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Prove :$$(e^x\sin x)^{(n)}=2^{\frac n2}e^x \sin (x+n\cdot\frac{\pi}4) \\ \forall x\in\mathbb R, \ n\ge1$$

Hint: Use the trig identity $\sin(a+b)$

Well the trig identity is equal to $\sin a \cos b +\cos a \ sin b$.

I tried to find a pattern as I derived, it goes like this:

$(e^x\sin x)^{(n)}=\\\begin{align}&(n=1) &(e^x\sin x+e^x\cos x)=A \\ & (n=2)& (A+(e^x\cos x -e^x\sin x)=B)\\ &(n=3)&A+B+B-A=2B\\ &(n=4)&...=-4A\\ &(n=5)&-4(A+B) \end{align}$

It seems periodic but I just don't see the rule that will make the two sides equal...

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Use $\sin x +\cos x =\sqrt2(\sin x \cos\frac\pi4+\cos x\sin\frac\pi4)$ starting in step $n=1$.

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  • $\begingroup$ Which is equal to the RHS only with n=1. I assume I need to continue with induction ? $\endgroup$ – GinKin Dec 30 '13 at 17:43
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    $\begingroup$ Yes of course. The next derivative contains $\sin(x+\frac π4)+\cos(x+\frac π4)$ and so on. $\endgroup$ – LutzL Dec 30 '13 at 17:47
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to find the $nth$ derivative of $$e^{az}\sin (bz+c)$$ differentiating this expression once $$D^1=e^{az}(a\sin (bz+c)+b\cos (bz+c))$$ let $a=r\cos \theta$ and $b=r\sin \theta$. you get $$D^1= e^{az}r\sin (bz+c+\theta)$$ if you observe you see $$D^n= e^{az}r^n\sin (bz+c+n\theta)$$ where $r=\sqrt{a^2+b^2}$ and $\theta=\arctan(\frac{b}{a})$ now substitute for $a,b,c$.

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  • $\begingroup$ Why is there an a in exponenet of e ? $\endgroup$ – GinKin Dec 30 '13 at 17:37
  • $\begingroup$ Also, this doesn't get you $2^{\frac n2}$... $\endgroup$ – GinKin Dec 30 '13 at 17:39
  • $\begingroup$ Yeah I noticed it's a very general solution. About the arctan, $atan \frac12\neq\frac\pi4$ (after checking in my calculator) also those $a,b$ won't give you $\sqrt{2}$ but $\sqrt{5}$ Also, Is this some useful/known identity ? $\endgroup$ – GinKin Dec 30 '13 at 19:52
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    $\begingroup$ @GinKin Suraj is incorrect, your case is $a=b=1,c=0$, which means his answer disagrees with you by a factor of $2^{n/2}$. $\endgroup$ – Alex Becker Dec 31 '13 at 1:18

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