# Thread: Maclaurin And Taylor Series

1. ## Maclaurin And Taylor Series

For example use sigma notation to write maclaurin seires ofr the function

e^x

use sigma notation to write taylor seires ofr the function

e^x x0=1

this is very easy function
my solution step for maclaurin is

1. find f , f', f'', f''', f'''' when x=0
2. then find p0, p1, p2, p3, p4
3. look at the regularity of them, write the sigma notation

same as taylor seires except first step is when x=1

I just want to know is this solution ok or not?
Do anybody have any other solution?

Thank you!

2. Originally Posted by soleilion
For example use sigma notation to write maclaurin seires ofr the function

e^x

use sigma notation to write taylor seires ofr the function

e^x x0=1

this is very easy function
my solution step for maclaurin is

1. find f , f', f'', f''', f'''' when x=0
2. then find p0, p1, p2, p3, p4
3. look at the regularity of them, write the sigma notation

same as taylor seires except first step is when x=1

I just want to know is this solution ok or not?
Do anybody have any other solution?

Thank you!
It is well known that $\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

3. Originally Posted by Mathstud28
It is well known that $\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$

yeah, I know
this is easy example
If I have a function is not that easy, not well know
How can I do?

4. Originally Posted by soleilion
yeah, I know
this is easy example
If I have a function is not that easy, not well know
How can I do?
ok for example how about $\displaystyle \frac{-1}{(1+x^2)^2}$? since we know that $\displaystyle \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2}$ and since we know using a simple substitution of $\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$ to get $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ so now using that knowledge we see that $\displaystyle \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}$

5. Originally Posted by Mathstud28
ok for example how about $\displaystyle \frac{-1}{(1+x^2)^2}$? since we know that $\displaystyle \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2}$ and since we know using a simple substitution of $\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$ to get $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ so now using that knowledge we see that $\displaystyle \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}$
I think d/dx [1/(1+x^2)]= -2x(1+x^2)^-2
and I dont understand since you are using a simple substitution

6. Originally Posted by soleilion
I think d/dx [1/(1+x^2)]= -2x(1+x^2)^-2
and I dont understand since you are using a simple substitution
Wow I am really out of it tonight...well i have no idea how why I did that but I will do another to make more sense...$\displaystyle \frac{-1}{(1+x)^2}=\frac{D\bigg[\frac{1}{1+x}\bigg]}{dx}=\frac{D\bigg[\sum_{n=0}^{\infty}(-1)^nx^{n}\bigg]}{dx}=\sum_{n=0}^{\infty}n(-1)^nx^{n-1}$

7. Originally Posted by Mathstud28
Wow I am really out of it tonight...well i have no idea how why I did that but I will do another to make more sense...$\displaystyle \frac{-1}{(1+x)^2}=\frac{D\bigg[\frac{1}{1+x}\bigg]}{dx}=\frac{D\bigg[\sum_{n=0}^{\infty}(-1)^nx^{n}\bigg]}{dx}=\sum_{n=0}^{\infty}n(-1)^nx^{n-1}$
If I do maclaurin and taylor problem
I'm still confusing right now.

8. Originally Posted by Mathstud28
ok for example how about $\displaystyle \frac{-1}{(1+x^2)^2}$? since we know that $\displaystyle \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2}$ and since we know using a simple substitution of $\displaystyle \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n}$ to get $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$ so now using that knowledge we see that $\displaystyle \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}$
He was just typing quickly, I belive what he ment is

let $\displaystyle f(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

Then $\displaystyle f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n-1}$

now to get the power series we wanted we would evaluate $\displaystyle f'(-x^2)$

so

$\displaystyle f'(-x^2)=\frac{1}{(1+x^2)^2}=\sum_{n=0}^{\infty}n(-x^2)^{n-1}=\sum_{n=0}^{\infty}n(-1)^{n-1}(x)^{2n-2}$

I hope this helps

9. Originally Posted by TheEmptySet
He was just typing quickly, I belive what he ment is

let $\displaystyle f(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

Then $\displaystyle f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n-1}$

now to get the power series we wanted we would evaluate $\displaystyle f'(-x^2)$

so

$\displaystyle f'(-x^2)=\frac{1}{(1+x^2)^2}=\sum_{n=0}^{\infty}n(-x^2)^{n-1}=\sum_{n=0}^{\infty}n(-1)^{n-1}(x)^{2n-2}$

I hope this helps

why 1/1-x=sigma x^n???????????????

10. Originally Posted by soleilion
If I do maclaurin and taylor problem
I'm still confusing right now.
yeah for example? say you wanted to do $\displaystyle arctan\bigg(\frac{3x}{11}\bigg)$...this is equal to $\displaystyle \frac{1}{11}\int{\frac{3}{1+\bigg(\frac{3}{11}x\bi gg)^2}}dx$ so this is equal to $\displaystyle \frac{3}{11}\int\sum_{n=0}^{\infty}(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n}dx=\sum_{n=0}^ {\infty}\frac{(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n+1}}{2n+1}$

11. Originally Posted by soleilion
why 1/1-x=sigma x^n???????????????
It is just a commonly known power series

12. Originally Posted by Mathstud28
It is just a commonly known power series
Maclaurin Series -- from Wolfram MathWorld

13. Originally Posted by Mathstud28
yeah for example? say you wanted to do $\displaystyle arctan\bigg(\frac{3x}{11}\bigg)$...this is equal to $\displaystyle \frac{1}{11}\int{\frac{3}{1+\bigg(\frac{3}{11}x\bi gg)^2}}dx$ so this is equal to $\displaystyle \frac{3}{11}\int\sum_{n=0}^{\infty}(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n}dx=\sum_{n=0}^ {\infty}\frac{(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n+1}}{2n+1}$
My HW question is not that hard

they are only cos(pi*x) or ln(1+x) or xsinx

If I do my way, I know how to fix them
but I see the solution mannual
they are different, I dont understand either

I just wanna know how to do this kind of problem
I want to have something like process, so I can follow it.

14. Originally Posted by soleilion
My HW question is not that hard

they are only cos(pi*x) or ln(1+x) or xsinx

If I do my way, I know how to fix them
but I see the solution mannual
they are different, I dont understand either

I just wanna know how to do this kind of problem
I want to have something like process, so I can follow it.
The first one you just insert $\displaystyle \pi{x}$ into the powers series of cos(x)...the second one you either know the formula or you use the fact that $\displaystyle ln(x+1)=\int\sum_{n=0}^{\infty}(-1)^{n}x^{n}dx$ and the third you take the regular power series for sin(x) and add one to the exponent of x

15. Originally Posted by Mathstud28
Thank you very much
I saw this website, it's very good
I understand #3 to #25
When I do this kind of problem, I use that method
For explicit form, I can not use that because my instructor want us to solve step by step

Page 1 of 2 12 Last