Expanding the power series

Question

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

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.

Answer 1

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 $7$.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.