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May 28th, 2019, 11:25 AM  #1 
Math Team Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 903 Thanks: 62 Math Focus: सामान्य गणित  Infinite series
I cannot remember the infinite series for $\displaystyle \sqrt {1  a}$ for $\displaystyle 0 < a <<1$

May 28th, 2019, 12:00 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,028 Thanks: 2259 
Can you remember this theorem?

May 28th, 2019, 06:27 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
Or even use Google.

May 29th, 2019, 01:05 AM  #4 
Senior Member Joined: Dec 2015 From: somewhere Posts: 729 Thanks: 98 
I think that you must apply complexanalysis 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. 
May 29th, 2019, 01:27 AM  #5  
Senior Member Joined: Oct 2009 Posts: 884 Thanks: 340  Quote:
Last edited by skipjack; May 29th, 2019 at 05:45 AM.  
May 29th, 2019, 06:12 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics 
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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