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.
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
@@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.
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
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.
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))
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 😁
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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.
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
Wow!
@@SyberMath Yeah, I love this stuff. Thanks for the problems.
@@paulortega5317 but what is original f(x) ?
@@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.
@@paulortega5317 It was brilliant but could not relate to the solution for what is the explicit function f(x).
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
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.
Did you check out www.youtube.com/@aplusbi? 😊
@@SyberMath #MostDefinitely
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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))
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 😁
😄
Got it!
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)
😊🎉😊👍👍👍😊🎉😊