July 11th, 2019, 04:10 PM  #1 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0  Infinite series
hi I found this infinite series in my calculations and I want to ask if it's have a name . I attached it to this thread. if anyone here knows anything about it ,please contact me. 
July 11th, 2019, 05:31 PM  #2 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
I don't know if that specific form has its own name, but it is an application of the Taylor series about x. We use it all the time for a lot of things, including coming up with different approximations of derivatives and choosing (or showing) their order of accuracy. 
July 11th, 2019, 05:55 PM  #3 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0 
this series have a relation between integration and derivatives of the function. is there a form for Taylor series like this one ?! also this series gives a different infinite series if you try it with any type of function. Last edited by Mohammad Hammad; July 11th, 2019 at 05:57 PM. 
July 11th, 2019, 06:07 PM  #4 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
Take the Taylor series expansion of $\displaystyle g(\xi)$ about x, and evaluate at $\displaystyle \xi = 0$. Now let $\displaystyle f(x) = g'(x)$. I believe you have what you wrote.

July 11th, 2019, 06:15 PM  #5 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0 
ok I will try to check that. thanks 
July 11th, 2019, 06:30 PM  #6 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0 
I tried that but I couldn't reach to the integration of the function. I am not expert in that also I know that Taylor series doesn't talk about the integration of the function. how it come that I can have the integration in Taylor series ?! 
July 11th, 2019, 07:35 PM  #7 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
Fundamental theorem of calculus. If f(x) = g'(x), then the integral is g(x)  g(0).

July 12th, 2019, 01:49 AM  #8 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0 
I didn't reach to this series from Taylor series. and if you check it you will find that it's ( +  +  ) and Taylor series is a summation and it's plus. I tried it and gives me a different answers . for example Taylor series for 1/((x+1)^2)= 1  2 x + 3 x^2  4 x^3 + 5 x^4  6 x^5 + 7 x^6  8 x^7 + 9 x^8  10 x^9 + 11 x^10 + O(x^11) but for this series it will give 1/((x+1)^2)=1  (2x/((x+1)^3)  (3x^2/((x+1)^4)  (4x^3/((x+1)^5) ... Last edited by Mohammad Hammad; July 12th, 2019 at 01:53 AM. 
July 12th, 2019, 06:37 AM  #9 
Member Joined: Jun 2019 From: USA Posts: 69 Thanks: 27 
Aha! There was a slight error in your original equation. It was off by an order in the derivatives. (In retrospect, this should have been obvious by looking at the "units.") See the derivation.

July 12th, 2019, 07:28 AM  #10 
Newbie Joined: Jul 2019 From: Jordan Posts: 8 Thanks: 0 
Amazing which program did you use to writ ethis expansion ? I like to use it in my calculations thanks also I didn't get where is the error ? Last edited by Mohammad Hammad; July 12th, 2019 at 07:30 AM. 

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