Simplifying a Fun Radical Expression

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[InnocentNeuron]

one of the ways could be to regard a=1/2 and b=sqrt(5)/2. then, realize that (a+b)^8 + (a-b)^8 is essentially a binomial expansion of (a+b)^8, where all the terms with odd powers of b are eliminated and the remaining terms are multiplied by 2. this results in an easily computable expression.

[udic]

It's also the Lucas series and we are looking for the 8th element Ln+1=Ln+Ln-1 where L1=1 L2=3
1, 3, 4, 7, 11, 18, 29, 47

[phee]

that's a lot simpler and quicker then my approach, which was summing the even terms of the binomial formula applied to (1+sqrt5)^8 (the odd terms cancelled out with the odd terms of (1-sqrt5)^8 ), doubling the result, then dividing by 2^8

[jimschneider]

Your method is a lot more general than what I did; I just exploited the fact that (1 +/- sqrt(5))/2 are the roots of x^2 - x - 1 = 0, and the connection with the Fibonacci sequence to determine that x^8 = 13 + 21 x, if x is (1 +/- sqrt(5))/2, and that gave me the value 2*13 + 21 = 47 (the 2*13 comes from the fact that we're adding 13 twice, and the 21 comes from the fact that we're adding 21/2 twice).

[ABhaim]

I took the long way. I knew there may be a more elegant solution but I did it just for fun.

I expanded the first term, not minding at the moment about the values of the coefficients. I then eliminated any other term I got and doubled the rest. Knowing that 8 choose 2 and 8 choose 6 are the same, I added them together, then I calculated all the factorials, multiplied, added and divided it all by 2 eight times.

And just to emphasize, I didn't use calculator.
Took me a while though 😅

[rkrk]

A nice solution could've been to use the Explicit form of the Fibonacci sequence : Fn=(A^n-B^n)/√5 with A the golden ratio and B its conjugate.

[shrey]

x^2-x-1 was the first thing that came to mind for me

[composerlmythomorphic]

Would use the relation p^2=p+1 to progressively reduce the power of the terms to linear terms in phi

[PlutoTheSecond]

I think the easiest way to solve this problem is to realize that x and y are just the roots of the quadratic equation a^2 = a + 1. This fact makes it easy to prove that a^n = F_n*a + F_{n-1} where F_0 = 0, F_1 = 1 and F_{n+2} = F_{n+1} + F_n (this is the Fibonacci sequence). Then this can be used to reduce the problem to the linear expression F_8*(x + y) + 2F_7. Since x + y = 1, we get F_8 + 2F_7 = 21 + 2*13 = 47.

[komilazimov]

it also could be solved using the fact that x^2-x-1=0, the roots (golden ratio) are the numbers of the expression, so: x^2 = x+1, and the roots relations are the (x_1) = -1/(x_2) in which [ (x_1)*(x_2) = -1] =>> need to find (x_1)^8 + (x_2)^8, and applying the fact that x^2 = x+1, we have : x^4 = (x^2)^2 = (x+1)^2 = (x^2)+2x+1=(x+1)+2x+1 = 3x+2, now to find x^8 = (x^4)^2 = (3x+2)^2 = 9(x^2)+12x+4 = 9*(x+1)+12x+4 = 21x+13 >>> so we have for each: [(x_1)^8] + [(x_2)^8] = [21*(x_1)+13 ] + [21*(x_1)+13] = 26 + 21*[(x_1)+(x_2)], from above given the relation of the roots we get: =>>> 26 + 21*[(x_1)- 1/(x_1)] or the same as : 26 + 21*[x- (1/x)] = 26 +21* [(x^2 - 1)/ (x) ] = 26 + 21*[ ((x+1) - 1)/ x ] =26+21*[ x/x ] = 26+21*1 = 47 ;

[jmart]

If you use the coefficients of the Tartaglia triangle 1, 8, 28, 56, 70, 56, 28, 8, 1 and you eliminate the coefficients corresponding to the odd powers of sqrt(5) doubling the others you get:
2+280+3500+7000+1250
Dividing the result by 2^8 you obtain 47.

[MichaelJamesActually]

in case you're wondering, the binomial theorem also works. doesn't take too long, esp if you know that all odd sqrt terms cancel.

[biswakalyanrath]

This can also be done by taking a quadratic eqaution with roots(root5+1/2 and root5-1/2 ) and then using newton sum

[WolfgangKais]

Another method, maybe not as fast:
From Vietas formula, because x + y = 1 and xy = -1, both x and y are solutions of u^2 = u + 1. Multiplying this equation with u^n, we get u^(n+2) = u^(n+1) + u^n, so the numbers u(n) := u^n obey to the linear recursion (“Fibonacci”) formula u(n+2) = u(n+1) + u(n) and so does any linear combination of x^n and y^n. Therefore, defining z(n) := 1*x^n + 1*y^n = x^n + y^n, we know that z(n+2) = z(n+1) + z(n). Starting with z(0) = x^0 + y^0 = 2 and z(1) = x + y = 1, we can successively conclude that
X^2 + y^2 = z(2) = z(1) + z(0) = 1 + 2 = 3,
X^3 + y^3 = z(3) = z(2) + z(1) = 3 + 1 = 4,
X^4 + y^4 = z(4) = z(3) + z(2) = 4 + 3 = 7,
X^5 + y^5 = z(5) = z(4) + z(3) = 7 + 4 = 11,
X^6 + y^6 = z(6) = z(5) + z(4) = 11 + 7 = 18,
X^7 + y^7 = z(7) = z(6) + z(5) = 18 + 11 = 29 and
X^8 + y^8 = z(8) = z(7) + z(6) = 29 + 18 = 47.

[mcwulf]

Quite quick to square (1+sqrt(5))/2 three times. Then the same with (1-sqrt(5))/2 which is the same but with the negative root 5. Then the answer is just the integer part.

[JohnRandomness]

I took the mostly-brute-force method, having played quite a bit with (8, m) in the recent past. The binomial coefficients are 1, 8, 28, 56, 70, 56, 28, 8, 1. I only used every other coefficient, because the signs on sqrt(5) caused odd powers of that to cancel. So we get (1 + 28*5 + 70*25 + 28*125 + 625)*2/128 = 94 last I checked. I might check again.

EDIT: I should divide by 256, right? So the answer should be 47.

[rockywei]

Is it a good hint to fast calculate Fibonacci number??

[michaelpurtell]

Did it with the Fibonacci golden ratio identities now looking forward to watching the video to learn of a perhaps more general approach

[lazaremoanang]

If I need to give an explanation, I just need to make the sum of pair powers twice: (1+140+1750+3500+625)×2=12032, when dividing by 2^8=256, it gives 47.

[siddharthabhattacharya]

WE can creeate a quadratic polynomisl whose roots are 1+-sqrt(5)/2 that is x^2-x-1 so we have to find the sum of 8th powers of the roots which is simple valuation by newtons sum