Maclaurin series for 1/(1+x²)

newton1 · Apr 10, 2003

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

FZ+ · Apr 10, 2003

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 · Apr 10, 2003

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 · Apr 10, 2003

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

Maclaurin series for 1/(1+x²)

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Maclaurin series for 1/(1+x²)

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