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I am supposed to find the nth order derivative of: $$\frac{1}{x^4+4}$$

I tried to resolve into partial fractions. But it didn't work out for me.

Edit- where I am stuck $$\frac{1}{x^4+4}=\frac{1}{(x-1+i)(x-1-i)(x+1+i)(x+1-i)}= \frac{A}{(x-1+i)} + \frac{B}{(x-1-i)} + \frac{C}{(x+1+i)} + \frac{D}{(x+1-i)} $$ where A,B,C,D are to to be found. I am unable to proceed further. If I am able to find these values, the rest is easy to handle.

I am a beginner in this subject. Please help.

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  • $\begingroup$ Do you need the whole derivative or just the derivative at a particular point? $\endgroup$ Commented Jan 14, 2021 at 8:28
  • $\begingroup$ @NinadMunshi Kindly elaborate your apprach as an answer. Thank you. $\endgroup$
    – S.S
    Commented Jan 14, 2021 at 8:30
  • $\begingroup$ I changed my question $\endgroup$ Commented Jan 14, 2021 at 8:31
  • $\begingroup$ @NinadMunshi The general nth derivative $\endgroup$
    – S.S
    Commented Jan 14, 2021 at 8:32
  • $\begingroup$ WolframAlpha can help sometimes. $\endgroup$ Commented Jan 14, 2021 at 9:17

2 Answers 2

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Hint

As the roots of $p(x)=x^4+4$ are $\xi_k = \sqrt 2 e^{i(\frac{\pi}{4} + k \frac{\pi}{2})}$ where $k \in \{0, 1, 2, 3\}$ you can write

$$\frac{1}{x^4+4} = \sum_{k=0}^3 \frac{a_k}{x-\xi_k}$$

where you have to find the $a_k$ using partial fraction decomposition. Based on the formula $a_k = \frac{1}{p^\prime(\xi_k)} = \frac{1}{4\xi_k^3}$, you get

$$\frac{1}{x^4+4} = \frac{1}{16}\sum_{k=0}^3 \frac{\xi_k}{x-\xi_k}$$ Then the $n$th-derivative of $\frac{a_k}{x-\xi_k}$ is easy to find. And you're done.

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  • $\begingroup$ Please have a look, I have added where I am actually stuck. $\endgroup$
    – S.S
    Commented Jan 14, 2021 at 10:30
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    $\begingroup$ You should look in detail at partial fraction decomposition wikipedia entry that I referenced in my answer. You need to compute $a_k$ and $a_k = \frac{1}{p^\prime(\xi_k)} = \frac{1}{4\xi_k^3}$. For example, what you name $A$ in your question is $\frac{1}{p^\prime(1-i)} = \frac{1}{4(1-i)^3}$ and so on for the other coefficients. $\endgroup$ Commented Jan 14, 2021 at 12:28
  • $\begingroup$ Thank you, I am trying to understand it. $\endgroup$
    – S.S
    Commented Jan 14, 2021 at 12:36
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One answer is as follows: \begin{align*} \biggl(\frac{1}{x^4+4}\biggr)^{(n)} &=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} B_{n,k}\bigl(4x^3,12x^2,24x,24,0,\dotsc,0\bigr)\\ &=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} x^{4k-n} B_{n,k}\bigl(\langle4\rangle_1, \langle4\rangle_2, \langle4\rangle_3, \langle4\rangle_4, \langle4\rangle_5,\dotsc,\langle4\rangle_{n-k+1}\bigr)\\ &=\sum_{k=0}^n \frac{(-1)^kk!}{(x^4+4)^{k+1}} x^{4k-n} (-1)^k\frac{n!}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{4\ell}{n}\\ &=\frac{n!}{x^n}\sum_{k=0}^n \frac{x^{4k}}{(x^4+4)^{k+1}} \sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{4\ell}{n}, \end{align*} where we used the formula \begin{equation}\label{binom-bell-factorial} B_{n,k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc,\langle\alpha\rangle_{n-k+1}) =(-1)^k\frac{n!}{k!}\sum_{\ell=0}^{k}(-1)^{\ell}\binom{k}{\ell}\binom{\alpha\ell}{n}. \end{equation}

A Reference

  1. Siqintuya Jin, Bai-Ni Guo, and Feng Qi, Partial Bell polynomials, falling and rising factorials, Stirling numbers, and combinatorial identities, Computer Modeling in Engineering & Sciences 132 (2022), no. 3, 781--799; available online at https://doi.org/10.32604/cmes.2022.019941.
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