# If $\sum a_n z^n = f(z) = \sum b_n z^n$, What Can Be Said About the Coefficients $a_n$ and $b_n$

I apologize for the cryptic title, this issue came up while obtaining a solution to my question here. I was given a power series:

$$\sum_{n=0}^{\infty}r(n)z^n = \frac{1}{(1-z)(1-z^2)(1-z^3)}$$

I was then asked to perform a partial fraction decomposition on the closed form and obtained:

$$-\frac{\frac{1}{6}}{(z-1)^3} + \frac{\frac{1}{4}}{(z-1)^2} - \frac{\frac{17}{72}}{z-1} + \frac{\frac{1}{8}}{z+1} - \frac{\frac{1}{9}e^{\frac{2\pi i}{3}}}{z-e^{\frac{2\pi i}{3}}} - \frac{\frac{1}{9}e^{\frac{-2\pi i}{3}}}{z-e^{\frac{-2\pi i}{3}}}$$

Now upon individually converting these fractions back into power series, in short, what I obtained is a power series with coefficients $\frac{(n+3)^2}{12} + x_n$ for $\frac{1}{3} \leq x_n \leq \frac{2}{3}$.

This was expected, since my book states that $r(n)$ is given by the integer closest to $\frac{(n+3)^2}{12}$. What I don't understand is how can we conclude such a close correspondence between $r(n)$ and $\frac{(n+3)^2}{12} + x_n$? Just because two power series converge to the same closed form function, can't their coefficients not still be very different for any particular $n$? I know Taylor series are unique, but I don't think my latter power series is a Taylor series, and even if the power series expansion of a function is always unique, it still has that extra $x_n$ added to or subtracted from each coefficient. What is going on here?

Edit: Ok so $r(n)$ is supposed to be the number of ordered triples $(x_1,x_2,x_3)$ such that $x_1 + 2x_2 + 3x_3 = n$. I formed this by employing the Cauchy product $(\sum z^n)(\sum z^{2n})(\sum z^{3n})$.

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When dealing with power series, one has that $\tilde A(x) = \sum a_n z^n=\sum b_n z^n=\tilde B (x)$ if and only if $a_n=b_n$ for every $n$, and conversely. – Pedro Tamaroff Jun 23 '12 at 0:25

You have two power series expansions of the same function, one with coefficients $r(n)$, one with coefficients $(1/12)(n+3)^2+x_n$. By the uniqueness of power series expansions, $r(n)=(1/12)(n+3)^2+x_n$. Unless, of course, you have edited your post several more times while I was typing this.

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ah good you were the one who helped me last time, well $r(n)$ is a number of partitions and thus was always a whole number, I thought.. – cuckmaster5000 Jun 23 '12 at 0:26
Yes, so, what's the problem? – Gerry Myerson Jun 23 '12 at 0:33
Oh so you're saying $r(n) = \frac{1}{12}(n+3)^2 + x_n$ is always an integer? I guess I should have checked that – cuckmaster5000 Jun 23 '12 at 0:35
It had better be an integer, otherwise we have proved mathematics is inconsistent.... – Gerry Myerson Jun 23 '12 at 0:46

If $f(z)=\sum a_nz^n$, then $a_n=n!f^{(n)}(0)$. So, answering your question from the title, $a_n=b_n$ for all $n$.

An even easier way to see it, though not really different is that if $f(z)=\sum a_nz^n=\sum b_nz^n$, then $\sum(a_n-b_n)z^n=0$, which of course forces $a_n=b_n$ for all $n$.

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Ah that clears that up for me, thanks Martin. – cuckmaster5000 Jun 23 '12 at 0:37
You are welcome! – Martin Argerami Jun 23 '12 at 1:06