But maybe I'm weird and this would not have helped at all... ¯\_(ツ)_/¯
Now this is understandable---thankyou !
I tested a cubic root with no success out of curiosity before seeing the geometrical explanation in the end of the video and now I both love this method and also got curious if there is a universal method to calculate "x root of y" and I hope this not too much of a nightmare to find out.
Remember the dark middle ages? Back when nearly nobody could write or read. We barely have memories and records from the dark ages. Look at humanity now and you realize we are heading in the same direction. Digitalization leads to no recording and humanity is slowly getting stupider. And yet ancient pottery is still found, even completely intact at times.
WWW VID THIS HELPED ME SOO MUCH IM STUDYING FOR THE SIENCE ACADEMY TEST SO I NEEDED TO KNOW THIS THANK YOU!!!!!!!
I love your channel
I will need to watch this again, but find it fascinating.
So in simpler terms the formula (a+b)^2= a^2 + 2ab + b^2 is used and since we take 'a' as the whole number that when squared results in a number closest to the number for which we seek to find the square root of and as for 'b' we take that as the fractional term whose square is negligible hence 'b^2' can be omitted in the expansion of (a+b)^2 = r with r being the number for which we wish to find the square root of. Resulting in a^2 + 2ab ~ r
1:54 I am sorry but what marks that this is a decimal number, did they have any decimal seperator?
like it can't be coincidence that the tablet has the aproximation but without a seperator asystem seems to complicated to use.
Its a convoluted way of explaining that you average the divisor and the result which gives the new divisor repeatedly until the values converge.
I asked a geometry teacher i had once if she could show me how to do it by hand, she got mad and told me to use a calculator. Looking forward to this
Hmmm. The only way I ever extracted a square root was guess-and-check.
For something like 97, you start with the nearest square to it which is 81, so 9 is the first digit. 2nd digit, guess pretty big, because 97's close to 100, so 9.7^2 = etc. Higher? Lower? Slowly, methodically, build each digit. Basically, at each step, find the least upper bound on the digit that keeps it less than 100 in the product.
This method seems more efficient, but I don't quite understand it.