# Mathematical Analysis

#### David8765

##### New member

I need help to proof how the series meets the equation

#### mmm4444bot

##### Super Moderator
Staff member
Please follow the forum's submission guidelines and post what you've tried or thought about so far. Thank you.

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#### topsquark

##### Full Member
I'll start you off, though, in case you are having trouble starting it. What are f'(x) and f''(x)?

-Dan

#### David8765

##### New member
I'll start you off, though, in case you are having trouble starting it. What are f'(x) and f''(x)?

-Dan
F’(x) and f’’(x) are the derivative of the function

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#### Anthobet

##### New member
I'll start you off, though, in case you are having trouble starting it. What are f'(x) and f''(x)?

-Dan
I also think you can simplify the summation using some arithmetic and a Maclaurin Series identity.

#### Subhotosh Khan

##### Super Moderator
Staff member
I also think you can simplify the summation using some arithmetic and a Maclaurin Series identity.
Can you calculate f'(x) and f"(x) for the given function?

To start off what is the first derivative of x2n?

#### David8765

##### New member
Can you calculate f'(x) and f"(x) for the given function?

To start off what is the first derivative of x2n?
2nx^(2n-1)

#### pka

##### Elite Member
\displaystyle \begin{align*}y &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{{2^{2n}}{{(n!)}^2}}}} \\ y' &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}n{x^{2n - 1}}}}{{{2^{2n - 1}}{{(n!)}^2}}}} \end{align*}

#### Subhotosh Khan

##### Super Moderator
Staff member
Can you calculate f'(x) and f"(x) for the given function?

To start off what is the first derivative of x2n?
Look at pka's answer in response #9.

Now tell us:

what would be the expression for f"(x) - or y"?