Science:Math Exam Resources/Courses/MATH103/April 2014/Question 07 (a)/Solution 1

Recall that

$e^{x}=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}.$

Note: If you have forgotten the series expansion of $e^{x}$ (at $0$), recall that, in general, such an expansion has the form

$f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}.$

For $f(x)=e^{x}$, we have $f^{(1)}(x)=f'(x)=e^{x}$ and it follows that $f^{(n)}(x)=e^{x}$ for all $n$. Thus, $f^{(n)}(0)=1$ and we get

$e^{x}=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}$

as above.

Now replacing $x$ by $-x^{2}$ in the above expansion yields

$e^{-x^{2}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}x^{2n}.$

Since only even powers of $x$ appear in this series, $a_{n}=0$ whenever $n$ is odd. In particular, $a_{11}=0$.

On the other hand, $a_{10}x^{10}=a_{10}x^{2n}$ for $n=5$. But the coefficient of $x^{2n}$ is ${\frac {(-1)^{n}}{n!}}$. For $n=5$, this yields $a_{10}={\frac {(-1)^{5}}{5!}}=-{\frac {1}{120}}$.