Mathematical Analysis

[David8765]

I need help to proof how the series meets the equation

 

[mmm4444bot]

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

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[David8765]

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

\(\;\)

I’m sorry. I am new member.

 

[topsquark]

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

-Dan

 

[David8765]

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
Thanks for your help.

 

[Anthobet]

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]

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]

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]

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

 

[Subhotosh Khan]

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"?