A Heptic Functional Equation

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[@stephenshefsky5201]

Assume f(z) is a 7th order polynomial. Then, let z = x + 1/x. We quickly find that even terms drop out, so f(z) = z^7 + a*z^5 + b*z^3 + c*z. Substituting z = x + 1/x and expanding, we have x^7 + 1/x^7 = x^7 + (7 + a)*x^5 + (21 + 5*a +b)*x^3 + (35 +10*a + 3*b +c)*x, from which we can determine a = -7, b = 14, c = -7.

[@thomaspickin9376]

Just tried to figure it expanding the binomials and getting rid of the highest power term each time: (x+1/x)^7 = x^7 + 1/x^7 + 7 x^5+ 21 x^3+ 35x + 35/x + 21/x^3 + 7/x^5 so we need -7(x+1/x)^5 to leave a x^3 power and so on. f(x) = x^7-7x^5+14x^3-7x f(x+1/x) = x^7+1/x^7

[@paulortega5317]

Let f(n) = x^n + 1/x^n f(0) = 2 Let f(1) = a Then f(n+2) = f(1)*(fn+1) - f(n) = a*f(n+1) - f(n) Then f(2) = a^2-2, f(3) = a^3- 3a, f(4) = a^4 - 4a^2 + 2,… f(7) = a^7 - 7a^5 + 14a^3 - 7a Or in general, f(n) = Σ (-)^k * (n/(n-k)) * (n-k)!/(k!(n-2k)!) * a^(n-2k) for k = 0 to n/2 or k = (n-1)/2 depending on whether n is even or odd

[@SyberMath]

Wow!

[@paulortega5317]

@@SyberMath Yeah, I love this stuff. Thanks for the problems.

[@wannabeactuary01]

@@paulortega5317 but what is original f(x) ?

[@paulortega5317]

@@wannabeactuary01 My objective was given x+1/x=f(1)=a, for f(n), what is the polynomial expansion in powers of a with a formula for the coefficients. Just like using n!/(k!(n-k)!) to calculate the coefficients for the nth row of Pascal's triangle.

[@wannabeactuary01]

@@paulortega5317 It was brilliant but could not relate to the solution for what is the explicit function f(x).

[@MisterPenguin42]

I like these and the exponential videos the best and I love whenever Lambert's function shows up. It's like the John Cena of math. The trig stuff and i stuff is growing on me; I just need to get more comfortable with complex numbers. There's a huge knowledge gap there.

[@SyberMath]

Did you check out www.youtube.com/@aplusbi? 😊

[@MisterPenguin42]

@@SyberMath #MostDefinitely

[@johnstanley5692]

Already a series for z(n)= (X^n+1/X^n): z(n+1)=z(n)z(1)-z(n-1) & z(2n)=z(n)^2-2. also z(0) = 2. Hence f(z(1)) = z(1)*(z(1)^6 - 7*z(1)^4 + 14*z(1)^2 - 7)

[@mathyfun]

Advertisement of king

[@MrGeorge1896]

I haven't seen the video yet and won't solve it now but as we all have seen your previous video with x^5 + 1 / x^5 it is the easiest way to reuse the solutions for x^3 and x^5 we found there and use the binomial expansion of (x + 1 / x) ^ 7. Reusing already achieved solutions is not cheating but an economical way to deal with it. 😊 Haha just watched the video and laughed out loud when you called it cheating. As if you had read my mind 😁

[@SyberMath]

😄

[@mikeparfitt8897]

I thought that this could be solved with the floor function, but it fails when t=±2 i.e. when x=±1 The Excel version is f(t) = SIGN(t) * (( FLOOR( ABS(t), 1) ^ 7) + 1/( FLOOR( ABS(t), 1) ^ 7))

[@scottleung9587]

Got it!

[@phill3986]

😊🎉😊👍👍👍😊🎉😊

[@DonEnsley]

f(x) = x⁷-7x⁵ +14x³-7x