Find the Generating Function from the sequence (0,1,0,3,0,5,...)

Question

Find the Generating Function from the sequence (0,1,0,3,0,5,...)

I can't conceptualize this one and I know it should be easy because it is just odds, but I am having a hard time figuring out how to cancel the even terms.

I was thinking like k$x^k$ but that doesn't cancel the evens.

Any help would be much appreciated.

Hint: if $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots$ then what is $f(x) - f(-x)$? — Daniel Schepler, Apr 12 '18 at 23:19

score 10 · Accepted Answer · answered Apr 12 '18 at 23:20

10

$G(x)=x+3x^3+5x^5+...=x(1+3x^2+5x^4+...)= xF'(x),$ where $$F(x)=x+x^3+x^5+...=\frac{x}{1-x^2}$$ I'm sure you can finish from here.

answered Apr 12 '18 at 23:20

saulspatz

53,131

3

That seems like a very complicated way of expressing (x mod 2) * x. – aroth Apr 13 '18 at 06:21

Donald Splutterwit · Answer 2 · 2018-04-13T01:16:17.367

2

\begin{eqnarray*} \sum_{i=0}^{\infty} (2i+1)x^{2i+1} =\frac{x}{1-x^2}+\frac{2x^3}{(1-x^2)^2} = \frac{x+x^3}{(1-x^2)^2}. \end{eqnarray*}

edited Apr 13 '18 at 01:16

answered Apr 12 '18 at 23:23

Donald Splutterwit

36,613
2
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What's up with the floating cube? Should that be $2x^3$ ? – mr_e_man Apr 13 '18 at 01:14
@mr_e_man You are right ... It is a mr_e why I did not put the $2x$ in ! – Donald Splutterwit Apr 13 '18 at 01:18

score -2 · Answer 3 · answered Apr 13 '18 at 05:40

-2

The answer is simple. Just iterate for x = 0 to Infinite And the xth term is

f(x) = (x%2)*x

Where % is modulus operator which returns the remainder of the two operands

Here x%2 cancels out all the even terms leaving all the odd terms behind

answered Apr 13 '18 at 05:40

Jagreet

17

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Yes, but that's not the generating function. The generating function is a formal power series. – saulspatz Apr 13 '18 at 06:48

Find the Generating Function from the sequence (0,1,0,3,0,5,...)

3 Answers3