## 8.4. Taylor Series

### Example 8.4.18 (b): Finding Taylor Series by Substituion

### Example

Find MacLaurin series for*f(x) =*

^{1}/_{1 + x2}The known series to start with is, of course, the Geometric series:

^{1}/_{1-x}= x^{n}

for *|x| < 1*. If we now substitute *-x ^{2}*
for

*x*we get:

^{1}/_{1+x2}=^{1}/_{1-(-x2)}= (-x^{2})^{n}= (-1)^{n}x^{2n}

and the series converges to the function for *|x| ^{2} < 1*,
i.e. for

*|x| < 1*.

It is interesting that the series only converges for
*|x| < 1* even though the function it represents is infinitely
often differentiable *for all* *x*. The reason for that is that
the function has a singularity at *x = i*, the imaginary unit, because
*1 + (i) ^{2} = 0*. That singularity is invisible when we
consider real numbers only, but is just on the border of a circle (of
convergence) centered at zero with radius 1. Thus, this invisible imaginary
singularity prevents the radius of convergence to be bigger than 1, even though
there is nothing wrong with our function on the real line.

Please note that this method has the big advantage that we don't have to
*prove* convergence and the substitution usually reveals the radius of
convergence quickly. But we do need to have an idea of which series to start
with, which can be quite tricky.

(-1)^{n}x^{2n}f(x) =^{1}/_{1+x2}