5
$\begingroup$

A practice problem reads:

Suppose $$\sum_{n=1}^{\infty} b_n x^n = \frac{x^3}{(x^4-1)^2}.$$

What could be an expression of $b_n$?

Some of the possible answers read $ 2^{3n}nx^{3n-1}, nx^{3n-1}, nx^{4n-1}$

There are a few other answers. I was just not sure how to start this problem?

$\endgroup$
  • $\begingroup$ Have you heard of Taylor series? That could be one place to start. $\endgroup$ – Matt Apr 27 '17 at 18:49
  • $\begingroup$ the expression depends on the value of $x$ because the radius of convergence of a geometric series is $1$. $\endgroup$ – Masacroso Apr 27 '17 at 18:51
  • 1
    $\begingroup$ Hint $\sum_{n=1}^{\infty} ny^{n} =\frac{y}{(1-y)^2}$ ... $y \rightarrow x^{\color{red}{?}}$ ... $\endgroup$ – Donald Splutterwit Apr 27 '17 at 18:53
4
$\begingroup$

HINT:

$$\int \frac{x^3}{(x^4-1)^2}\,dx=\frac{1}{4(1-x^4)}+C$$

Expand $\frac{1}{1-z}$ in a geometric series and set $z=x^4$?

$\endgroup$
  • $\begingroup$ Does that mean I can say $$\int \sum_{n=1}^{\infty} b_n x^n = \frac{1}{4(1-x^4)} + C$$ could be written as a power function $$ \sum z^n, $$ where z = x^2? $$ $\endgroup$ – Haim Apr 27 '17 at 19:59
  • $\begingroup$ Note quite. Note $z=x^4$. $\endgroup$ – Mark Viola Apr 27 '17 at 20:08
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola May 1 '17 at 3:48
2
$\begingroup$

Another variation is based upon the binomial series expansion. \begin{align*} (1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1 \end{align*}

We obtain \begin{align*} \frac{x^3}{(x^4-1)^2}&=x^3\sum_{n=0}^\infty\binom{-2}{n}(-x^4)^n\tag{1}\\ &=x^3\sum_{n=0}^\infty\binom{n+1}{1}x^{4n}\\ &=\sum_{n=0}^\infty (n+1)x^{4n+3}\tag{2}\\ &=x^3 + 2 x^7 + 3 x^{11} + 4 x^{15} + 5 x^{19} + 6 x^{23}+\cdots \end{align*}

In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

We conclude according to (2) the representation \begin{align*} \sum_{n=1}^{\infty} b_n x^n=\sum_{n=0}^\infty (n+1)x^{4n+3} \end{align*} implies for $n\geq 0$ \begin{align*} b_{4n+k}= \begin{cases} n+1&\qquad k=3\\ 0&\qquad k\neq 3 \end{cases} \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.