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I've received this problem to solve, and frankly, I'm not sure how to do it. I know what the Geometric Series looks like, and for a (limited) number of examples, I know how it can be used to determine the Taylor series. But for this problem, because the function is not a (in my opinion) conventional one, I'm not sure how to do it. Without further ado, here's the problem:


Determine the Taylor series of $f(x)=A/(x-B)^4$ at the point $x=c($$\ne$$B)$ using the Geometric series.



What I have tried so far is this:

$A(\frac{1}{x-B})^4 = AB^4(-\frac{1}{1-\frac{x}{B}})^4 = -AB^4(1+\frac{x-c}{B}+\frac{(x-c)^2}{B^2}+$...$)^4$


If I take the $n^{th}$ term, I end up with $$\frac{(x-c)^{n}}{B^{n}}$$

Then, the Taylor expansion would look something like this:

$-AB^4(1+\frac{x-c}{B}+\frac{(x-c)^2}{B^2}+$...$+\frac{(x-c)^{n}}{B^{n}})^4$

I would now like to rewrite this to something that doesn't look like this, though my professor told me you could also write a Taylor series like this.

Can someone confirm if my answer is correct, and add suggestions where needed? I really want to know if what I've done so far is correct and how it can lead to the final answer.

Thanks in advance!

Edit: I still have no clue how to get to the final answer from what I've written down. Can someone help me?

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    $\begingroup$ Are you demanding us to do it ASAP? $\endgroup$ – Juniven May 2 '17 at 17:28
  • $\begingroup$ Try to solve some easier examples first. Two of them will be enough. First, try to adapt $A/(x-B)$ directly to the form of the sum of the geometric series $N/(1-X)$ with $X$ anything you need, including $M(x-c)$ and $N$ some factors you have to carry. Later, consider that $A/(B-x)^2$ is the derivative of the sum for the geometric series... $\endgroup$ – Rafa Budría May 2 '17 at 17:49
  • $\begingroup$ "(if possible) ASAP" is a pleonasm, please rephrase. $\endgroup$ – Yves Daoust May 2 '17 at 18:01
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We have $x-B = (x-C)-(B-C)$, so $$\frac1{x-B}=\frac1{(x-C)-(B-C)} = -\frac1{B-C}\cdot\frac1{1-\frac{x-C}{B-C}} = -\frac1{B-C}\cdot \sum_{n=0}^\infty \left(\frac{x-C}{B-C}\right)^n$$ whenever $|x-C|<|B-C|$. To get the series for $\dfrac1{(x-B)^4}$, we should differentiate this series $3$ times, getting \begin{align*} (-1)(-2)(-3)\frac1{(x-B)^4} &= -\frac1{B-C}\sum_{n=3}^\infty n(n-1)(n-2)\left(\frac{x-C}{B-C}\right)^{n-3} \\ &= -\frac1{B-C}\sum_{n=0}^\infty (n+3)(n+2)(n+1)\left(\frac{x-C}{B-C}\right)^n. \end{align*} That is, $$\frac1{(x-B)^4} = \frac 1{6(B-C)}\sum_{n=0}^\infty (n+3)(n+2)(n+1)\left(\frac{x-C}{B-C}\right)^n.$$

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With not a lot of effort by convolution you get ASAP: $$\frac {a}{(-b + x)^4} =a\left(\frac {1}{-b + x}\right)^4= a\sum_{n=0}^∞ \frac{1}{6} x^n ( b^{-4 - n} (1 + n) (2 + n) (3 + n))$$

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  • $\begingroup$ I would prefer a longer solution or suggestions as to how to improve my answer. I cannot see if this answer corresponds to what I've written down. $\endgroup$ – Adam Warlock May 6 '17 at 8:39

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