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Find $\large\color{maroon}{a^n}$ for the following generating function,

$$\color{green}{\begin{align*} \frac{1}{1-2x^2} \end{align*}}$$

$\large\color{maroon}{a^n}$ = closed form of the $nth$ term in the corresponding sequence.
asked in Combinatory by Veteran (57.4k points)
retagged by | 292 views
<1,2,4,8....> ??

This is what you wanted ??
possibly a closed form

1 Answer

+3 votes
Best answer

$\frac{1}{1-2x^{2}}$ = $\frac{1}{1-\sqrt{2}x}$ * $\frac{1}{1+\sqrt{2}x}$    ................(1)

 

Now, we know that,

$\frac{1}{1-ax}$ = 1+ ax + a2x + a3x3  +..................         ..........(2)

$\frac{1}{1+ax}$ = 1- ax + a2x - a3x3  +..................         ...........(3)

 

Now, from (1) , we observe that the result is the product of 2 functions.

 

If the coefficients of function 1 are a0, a1, a2................

and that of function 2 are b0, b1, b2................

then the result of product of two functions is a function with terms having coefficients rk corresponding to ''k'th term of the sequence. The coefficients can be calculated as follows:

rk = akb0 + ak-1b1 + ak-2b2.........................+ a0bk

 

From (2) and (3), we get,

r0 = 1

r1 = 0

r2 = a2

r3 = 0

........

 

We observe that when 'k' is odd, rk = 0. If 'k' is even, rk = ak

Thus, the terms of the sequence are {1, 0, a2, 0, a4, 0, ........................}

 

Now, substitute $\sqrt{2}$ for 'a' to get the actual sequence.

answered by Veteran (18k points)
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nice answer.


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