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It is given that $a_n$ is generated by the following function:
$${1+6x} \over {(1+5x)(1-x)}$$

What is the generating function of:
$$\left\{ {\matrix{ {{a_n},\mathbb{N}_{even}} \cr {0,\mathbb{N}_{odd}} \cr } } \right.$$

I'm guessing it's a technical issue, which i am lacking of.
I'll be glad for an explanation.

Thanks!

AnnieOK
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1 Answers1

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Hint: $\dfrac{f(x)+f(-x)}{2}$.

André Nicolas
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  • it has something with $\sinh(x)$ maybe? I recall it generates somewhat similar series. I am not quite sure how to use your hint. – AnnieOK May 23 '14 at 18:50
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    It is only vaguely related. In general, if you have $f(x)=a_0+a_1 x+a_2x^2+a_3x^3+\cdots$, then $a_0+a_2x^2+a_4x^4+\cdots$ is $(1/2)[f(x)+f(-x)]$. For plug in $-x$ for $x$ and add. The odd terms cancel. It is similar to the way we get the series for $\cosh(x)$ from the series for $e^x$ and $e^{-x}$. In your case, the answer would be $\frac{1}{2}\left[\frac{1+6x} {(1+5x)(1-x)}+ \frac{1-6x}{(1-5x)(1+x)}\right]$, which we can simplify further if we wish. – André Nicolas May 23 '14 at 18:58
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    In your case, the desired generating function simplifies to (I think) $\frac{1-29x^2)}{(x^2-1)(25x^2-1)}$. You could also use partial fractions to simplify the original generating function before doing the $(1/2)(f(x)+f(-x))$ bit. – André Nicolas May 23 '14 at 19:14
  • You are welcome. To eliminate the even terms, we would have used $(1/2)(f(x)-f(-x))$. There are similar tricks using complex numbers to "eliminate" other terms of the generating function. – André Nicolas May 23 '14 at 20:14