How to find a square root
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- Опубликовано: 15 мар 2018
- Here is the original π video:
• Calculating π by hand:...
Watch the long version of the extra working out here:
• Calculating π by hand:...
My calculation gave the square root of 10,005 to be 100.02499687578103
Compared to the true value of 100.0249968757810059447921878763577780015950243686963146571...
CORRECTIONS
- None yet. Let me know if you spot anything!
Thanks to my Patreon supporters who enable me to spend a day doing a lot of maths by hand. Here is a random subset:
David McCarthy JR
Derek Chandler
Mauro Cioni
Kevin Petrychyn
James Tanner
Support my channel and I can make more videos:
/ standupmaths
Music by Howard Carter
Filming and editing by Trunkman Productions
Design by Simon Wright
MATT PARKER: Stand-up Mathematician
Website: standupmaths.com/
Maths book: makeanddo4D.com/
Nerdy maths toys: mathsgear.co.uk/ 
Развлечения
Tripping over a root while discussing roots... priceless 0:26
"I was able to find the root"... while finding root on the ground xD
And talking about squares in a square.
I can't believe I missed that. :O Amazing xD
Yea, what are the chances?
... That he didn't measure one take, estimate how fast he walks, and then trip over the root at the punch line? :D
Apparently the best way to find the root is to trip over it.
The dedication to do an opening bit in the woods just so you could trip on a root at the right time is an admirable commitment to comedy.
Start = root
End = square
You have a great sense of humour Mr Parker.
Start=(Matt) Parker
End=square
Nice observation !
What root? There is a square but there is no root
@@farciarzfunny5326 in start of the video, he trips on a root
0:27 goes to the middle of the forest just ro make a root pun. 0.0
Ahh, a Parker Square Root
🔥🔥🔥🔥
Actually, I was gonna comment that on the original video but somehow forgot. So thanks. :)
Parker root
For goodness sake
Oh so it’s just what I watched 2 days ago but again. *watches again anyways*
We're just suckers for math.
Because it's still astonishing
I did that too
I had a Dynamics professor, Dr. Penrod, at Auburn Univ. who could work out roots and powers simultaneously (e.g. 2.68^2/3) and give an answer to about 3 or 4 decimal places. It was absolutely amazing.
The root joke was levels of comedic genius never before witnessed by the human race
Did you stumble (0:27 s) over the root on purpose? Maybe because you end the video on the city square. ;)
lol
Yeah, the square answers the question.
Definitly on purpose. The real question is: How many takes to nail it that well? :-)
Yes, it look very Chaplin profesional!
I'm sure he did. Matt always goes the extra (Square) Mile.
0:27 - Looks like you found a different type of root :)
I wonder how many takes were needed to trip at exactly the right time. :-)
He tripped over a root when talking about square roots
in high school i once overestimated a really simple square root on a test, and couldn't figure it out. so to make sure i never got stuck on any root again, i learned an iterative method by heart which only uses simple operators (so i could do it on paper, albeit very slowly). naturally, we never got any square roots on tests after that
mfwplayinggames My school was weird. Before we got to algebra, we learned the iterative method for square roots. I'd have to solve 6 a day, first with four digit number, then with six digit ones. Though i eventually caught on that they were all exact roots, so that made it faster.
When I find a square root, I sketch a big parabola and then estimate distances using an elaborate system of paperclips and hourglasses.
Keep it up Matt. Always excited when I see that you uploaded a new video.
00:26
“Find the square ROOT (trips over tree root, notices that he accidentally made a Parker pun, and smiles)
I predict that the frequency at which Matt's hair is going to change will not be random but rather contain some sort of coded message.
Volbla He did say "seemingly", I find that suspicious.
Hair = 1, bald = 0, and we've got a binary code using follicles.
0:27 when he was saying “root” and literally tripped over the root.
thanks for the explanation I've often wondered, and had studens asking me, this seems like a good way to explain what is going on with root squares.
Pi-head was the best one. thx for brightening up my day with the reflection of the sun on your perfectly round head. ^^ Big thumbs up
I wouldn't mind if my hair made an hilarious reappearance.
n ** 0.5
IKR
Beethovenethylenenium 64 whoosh
"Find the ROOT-" trips on actual tree root. Love it.
I looked and immediately wondered whether what you did or finding the solutions to x^2+2xy+y2=10005 is easier. After attempting the latter it's still a bit of guess and check but I was able to place some restraints on the possible value of y that were helpful in increasing the accuracy by any amount desired.
The method I learned to do square roots looks a lot like long division but it is done two digits at a time.
For the summer, I'd gotten a pretty bleak hair cut as well (if that's the way to use the word 'bleak' )
But after watching your video, I am pondering over the option of going full blown shiny head
Really like this, wanted to know how to do this for a long time!
Newton's method for √N:
Next term = xₙ/2 + N/2xₙ
The first one has a specific name :Heron's method after a Greek mathematician
My method:68x(39484)[55]^12+13
@@INT41O
Babylonian Method For Square-Root(s).
The way I was shown to do square roots by hand is using this formula (x+(r/x))/2 where r is the number you want the square root of and x is a number close to √r. If you plug the answer you get back into the formula as x, the result you get gets closer and closer to √r.
ex. (2+(5/2))/2=(9/4)=2.25
((9/4)+(5/(9/4)))/2=161/72≈2.2361111111...
((161/72)+(5/(161/72)))/2=51841/23184≈2.2360679779...
which is a good approximation for √5, what I was trying to find: 2.2360679775...
There is a sort of long division method to find square roots that works in binary very easily:
1. write the number in digit pairs left and right of the point, adding a zero before the first digit if odd number of digits before point, appending a zero after the ladt digit idmf an odd number of digits after the point. (extra pairs of zeros can be added after the nunber after rhe point.)
2. Do a long division taking the pairs of digits of the nunber (dividend) as a unit and creating a new divisor at each step:
2.1. double the answer so far and multiply by 10 (the base).
2.2.. Now find a digit to put in the ones place to create a divisor that when multiplied by this digit is the largest result that does not exceed the "working" digits (of the division)
2.3. Put this digit over the pair of digits that form the lat pair of digits of the "working" number.
2.4. Multiply the divisor by this digit and subtract (like long division)
2.5. Bring down next pair of digits to continue the division.
3. Continue the "division" until all pairs of digits have been used and result of last subtraction is zero, or until enough precision has been reached.
In binary the double is a shift left 1 bit, and multiply by 10 (the base - in decimal this is also 2) shift left anothe bit; only possible values for the units bit are 1 and 0, so try 1 - put a 1 in the units place and if divisor is less than "working" number can subtract and a 1 goes in the answer; otherwise a 0 goes in answer and next 2 bits of dividend are appended to the end of the working number. - highly efficient for a digital computer.
For 10005 the division is:
10005 -> 01 00 05 . 00 00 00 00
Put a decimal point above the point in the "dividend"
first pair is 01; answer do far is 0 -> 0_ so need units digit 1; 1 in answer above 01, 01×1=01 subtract to get 0 and bring down next pair 00 to make the "working" number 0 00
As this is zero can put a 0 above the 00 and bring down the next pair 05 to make the working number 0 00 05
Double answer and multiply by 10 gives 20_ which is greater than 5 so put 0 above 05. Bring down next pair 00 to make working number [0 00] 05 00
Double etc answer -> 200_ again 0 in answer; bring down next pair for 05 00 00
Double etc -> 2000_ this time can put a 2 in the units place -> 20002, multiply by that 2 to get 40004 and subtract to get 99 96, bring down 00
Double etc -> 20 00 4_ this time a 4 to give 20 00 44 × 4 = 80 01 76 to subtract = 19 94 24; bring down 00
Double etc -> 2 00 04 8_ units = 9 to give 2 00 04 89 x 9 = 18 00 44 01 to subtract = 1 93 79 99
and repeat as required.
(The "division" is easier to see (and do) when written out as a long division with the new divisor written down each time along side each "working" number, extending the vertical line of the divisiom "bus stop shelter" down the page.)
4k maths video, just what I needed XD
"find the root" he says as he trips over a root. Great pun. A+
There is an issue in your correction at 2:55 . When you subtract the area of x-square. You first cut off 100+x times y. With this cut you leaf not enough area for the second cut. So, your new estimate is also overvalued and it’s overvalued by y square, because you can’t cut this square of. In the next step you need to subtract the new estimate by y square. And so on, and so on.
Your method reminds me of using log tables (not logs for square roots before anyone says anything, but my log table book which had all the lists of numbers to work out things like cos, sin, sq rt, sq, logs, nat logs etc etc etc). I still have my book - used it for my O levels. My little sister (she's 50 soon lol) was in the first year to be allowed to use calculators. My year and the one between us were absolutely horrified that they'd be _given_ the answers! Where was the working out?! Ah, time flies, eh?
Btw, hubby's got the same haircut. Had it for years. Much neater than a comb over lol!
so ez yet so helpfull i love it thank you!
"How I was able to find the root..." *Trips on a root*
If you do it geometrically and algebraically with the binomial theorem I think it is quite straight forward. The advantage is that you can see whether the number is a perfect square or not. But at the same time you have to do all the sub-calculations increasing with the growing answer 😅 just giving you a single digit every time. Therefor Herons method remains my favorite.
I like the new doo too. Matt's eyebrows now take center stage and rule the school.
Subtle root placement; I appreciate that cheeky pun delivered through physical prop comedy. Thank you for that.
Your sense of humor is amazing. Heisenberg haircut looks badass.
0:24 "there's a whole section on how I was able to find the roo...*trips on root*...t"
“How I was able to find the root”
Trips on a tree root.
At the end you were filming in Trafalgar SQUARE. Haha nice!
Hey man great video and you look way better short/bald like that!
Nice pun with that root
Yay, he's back! Math dad.
You could use Newton's Method to approximate the root of the polynomial x^2-n. Successive iterations of the method would yield better and better approximations of root n.
3 years later BUT what he's doing is actually the exact same as Newton's method. He's using the derivative of x^2 which are the two rectangles (2x)
@@TobyBW I realised the same thing. But correctly Newton's method is stated as recursive function and Matt never wrote that down.
It is basically just the taylor series expansion
(1+x)^m=1+mx+m(m-1)/2 x^2...
For the special case of m=1/2, x=0.0005 and a scaling factor of 100
(10005)^(1/2)=100* (1+.0005)^(1/2)
But this is a nice intuitive geometric illustration! Great job!
hyh2012 exactly what I was thinking!
That ROOT totally made my day
I found a video not so long ago telling how to algebraicly (spelling no?) calculate the square root, digit by digit, without approximation.
1. Start off by having your number, in this case 10005.
2. Now write down the closest square root integer that's closest to your number without going over. In this case 100.
3. Next up, write down 100 at the start of the result line, and add a period. While also subtracting the square (10000) from 10005 and remembering that. In this case it's 5.
4. Next up, double the result, but without the decimal place, and remember the integer, in this case 200.
5. Now if you have decimal places of your starting number, bring down 2 of them and place them after your subtracted number. In this case we have 500.
6. We now do the same as we did on #2, but we take the closest integer x without going over where we can multiply 200_ * _ where the _ gets replaced by the digit x. In this case we can't so it's 0.
7. Once again we add 0 to the result, and we subtract from the number above so we still have 500.
8. Just like at #4, we double the result without the decimal place, and remember it, in this case 2000.
9. We now bring down 2 additional decimal places just like #5 and we end up with 50000.
10. Now repeat from #6 over and over and we will end up calculating digit by digit what the exact square root of the number is.
To show an example, I will continue doing a few of the ones following.
We need 2000_ * _ < 50000. In this case it's 2, since 20002 * 2 < 50000. The result is 40004.
We subtract 40004 from 50000 and get 9996 while we also add 2 to the result.
We turn 100.02 from the result into 10002 and then 20004, and we also carry down 2 decimals to have 999600.
We do 20004_ * _ < 999600 and we get that it's 4 since
200044 * 4 = 800176 which is just as far as we can go.
While the trade off of using this method is that the numbers used in the formula gets insanely huge quickly, it does still mean that if you're able to perform them, the answer will always be accurate, compared to the more used techniques which indeed is easier but they're not exact.
Credits to RUclips user Davidson1956 who made a video on this 10 years ago.
I appreciate the pun. You're at Piccadilly circus, which is actually a square and you speak about interest in finding a square, while being at circus that's actually a square.
That's a good one
I remember when I ran across the technique of finding a square root by hand. I was fascinated. I didn't understand why it worked (and still don't, I'm a humanities guy), but it was fascinating.
Damn, I thought the video will be about botanical research...
Nailing that grothendieck look!
This is awesome
3:25 Aside from the whole nicely approximating errors thing, it's nice to know that I'm not the only person who does the weird hovering raptor arm thing.
I remember i used the same method to square numbers 10-20 with a simmilar method. I was like 7and somehow found this out.
You could also do a taylor expansion from the closest perfect square!
4:12
"Another
overlycomplicated
VDt5Dvision....."
Masterpiece
"...a whole section about how I was able to find the root..." Right as you trip over a root. I'm not sure if that was intentional or accidentally perfect. Either way, I loved it.
Very interesting!!!!
Cue-ball look is a good look.
Haircut's looking great.
Back in the day when desktop PCs executed 16-bit software at about 20MHz, without hardware floating point, where integer multiplication was slow and division was ridiculously slow, I implemented efficient code to do square roots.
I used Newton's Method of approximations. If you have not covered that, you might consider it, especially with a focus on using only simple operations (add or subtract, multiply or divide by 2, bitwise logical operations like AND and OR, compare).
In my application (Lambert shading) the value would not change much from one pixel to the next, so the last result is used as an approximation to start with. For successive pixels being rasterized, it could adjust it with a single iteration of Newton's method, and skip the overhead of the looping.
It’s a bit of a Parker sphere, though, tbh. (This video is going into my pocket as a demo of successive approximation techniques. Thanks for making it! In my historical math elective in undergrad, I did a small survey essay on ancient square root methods, which gave me a solid appreciation for this sort of thing. If you haven’t read The Crest of the Peacock, you really should!)
Very nice graphical method and a beautiful illustration of what (I think???) amounts to a Taylor expansion without resorting to calculus! Another nice method might be the Newton-Raphson method but again, needs calculus.
this is Newton's method (which happens to coincide with solving it via Taylor expansion but only using first order approximation)
I saw a video of different way of approximating sqrt using the function y=x^2-c. Sqrt (c) is a solution. You start with a guess. And you calculate the slope at your guess and make a line from your guess with that slope. Find the 0 of that line. This is your new guess. This method doubles the number of digits at every step. Seems like your method also doubles the digits at every step.
Here I was beginning to think I was weird for using an estimation algorithm I learned a long time ago from a website about bit manipulation in C-like programming languages. I usually carry out the estimation in my head. Turns out it's actually very similar to this estimation technique you demonstrate here. Just adjusted for binary since I work just as well in binary as I do in decimal and conversion between them is simple enough. That and it's how I first learned of the algorithm before learning what it was approximating.
/r/iamverysmart
For next pi day, can we get a video of you manually approximating the pi root of pi?
I do like the "coincidence" of saying "finding the root" and almost stumbling on a root!
Calculating a square root is a fairly simple process. It is similar to long division except you pair the dividend and do some additional calculations on each subsequent subtrahend. I learned it in primary school back when calculators were the size of buildings. We were still faster.
Another iterative technique that requires less thinking to do right is to take whatever estimate x that you have, and calculate (x+10005/x)/2. For a first guess of 100, it gives 100.025, and starting at 100.025 it gives too many correct digits for my phone calculator to tell the difference. (You still have to divide by 100.025, so calculation wise it may not be easier, but there are fewer things to get wrong). The idea is that what you're really after is the geometric mean of x and 10005/x, so you use the arithmetic mean instead, which is a little bit bigger. But with a good starting point, x and 10005/x will be quite close, and then the arithmetic mean and geometric mean are really close to one another.
Could we do a little binomial expansion? Factorials and fancy brackets would look good on a whiteboard.
I love how he tripped on a root while filming this
0:26 "I was able to find the *ROOT*" as he trips on a root lol
Nice touch ending in a square.
Why don't you cover the method for determining an integer root by hand some time? The one where you break it up into groups of digits that match the base (2 for square, 3 for cube, 4, 5, etc.) and work it sort of like long division but with what gets subtracted being determined with the Pascal 's Triangle numbers.
0:28 "I was able to find the root" *trips over root*
0:27 This root almost got you ! ;)
“Find the root”
Promptly trips over a root.
Wouldn't a Taylor expansion of 100*sqrt(1+x) be easier to compute by hand?
The expansion is wrote down rather easily and all divisions are pretty simple in comparison with yours (because you insert x=5/10000 into it and divide only by integers).
It might converge a bit slower though.
Ahahahaha he tripped over a *root* at the start of the video and ended the video in a *square*. Very clever
The Aussie crowd is going "Is this really how a square gets a root in London? Is it the numbers demo or just the haircut?"
"...how I was able to find the roOT!" Falls over root.
Liking the new haircut, Matt! :) Suits you
i just love how he tripped over a root as he was saying square root xD
"Of how I was able to find the-"
*Trips over root*
"Root"
0:26 I see what you did there clever cookie.
I'm amazed that using "y" you were able to get that much more precision than you did using only "x".
Shaven head looks good!
Don't assume your head is a sphere.
My favorite algorithm is a recursive one. It involves arrangement of the equation into the form f(x) = x, then plugging the estimate for x into f(x) and solving for a new estimate of x. From your intitial estimate of .025 I was able to come up with .024996875 relatively quickly compared to the amount of work you put in. However, the next interaction didn't buy me much for a very long hand multiplication. I suspect another iteration would be required to make the work pay off.
What's the algorithm name?
@@zfolwick It sounds like the successive approximations method, however I haven't found a way to approximate a square root using it.
@@zfolwick newton raphson. It really makes sense if you think about it, even algebraically.
To find the square root of c, use the recursive formula:
x_(n+1) = 1/2(x_n + c/x_n)
This works since x/sqrt(x) = sqrt(x). When you find the average, it’ll approximately double the accuracy. There are geometric proofs involving some basic calculus and analytical geometry online too.
"... to find the root.." *almost trips on root*
While this explanation works well for a square, I think that the general approach of Newton's method. In this particular case, you're doing exactly the same computations, but the use of newton's method is more versatile when a geometric interpretation is not so clear. In fact, for working out by hand, this newton's method approach gives you fewer digits in your division problem ( 625 vs 3125 and 20005 vs 100025).
y(x):=x^2- A
"route", with a "root", to a "square". masterful
It was your approximation of sqrt(10 005) that limited the precision of your approximation of pi.
Using the value you got here for the Chudnovski algorithm first approximation would have given you about the same precission as you got for the second approximation - with a lot less work.
Awesome.
Matt should try calculating pi using his head one day
Hm, I tried playing along with a random number (92737 for anyone curious) and I got a pretty good approximation using the method in the first video.
Then I reverted to your number (10005) to try and follow along with your calculations and also go further (not just removing y once, but doing it again) but I cheated and used wolframalpha because why not.
However, I must've done something wrong, because not only did I get a different result than you presented at the end of your video after the first removal of y, but after removing y again, I ended up with a length less than the actual root, and further away from the root than I was beforehand.
Is this the kind of algorithm that approximates closer and closer for each y-interaction and I just did it wrong, or can it fluctuate between closer and further away from the real truth depending on how many times you apply the 'calculate y and remove/add it'-method?
for this specific case (calculating the square root) the result should always be getting closer
moar square root puns! :)
There is a simpler method which involves no estimating or rounding. Search "afjarvis square root" and there's a pdf file. It can even be done mentally, since there are only two variables to keep track of which change with every step of the algorithm.