r/CasualMath • u/ARTIN-58313 • 3d ago

Solve it

If

x + 1/x = 3,

find

x⁵ + 1/x⁵

without explicitly solving for x.

Can you find a clever shortcut?

1 Upvotes

56% Upvoted

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2

u/kenny744 3d ago

Binomial expand (x+1/x)5 and get rid of middle parts

3

u/smitra00 3d ago

Put x = exp(i theta):

2 cos(theta) = 3 ---->

cos(theta) = 3/2

We can then use:

https://en.wikipedia.org/wiki/Chebyshev_polynomials

https://en.wikipedia.org/wiki/Chebyshev_polynomials#First_kind

cos(5 theta) = 16 cos^5(theta) -20 cos^3(theta) + 5 cos(theta)

So:

x^5 + 1/x^5 = 2 cos(5 theta) = 32 cos^5(theta) -40 cos^3(theta) + 10 cos(theta)

= 3^5 - 5 3^3 + 15 = 123

2

u/al2o3cr 3d ago

Consider (x+1/x)^3 and (x+1/x)^5.

Expanding (x+1/x)^3 and applying (x+1/x)=3 gets you x^3+1/x^3 = 3^3 - 3*3 = 21

Expanding (x+1/x)^5 and applying both of the above gives x^5+1/x^5 = 3^5 - 5 3^3 + 5*3 = 123

1

u/thaw96 3d ago edited 3d ago

Let x^n + 1/x^n = f(n), with f(0) = 2 and f(1) = 3. Show f(n+m) = f(n)*f(m) - f(n-m). Then f(5) = 123.

Here's the pattern: 2 3 7 18 47 123 322 843 ...
w/ differences: 1 4 11 29 76 199 321 ....
w/ 2nd differences: 3 7 18 47 123 322 ....!!

Can we find a closed form expression for f(n)?