Discover the Power of Maclaurin's Series: Solving 1/(1+x^2) for -1 > x > 1

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The discussion focuses on deriving the Maclaurin series for the function 1/(1+x^2) within the interval -1 > x > 1. It starts with the formula for Maclaurin's series, which involves evaluating the function and its derivatives at zero. The series expansion results in 1-x^2+x^4-x^6+...+(-1)^n(x^2n)+..., demonstrating the alternating nature of the series. Additionally, the geometric series formula is mentioned as a simpler method to arrive at the same result by substituting y with -x^2. This highlights the versatility of series expansions in solving complex functions.
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1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...
 
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It's basically Maclaurin's series.

f(x) = f(0) + x*f'(0) + (x^2)/2! * f''(0) +...

Which derives the binomial series:
(1+a)^n = 1+ na + n(n-1)/2! * a^2 + n(n-1)(n-2)/3! * a^3 ... As long as |a|<1

Substitute a = x^2 and n = -1 et viola!
 


Originally posted by Newton1
1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...

If you do not want to go all the way to the general theory of series of powers (Taylor and MacLaurin series), you can simply use the result of the geometric series

1/(1-y)=1+y+y^2+...+y^n+...

which can be obtained with basic arithmetic arguments and substistute y with -x^2.
 
Yes. Because, if you abbreviate the right hand side as S, then obviously
y*S = S-1
 

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