## Example 8.4.18 (a): Finding Taylor Series by Definition

*a*to find the

_{n}=*a*.

_{n}
This in theory will always give you the right coefficients, but practically
speaking there are two problems: first you need to determine a *general*
formula for the *n*-th derivative, and second you still need to prove
that the series you obtain really converges to the original function.

### Example

- Find MacLaurin series for
- Guess what the
**general binomial series**for*(1+x)*might be.^{a}

A MacLaurin series is just a Taylor series centered at *c=0*.
If the function had a Taylor series, the coefficients would be

a_{n}=

Let's compute the derivatives:

f(x) = (1+x)^{1/2}f(0) = 1f '(x) =^{1}/_{2}(1+x)^{1/2 - 1}f '(0) =^{1}/_{2}f ''(x) =^{1}/_{2}(^{1}/_{2}-1) (1+x)^{1/2 - 2}f ''(0) =^{1}/_{2}(^{1}/_{2}-1) =^{-1}/_{4}f '''(x) =^{1}/_{2}(^{1}/_{2}-1)(^{1}/_{2}-2) (1+x)^{1/2 - 3}f '''(0) =^{1}/_{2}(^{1}/_{2}-1)(^{1}/_{2}-2)=^{3}/_{8}

and so on, so that the general term would be:

for n > 1

This is not that useful (because of the ...) but after some thought we come up with:

f^{ (n)}(0) =

or, since we really need the *a _{n}* we have:

a_{n}=

Thus, we claim

= a_{n}x^{n}= x^{n}

To finish this example we need to show that (a) the power series converges, and (b) it converges to the function. For the radius of convergence we compute, as usual:

r = | a_{n}/a_{n+1}| = = |^{n+1}/_{1/2 - n}| = 1

Thus, our series above converges uniformly and absolutely for
*|x| < 1*. But as we know, that does not automatically imply that
it converges *to our function*, so we still need to show that the Taylor or
Lagrange remainder goes to zero.

**Case 1:** *1 > x > 0*: Let's take the Lagrange remainder
with *c=0*:

| R_{n+1}(x) | =

with *a _{n}* as before and

*t*between

*0*and

*x*. We also know that

*0 < t < x*so that:

| R_{n+1}(x) | < |a_{n+1}| x^{n+1}(1+t)^{1/2}

But we already know that the series
*
a _{n} x^{n}*
converges, so that by the Divergence test

*a*. But then the remainder must indeed converge to zero.

_{n}x^{n}= 0**Case 2:** *-1 < x < 0* This time we'll use the regular
remainder:

| R_{n+1}(x) | =^{1}/_{n!}| (x-t)^{n}f^{(n+1)}(t) dt | =

=^{1}/_{n!}| (x-t)^{n}(n+1)! a_{n+1}(1+t)^{1/2-n-1}dt | =

= (n+1) | |a_{n+1}| | (^{x-t}/_{1+t})^{n}(1+t)^{-1/2}dt |

(n+1) | |a_{n+1}| |^{x-t}/_{1+t}|^{n}(1+t)^{-1/2}dt

with *a _{n}* as before and

*x < 0*. But if we consider the function

*f(t) =*for

^{x-t}/_{1+t}*x < t < 0*we can take the derivative to see it is monotone decreasing, with maximum value

*f(x) = 0*on the left side of the interval at

*t = x*. But then the function

*|f(t)|*must have its maximum on the

*right*side at

*t = 0*with value

*|f(t)| = |x|*. Thus:

| R_{n+1}(x) | (n+1) | |a_{n+1}| |^{x-t}/_{1+t}|^{n}(1+t)^{-1/2}dt

(n+1) | |a_{n+1}| |x|^{n}(1+t)^{-1/2}dt =

= (n+1) | |a_{n+1}| |x|^{n}C

for some constant *C*. As before we can now use the Ratio test to
show that the series
*
(n+1) | |a _{n+1}| |x|^{n}*
converges for

*|x| < 1*so that by the Divergence test

*(n+1) | |a*, which finishes the proof.

_{n+1}| |x|^{n}= 0Some proof !!! But it illustrates nicely the pros and cons of using derivatives to determine the Taylor coefficients.

To take a stab at the general case, recall our answer:

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

We simply replace *1/2* by *a* to get the general
series for *(1 + x) ^{a}* centered at

*c = 0*:

### Theorem: The Binomial Series

The general binomial series for*(1+x)*, where

^{a}*a*is not an integer, is:

The series converges uniformly and absolutely for(1 + x)^{a}= x^{n}

*|x| < 1*

The proof is similar to the special case above ...