## Example 8.4.4 (b): Taylor Series

If the given function had a convergent Taylor series, what would it be:

*f(x) = e*around^{x}*c = 0*and*f(x) = e*around^{x}*c = 1**g(x) = cos(x)*around*c = 0*and*g(x) = cos(x)*around*c = Pi/2**h(x) = sin(x)*around*c = 0**k(x) =*around^{1}/_{1+x}*c = 2*

In each case we need to find
for the
given function at the specified center. We therefore compute derivatives
until we detect a pattern. Once we see the pattern we have the general
coefficient *a _{n}* and we can write down our
Taylor series.

**1. f(x) = e^{x}**

The pattern is clear ... it remains to divide by

c=0c=1f(x) = e^{x}f(0) = 1f(1) = ef '(x) = e^{x}f '(0) = 1f '(1) = e... f^{(n)}(x) = e^{x}f^{(n)}(0) = 1f^{(n)}(1) = e

*n!*to get the

*a*:

_{n}

T_{f}(x, 0) =^{1}/_{n!}x^{n}T_{f}(x, 1) =^{e}/_{n!}(x-1)^{n}

**2. g(x) = cos(x)**

The pattern now repeats so that:

c=0c=Pi/2f(x) = cos(x)f(0) = 1f(Pi/2) = 0f '(x) = -sin(x)f '(0) = 0f '(Pi/2) = -1f ''(x) = -cos(x)f ''(0) = -1f ''(Pi/2) = 0f '''(x) = sin(x)f '''(0) = 0f '''(Pi/2) = 1...

T_{f}(x, 0) = 1 + 0 x - 1/2! x^{2}+ 0 x^{3}+ 1/4! x^{4}+ ... =T_{f}(x, 1) = 0 - 1/1! x + 0 x^{2}+ 1/3! x^{3}+ 0 x^{4}+ ... = -

**3. h(x) = sin(x)**

The pattern now repeats so that:

c=0f(x) = sin(x)f(0) = 0f '(x) = cos(x)f '(0) = 1f ''(x) = -sin(x)f ''(0) = 0f '''(x) = -cos(x)f '''(0) = -1...

T_{f}(x, 0) = 0 + 1 x + 0 x^{2}+ 1/3! x^{3}+ 0 x^{4}+ ... =

Note that if we substitute *x - Pi/2* for *x* and add a
minus sign, we get the second series in the second example. That actually
proves a trig identity - which one ?

**4. k(x) = ^{1}/_{1+x}**

c=2f(x) =^{1}/_{1+x}f(0) = 1/3f '(x) =^{-1}/_{(1+x)2}f '(0) = -1/3^{2}f ''(x) =^{1*2}/_{(1+x)3}f ''(0) = 2!/3^{3}f '''(x) =^{-1*2*3}/_{(1+x)4}f '''(0) = -3!/3^{4}... f^{(n)}(x) =^{(-1)n n!}/_{(1+x)n+1}f^{(n)}(0) =^{(-1)n n!}/_{3n+1}

Thus

Note that we can arrive at the same answer using our - by now familiar - geometric series and some appropriate substitution:

T_{f}(x, 2) =^{(-1)n}/_{3n+1 (x-2)n}