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Expand given function $f$ as Taylor series around $c=3$ $$f(x) = \frac{x-3}{(x-1)^2}+\ln{(2x-4)} $$

and find out open interval at what that series is convergent. What is radius of convergence?

This is what i have so far. We know that $\ln{(1+x)} = \sum_{n=1}^{\infty} \frac{x^n}{n}$ and $\frac{1}{1-x}=\sum_{n=1}^\infty x^n ,|x| \lt 1$

We can rewrite $\ln{(2x-4)}$ as $\ln{(-\frac{1}{4})}+\ln{(-x/2+1)}$ and then expand last expression, but i don't know what to do with $\ln{(-1/4)}$. We can can split given fraction onto partial fractions as $\frac{x-3}{(x-1)^2} = \frac{1}{x-1}-\frac{2}{(x-1)^2}$. First fraction we can expand easily, but i don't know how to expand fraction with binomial as denominator.

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We have that for $|z|<1$, $$\frac{1}{(1-z)^2}=\left(\frac{1}{1-z}\right)'=\left(\sum_{n\geq 0} z^{n}\right)'=\sum_{n\geq 1}n z^{n-1}.$$ Let $t=x-3$, then for $t\in (-1,1]$ $$\frac{x-3}{(x-1)^2}+\ln{(2x-4)}=\frac{t}{(t+2)^2}+\ln(2)+\ln(1+t)\\ =\frac{t}{4(1-(-t/2))^2}+\ln(2)+\ln(1+t)\\ =\frac{t}{4}\sum_{n\geq 1}n (-t/2)^{n-1}+\ln(2)-\sum_{n\geq 1}\frac{(-t)^n}{n}\\ =\ln(2)+\sum_{n\geq 1} t^n\cdot(-1)^{n-1}\left(\frac{n}{2^{n+1}}+\frac{1}{n}\right). $$ Note that the radius of convergence is $1$, that is the smaller radius between the radius of the series of $\ln(1+t)$, that is $1$, and the radius of the series of $\frac{1}{(1-(-t/2))^2}$, that is $2$.

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  • $\begingroup$ But, what are you gonna do with $t$ in numerator in that fraction, i think we have to split it in partial fractions $\endgroup$ – Rush ThaMan Aug 24 '16 at 18:19
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In general, for integers n>1:

$$ \frac{n!}{(1-x)^{n+1}}=\sum_{r=0}^{\infty}(r+1)(r+2)...(r+n)x^r =\sum_{r=0}^{\infty} \frac 1r r^{(n+1)}x^r $$ where $r^{(n)} $ denotes rising factorial.

This can be easily derived using the binomial theorem. This is the only hurdle I can see in this question - this is a useful identity to remember.

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$$f(x)=\frac{x-1-2}{(x-1)^2}+\ln2+\ln(x-2)=\ln2+\frac1{x-1}-\frac2{(x-1)^2}+\ln(x-2)$$ and $f(3)=\ln2$.

Then

$$f'(x)=-\frac1{(x-1)^2}+\frac4{(x-1)^3}+\frac1{x-2}.$$

Next $$f''(x)=\frac2{(x-1)^3}-\frac{12}{(x-1)^4}-\frac1{(x-2)^2}.$$

The general pattern isn't difficult to find out,

$$f^{(k)}(x)=(-1)^k\left(\frac{k!}{(x-1)^{k+1}}-2\frac{(k+1)!}{(x-1)^{k+2}}+\frac{(k-1)!}{(x-2)^k}\right).$$

Finally, for $k>0$,

$$a_k=\frac{f^{(k)}(3)}{k!}=(-1)^k\left(\frac1{2^{k+1}}-2\frac{k+1}{2^{k+2}}+\frac1k\right).$$

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  • $\begingroup$ I don't think these types of problems are solved this way. Although, its nice to see other way to look at it. $\endgroup$ – Rush ThaMan Aug 24 '16 at 19:23

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