A question about the series $\frac{\arctan(x)}{e^x}$

Question

I just see a demonstration on how to find the Maclaurin expansion for $\dfrac{\arctan(x)}{e^x}$

The part I don't understand I circle in red.

How does he obtains $c_0=0$? and then $c_0+c_1=1$ then $c_1+c_2=0$

And then for the third degree, why is there an $\frac{1}{2}$ that involves here?

I am very interested in this technique of find the unknown coefficients of an unknown series based on two known series.

Answer 1

You can see that on the RHS, there is no constant. But on the LHS, $c_0\cdot1$ is a constant, which means $c_0=0$.

Similarly, you could see, on the RHS, the term of degree $1$ is $x$ with coefficient $1$, and on the LHS the only way to get that term is $c_1x\cdot1+c_0\cdot x=(c_1+c_0)x$. That will give us $c_1+c_0=1$.

If you do this for terms on the RHS of higher degrees, you will get a system of equations of $c_i$'s and then you will be able to solve each of them (though not 100% rigorous since this solution didn't demonstrate a general equation for the term of arbitrary degree $n$).

Answer 2

When you multiply the series on the left you want to get the series on the right.

You count the constants on both sides, you have $(C_0)(1)$ on the left side but no constant on the right side so $C_0 =0$

Now you count how many $x$ you have on each side.

On the left side you have $(C_0+C_1)x$ and on the right side you have $x$, so $C_0+C_1 =1$ and so forth.

Answer 3

So I post this up as a question again:

In this series, the constant term is one and not zero, why is it the case?

Answer 4

It is well known that \begin{equation*} \arctan x=\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{2k+1}, \quad |x|<1 \end{equation*} and \begin{equation*} \operatorname{e}^{-x}=\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}, \quad |x|<\infty. \end{equation*} Therefore, we obtain \begin{align*} \frac{\arctan x}{\operatorname{e}^{x}} &=(\arctan x)\operatorname{e}^{-x}\\ &=\Biggl[\sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{2k+1}\Biggr] \Biggl[\sum_{k=0}^\infty(-1)^k\frac{x^k}{k!}\Biggr]\\ &=x\Biggl[\sum_{k=0}^{\infty}\frac{1}{k+1}\cos\biggl(\frac{k\pi}{2}\biggr)x^k \Biggr] \Biggl[\sum_{k=0}^\infty\frac{(-1)^k}{k!}x^k\Biggr]\\ &=\sum_{k=0}^\infty\Biggl[\sum_{\ell=0}^k\frac{1}{\ell+1}\cos\biggl(\frac{\ell\pi}{2}\biggr) \frac{(-1)^{k-\ell}}{(k-\ell)!}\Biggr]x^{k+1}, \quad |x|<1. \end{align*}