# Use power series operations to find the Taylor series at x=0 for the following function. (5x^2)/2-5+5cosx.

Use power series operations to find the Taylor series at x=0 for the following function.
$\frac{5{x}^{2}}{2}-5+5\mathrm{cos}x$.
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Alara Mccarthy
Two Taylor series at x=0 can be added term by term, giving rise to another Taylor series. The radius of convergence is going to be the minimum of the radii of convergence of the two (Taylor) power series. In our case, the term $\left(\frac{5{x}^{2}}{2}-5\right)$ is already a Taylor series at x=0. The Taylor series for cosxcosx at x=0 is given by
$1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots +{\left(-1\right)}^{n}\frac{{x}^{2}n}{\left(2n\right)!}+\dots$
Adding term by term the two Taylor series, we get:
$\left(-5+1\right)+\left(\frac{5}{2}-\frac{1}{2}\right){x}^{2}+\frac{{x}^{4}}{4!}=\frac{{x}^{6}}{6!}+\dots +{\left(-1\right)}^{n}\frac{{x}^{2}n}{\left(2n\right)!}+\dots$
which is equal to
$-4+2{x}^{2}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+\dots +{\left(-1\right)}^{n}\frac{{x}^{2}n}{\left(2n\right)!}+\dots$