## 8.4. Taylor Series

### Proposition 8.4.11: Taylor Series for the Exponential Function

We know - assuming the assumptions of Taylor's theorem are satisfied (which they are) - that:

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

where I can pick my preferred form of the remainder. Let's pick the regular
remainder with *c = 0*:

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

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

But for a fixed *x* we have
already shown before that for
any fixed number *x*

^{|x|n}/_{n!}= 0

which finishes the proof.

### Fun Facts:

*e*(Euler's Formula)^{ix}= cos(x) + i sin(x)*(cos(x) + i sin(x))*(De Moivre's Formula)^{n}= cos(nx) + i sin(nx)*e*and^{x}= e^{x}*e*^{x}= e^{x}

**Fact 1:**The

*i*in Euler's Formula is the imaginary unit, i.e. that symbol whose square is -1:

*i*. Incidentally, that implies:

^{2}= -1

i^{2}= -1i^{3}= i^{2}i = -1 i = -ii^{4}= i^{2}i^{2}= (-1) (-1) = 1i^{5}= i^{4}i = ii^{6}= i^{4}i^{2}= -1- ...

If we now substitute *ix* into the series expression for the
exponential function we get:

e^{ix}= 1 + (ix) + 1/2! (ix)^{2}+ 1/3! (ix)^{3}+ 1/4! (ix)^{4}+ 1/5! (ix)^{5}+ 1/6! (ix)^{6}+ 1/7! (ix)^{7}+ ...

= 1 + ix - 1/2! x^{2}- i 1/3! x^{3}+ 1/4! x^{4}+ i 1/5! x^{5}- 1/6! x^{6}- i 1/7! x^{7}+ ...

= 1 - 1/2! x^{2}+ 1/4! x^{4}- 1/6! x^{6}+ ... + ix - i 1/3! x^{3}+ i 1/5! x^{5}- i 1/7! x^{7}+ ...

= + i

= cos(x) + i sin(x))

assuming we know the series expansions for *sin* and
*cos*.

**Fact 2:** De Moivre's Formula
*(cos(x) + i sin(x)) ^{n} = cos(nx) + i sin(nx)* is now
merely an application of the Euler's Formula together with the properties
of the exponential functions:

(cos(x) + i sin(x))^{n}) = (e^{ix})^{n}= e^{inx}= e^{i (nx)}= cos(nx) + i sin(nx)

**Fact 3:** We may have proved that elsewhere, so this would be
a good review exercise.