$$F(p) = \frac{\cos p + pe^p}{p^{10} + \dfrac{1}{p^2}}$$

O.K. What a simple function! /s

By quotient law we have:

$$\frac{dF}{dp}=\frac{(-\sin(p) + pe^p + e^p)(p^{10} + p^{-2}) - (\cos(p) + pe^p)(10p^9 - 2p^{-3})}{(p^{10} + p^{-2})^2}$$

Some fractional simplification later:

$$\frac{(\sin(-p) + pe^p + e^p)(\frac{p^{12} + 1}{p^2}) - (\cos(p) + pe^p)(\frac{10p^{12} - 2}{p^3})}{p^{20} + p^{-4} + 2p^{12}}$$

Still very long, still not very jolly. No matter how much I think I can't think of a way to simplify it further. Can anyone prompt me? (Emphasis on prompt, please don't give me the entire solution).

Thanks in advance!

  • $\begingroup$ For instance, multiply numerator and denominator by $p^4$, so that you cancel the fractions in the numerator and you remove negative exponents in the denominator. $\endgroup$ – AugSB Mar 11 '16 at 17:37

Rewrite it as $$ F(p)=\frac{p^2\cos p+p^3e^p}{p^{12}+1}=\frac{f(p)}{g(p)} $$ and apply the quotient rule: \begin{align} f'(p)&=2p\cos p-p^2\sin p+3p^2e^p+p^3e^p\\[4px] g'(p)&=12p^{11} \end{align} Therefore \begin{align} f'(p)g(p) &=(2p\cos p-p^2\sin p+3p^2e^p+p^3e^p)(p^{12}+1)\\[4px] &=2p(p^{12}+1)\cos p-p^2(p^{12}+1)\sin p+(3p^2+p^3)(p^{12}+1)e^p \end{align} and $$ f(p)g'(p)= 12p^{13}\cos p+12p^{14}e^p $$ Now, just put together the pieces

$f'(p)g(p)-f(p)g'(p)=2p(1-5p^{12})\cos p-p^2(p^{12}+1)\sin p+p^2(3+p-9p^{11}+p^{12})e^p$

and, finally,

$F'(p)=\dfrac{2p(1-5p^{12})\cos p-p^2(p^{12}+1)\sin p+p^2(3+p-9p^{11}+p^{12})e^p}{(p^{12}+1)^2}$

  • $\begingroup$ When I used @AugSB's comment, I got: $$\dfrac{(p^{14} + p^2)\sin(-p) + (10p^{13} - 2p)\cos(p) + (11p^{14} + p^{15} + p^3 - p^2)e^p}{p^{24} + p + 2p^{16}}$$ $\endgroup$ – naiveai Mar 12 '16 at 10:20
  • $\begingroup$ @eshansingh1 Why multiplying by $p^4$? Just $p^2$ suffices (which is essentially the same I did). $\endgroup$ – egreg Mar 12 '16 at 10:21
  • $\begingroup$ Yes, but what I'm more worried about is the disagreeing coefficient for $e^p$. I'm not sure how to check this. Forgive me, I'm not really a math person. $\endgroup$ – naiveai Mar 12 '16 at 10:24
  • $\begingroup$ @eshansingh1 Such computations are always a mess and it's quite easy making some small mistake. I used all the care I could in my answer, but some mistake could have sneaked in nonetheless. $\endgroup$ – egreg Mar 12 '16 at 10:25
  • $\begingroup$ Yeah, you're right. Let me just have a look through it again. $\endgroup$ – naiveai Mar 12 '16 at 10:27

Rewrite your function as $$ F(p) = \mathrm{Re}\left[p^2(e^{\mathrm{i}p} + pe^p)(p^{12}+1)^{-1}\right]\ , $$ where Re stands for real part. Differentiating (which is fast using product and chain rule), you will get a sum of three reasonable (at least to my taste) fractions.

  • $\begingroup$ Sorry if this is a bit of a stupid question, but... where did the cosine go in your rewrite? And where did $i$ come from? $\endgroup$ – naiveai Mar 12 '16 at 3:15
  • $\begingroup$ @eshansingh1 Ok, I assume you haven't seen Euler's formula before en.wikipedia.org/wiki/Euler's_formula ... in this case, ignore my answer. $\endgroup$ – Pierpaolo Vivo Mar 12 '16 at 9:43
  • $\begingroup$ Yep, though it is very interesting. Thanks for that! $\endgroup$ – naiveai Mar 12 '16 at 10:06
  • $\begingroup$ You're welcome. $\endgroup$ – Pierpaolo Vivo Mar 12 '16 at 12:16

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.