Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let the power series to be defined as :

$$\sum\limits_{n=0}^\infty a_nx^n$$

How to determine $a_{n}$ in such cases?

$$\begin{eqnarray}&a)&\quad \sum\limits_{n=0}^\infty \frac {n^{n}} {n!} x^{3n}\\ \\ &b)&\quad \sum\limits_{n=0}^\infty \frac {n}{2^{n}(x+2)^{2n}}\end{eqnarray}$$

share|improve this question
    
Could you check (b)? It makes little sense at present... –  Did Dec 9 '12 at 22:13
    
@did why do you think so? –  0x6B6F77616C74 Dec 10 '12 at 0:44
1  
For one thing, because one divides by zero. –  Did Dec 10 '12 at 5:44
    
@did fixed :--) –  0x6B6F77616C74 Dec 10 '12 at 22:37
    
Still in b), are you sure you mean $\frac{n}{2n}$? –  Did Dec 11 '12 at 6:00
add comment

2 Answers

up vote 5 down vote accepted

Just read off the coefficients. For (a), for instance, you have

$$a_n=\begin{cases} \frac{k^k}{k!},&\text{if }n=3k\\\\ 0,&\text{otherwise}\;. \end{cases}$$

share|improve this answer
    
Your answer is incorrect (despite three upvotes so far). Your formula yields $a_3=9/2$ but $a_3=1$. –  Did Dec 9 '12 at 22:13
    
@did: Aargh. Mindless. Fixed now; thanks. –  Brian M. Scott Dec 9 '12 at 22:17
add comment

The two problems are so different from each other that it leaves me wondering if they're even supposed to be part of the same exercise, or if they may be mis-transcribed in some fashion. That said, here's one approach to solving (b):

a) Expand out $\displaystyle\frac{1}{(x+2)^{2n}}$ into a power series in $x$ using the Binomial Series : $$\begin{eqnarray}\frac{1}{(x+2)^{2n}} &=& \frac{1}{\left(2(\frac{x}{2}+1)\right)^{2n}} \\ &=&\frac{1}{2^{2n}}\frac{1}{\left(\frac{x}{2}+1\right)^{2n}} \\ &=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\binom{i+2n-1}{i}\left(-\frac{x}{2}\right)^i \\ &=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i \end{eqnarray}$$

b) Plug this expansion into your original power series: $$\begin{eqnarray}\sum_{n=0}^\infty\frac{n}{2^n\left(x+2\right)^{2n}} &=&\sum_{n=0}^\infty\frac{n}{2^n}\frac{1}{\left(x+2\right)^{2n}}\\ &=&\sum_{n=0}^\infty\frac{n}{2^n}\left(\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\ &=&\sum_{n=0}^\infty\left(\frac{n}{2^{3n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\ \end{eqnarray}$$

c) Exchange the orders of summation so that the sum over $i$ is your outer summation; you'll be left with something that looks like $\displaystyle\sum_{i=0}^\infty\left(\sum_{n=0}^\infty b_{n,i}\right)x^i$ where $b_{n,i}$ is some complicated expression in $n$ and $i$. Your coefficients $a_i$ are then just $\displaystyle\sum_{n=0}^\infty b_{n,i}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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