Question

\[d/dx \sum_{o}^{\inf} [(x^(2n)))/n!]\] I will award a medal if you help me get the right answer!!

Answer 1

\[d/dx \sum_{n=o}^{\inf} [(x^(2n)))/n!]\]

Answer 2

Derivative of a sum is just the derivative of the individual terms, right?
So you can interchange the sum and derivative because we are essentially taking:
\( \dfrac{d}{dx} \left( f_1 (x) + f_2 (x) + f_3 (x) + \cdots \right) = \dfrac{d}{dx} f_1 (x) + \dfrac{d}{dx} f_2 (x) + \dfrac{d}{dx} f_3(x) + \cdots \)

Answer 3

okay so that leaves us with what?

Answer 4

\( \displaystyle \dfrac{d}{dx} \sum_{n=0}^{\infty} x^{2n} / n! = \sum_{n=0}^{\infty} \dfrac{d}{dx} \left( x^{2n} / n! \right) \)

Answer 5

okay

Answer 6

Then take the derivative, using the power rule. d/dx(x^a) = a x^(a-1). d/dx (x^(2n)) = ?

Answer 7

2nx^(2n-1)

Answer 8

Yep, looks good.
So we would now have:
\( \displaystyle = \sum_{n=0}^{\infty} 2n x^{2n - 1} / n! \)

There may be some minor simplifications (n=0 gives us 0, so it can be lost. n and n! = n(n-1)! have common factors to cancel), but that may be done as you feel necessary here.

Answer 9

okay so how do we find what it equals?

Answer 10

Like, we want to find the actual function with that series expansion?

Answer 11

the answer options are as follows a) -sin(x^2) b) -2xsin(x^2) c) cos(x^2) d) \[e^x^2 \]

Answer 12

or e) 2xe^(x^2)

Answer 13

d) was supposed to say e^(x^2)

Answer 14

any idea?

Answer 15

Hm.. if you know the expansion for e^x:
\( \displaystyle e^{x} = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} \)
which when we have \(x \to x^2\), this becomes what we started with:
\( \displaystyle e^{x^2} = \sum_{n=0}^{\infty} \dfrac{x^{2n}}{n!} \)
When we take this function's derivative (which is equivalent to above, this is d/dx(x^2) e^(x^2) = 2x e^(x^2).

Otherwise we can put the derivative expression in the same terms as the e^(x^2) expansion by making those simplifications I mentioned above:
\( \displaystyle \sum_{n=0}^{\infty} 2n \dfrac{x^{2n-1}}{n!} = \sum_{n=1}^{\infty} 2 \dfrac{x^{2(n-1)+1}}{(n-1)!} = 2x \color{blue}{\sum_{n=0}^{\infty} \dfrac{x^{2n}}{n!}} \)

Answer 16

okay, so do we get d as the answer then?

Answer 17

I condensed a lot of steps in that last equation, but if anything is unclear I can elaborate that.

Answer 18

No, i understand the steps you took!

Answer 19

(d) is just e^(x^2), but that was just the original thing we started with. What we took was the derivative of the e^(x^2) expansion, so it is the equivalent of taking the derivative of e^(x^2).

Answer 20

so would it be e then?

Answer 21

d/dx ( e^(x^2) ) is just a chain rule application, we take the derivative of the x^2 times the original function (because exponential). 2x * e^(x^2), would be (e) yes.

Answer 22

Okay, thanks so much for your help!

Answer 23

Yup, glad to help! :)