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May 28th, 2019, 11:25 AM   #1
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Infinite series

I cannot remember the infinite series for $\displaystyle \sqrt {1 - a}$ for $\displaystyle 0 < a <<1$
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May 28th, 2019, 12:00 PM   #2
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Can you remember this theorem?
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May 28th, 2019, 06:27 PM   #3
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Or even use Google.
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May 29th, 2019, 01:05 AM   #4
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I think that you must apply complex-analysis theorems.
Or use $\displaystyle sqrt(x) =x/sqrt(x)$.
I'm not sure, but try exponential complex variable.
Expand the function with infinite product and then add the square root to each term.
If not, then someone else I hope will assist.

Last edited by skipjack; May 29th, 2019 at 05:45 AM.
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May 29th, 2019, 01:27 AM   #5
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Quote:
Originally Posted by idontknow View Post
I think that you must apply complex-analysis theorems.
Or use $\displaystyle sqrt(x) =x/sqrt(x)$.
I'm not sure, but try exponential complex variable.
Expand the function with infinite product and then add the square root to each term.
If not, then someone else I hope will assist.
Look at the previous replies lol...

Last edited by skipjack; May 29th, 2019 at 05:45 AM.
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May 29th, 2019, 06:12 PM   #6
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Which infinite series? There are infinitely many infinite series for this function (and of course any other smooth function). Do you mean Taylor series?
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