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How would I write the power series $$1+3x^2+8x^4+21x^6+\cdots$$ as a power series representation (something neat similar to $\frac{1}{1-x}$)?

This reminds me of the power series $1+x^2+x^4+x^6+\cdots$ where the power series representation for that is $\frac{1}{1-x^2}$, but how would I add the Fibonacci numbers as coefficients into that?

Hints only!!

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  • $\begingroup$ Can you please tell us the sequence of coefficients? It's not obvious to me. $\endgroup$ – flawr Dec 20 '14 at 23:40
  • $\begingroup$ The coefficients are every other Fibonacci number. $\endgroup$ – JimmyK4542 Dec 20 '14 at 23:40
  • $\begingroup$ @flawr: Originally it was $x^2+3x^4+8x^6+21x^8+...$, but I factored out $x^2$ $\endgroup$ – Mathy Person Dec 20 '14 at 23:41
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Let $f(x) = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$.

You probably already know a closed form for $f(x)$.

Then, $f(-x) = 1-2x+3x^2-5x^3+8x^4-13x^5+21x^6-\cdots$.

Do you see how to get the series you want from $f(x)$ and $f(-x)$?


To get a closed form for $f(x)$ try combining the following equations in a way that leaves a finite number of terms on the right side:

$f(x) \ \ \ \ = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$

$xf(x) \ \ = \ \ \ \ \ \ \ 1x+2x^2+3x^3+5x^4+ \ \ 8x^5+13x^6+\cdots$

$x^2f(x) = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1x^2+2x^3+3x^4+ \ \ 5x^5+ \ \ 8x^6+\cdots$

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  • $\begingroup$ $f(x)+f(-x)$ and factoring out 2 will yield the desired result. However, I don't know the closed form for f(x)...can you give me a hint on how to find that? EDIT: The coefficients remind me of the Fibonacci numbers a bit. $\endgroup$ – Mathy Person Dec 20 '14 at 23:47
  • $\begingroup$ maybe multiply the second equation by x as well ? $\endgroup$ – Mathy Person Dec 20 '14 at 23:51
  • $\begingroup$ I can't seem to make it finite. If you add all three up, then it results in $1+3x+6x^2+10x^3+16x^4+....$...will that help at all? $\endgroup$ – Mathy Person Dec 20 '14 at 23:56
  • $\begingroup$ You can add and/or subtract equations. Remember that the fibonacci numbers satisfy $F_{n+1}-F_{n}-F_{n-1} = 0$. $\endgroup$ – JimmyK4542 Dec 20 '14 at 23:59
  • $\begingroup$ I think I got it! $f(x)-xf(x)-x^2f(x) = 1+x$ -> $f(x) (1-x-x^2) = 1+x$ -> $f(x) = \frac{1+x}{1-x-x^2}$ $\endgroup$ – Mathy Person Dec 21 '14 at 0:01
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Hint: Show that if $A(x) = a_0 + a_1 x + a_2 x^2 + \cdots $, then:

$$\frac{A(x) + A(-x)}{2} = a_0 + a_2 x^2 + \cdots $$

Can you apply this to your series?

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Hint 1: What is the generating function for the Fibonacci numbers?

Hint 2: How would you get every other term of a power series? (Big hint: What does $f(x)\pm f(-x)$ do?)

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  • $\begingroup$ generating function for the Fibonacci numbers is $x+x^2+2x^3+3x^4+5x^5+8x^6+....$ $\endgroup$ – Mathy Person Dec 20 '14 at 23:45
  • $\begingroup$ As a rational function. $\endgroup$ – marty cohen Dec 20 '14 at 23:47
  • $\begingroup$ $\frac{x}{1-x-x^2}$ $\endgroup$ – Mathy Person Dec 20 '14 at 23:48
  • $\begingroup$ What are f(x) and f(-x)? $\endgroup$ – Mathy Person Dec 20 '14 at 23:59
  • $\begingroup$ f(x) is the power series for the fibbies. $\endgroup$ – marty cohen Dec 22 '14 at 1:17

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