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Let me bring out a very old question first:

$$ \displaystyle {f\left(x\right)=\frac{x^{2}+3}{x^{6}\left(x^{2}+1\right)}} $$

(Source: Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $)

I think by now everyone knows that we can rewrite the function as:

$$ f\left(x\right)=\frac{a_{2}}{x^{2}}+\frac{a_{4}}{x^{4}}+\frac{a_{6}}{x^{6}}+\frac{bx+c}{x^{2}+1} $$

By using the Heaviside cover-up method, it is easy to obtain $a_{6}$, $b$ and $c$ (which is $3$, $0$ and $-2$ respectively). However, when it comes to other coefficients, it is (almost) impossible to use this method. (No, I do not want to use other methods like subtracting the found fractions from $f(x)$, when it comes to other very long scenarios, it can be very difficult to simplify). So, I started searching for an alternative method that at least a handheld calculator could solve quickly, and I found this formula for those coefficients.

$$a_{k}=\frac{1}{\left ( k-1 \right )!}\left\{ \frac{d^{k-1}}{dx^{k-1}} \left [ \left ( x+x_{0} \right )^{n}f\left ( x \right ) \right ]\right\}\left|\begin{matrix}\\_{x=-x_{0}}\end{matrix}\right. (2 \leq k \leq n-1)$$

Source:

My question is "How can anyone come up with this???" Surely, it makes sense for the highest order since you just multiply everything with that term and calculate the limit, but it doesn't make sense how differentiation can lead you straight up to the coefficient of lower orders. For example, in the problem I proposed initially, let's say I want to find $a_{4}$ first. First, multiply everything by $x^{6}$, then I try to differentiate the RHS, twice, plug in $x = 0$, and all fractions magically equal 0, except for that specific $a_{4}$ coefficient. How does this happen?

TargetVN
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  • What is $A_k$ in your passage. Is it the coefficient of $(x+x_0)^{-k}$ as it seems not consistent with your example above. – Angae MT Jul 24 '24 at 04:27
  • You can check the original sources, it's basically what you just said

    (should've made it a instead of A)

     – TargetVN Jul 24 '24 at 04:59
  • By the way, I have edited the examples a bit so it can relate with the formula more – TargetVN Jul 24 '24 at 05:10
  • Do you know how to express the coefficients of $F(x)=\sum_{0}^{\infty} a_n (x-x_0)^n$ as derivatives of $F(x)$? If so, you should be able to see easily how this applies to your $x^6 f(x)$. It's not difficult to see "how somebody came up with this". – ancient mathematician Jul 24 '24 at 13:39
  • After a bit of thinking, that seems very similar to Taylor Series... Oh! So that's basically the whole concept! It's just taking the coefficients from a Taylor Series for f(x)! – TargetVN Jul 24 '24 at 14:42

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