The Maclaurin series for f(x)=cos⁡(x)\displaystyle {f{{\left({x}\right)}}}={\cos{{\left({x}\right)}}} is
∑n=0∞(−1)n(x)2n(2n)!=1−x22!+x44!−x66!+…\displaystyle {\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}^{{n}}{\left({x}\right)}^{{{2}{n}}}}}{{{\left({2}{n}\right)}!}}={1}-\frac{{x}^{{2}}}{{{2}!}}+\frac{{x}^{{4}}}{{{4}!}}-\frac{{x}^{{6}}}{{{6}!}}+\ldots

By transforming the series for cos⁡(x)\displaystyle {\cos{{\left({x}\right)}}}, find the first 4 nonzero terms of the Maclaurin series for 1−cos⁡(15x)x\displaystyle \frac{{{1}-{\cos{{\left({15}{x}\right)}}}}}{{x}}.



T(x)=\displaystyle {T}{\left({x}\right)}= + + + + ...
       

Written compactly, this series is

 T(x)=∑n=0∞\displaystyle {T}{\left({x}\right)}={\sum_{{{n}={0}}}^{{\infty}}}   


Using your new series lim⁡→0T(x)=\displaystyle \lim_{{\rightarrow{0}}}{T}{\left({x}\right)}=  



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