Find the coefficient of $f(x)=e^x \sin x$ for the 5th derivative $f^{(5)}(0)$ using maclaurin series
the maclaurin polynomial is supposed to be $M_5(0)=f(0)+ \frac{f'(0)}{1!}+…+ \frac{f^{(5)}(0)}{5!}$
We know that $e^x=1+x+ \frac{x^2}{2!}+…$
we also know $\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}+…$
the answer is supposed to be $-4$ and the solution it shows is just $(\frac{1}{5!}- \frac{1}{12}+\frac{1}{4!})x^5$ and then $(\frac{1}{5!}- \frac{1}{12}+\frac{1}{4!})=\frac{f^{(5)}(0)}{5!}$
but I do not understand how they got to this? I assumed they took the expansion of each $e^x$ and $sinx$ till the fifth order and multiplied but that didn't work for me
there has to be a simple way and I am unable to figure it
Thanks for any tips and help!
Best Answer
You have that
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\text{higher order terms}$$
and
$$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+\text{higher order terms}.$$
If we multiply these together and just look at the resulting $x^5$-term, we get
$$e^x\sin x=\text{lower order terms}+\left(\frac{1}{4!}\cdot1-\frac{1}{2!}\cdot\frac{1}{3!}+1\cdot\frac{1}{5!}\right)x^5+\text{higher order terms}.$$
Now since the factor in front of the $5$'th order term in the expansion corresponds to $\frac{1}{5!}f^{(5)}(0)$ (with $f(x)=e^x\sin x$), we get that
$$f^{(5)}(0)=5!\left(\frac{1}{4!}\cdot1-\frac{1}{2!}\cdot\frac{1}{3!}+1\cdot\frac{1}{5!}\right)=-4.$$