# Find the Taylor polynomial of degree n=4 for each function expanded about the given value of x_0 f(x)=cos(x), x_0 =0

Find the Taylor polynomial of degree n=4 for each function expanded about the given value of ${x}_{0}$.
$f\left(x\right)=\mathrm{cos}\left(x\right),{x}_{0}=0$
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$f\left(x\right)=f\left(x0\right)+\frac{xf\ast \left(x0\right)}{1!}+\frac{{x}^{2}f\ast \left(x0\right)}{2!}+...$
$f\left(x\right)=\mathrm{cos}0=1$
${f}^{\prime }\left(x\right)=-\mathrm{sin}x⇒{f}^{\prime }\left(0\right)=-\mathrm{sin}0=0$
${f}^{″}\left(x\right)=-\mathrm{cos}x⇒{f}^{″}\left(0\right)=-\mathrm{cos}0=-1$
${f}^{‴}\left(x\right)=\mathrm{sin}x⇒{f}^{‴}\left(0\right)=\mathrm{sin}0=0$
${f}^{\left(iv\right)}\left(x\right)=\mathrm{cos}x⇒{f}^{\left(iv\right)}\left(0\right)=\mathrm{cos}0=1$
${f}^{\left(v\right)}\left(x\right)=-\mathrm{sin}x⇒{f}^{\left(v\right)}\left(0\right)=-\mathrm{sin}0=0$
${f}^{\left(vi\right)}\left(x\right)=-\mathrm{cos}x⇒{f}^{\left(vi\right)}\left(0\right)=-\mathrm{cos}0=-1$
so on
$f\left(x\right)=1-\frac{{x}^{2}}{2!}+\frac{{x}^{4}}{4!}-\frac{{x}^{6}}{6!}+....=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}{x}^{2n}}{\left(2n\right)!}$