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So I have to find the power series representation for $f(x) = \ln (3-x)$.

I attempted the following:

$$\ln(3-x) = \int {- \frac{1}{3-x} dx}$$ $$ = - \int { \frac{1}{1-(x-2)} dx}$$ $$ = - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$ = \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$ = \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K $$

Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.

However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$

Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.

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    $\begingroup$ It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). $\endgroup$ – Paolo Leonetti Aug 15 '15 at 12:24
  • $\begingroup$ @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? $\endgroup$ – Aaron Aug 15 '15 at 12:36
  • $\begingroup$ In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? $\endgroup$ – Paolo Leonetti Aug 15 '15 at 12:36
  • $\begingroup$ @PaoloLeonetti is that because we are allowed to do term-by-term integration? $\endgroup$ – Aaron Aug 15 '15 at 12:41
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    $\begingroup$ " the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) $\endgroup$ – Did Aug 16 '15 at 15:59
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Hint. Your route is OK, but you should rather start with $$ \ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3 $$ then follow the same path to obtain the right answer.

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    $\begingroup$ Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? $\endgroup$ – Aaron Aug 15 '15 at 12:36
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Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.

Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$

$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$

We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.


EXPANSION AROUND $x=0$

Approach 1: Using the approach outlined in the posted question, we find that

$$\begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int_2^x\frac{1}{1-(t-2)}dt\\\\ &=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align}$$

which converges for $-1\le x<3$ and diverges otherwise.


Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$

Therefore, we can write the series representation as

$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$

which converges for $-1\le x<3$ and diverges otherwise as expected!


EXPANSION AROUND $x=3$

Approach 1: Using the approach outlined in the posted question, we find that

$$\begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n} \end{align}$$

which converges for $-3\le x<3$ and diverges otherwise.


Approach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$.

Therefore, we can write the series representation as

$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$

which converges for $-3\le x<3$ and diverges otherwise as expected!

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    $\begingroup$ Please let me know how I can improve my answer. I really just want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 17 '15 at 15:10

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