$\dfrac {\mathrm d^{1000}}{\mathrm d x^{1000}}(x^{2}+4)e^{2x}$ using Maclaurin series?

Question

Find the maclaurin series for $f(x)=(x^{2}+4)e^{2x}$ and use this to calculate the 1000th derivative for $f(x)$ at $x=0$.

Im not sure how to start this, any help is appreciated thanks!

Answer 1

Hint: $f(x)=x^2e^{2x}+4e^{2x}$. Now it is easy to find the Maclaurin series, then we can differentiate term-by-term.

Answer 2

$$ e^x=1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}+\ldots $$ $$ e^{2x}=1+(2x)+\frac{(2x)^2}{2!}+\ldots+\frac{(2x)^n}{n!}+\ldots $$ $$ (x^2+4)e^{2x}=\sum_{n=0}^\infty(x^2+4)\frac{(2x)^n}{n!}=\sum_{n=0}^\infty\frac{2^n}{n!}x^{n+2}+\sum_{n=0}^\infty \frac{2^{n+2}}{n!}x^n$$

Now the 1000th derivative at $x=0$ is just the coefficient at $x^{1000}$ times $1000!$ which is $$ 1000!\left[\frac{2^{998}}{998!}+\frac{2^{1002}}{1000!}\right] $$

Answer 3

Here is the derivation

$$ f(x)=(x^2+4)e^{2x}=\sum_{k=0}^{\infty} \frac{2^kx^{k+2}}{k!} + \sum_{k=0}^{\infty} \frac{2^{k+2}x^{k}}{k!}$$

$$ = \sum_{k=2}^{\infty} \frac{2^{k-2}x^{k}}{(k-2)!} + \sum_{k=2}^{\infty} \frac{2^{k+2}x^{k}}{k!} +4+8x $$

$$= 4+8x+\sum_{k=2}^{\infty} \left(\frac{2^{k-2}}{(k-2)!}+\frac{2^{k+2}}{k!}\right)x^k $$

$$ \implies \frac{f^{(k)}(0)}{k!} = \frac{2^{k-2}}{(k-2)!}+\frac{2^{k+2}}{k!},\quad k\geq2 $$

$$ \implies f^{(k)}(0) = k! \left(\frac{2^{k-2}}{(k-2)!}+\frac{2^{k+2}}{k!}\right),\quad k\geq2 $$

$$ \implies \frac{{2}^{k}}{4} \left( {k}^{2}-k+16 \right) $$

Substituting $k=1000$ in the above formula gives the desired result.