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

[newton1]

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

 

[FZ+]

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!

 

[dg]

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.

 

[arcnets]

Yes. Because, if you abbreviate the right hand side as S, then obviously
y*S = S-1