Re: Maclaurin's Series

In article
<a158f245-f019-4dfa-a7ac-a57b8ea0e646@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mitch@xxxxxxxxxxxxxxxxxx wrote:

The Maclaurin series for ln(1+x) is:

x - (x^2/2!) + (x^3/3!) - (x^4/4!) + .....

And this is convergent for -1 < x <= 1 (as can be seen by exmaining
the radius of convergence of the given series where nth term is (-1)^(n
+1) * x^n / n! (n = 1,2,3......)

Okay,

Using the above it a easy to see that for instance the Maclaurin
series for ln(1+2x) is convergent for -0.5 < x <= 0.5 because (by the
series for ln(1+x) ) it is convergent for -1 < 2x <= 1

and so on, similarly if we wanted the convergence criteria for the
Maclaurin series of ln(Sqrt(1-x-2x^2)) we could write the function as
ln(Sqrt[(1-2x)(1+x)]) = 0.5 * [ln(1-2x) + ln(1+x)] ans so we get (by
comparison with the above) -0.5 <= x < 0.5.

In a similar way how can we establish the convergence criteria for the
Maclaurin expansion of ln(1+x+x^2).

Can we not say that it is -1 < x + x^2 <= 1 ?

No. That would be the answer for an expansion in powers of (x + x^2),
but Maclaurin is an expansion in powers of x.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.