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.

$$g_2(\epsilon^1 \phi_1+ \epsilon^2 \phi_2+ \epsilon^3 \phi_3+\cdots)^2+ g_3(\epsilon^1 \phi_1+ \epsilon^2 \phi_2+ \epsilon^3 \phi_3+\cdots)^3+g_4(\epsilon^1 \phi_1+ \epsilon^2 \phi_2+ \epsilon^3 \phi_3+\cdots)^4+\cdots$$

I want to expand the above equation in such way to get coefficients of $\epsilon^7$. How long I have to do? Is there any way to get quick solution?

Edit: I need the coefficients of $\epsilon^7$ from the above equation i mentioned. The dots (...) mean the equation goes on with increasing dummy index and power index.

share|improve this question
    
Is there $g_1(\epsilon\phi_1+\epsilon^2\phi_2+\epsilon^3\phi_3 + \ldots )$ term as well? –  Kaster Mar 4 '13 at 20:03
    
I have no idea how you can relate the $g_k$ with its arguments, so have no idea how to begin to answer. –  copper.hat Mar 4 '13 at 20:03
add comment

5 Answers

up vote 3 down vote accepted

When multiplying two formal power series (not worrying about convergence), we have that the coefficient of $x^n$ in the product of $$ (a_1x^1+a_2x^2+a_3x^3+\dots)(b_1x^1+b_2x^2+b_3x^3+\dots) $$ is $$ \sum_{k=1}^{n-1}a_kb_{n-k} $$ This can be used to inductively compute the coefficients of each power of $x$ in $$ \left(a_1x^1+a_2x^2+a_3x^3+\dots\right)^n $$ This should be enough to compute the coefficient of $\epsilon^7$ as long as all subseries are included up to $$ g_7\left(\epsilon^1\phi_1+\epsilon^2\phi_2+\epsilon^3\phi_3+\dots\right)^7 $$ and each of those include all terms with powers of $\epsilon$ less than or equal to $7$.

share|improve this answer
add comment

You will need as many terms as the power, so in this case you should go to $g_7$. The reason is that your functions contain at lowest order $\epsilon^1$ so expanding the addition will lead to terms at the outer power (i.e. $n$ in $(...)^n$) and higher.

share|improve this answer
    
Can you do that for me? –  mathphy Mar 4 '13 at 20:54
    
Nope, sorry. But since this is homework I think you will have to do with hints and pointers instead of full answers –  Michiel Mar 4 '13 at 21:25
    
But no one couldn't give a easy way. –  mathphy Mar 4 '13 at 21:27
add comment

Consider how there can be a contribution to $\epsilon^7$ for example from the $g_4(\ldots)^4$ term:

You need foru factors together containing $\epsilon^7$, that is you need four (natural) exponents that sum to $7$. The possibilities are $$\tag17=1+1+1+4=1+1+2+3=1+2+2+2,$$ so we obtain summands $g_4\phi_1\phi_1\phi_1\phi_4\epsilon^7$, $g_4\phi_1\phi_1\phi_2\phi_3\epsilon^7$, $g_4\phi_1\phi_2\phi_2\phi_2\epsilon^7$. But: the summands in the pratiotions given in $(1)$ may occur in any order and the number of orders can be obtained by combinatorics. For example $1+1+1+4$ can be written in foru different orders, depending on where the $4$ is; $1+1+2+3$ can be written in $4\cdot 3=12$ orders as we can place the $3$ anywhere and then the $2$ anywhere else. In summary, we obtain $$ g_4(4\phi_1^3\phi_4+12\phi1^2\phi_2\phi^3+4\phi_1\phi_2^3)\epsilon^7.$$ Do the same with all powers up to $g_7(\ldots)^7$ and add. More is not needed because every summand in expanding $g_k(\ldots)^k$ will have $\epsilon^k$ and above.

share|improve this answer
add comment

Use geometric power series & Pascal's Triangle (Binomial Theorem).

First you convert your series to make it look like $(a+b)^n$ as much as possible, especially the higher power terms (see example).

Second, you use Binomial Theorem, $(a+b)^n = \sum^{n}_{0} {n \choose k}a^{n-k}b^{k}$.

For example, $(1+x+x^2+x^3+x^4)^3 = (\frac{1-x^5}{1-x})^3$, now you can use $(1-x^5)$ in the second step. Consider $\frac{1}{1-x}$ separately to match the exponents, and manipulate backwards to get the original series.

share|improve this answer
1  
Could you provide a bit more detail? As is, this is just a comment. –  robjohn Mar 4 '13 at 21:03
    
Can you do a little? –  mathphy Mar 4 '13 at 21:04
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  gnometorule Mar 4 '13 at 21:17
    
I'm not asking to give a completesolution but show a little bit solution. I did the lower power expansion but problem on the higher power. –  mathphy Mar 4 '13 at 21:23
    
What is the problem on th ehigher power? –  Hagen von Eitzen Mar 4 '13 at 21:28
show 3 more comments

Just check which terms of each power give rise to terms up to $\epsilon^7$ (for the square, need to go to $\epsilon^4$; for the cube, to $\epsilon^3$, ...). Feed the truncated stuff to your neighborly computer algebra package (like Maxima ow Wolfram Alpha) and cut the resullting mess at the right point.

If the $g_k$s come from a known function, ditto the $\phi_i$s, just ask for a power series expansion.

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.