The Hardest Exam Question | Only 6% of students solved it correctly
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- Опубликовано: 6 окт 2024
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Quicker method: since x^10 is (x^5)², and x^5 is (x²)(x³), (√2-1)^10 is the square of (√2-1)² that is 3-2√2 times (√2-1)³ that is 5√2-7, which is 29√2-41, and its square is 3363-2378√2.
Same method got in 3 min dude
what I did too...
Much saner approach.
Don't use the times sign when doing algebra, especially among x-variables. Use grouping symbols such as parentheses: (x^2)(x^3)
(√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)^10 = (√2 - 1)²(√2 - 1)^8 = (3 - 2√2)(577 - 408√2) = 3363 - 2378√2
This is so much easier indeed.
That's how I did it. Didn't take 17 minutes, and I didn't use a calculator (except to check it at the end)
Simple...how to make a simple problem complex foolishly and create a video is what it's all about.
Since 10 is pretty small(small enough to calculate) you could do it with binomial theorem
That's how I did it! I got the answer, but it took some time. I think the method shown in the video is faster.
Yeah man , i dont understand why they make simpler problems look like its super tough. Here wlso 2nd term of binomial is 1, so the expansion becomes more easily right?
@@sidharthpatra4751 I took it as [-(1-√2)]^10 so I got 1 as the first term, I don't have much experience with binomial equations, but I think that's faster?
I really enjoyed this video! But in this case, since the square root turns into a whole number when squared, I think it is actually easier just to power through using the original numbers, rather than defining 𝑥 = √2 - 1 and then substituting it back in the end:
x¹⁰ = x⁸ ⋅ x² = [(x²)²]² ⋅ x²
So, (√2 − 1)¹⁰ = {[(√2 − 1)²]²}² ⋅ (√2 − 1)²
STEP 1: (√2 − 1)² = 3 − 2√2
STEP 2: (3 − 2√2)² = 17 − 12√2
STEP 3: (17 − 12√2)² = 577 − 408√2
STEP 4: (577 − 408√2) ⋅ (3 − 2√2) = 3363 − 2378√2
Pascal triangle... 2 minutes, exact solution
Let a be (SQRT(2)-1) then the equation can be factored into a² *( a²)² * (a²)²= a² * (a²)²)²
a² is easy from 2nd binomial formula, squaring this a 2nd and 3rd time is not much more difficult. Now multiply 1st and 3rd term and you're finished.
This is probably what most of us started with.
You needed a calculator anyway, so instead of “simplifying” over 17 minutes and the using a calculator, you could have used a calculator from the beginning and arrived to same result in less than 15 seconds ;)
But I do love the arithmetics - always good to refresh :)
Which would have been marked wrong. The question was seeking the exact answer, not a calculator approximation. I know he fiddled about with a calculator at the end, but the problem was solved before that.
Beautiful question! Thank you!
Demonstrates the incredible accuracy of modern calculators. The difference of too very similar values will show any errors propagation.
Thanks man.
Engineer says 1.5E-4
My brother is an Engineer, and I admire him and his fellow Engineers for making things I could never make; but Mathematics? No.
I solved it using Pascal's Triangle, fun problem
Evaluate : (√2 - 1)¹⁰.
There are a number of ways of doing
this
I shall use the Binomial Theorem
Firstly I am going to do some
preparation
In order to reduce repetition and
tedium.
Now, in the Binomial Expansion of
(√2 - 1)¹⁰.
There are 11 terms.
Each term has the following structure
(binomial coefficient) x (a power of - √2 x (a power of 1)
The powers of 1 are each equal to 1
So each term is now (a binomial coefficient) x (a power of √2 )
Every term with an even power of 2 is rational these powers are (L₁)
(-√2)¹⁰=2⁵=32, (-√2)⁸=2⁴=16, (-√2 )⁶=2 ³=8, (-√2 )⁴= (2)²=4, (-√2 )²= 2
Every term with an odd power of √2 is irrational
and can be written as a multiple of - √2
Thus (-√2)⁹= (-√2)⁸⋅〈-√2 )=(2)⁴.-√2 =-16√2 the complete list of these, (L₂)
- 16√2, - 8√2, - 4√2, - 2√2, - √2.. ... (L₂)
32,16,8,4,2,1................ ..... ...(L₁)
The Binomial Coefficients for power 10 are
1,10. 45, 120, 210, 252, 210, 120,45,10, 1.......... (L₃).
(These form the 10th row of Pascal'sTriangle, or direct calculations
using combinations).
Now we can calculate the value of each term by multiplying each
corresponding values in lists L₁, L₂ and L₃.
( 1x32 + 45x16+210x8+210x4+45x2+1)-
√2 (10x16+120x8+252x4+120×2+10x1
Giving (32+720+1680+840+90+1)
- (160+960+1008+240+10)√2
=3363 - 2378√2
I've done binomial expansion for this one... the other way is obviously a nice shortcut.
A binomial theorem is the common solution to that expression, but your solution is so amazing. Thank you sir
I'll take your word for it.
But the final answer still takes 6 computer operations to calculate a real number, just like the original. So is it really simplified or just a different format?
Well, for the expression (v-1)^10 = 3363 - 2378*v, the former requires one subtraction and then exponentiation which takes up log(n) multiplications. The latter is one multiplication and one subtraction. It's clearly a much easier expression to compute.
@@AvihooI Actually 3363-2378*v^0.5; It takes the same amount of time to do e^(0.5 * ln v) as for e^(10 * ln v); But if we did it all in assembly, I guess the simplified version is actually faster.
@@JaimeWarlock yeah I mean, if you formulate this problem using a logarithm and exponentiation which can be approximated using a Taylor expansion then indeed the simplifcation becomes redundant.
This is why it depends on whatever computation model/system you're dealing with and what exactly is the problem you're trying to simplify. If the only tools you have are addition and multiplication (and subtraction and division) then the simplified form is better computationally.
Btw I just noticed that in what I wrote, I forgot to mention that v = sqrt(2)... that expansion is obviously only true when that is the case.
Chefe raiz cuadrada de 2 = (1,4142......) es un numero irracional, asi que si le resto 1 el resultado sigue irracional. Al ser elevado a la potencia 10 la parte entera tendra 10 digitos y la decimal infinitos ...
Really close to 1/6726. What's the simplest way to get this answer algebraically?
Let a=√2-1 and b=√2+1. ab=1; a+b=2√2. Using ab=1 without explicitly writing in the expansions below.
(a+b)^2=a^2+b^2+2. a^2+b^2=6
(a^2+b^2)^2=a^4+b^4+2. a^4+b^4=34
(a^2+b^2)(a^4+b^4) = a^6+b^6+(a^2+b^2). a^6+b^6=6*33
(a^4+b^4)^2=a^8+b^8+2. a^8+b^8=34^2-2.
(a^2+b^2)(a^8+b^8)=a^10+b^10+(a^6+b^6)
a^10+b^10 = 6*(34^2-2)- 6*33 = 6*(34*33-1) = 6726.
Note that a
@@FatihKarakurt I love you. Please continue to exist!
(2 ➖ 1)^2^5 (1 ➖ 1)^2^5^1 ( ➖ 1)2^1^1 ()^2^1(x ➖ 2x+1).
(√2-1)^10
= (√2-1)^10 ×(√2+1)^10 / (√2+1)^10
= (2-1)^10/ (√2+1)^10
= 1 / (√2+1)^10
= (√2+1)^-10
Bringt genau gar nichts, war aber innovativ 😅😅😅
What kind of pen are you using?
I like at minimum 4 digits maybe 5.
X= (√2 - 1) ? X = (√2 - 1)^10
Hang on. Right at the end you used a calculator. So why all other stuff ?
Nice❤❤🎉
Very good.
Hmm, am I wrong if i think that should be two results? √2 can be ~ 1,4142 and -1,4142. If you put 1,4142 as a result of √2 your answer is correct, but if you use -1,4142 as a result of √2 then not. Thank you for answer.
Funny, where do they teach such science? The result is 3263 - 2378*√2 , so you may insert whatever you think that √2 represents. But in mathematics it is usually a positive number, which square equals 2.
No, the square root function is defined as the unique non negative solution.
✅
This is a stupid problem!
Easy way :
√2 = 2 ^1/2
2^1/2*10 - 1
2^5 - 1
32-1 = 31 Ans.
But i haven't learned the binomial theorem that's why I'll prefer this method
You cannot distribute the 10 to the Equation if in the brackets theres subtraction or addition
since
(X±Y)^2=X²±2XY-Y²
If it was
(XY)^2=X²*Y²
If it was
(X/Y)^2=X²/Y²
The easy way is often the wrong way.
You are missing needed grouping symbols around the fractional exponent: 2^(1/2)