1/(1+x^2)=1-x^2 + x^4-x^6+....+(-1)^n(x^2n)+.... -1 > x > 1
how to get this??.....
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!
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
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
Separate names with a comma.