Generating function

Question

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.

Comments

Magma

Can u give me correct formula??

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(a+x)^-n =  $\binom{n-1}{n-1}$ a^-n  - $\binom{n}{n-1}$ a^-(n+1)  x + $\binom{n+1}{n-1}$ a ^-(n+2) x² -  $\binom{n+2}{n-1}$ a ^-(n+3) x³ ........

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Is this wrong??

Answer 1

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

Now, we know that,

$\frac{1}{1-ax}$ = 1+ ax + a²x²  + a³x³  +..................         ..........(2)

$\frac{1}{1+ax}$ = 1- ax + a²x²  - 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, a_{2................}

and that of function 2 are b₀, b₁, b_{2................}

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

r_k = a_kb₀ + a_k-1b₁ + a_k-2b_{2.........................}+ a₀b_k

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

r₀ = 1

r₁ = 0

r₂ = a²

r₃ = 0

........

We observe that when 'k' is odd, r_k = 0. If 'k' is even, r_k = a^k

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

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

Comments

nice answer.

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Good answer

Answer 2

Another approach using Extended Binomial Theorem

$G(x) = \frac{1}{(1-2x^{2})} = (1-2x^{2})^{-1} $

from the Extended Binomial coefficient, (Ref)

$a^{n}$ = $\binom{- 1}{n}* (-2{x}^{2})^{n}$                        ,   ( where $\binom{n}{r}$  is binomial coefficient $\forall n \epsilon \mathbb{R}$ )

      = $\binom{- 1}{n}* (-1)^n*(2{x}^{2})^{n}$

      = $ (-1)^{n} * \binom{1+n -1}{n} * (-1)^n*(2{x}^{2})^{n}$

      = $ \binom{n}{n} * 2^{n} x^{2n}$ 

      = $\frac{n!}{{n!}*{0!}} * 2^{n} x^{2n} $

$a^{n}$ = $2^{n} x^{2n}$

then G(x) = $2^{0} x^{0} + 2^{1} x^{2} + 2^{2} x^{4}+ 2^{3} x^{6} + . . . $

                 = $1 + 2x^{2} + 4 x^{4}+ 8 x^{6} + . . . $

Sequence of {$a^{n}$} = $\left \langle a^{0}, a^{1}, a^{2}, a^{3}, a^{4}, . . .  \right \rangle$  is given as 

{$a^{n}$}  = $\left \langle 2^{0}, 0, 2^{1}, 0, 2^{2}, . . .  \right \rangle$ = $\left \langle 1, 0, 2, 0, 4, . . .  \right \rangle$

Hence,

$a^{n} $= $\left\{\begin{matrix}
2^{\frac{n}{2}} &,n = 2k \mathit{(even)}\\ 
0   &   , n = 2k+ 1\mathit{ (odd) }
\end{matrix}\right.$