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.

Answer 1

oh gosh not this stuff again

Answer 2

I already found the Maclaurin series and I got:

\[\sum_{n=0}^{\infty}\frac{ (x^2+4)(2x)^n }{ n! }\]

Answer 3

Now how do I compute the 1000th derivative?

Answer 4

just plug 1000?

Answer 5

Because you obviously don't differentiate that 100 times :P .

Answer 6

Now that, I am not sure about.

Answer 7

Because that's a summation. It's not the derivative.

Answer 8

Ohh wait... It's a maclaurin series right? SO I already took the derivative.

Answer 9

Ohh... What a dumb question...

Answer 10

Thanks lol.

Answer 11

But, isn't that 0?

Answer 12

because we have :

\[\sum_{n=0}^{\infty}\frac{ (x^2+4)(2x)^n }{ n! }\]

So if we plug in 0 we get 0 for everything.

Answer 13

oh no wait

Answer 14

u know that f^1000(0) / n! = the coefficient of the term containing x^1000 term in the power series

Answer 15

Did you mean f(1000)/n! the coefficient of the term containing x^1000 term in the power series?

Answer 16

sithandgiggles is here he is pro

Answer 17

That he is my friend.

Answer 18

I don't think that's the right Mclaurin series for your function... I've already answered a similar question asking for the Mclaurin series of that same function. I spent some time looking for a pattern for the n-th derivative, and it doesn't look anything like what your series suggests. Give me a sec to look for it.

Answer 19

Well here is another one I found:

\[\sum_{k=0}^{\infty}\frac{ (2^k)(x^k+2) }{ k!}+\sum_{k=0}^{\infty}\frac{ 2^{k+2}x^k }{ k!}\]