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July 11th, 2019, 04:10 PM   #1
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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.
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July 11th, 2019, 05:31 PM   #2
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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.
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July 11th, 2019, 05:55 PM   #3
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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.
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July 11th, 2019, 06:07 PM   #4
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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.
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July 11th, 2019, 06:15 PM   #5
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ok
I will try to check that.
thanks
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July 11th, 2019, 06:30 PM   #6
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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 ?!
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July 11th, 2019, 07:35 PM   #7
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Fundamental theorem of calculus. If f(x) = g'(x), then the integral is g(x) - g(0).
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July 12th, 2019, 01:49 AM   #8
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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.
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July 12th, 2019, 06:37 AM   #9
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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.
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July 12th, 2019, 07:28 AM   #10
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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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