A tricky problem from Harvard University Interview
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- Опубликовано: 1 ноя 2024
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The fast, direct, straightforward method:
(√2 - 1)¹²
= [[ (√2 - 1)²*(√2 - 1) ]²]²
= [[ (3 - 2*√2)*(√2 - 1) ]²]²
= [[ 5*√2 - 7 ]²]²
= [ 99 - 70*√2 ]²
= 9801 + 9800 - 140*(100 - 1)*√2
= 19601 - 13860*√2
The slow, complicated, error-prone method:
Let x = √2 - 1 . We find:
(x + 1)² = 2
x² + 2*x + 1 = 2
x² = 1 - 2*x .
x¹²
= [[ (x²)*x ]²]²
= [[ (1 - 2*x)*x ]²]²
= [[ x - 2*x² ]²]²
= [[ 5*x - 2 ]²]²
= [ 25*x² - 20*x + 4 ]²
= [ 29 - 70*x ]²
= 841 + 4900*x² - 4060*x
= 5741 - 13860*x
= 19601 - 13860*√2
This is faster than what was done in the video. But both evaluate to the same value of about 0.0000255089026236...
At 15:00, you make the mistake of saying 0.000025508 is the exact value.
Im starting to relearn maths,.can you explain more in detail how the 12 power was fractioned in the second step
@@Bisinski Oahu worked the square out, then used that to work the cube out. (3- 2sqrt(2)) * (sqrt(2)-1)= 5sqrt2- 7= approx 0.07106
Then squared the cube to get the sixth power. Lastly squaring the sixth power to get the required twelfth power.
I recommend finding the calculator on your phone or computer and increasing skill with that as well as with the numeracy and algebra. Enjoy your persistence as well as the corrct result
I like ur first way better
Combine the first two steps using the cubic identity:
(a - b)³ = a³ - 3a²b + 3ab² - b³
So (√2 - 1)³
= (√2)³ - 3.(√2)² + 3.√2 - 1
= 2√2 - 6 + 3√2 - 1
= 5√2 - 7
Then just square twice and you’re done!
yep. all those simplifications when you substitute x² are not superior to the simplification that happens when you see a√2 × b√2. and this way you're directly keeping track of square->cube->6th power->12th power by only ever multiplying two sums of two terms ... instead of praying that you didn't make an error in the intermediate working. i kept waiting for the elegant trick and there was none. the fundamental error in this example was feeding our intuition that a straightforward approach would get out of hand with too many terms, when in fact the even powers of √2 keep being consolidated with the integers. all you need to know is that it *won't* look the same as multiplying (a+b) twelve times
Engineer's answer: √2 - 1 is almost 0.5. So, (√2 - 1)¹² is almost zero 😂.
would say though that 0.5 is 1/2 which is 2^-1, so the answer is approximately (2^-1)^12. We can calculate that using powers of 2 fairly easily, so that is also approximately (2^12)^-1. This might be more what an engineer would say, but I get the joke and surprisingly they are not wrong.
@kennethgee2004: But a good engineer wouldn't say that because 2⁻¹² is an order of magnitude off from (√2 - 1)¹² .
@@oahuhawaii2141
🤣😂🤣🤣🤣😂👍
Ответ любителя: (√2-1)¹² более красиво и кратко выглядит, чем 19601-13860√2. 😊
And it's true
(✓2 - 1)^2 = 3 - 2✓2
(✓2 - 1)^4 = (3 - 2✓2)^2 = 17 - 12✓2
(✓2 - 1)^8 = (17 - 12✓2)^2 = 577 - 408✓2
(✓2 - 1)^12 = (✓2 - 1)^8 x (✓2 - 1)^4 = (577 - 408✓2)(17 - 12✓2) = 577 x 17 + 408 x 24 - (577 x 12 + 408 x 17)✓2
This is more straightforward and likely faster.
This is exactly how I did it. Doing it this way is much much faster.
Good approach, and maybe we could more easily calculate 12th degree by cubing the fourth?
@@SergeyPricka It is easier to multiply the 8th power by the 4th power. 8+4=12. This is because one needs to calculate the 4th power anyway.
@cyruschang1904 ...
In the penultimate line you mistakenly wrote 480x24 instead of 408x24...
Exactly. This is turning a straight forward piece of arithmetic into click bait.
Lovely efficient solution. And super-clearly presented.
Beautiful, thank you. I am an engineer, I love math, and every time I watch you solving a problem I learn something new.
Many thanks!
binomial theorem exists
My first two thoughts were binomial theorem or just calculating (((sqrt(2) - 1)^2)^2)^2 * (sqrt(2) - 1)^2)^2
Yes, it does.
Does anyone else find these videos really tedious cos of the amount of unnecessary duplication? Why write out an entire complicated expression again merely to move a digit from one side of the equals sign to the other and change the sign or to subtract a "1" from both sides or why write out "(x^2)^2" in an entire duplicated line with the only difference being that it's then written as "x^2.x^2" and then a further duplication of the entire exact same expression merely to finally express it as "x^4"? Please have mercy on us, for goodness sake.
Double tap on the right side of the video to move forward by 5 second intervals, or press the right arrow. The steps are there if you need them...
@@todd8155 i run these math videos, (that are usually designed for someone who just learnt how to spell math) at twice the speed. 😄
Yes, but then I remember how many stupid people there are out there who don't understand what happened in a trivial step.
This is what non math people think what hard maths are.
He seems to think that it is higher mathematics, and that his "trick" is so good that he can bring the (essentially) same problem again and again.... (today, yesterday, 1 month ago, 7 months ago, 11 months ago).
so agreed
Hard math is.
@@aspenrebel I would love to see abstract algebra on this channel
@@xl000 pi r squared. No! Pie are round.
Ohh dear, you can easily reduce it to 17 -12√2 to the power of 3 and then use newton binomial series which will be way faster
The exponent of 12 factors as 2*2*3. That can be permuted in 3 ways. We can see that squaring big binomials is easier than cubing them. The former likely can be done in my head, whereas the latter likely needs scratch paper. I opt to cube first with small numbers and then square twice.
BTW, they're easier to compute when grouped in odd and even powers for the element with a square root:
(x - y)² = (x² + y²) - 2*x*y
(x - y)³ = x*(x² + 3*y²) - (3*x² + y²)*y
The easiest way:
(√2 - 1)¹² { 12 = 3*2*2 }
= (((√2 - 1)³)²)² { Reference the cubic formula … }
= ((5*√2 - 7)²)² { Square this in my head }
= (99 - 70*√2)² { Next square isn't as easy ... }
= ((100-1) - 70*√2)² { Make it manageable }
= (100-1)² + 70²*2 - 2*(100-1)*70*√2
= 19601 - 13860*√2
≈ 0.0000255089026236... { Calculator verified }
The harder way:
(√2 - 1)¹² { 12 = 2*3*2 }
= (((√2 - 1)²)³)² { Square this in my head }
= ((3 - 2*√2)³)² { Reference the cubic formula … }
= (99 - 70*√2)² { Same as above }
= ...
= 19601 - 13860*√2
The hardest way:
(√2 - 1)¹² { 12 = 2*2*3 }
= (((√2 - 1)²)²)³ { Square this in my head }
= ((3 - 2*√2)²)³ { Square this in my head }
= (17 - 12*√2)³ { Reference the cubic formula … }
= 17*(289 + 3*288) - (3*289 + 288)*12*√2 { Ugh! }
= 17*(4*288 + 1) - (4*288 + 3)*12*√2
= 17*1153 - 1155*12*√2
= 11530 + 8071 - 4620*3√2
= 19601 - 13860*√2
build pascal triangle up to power 6 => 1, 6, 15, 20, 15, 6, 1
multiply them accordsingly to powers of -sqrt(2): 8, -4*sqrt(2), 4, -2*sqrt(2), 2, -sqrt(2), 1 and add together
=> (8+15*4+15*2+1) - (6*4+20*2+6)*sqrt(2) =>99-70*strt(2), this is (sqrt(2)-1)^6
^2 of it -> (99*99+70*70*2) - 99*70*2*sqrt(2)
=> 10601 - 13860*sqrt(2)
Hi. This is fairly straightforward and able to be solved in a few minutes even without a calculator:
1. square the expression
2. square that expression
3. cube that expression (by squaring and multiplying the result by the original expression)
Cheers :)
That's the way I've done it too... But it does become cumbersome to calculate coefficients by hand :)
That's exactly the thought I had after two or three seconds. I jumped through the video just to see if there is a more simple solution. After the second or third stop, I was only laughing!
Let x = √2 - 1
x² = (√2 - 1)² --> x² = 3 - 2√2
x³ = x*x² = (√2 -1)*(3 -2√2) --> x³ = 5√2 -7
x⁶ = (x³)² … x⁶ = (5√2 -7)² = 50 -70√2 +49 --> x⁶ = 99 -70√2
x¹² = (x⁶)² = (99 -70√2)² = 99*99 -2*99*70√2 +4900*2
x¹² = 19601 -13860√2
SOLUTION :
x = √2 - 1
==> x² + 2x -1 = 0
==> x² = -(2x-1)
==> x⁴ = -(12x - 5)
==> x⁶ = -(70x - 29)
==> x¹² = -13860x + 5741
Applying
x¹² = -13860√2 +19601
Very, very excellent solution procedure.
I have no idea what you did there.
It seems that your answer would give an negative solution? Not possible in x¹².
You have a mistake in x^12, it should be
x^12 = (x^6)^2 = 4900x^2 - 2*29*70x + 841 = -9800x + 4900 - 4060x + 841 = -13860x + 5741 = -13860V2 + 19601
@@NickDos-r7f ... Edited, thanks!
There is a much simpler way by raising the sqrt(2)-1 to the 2nd, 2nd and 3rd power, since 2·2·3=12. this results in two times binomial (a-b)² and one (a-b)³. Done.
I'd rather start with 3, then twice 2. But the idea is the same.
(V2-1)^3 = (V2)^3 - 3(V2)^2 + 3V2 - 1 = 2V2 - 6 + 3V2 - 1 = 5V2 - 7
(V2-1)^6 = (5V2-7)^2 = 25(V2)^2 - 2*7*5V2 + 49 = 50 - 70V2 + 49 = -70V2 + 99
(V2-1)^12 = (-70V2+99)^2 = 4900*(V2)^2 - 2*99*70V2 + 99^2 = 9800 - 13860V2 + 9801 = -13860V2 + 19601
You may not even even use the cubic expansion 2.2.(2+1) = 12. (17-12*sqrt(2))^2*(17-12*sqrt(2))=19601-13860*sqrt(2)
@EnginAtik: Well, I can't cube (17 - 12*√2) in my head! It's much easier to square for x², multiply for x³ = x*x², and square twice for x⁶ and then x¹² . The coefficients will be smaller when the "cubing" is done early; squaring isn't as bad.
(((√2 - 1)³)²)² { cube it in my head }
= ((5*√2 - 7)²)² { square it in my head }
= (99 - 70*√2)² { must put down partial results }
= 9801 + 9800 - (100-1)*140*√2 { manageable }
= 19601 - 13860*√2
I was thinking the same just break the power into 2.2.3 and apply formula I got the same ans ❤👍
Wonderful guys. All have unique ideas. Enjoyed .
and hence in its turn we obtain a rational approximation for √2 ~ 19601/13860
Cool! Somebody tell Pythagoras.
Surely anyone interviewing at Harvard for Maths has knowledge of the binomial theorem? It’s a trivial solution.
Well, if you can do error-free work for (x + y)¹² , then go for it! Most folks will make an error or two in the process.
Calculate {0,1}.{{-1,2},{1,-1}}^12.
Diagonalizate the matrix.
The thing I appreciate the most about your videos is that you go step by step and you don't skip any steps. Thank you for that.
How many of us are here just to see how badly this guy fumbles an easy problem?
More then one.
With the matrix X = (-1, 2; 1, -1), representative for x = (-1 + sqrt(2)), just calculate X^12 (1, 0)^t = (19601, -13860)^t, representing x^12 = 19601 - 13860 sqrt(2).
Fun fact: 1/x = 1 / (sqrt(2) - 1) = sqrt(2) + 1. Thus (1/x)^12 = R^12 (1, 0)^t = (19601, 13860)^t with R = (1, 2; 1, 1)!
That means also x^12 = 1 / (19601 + 13860 sqrt(2)) = 1 / 39201.9999744910973763914... ~= 1 / 39202 (double 19601).
(Sqrt[2]-1)^12=19601 - 13860Sqrt[2]
I propose a general solution for every power n.
x = √2 - 1
=> x² + 2x -1 = 0 (or x² = -2x +1) (1)
=> x^(n+2) = (-2) * x^(n+1) + x^n, where n = 0,1,2,3,4 ......
Let x_n be equal to x^n, then it is easy to prove that x_n = a_(n-1)*x + a_(n-2), where sequence a_n satisfies a reccurrent equation a_(n+2) = (-2) * a_(n+1) + a_n.
From (1) we get a_1 = -2 and a_0 = 1.
So, a_10 = 5741 and a_11 = -13860.
Thus, x^12 = x_12 = a_11 * x + a_10 = -13860 * (√2 - 1) + 5741 = -13860√2 + 19601.
So, it is also easy to calculate a much higher power of x using this method in general case. The only thing needed is to calculate the corresponding members of the seaquence.
When 0.414 is raised to the power of 12, it is almost Zero.
What is he calculating,,?
Almost zero is not equal to zero. Precise answer is required.
I would use the binomial theorem (i.e. Pascal's triangle).
Nice!
Good to see, that Algebra has not changed by war!
Brilliant. This line of attack of a problem like this, I haven't seen before.
When I think of simplifying, I think of some answer which can be computed “faster” and “easier” than the original.
This “solution” seems to make the computation even more complex and “longer”
Even with calculator its easier to approximate the original, than the solution.
Not sure it’s correct to say that the expression was simplified.
I would ague that the original form was more simple than the result.
Just use Pascal's triandle to get the binomial coefficients for the sixth power say. Then it is straightforward to solve by squaring.
Better to cube, then square twice. And if the cube isn't easy, then square and then multiply.
It works and is faster
This is the shortest way. Sixth power, then squared.
This is too long,
(2^o.5 - 1)^2 = 3-2*2^o.5
For another square = 17-12*2^o.5
Using cubic formula= 17^3 + 3*17*12^2*2 - (3*17^2*12+12^3*2) 2^o.5
Alternatively [sin 5pi()/24 * cos pi()/12]^12
Good work and keep it up
Great approach
Direct multiplication (...)² -> (...)³ -> (...)⁶ -> (...)¹² will be simplest.
An exemplary lesson in how to lay out a solution. Neat, organized, detailed, clear - the way it 'sposed to be. Which, when done, is itself a great aid to thinking.
Although I’m not keen on his use of the phrase ‘cancels out’.
Just to level down the barbs here ...
1. None of the so called better methods (below) are really any more than microscopically better, if at all; bold claims.
2. Given the size and "irregularity" of those numbers, there is going to be a lot of ugly hand arithmetic no matter how you do it (this includes diagonalisation, solving recurrence relations, tricks etc.)
3. The approach given is quite novel (not entirely original I agree but not what most would do first) and in some similar problems may actually be a HUGE time saver so is a technique worth seeing; thanks to the author for sharing.
4. The discussion at the end is significant. (v2 - 1)^n -> 0 as n-> inf, so what those seemingly meaningless numbers are really telling us, are better and better rational approximations for v2. In this case (n=12) the ratio 19601/13860 is accurate to 9 sig figs ... pretty good!
5. continuing the last point, if you calculate the decimal expansion of 19601/13860 you obtain 1.414(213564)* where the * indicates recurring pattern. Now why those six digits and is it a coincidence that it is the first 6 positive integers?
So this guy has shared an idea and opened up a new vista for many of us wanting to explore and understand mathematics, and those being critical should be ashamed.
X is equals to root of 2 minus 1 whole power 12. As root of 2 plus 1 whole power12 is its conjugate. Taking one as x and other as 1 over x gives these two terms' sum as x plus 1 over x let this some be 't'. Now, we can easily calculalate x minus one over x by the relation x minus one over x equlas to whole square root of x plus one over x whole square minus four .Let it be '1 over t'.Now calculating sum of t and 1 over t gives us 2x which further gives x.After calculation answer is square root of two plus one raised to whole power 12.All the game is hidden in overseeing conjugates.
this leads to a fraction that approximates the square root of two.
This is what’s interesting about this result
x = √2 - 1
x^2 + 2x + 1 = 2
x^2 = 1 - 2x
x^4 = 1 - 4x + 4x^2 = 1 - 4x + 4(1 - 2x) = 5 - 12x
x^8 = 25 - 120x + 144(1 - 2x) = 169 - 408x
x^12 = (5 - 12x)(169 - 408x) = 845 - 4068x + 4896(1 - 2x) = 5741 - 13860x
= 5741 - 13860(√2 - 1)
= 19601 - 13860√2
19601/13860 is a good approximation to √2
x^2 = 3 - 2√2
x^4 = 17 - 12√2
x^8 = 577 - 408√2
x^12 = (17 - 12√2)(577 - 408√2) = 19601 - 13860√2
Repeated squarings are faster.
X^2=3-2root(2)
X^4=17-12root(2)
X^8=577-408root(2)
Now foiling the last two gives the answer.
But your last step is going to be hard, which is why you left it out. It's better to cube first, then square twice. And if the cube isn't easy, do the square and then multiply.
(√2 - 1)¹²
= (((√2 - 1)³)²)² { cube it in my head }
= ((5*√2 - 7)²)² { square it in my head }
= (99 - 70*√2)² { must put down partial results }
= 9801 + 9800 - (100-1)*140*√2 { manageable }
= 19601 - 13860*√2
lf looking for an approximate value you can do with much less arithmetic and numerical calculation.
let a=sqrt(2) - 1 and b=-sqrt(2)-1
we have ab=-1 and a+b=-2
(a+b)^2=a^2+b^2+2ab. =>. a^2+b^2=6
(a^2+b^2)^2=a^4+b^4+2a^2b^2. => a^4+b^4=36-2=34
(a^4+b^4)^3=a^12+b^12+3a^4b^4(a^4+b^4) => a^12+b^12 = 34^3-3*34 = 34(34^2-3) ~ 39000
note that a
Actually, if you plug the original expression into your calculator and then plug in the derived formula you will not get the same answer because of rounding error. There is no such thing as an exact answer on a calculator when the expression involves an irrational number (and even most rational numbers have no exact representation on a calculator). So each method of calculating the answer when plugged into a calculator only comes up with an approximate answer. In this case the difference is significant if your calculator only has about 9 or 10 digits of precision.
So which answer is closer to the truth, and why?
If instead we let x=1/(sqrt(2)-1)=sqrt(2)+1 and use the method demonstrated in this video we get the formula:
x^12=19601+13860sqrt(2)
(sqrt(2)-1)^12=(1/x)^12=1/(19601+13860sqrt(2))
If you plug this formula into your calculator the answer should be a much better approximation than the formula derived in the video. Can you see why?
(19601 - 13860*√2) vs (19601 + 13860*√2)⁻¹
My calculator returns the same result, since it has great precision. However, it's always good to be aware of finite precision in calculations and alter the computation to avoid problems, such as subtracting 2 numbers that are very close to each other. That's why my old HP has the [eˣ - 1] function, which is targeted for x near 0.
@@oahuhawaii2141 Possibly your calculator is doing 64 bit arithmetic but only displaying 9 or 10 digits. In that case the difference will be in the digits that aren't displayed. If you subtract one result from the other you will probably get a non-zero answer. The size of the difference should give you a clue as to how many digits of precision your calculator has.
Can’t you just use binomial theorem or even pascal’s triangle for this question?
How?
Yes
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice:
1:3:3:1
1:2:1
1:2:1
Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice:
1:3:3:1
1:2:1
1:2:1
Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
Yes, you can get the coefficients and powers for (x + y)¹² . But it's much easier to apply (x + y)³ first, then use (x' + y')² twice:
1:3:3:1
1:2:1
1:2:1
Cubing last is a big mess, so doing it early makes the numbers manageable. That is, I'd rather do (((√2-1)³)²)² instead of (17-12*√2)³ .
Assalomu alaykum. Rahmat sizga qiziqarli matematika uchun. Salomat bo‘ling
This is insane
Because you can say that
(SR(2) -1)^n = a-b.SR(2)
Where a and b are Naturals
And when n -> +infinite
The left member limit is 0
So there is a couple (a,b) where
0 = a -b.SR(2)
SR(2) = a/b
So SR(2) is rational :-)
There is a really simple way to calculate coefficients of the binomial theorem as you go without resorting to drawing Pascal's triangle.
Yes, demonstrate this for the 12th power:
(x + y)¹² = ???
@@oahuhawaii2141 Sure, I'd be happy to explain! Bear with me, since RUclips comments aren't the best for math.
1. *Binomial expansions are symmetric*, so we only need to figure out half the coefficients.
2. *Exponents of x decrease by 1* each term, while *exponents of y increase by 1*.
3. The *first coefficient is always 1*.
Let's start with x^12.
The next term will have the form Cx^11y.
- Multiply the coefficient and exponent of x in the previous term: 1 * 12 = 12.
- Divide by the exponent of y in the current term: 12 / 1 = 12.
So the second term is 12x^11y.
For the third term Cx^10y^2:
- Multiply the previous coefficient by the exponent of x: 12 * 11 = 132.
- Divide by the exponent of y: 132 / 2 = 66.
Now you have 1x^12 + 12x^11y + 66x^10y^2.
Keep going:
- 66 * 10 / 3 = 220
- 220 * 9 / 4 = 495
- Then 792 and 924.
At this point, you have:
x^12 + 12x^11y + 66x^10y^2 + 220x^9y^3 + 495x^8y^4 + 792x^7y^5 + 924x^6y^6.
From here, the coefficients mirror, so the final expansion is:
x^12 + 12x^11y + 66x^10y^2 + 220x^9y^3 + 495x^8y^4 + 792x^7y^5 + 924x^6y^6 + 792x^5y^7 + 495x^4y^8 + 220x^3y^9 + 66x^2y^10 + 12xy^11 + y^12.
I just use Pascal's triangle which leads to a one liner: (sqrt(2))¹²+1¹²+12*(sqrt(2)¹¹+1¹¹)+... this will leads to the next line with terms of Integer*sqrt(2) and Integer*sqrt(2)^(even number).. 😂
just calculate it straight up. Yes when you know the answer and can notice all the correct simplifications(copying from someone else work) it becomes slightly shorter, but...
Are calculators not allowed? According to mine (T^exas Instruiments TI-36 SOLAR) - about 30m years old ...
sqrt(2) = 1.414213562
sqrt\(2) - 1 = 0.414213562
0.414213562^12 = 0.000025508 ... which is probably close enough for most practical purposes.
Picking a randomly chosen way out of a myriad of possibilities to reorganize the expression with no clear plan up front where it might lead to. Below are several clear cut ways shown leading straight to the solution.
Just keep on squaring terms.Just as easy and intuitive.
Intéressant. Cela donne une bonne méthode pour trouver des valeurs approchées de √2 à l'aide de fractions 👍
Thank you guy
1 = sqrt(1) so sqrt(2) - sqrt(1) = sqrt(1) = 1 🤷🏻♂️
This video is painful. While factual, the presenter did not explain how, where, and why he was going. Waste of precious time in life.
Perhaps no one can make it more bigger than as it is solve🤔🤔🤔.
This type of solution done by self reduce anxiety and stress !!!
Taking the longer route I see. When you had x^2 could have substituted that into x^12 as 12=2*6 and have that expression to the power of 6. 6=2*3 so you just had to power that to 2 quite simply and then 3.
Давно на пенсии, но так полезно пошевелить мозгами. Я наслаждаюсь вашими уроками. Спасибо. ❤
Much simpler:
(√2-1)^12 = x
multiply both sides by (√2+1)^12
(2-1)^12 = 1 = x * (√2+1)^12
hence:
x = 1 / (√2+1)^12
I had to check- 19,601 and 13,860 have no common factors.
But the ratio is close to √2 .
Convert to polar and use Demoivres theorem. Much easier.
С таким же решением я предложу решить квадрат бесконечности...
Квадрат бесконечности это бесконечность
hmm but this is also a power rule situation is it not? I mean (sqrt(2)-1)^12= (((sqrt(2)-1)^2)^2))^3. In this look we can easily square the expression and reach a^2-2ab+b^2, which is 2 - 2*sqrt(2) - 1. Simplify to 1-2*sqrt(2). We square that again and then cube it. hungry to having issues following the formula, but i hope you get the gist of it. and then that should be a much simpler form.
where you wrote "which is 2 - 2*sqrt(2) - 1" it should end in "+1" not "-1"
@@linsqopiring6816 thank you. i was hungry at the time posting so was messing up the math badly. The idea though was that you could use power rules evaluate this without getting crazily big numbers.
@@kennethgee2004 No problem, and yea I think your way is less work. But I'm glad to see the approach taken in the video because it's an interesting system and it's good to know different ways to do something.
It's much easier to square for x², multiply for x³ = x*x², and square twice for x⁶ and then x¹² . The coefficients will be smaller when the "cubing" is done early; squaring isn't as bad.
(((√2 - 1)³)²)² { cube it in my head }
= ((5*√2 - 7)²)² { square it in my head }
= (99 - 70*√2)² { must put down partial results }
= 9801 + 9800 - (100-1)*140*√2 { manageable }
= 19601 - 13860*√2
(2^1 ➖ 1)3^4,(2^1 ➖ 1)^3^2^2 (1^1 ➖1)^1^1^2 1^2 (x ➖ 2x+1). Avery engaging maths problems .
To present the 19600.0000745 at the end, you must have either looked up or used a calculator to multiply 13860 times the √2. If you are going to use a calculator for the √2, why not just subtract 1 from √2 and then take that to the 12th power.
he was exploring at that point rather than showing how to answer the original qu sans calculator. That's fair enough. How often have I done a really tricky definite integral to find a closed form solution, then to enthusiastically reach for my calculator to see if I am right!!!!
Oh my god. This is probably the only YT math I knew how to solve xD
Interesting is the relationship of this series of numbers to diophantine equations x² = 2y² + 1 and x² = 2y² - 1
I was going to do a calculator check, but you beat me to it! What a great question and a cool way of solving it.
The term in parentheses is the solution to a quadratic equation. That implies the square of x is a constant times x plus another constant. But x^3 is the same first constant squared plus the constant term times x. With convenient expressions for x^2 and x^3 in only linear and constant terms, a 3 step iterative process gives the solution. I'd not think this is beyond the skill of A student in an introductory algebra course. Of course the binomial theorem would do it as well. I'm skeptical this comes from a Harvard interview.
Cube first, then square twice.
Do you think that is easier than just calculate it?
(2^.5-1)²=3 - 2V2
(2^.5-1)⁴=17 - 12V2
(2^.5-1)⁸=577 - 408V2
(2^.5-1)**12=9809 - (17x408+12x577)V2 + 12x408x2=19601 - 13860V2
Egg zackly.
I agree with you but you forgot to calculate (add) 17x577 in the last line ;-)
Congratulations, your application for Harvard has been approved!
@@linsqopiring6816 Thank you. I am 76 , so i will go right away.
@@mastnejbucek3411 17x577=9809 and adding 12x408x2 makes 19601
I understand why Korean student have advantages. they train those operations in quick.
very interesting trick
It is tending to Zero.
0.414 raised to power 12 is nearly Zero
square root of 2 = 1.4 - 1 = 0.4. take that to power 12, it is close to zero.
Since the answer is pretty straightforward, wouldn't it had been better to simply calculate the result..?
I liked the approach very much.
He gets to the heart of the problem and doesn't just want to solve it quickly.
I did not used substitution for this test. Instead i did this (sqrt(2)-1)^12=((sqrt(2)-1)²(sqrt(2)-1)²)^3 after developing this expression i found the right solution.
It's easier to cube first, then square twice:
(((√2 - 1)³)²)²
If you don't remember how to cube, just square and then multiply:
(((√2 - 1)²*(√2 - 1))²)²
This is less tedious and less error-prone.
Выражение в скобках меньше 1. Приблизительно 0,41. А в 12 степени вообще мизер. К чему вся эта писанина с неверным ответом?
From using binomial theorem we can easily solve this question and the answer becomes from binomial theorem is (-12√2)
It’s in my head.
It's easier to use the Binomial Combination Expansion
I did not take much time to calculate as x^6=(×^2×x^1)^2
X^12= (×^6)^2
Except for the last step,Idid not have to handle large nos
Yes, cube first, then square twice.
And that folks is how I met your mother. Any questions? 😎
One should use trigonometry and remember PI is useful
It is faster to multiply directly……
Why not use the binomial expansion? Seems like that would have been just about as quick.
I think that calculating ^2 / {(√2-1)^2}2 is faster than using the X.
I noticed some peoples are using V or sqrt instead of √ . (^-^)
This solution is probably as cumbersome as evaluating the expression using the binomial expansion.
yes but it's marginally better and it is quite a nice way to go about it
There is NO specific test for Harvard or any other school!!Its the SAT or ACT!!
What is the thing that you are continiously saying as "it is really great"?
Use Binomial Theorm
The method described has some interest, I suppose. But to get the answer, we have to rely on a calculator to find a very small difference between 2 large numbers. If a calculator were available anyway, we could have used it straight forwardly without doing all this manipulation. It is not clear at all what insight has been gained.
Why should that equation be a problem?
I just take out my calculator and put in the numbers for the result.🤔
Wouldn't it be faster to draw pascal triangle to 12th layer and solve this without x?
Talk about doing difficult for no apparent reason...
First do the problem in the parentheses. Calculator is easy. Few steps and done. With an actual number as outcome.
Not how I did it in my head. Use conjugate multiplication. X(srt 2+1)^12=2-1=1. Then x=1/((srt 2+1)^12).
(A+b)(a-b)= a^2-b^2.
This is not higher mathematics, but school's Olympiad mathematics
12=3*2*2
Можно сначала возвести в куб, а потом два раза в квадрат
И это не 18 минут как на видео, а от силы 5 минут
Binomial Theorem Left The Chat 💀💀
Quicker to calculate the square, then the 4th power then the 12th power. By the time you are half on your theory, I have already got the answer.
Your method of square, square, cube isn't as easy as cube, square, square. Try them out.
@@oahuhawaii2141 not much different in terms of complexity if you know binomial expansion.
WRONG....0 is the answer
Factor out the sqrt of 2 and it leaves you with 1-1 in the parenthesis which is equal to 0....0 times sqrt of 2 is 0.....so 0 is the answer
It's a lot simpler to square -> square -> cube. It took me almost 5 min.
(√2-1)¹² = ((√2-1)²)⁶ = ((3-2√2)²)³ = (17-12√2)³ =
= 17³ - 3 * 17² * 12√2 + 3 * 17 * (12√2)² - (12√2)³ =
= 289*17 - 3*289*12√2 + 3*288*17 - 288*12√2 =
= (289 + 864)*17 - (867 + 288)*12√2 =
= 1153*17 - 1155*12√2 =
= 19601 - 2310*6√2 =
= 19601 - 13860√2