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I was trying to solve a differentiation question but unable to understand .

My question is :

find the $n^{th}$ derivative of $1/(1+x+x^2+x^3)$

I know that if we divide the numerator by denominator then the expression would be :

$1- x(1+ x^2 + x )/(1+x+x^2+x^3)$

But now how to find the nth derivative?

Please somebody explain this..

Thanx :-)

Umberto
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Gaurang Tandon
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  • First you should review how to derivate $1/f(x)$. After that things will be easier. Have you tried to derivate something? Can we see what you have done? – Umberto Jan 13 '14 at 09:45
  • My first two thoughts are: (1) compute its Taylor series (by doing arithmetic with other, known Taylor series), and (2) simplify the function (e.g. partial fractions). –  Jan 13 '14 at 09:49

4 Answers4

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HINT:

Use Partial Fraction Decomposition

$$\frac1{1+x+x^2+x^3}=\frac1{(x+1)(x^2+1)}=\frac A{x+1}+\frac B{x+i}+\frac C{x-i}$$

lab bhattacharjee
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Hint: Note that $(1-x)(1+x+x^2+x^3)=1-x^4$, so $$\frac1{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\frac{1-x}{(1-x^2)(1+x^2)}=\frac{1}{(1+x)(1+x^2)}.$$ (After the fact, this factorization is easy to see directly.)

Harald Hanche-Olsen
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My answer is as follows: \begin{align*} \frac1{1+x+x^2+x^3}&=\frac{1}{2 (x+1)}+\frac{1}{2(x^2+1)}-\frac{2 x}{4 (x^2+1)}\\ &=\frac{1}{2 (x+1)}+\frac{1}{2(x^2+1)}-\frac{1}{4}\bigl[\ln\bigl(x^2+1\bigr)\bigr]',\\ \biggl(\frac1{x+1}\biggr)^{(n)}&=\frac{(-1)^nn!}{(x+1)^{n+1}},\\ \biggl(\frac1{x^2+1}\biggr)^{(n)}&=\frac{n!}{(2x)^{n}}\sum_{k=0}^{n}(-1)^k\binom{k}{n-k}\frac{(2x)^{2k}}{(1+x^2)^{k+1}}\\ &=(-1)^nn!\frac{\sin[(n+1)\operatorname{arccot}x]}{(1+x^2)^{(n+1)/2}}\\ &=\frac{(-1)^{n}}{(1+x^2)^{n+1}} \begin{vmatrix} 2x & 1+x^2 & 0 & \dotsm&0& 0& 0\\ 2 & 4x & 1+x^2 & \dotsm& 0&0& 0\\ 0 & 6 & 6x & \dotsm& 0& 0&0\\ \dotsm & \dotsm & \dotsm & \ddots&\dotsm& \dotsm& \dotsm\\ 0 & 0 & 0 & \dotsm&2(n-2)x& 1+x^2& 0\\ 0 & 0 & 0 & \dotsm&(n-2)(n-1)&2(n-1)x & 1+x^2\\ 0 & 0 & 0 & \dotsm&0& (n-1)n& 2nx \end{vmatrix},\\ \bigl[\ln\bigl(x^2+1\bigr)\bigr]^{(n+1)}&=\sum_{k=1}^{n+1}\frac{(-1)^{k-1}(k-1)!} {(x^2+1)^{k}} B_{n+1,k}(2x,2,0,\dotsc,0)\\ &=\sum_{k=1}^{n+1}\frac{(-1)^{k-1}(k-1)!} {(x^2+1)^{k}} 2^k B_{n+1,k}(x,1,0,\dotsc,0)\\ &=\sum_{k=1}^{n+1}\frac{(-1)^{k-1}(k-1)!} {(x^2+1)^{k}} 2^k \frac{1}{2^{n-k+1}}\frac{(n+1)!}{k!}\binom{k}{n-k+1}x^{2k-n-1}\\ &=\frac{(n+1)!}{(2x)^n}\sum_{k=1}^{n+1}\frac{(-1)^{k-1}} {(x^2+1)^{k}} \frac{1}{k}\binom{k}{n-k+1}(2x)^{2k-1}. \end{align*}

References

  1. Show that $\frac{d^n}{dx^n}\left[ \frac{1}{1+x^2} \right] = \frac{(-1)^nn!}{(x^2+1)^{\frac{n+1}{2}}}\sin[(n+1)\arctan x]$
  2. https://math.stackexchange.com/a/4418636
  3. F. Qi and B.-N. Guo, Explicit formulas for special values of the Bell polynomials of the second kind and for the Euler numbers and polynomials, Mediterr. J. Math. 14 (2017), no. 3, Article 140, 14 pages; available online at https://doi.org/10.1007/s00009-017-0939-1.
  4. F. Qi, D.-W. Niu, D. Lim, and Y.-H. Yao, Special values of the Bell polynomials of the second kind for some sequences and functions, J. Math. Anal. Appl. 491 (2020), no. 2, Article 124382, 31 pages; available online at https://doi.org/10.1016/j.jmaa.2020.124382.
  5. Feng Qi, Determinantal expressions and recursive relations of Delannoy polynomials and generalized Fibonacci polynomials, Journal of Nonlinear and Convex Analysis 22 (2021), no. 7, 1225--1239.
qifeng618
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Find the first derrivative and then the second and then the third, you might see a pattern emerge, then write down the general equation that follows that pattern the nth derrivative

Abel
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