# Use the definition of Taylor series to find the Taylor series, centered at c, for the function. f(x)=cos x, c=frac{-pi}{4}

Use the definition of Taylor series to find the Taylor series, centered at c, for the function.
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The Taylor polynomial for a function f(x) is
$f\left(x\right)=f\left(c\right)+{f}^{\prime }\left(c\right)\left(x-c\right)+\frac{{f}^{″}\left(c\right)}{2!}\left(x-c{\right)}^{2}+\frac{{f}^{‴}\left(c\right)}{3!}\left(x-c{\right)}^{3}+...+{f}^{\left(n\right)}\frac{c}{n!}\left(x-c{\right)}^{n}$
Now is
$f\left(c\right)=\mathrm{cos}\left(\frac{\pi }{4}\right)$
$=\frac{1}{\sqrt{2}}$
${f}^{\prime }\left(c\right)=-\mathrm{sin}\left(\frac{\pi }{4}\right)$
$=-\frac{1}{\sqrt{2}}$
${f}^{″}\left(c\right)=-\mathrm{cos}\left(\frac{\pi }{4}\right)$
$=\frac{-1}{\sqrt{2}}$
${f}^{‴}\left(c\right)=\mathrm{sin}\left(\frac{\pi }{4}\right)$
$=\frac{-1}{\sqrt{2}}$
Then is
$\mathrm{cos}x=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}\left(x-\frac{\pi }{4}\right)}-\frac{\sqrt{2}}{\left(2!\right)\left(x-\frac{\pi }{4}{\right)}^{2}}+\frac{\sqrt{2}}{\left(3!\right)\left(x-\frac{\pi }{4}{\right)}^{3}}+...+\frac{{f}^{n}\left(c\right)}{\left(n!\right)\left(x-\frac{\pi }{4}{\right)}^{n}}$
$=\frac{1}{\sqrt{2}}+\frac{-1}{\sqrt{2}\left(x-\frac{\pi }{4}\right)}+\frac{-1\left(2\right)}{\left(2!\right)\left(x-\frac{\pi }{4}{\right)}^{2}}+\frac{\sqrt{2}}{\left(3!\right)\left(x-\frac{\pi }{4}{\right)}^{3}}+...+\frac{{f}^{n}\left(c\right)}{\left(n!\right)\left(x-pi/4{\right)}^{n}}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2}}\frac{\left(-1{\right)}^{\frac{n\left(n+1\right)}{2}}}{n!}\left(x-\frac{\pi }{4}{\right)}^{n}$
Result
$\mathrm{cos}x=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{\sqrt{2}}\frac{\left(-1{\right)}^{\frac{n\left(n+1\right)}{2}}}{n!}\left(x-\frac{\pi }{4}{\right)}^{n}$