Taylor and Maclaurin series

[liubava]

Hello, everyone! I am trying to represent this function as Maclaurin series, however I am a bit stuck with finding the n-th order derivative of f(x). Can someone please help me?

 

[Deleted member 4993]

View attachment 25738
Hello, everyone! I am trying to represent this function as Maclaurin series, however I am a bit stuck with finding the n-th order derivative of f(x). Can someone please help me?

What is f'(x)?

What is f"(x)?

 

[liubava]

What is f'(x)?

What is f"(x)?

f'(x) = e^x - e^(-x)
f''(x) = e^x + e^(-x)
f'(0) = 0
f''(0) = 2
I am okay with finding derivatives, I just wonder how I can represent exactly the n-th order derivative as a general formula

 

[liubava]

f'(x) = e^x - e^(-x)
f''(x) = e^x + e^(-x)
f'(0) = 0
f''(0) = 2
I am okay with finding derivatives, I just wonder how I can represent exactly the n-th order derivative as a general formula

here is what I am talking about

 

[Deleted member 4993]

f'(x) = e^x - e^(-x)
f''(x) = e^x + e^(-x)
f'(0) = 0
f''(0) = 2
I am okay with finding derivatives, I just wonder how I can represent exactly the n-th order derivative as a general formula

Continue on → f"' & f""

and so on .....

Do you see a pattern?

 

[liubava]

it's 0, 2, 0, 2 ....
0 - for odd n
2 - for even n
but still, I have no idea how to show it as a formula

 

[liubava]

it's 0, 2, 0, 2 ....
0 - for odd n
2 - for even n
but still, I have no idea how to show it as a formula

so, after several attempts, I end up with 2* x^2n / (2n!). Is it correct?

 

[skeeter]

[MATH]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^n}{n!} + ...[/MATH]
[MATH]e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ... + \frac{(-1)^nx^n}{n!} + ...[/MATH]
------------------------------------------------------------------

[MATH]e^x+e^{-x} = 2\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ... + \frac{x^{2n}}{(2n)!} + ... \right)[/MATH]

 

[liubava]

[MATH]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^n}{n!} + ...[/MATH]
[MATH]e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ... + \frac{(-1)^nx^n}{n!} + ...[/MATH]
------------------------------------------------------------------

[MATH]e^x+e^{-x} = 2\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ... + \frac{x^{2n}}{(2n)!} + ... \right)[/MATH]

Thank you very much!!! I understood the logic

 

[HallsofIvy]

You could also note that \(\displaystyle e^x- e^{-x}= 2\frac{e^x- e^{-x}}{2}= 2 sinh(x)\)
and use the fact that \(\displaystyle sinh'(x)= cosh(x)\) and \(\displaystyle cosh'(x)= sinh(x)\).