0
$\begingroup$

How would you find the Cauchy product of two power series of different powers? For example, I want to find the Cauchy product of the two series $ \exp(x) = \sum_{k=0}^{\infty}\frac{x^k}{k!}$ and $ \cos(x) = \sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!}$ directly. I tried writing $x^{2k}$ as $(x^2)^k$, but I'm not sure if I can still use the Cauchy product definition in this form.

$\endgroup$
0
$\begingroup$

If you see the power series of the cosine function as$$\require{cancel}1+0\times x-\frac1{2!}x^2+0\times x^3+\frac1{4!}x^4+\cdots,$$then you have\begin{align}\exp(x)\cos(x)&=\left(1+x+\frac1{2!}x^2+\frac1{3!}x^3+\cdots\right)\left(1+0\times x-\frac1{2!}x^2+0\times x^3+\frac1{4!}x^4+\cdots\right)\\&=1+x+\cancel{\left(-\frac1{2!}+\frac1{2!}\right)}x^2+\left(-\frac1{2!}+\frac1{3!}\right)x^3+\cdots\\&=1+x-\frac13x^3-\frac16x^4+\cdots\end{align}As far as I know, there is no simple expression for this series.

| cite | improve this answer | |
$\endgroup$

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.