Answer is: option4
\( e^{-x} \)Solution:
\[ f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!} =1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots \]
Since this has both odd and even powers of \(x\), it is neither \( \sin x \) nor \( \cos x \).
Compare with:
\[ e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots \]
\[ e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots \]
Therefore, \[ f(x)=e^{-x} \]
