I'm trying to find Taylor approximation for the function: $$x^4e^{-x^3}$$

I started taking the first, second, third, etc. derivatives but the expression for it seems to explode with terms. I was just wondering is there a trick for this one or do I just have to use brute force in order to discover the pattern? :)

thnx for any help


Notice $$\exp(x) = \sum \frac{x^n}{n!} \implies \exp(-x^3) = \sum\frac{(-1)^nx^{3n}}{n!} \implies x^4 \exp(-x^3) = \sum\frac{(-1)^nx^{3n+4}}{n!} $$

  • 1
    $\begingroup$ +1 @Citizen Thank you for your help :) I thought I might have to use the expansion for $e^x$ for this somehow :) thnx again! $\endgroup$ – jjepsuomi Oct 9 '13 at 8:31
  • $\begingroup$ The first equality IS the Taylor expansion of $e^x$ around $0$. $\endgroup$ – Laurent LA RIZZA Mar 17 '15 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.