Question Regarding The Power Series For $e^x$ Currently I'm reading Higher Engineering Math by John Bird and under exponential function he talks about obtaining the value of $e$.
He begins by saying 

The value of $e^x$ can be calculated to any required
  degree of accuracy since it is defined in terms of the following power series:
  $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots $$
  The series is said to converge upon substituting $x = 1$ which gives $e = 2.7183$ correct to 4 decimal places. 

My question is what is this "power series" and where did it come from?
How can one define $e^x$ in terms of the above power series?
 A: The power series you wrote down is the Taylor series of $e^x$ at $a=0$. Recall that the Taylor series of a function $f(x)$ at $a$ is given by 
$$f(x)=\sum_{k=0}^\infty\frac{f^{(k)}(a)}{k!}(x-a)^k.$$
Now if $f(x)=e^x$ and $a=0$, we have $f^{(k)}(0)=e^x\big|_{x=0}=1$. This implies that the Taylor series of $e^x$ at $0$ is given by
$$e^x=\sum_{k=0}^\infty\frac{1}{k!}x^k=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$
A: The said series is the Taylor expansion of the exponential function in $x_0=0$, i.e. is the function series

$$\sum_{i=0}^\infty \frac{d}{dt}(e^t)_{t=0}\frac{x^n}{n!}=
     \sum_{i=0}^\infty e^0\frac{x^n}{n!}=\sum_{n=0}^\infty \frac{x^n}{n!}$$

One can define the function $e^{-} \colon \mathbb R \to \mathbb R$ exactly through this function series: in practice one can define $e^x$ as $\lim_{N \to \infty} \sum_{n=0}^N \frac{x^n}{n!}$ (this is the way as it's defined complex and matrix exponential). 
Hope this may answer your question.
A: Let 
$$
f(x) = e^x
$$
and
$$
g(x) = 1+x+\frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \frac{x^5}{120} + \cdots
$$
Are they the same function?  Notice that $f'(x)=f(x)$ and $g'(x)=g(x)$, and $f(0)=1$ and $g(0)=1$.  So they both satisfy the same first-order differential equation and the same initial condition.  Is that enough to imply they're the same?
According to the conventional methods of solving differential equations taught to (for example) engineering students, the answer would have to be "yes", since one would proceed like this:
$$
\begin{align}
f\;' & = f \\  \\
\frac{df}{dx} & = f \\  \\
\frac{df}{f} & = dx \\  \\
\int\frac{df}{f} & = \int dx \\  \\
\log_e |f| & = x + C\qquad\longleftarrow\text{remember this step} \\  \\
f & = e^{x+C} = e^x\cdot(\text{positive constant})  \\  \\  \\
f & = e^x\cdot\text{constant}.
\end{align}
$$
If $f(0)=1$, that determines what number the "constant" is.
Now look at the step labeled "remember this step".  That says if two functions have the same derivative, then they differ by a constant.  If that's true, then that answers the "theoretical" question above.  That it is true is a consequence of the mean value theorem.
A: Let's start with a simpler task:
We wish to approximate the function $f$ with simpler functions, namely  polynomial functions.
In fact, we will construct an approximation scheme that gives a sequence of polynomial functions that approximate $f$.
The sequence will be denoted by $P_0(x)$, $P_1(x)$, $P_2(x)$, $\ldots\,$, where each $P_n(x)$ is a polynomial of degree $n$.
Moreover,  the scheme will be simple in the sense that, given $P_n$ and $P_m$ with $n<m$, the coefficients $x^i$ are identical for $i\le n$.  
We decide to define $P_n(x)$ to be the polynomial of degree $n$ that has the same value as $f$ at $x=0$ and has the same derivatives as $f$ at $x=0$ up to the $n^{\rm th}$-derivative.
That is, we demand 
$$P_n(0)=f(0),\quad\text { and }\quad P_n^{(k)}(0) = f^{(k)}(0) \text{ for }k=1, 2, \ldots\,, n.$$
So, $P_n(x)$ has the form
$$
P_n(x) =  a_0+ a_1x+ a_2x^2+ a_3x^3+a_4x^4+\cdots+a_n x^n;
$$
where
$$
 P_n^{(k)}(0) = f^{(k)}(0) \quad{\rm for}\quad k=1, 2, \ldots\,, n.
$$
We can solve for the unknown constants without too much difficulty.  We first write down the derivatives of $P_n(x)$:
$$
\eqalign{
P_n(x) &=  a_0+ a_1x+ a_2x^2+ a_3x^3+a_4x^4+\cdots+a_n x^n  \cr
P_n'(x) &=  a_1+ 2a_2x+ 3a_3x^2+4 a_4x^3+\cdots+n a_n x^{n-1}  \cr
P_n''(x) &=  2a_2+ 3\cdot2 a_3x+4\cdot3 a_4x^2+\cdots+n(n-1) a_n x^{n-2}  \cr
P_n'''(x) &= 3\cdot2a_3+ 4\cdot3\cdot2a_4x+\cdots+n(n-1)(n-2) a_n x^{n-3}  \cr
&\ \ \vdots\cr
P_n^{(n)}(x) &= n! a_n \cr
}
$$
Now, we set $f^{(k)}(0) = P_n^{(k)}(0)$ and solve for the unknown coefficients:
$$
\eqalign{
f(0)=P_n(0)&=a_0\cr
f'(0)= P'_n(0)&=a_1\cr
f''(0)=P_n''(0)&=2! a_2\cr
f'''(0) = P_n'''(0)&=3!a_3\cr
&\vdots \cr
f^{(n)}(0)=P_n^{(n)}(0)&=n! a_n\cr
}
\eqalign{
&\Rightarrow\vphantom{f v(0)}\cr
&\Rightarrow\vphantom{f'(0)}\cr
&\Rightarrow\vphantom{f'(0)}\cr
&\Rightarrow\vphantom{f'(0)}\cr
\ \ \ \ &\ \ \vdots\cr
&\Rightarrow\vphantom{f'(0)}\cr
}
\eqalign{
a_0 &= f'(0) \cr
a_1 &=f'(0)\cr
a_2 &= f''(0)/2! \cr
a_3 &=f'''(0)/3!\cr
&\vdots \cr
a_n &=f^{(n)}(0)/n! \cr
}
$$
Thus,
the  Taylor Polynomial of degree $n$ for the function $f$ is
$$\tag{1}
P_n(x) = f(0)+{f'(0)\over 1!}x+{f''(0)\over 2!}x^2+{f'''(0)\over3!} x^3+\cdots+{f^{(n)}(0)\over n!}x^n
$$
Note that to compute a Taylor Polynomial, one need only find the values 
$$\def\hfil{}{\hfil f(0),\  f'(0),\  f''(0),\  \ldots\,, \  f^{(n)}(0) \hfil}
$$
and substitute into  formula (1).

 Let's find the Taylor Polynomial for $f(x)=e^x$:

 Example:  Find the Taylor Polynomial of degree $n$ for
$f(x)=e^x$. 
We first evaluate the derivatives of $e^x$ at $x=0$.
Since ${d^k\over dx^k }e^x = e^x$ for any $k$, we have $f^{(k)}(0)=e^0=1$ for all $k$.
Thus
for  any $n$:
$$
P_n(x) =1+x+{x^2\over 2!}+{x^3\over 3!}+{x^4\over 4!}+\cdots +{x^n\over n!}.
$$
In particular:
$$
\eqalign{
P_0(x)&= 1  \cr
P_1(x)&= 1+x  \cr
P_2(x)&= 1  +x+{x^2\over2}\cr
P_3(x)&= 1  +x+{x^2\over2}+{x^3\over6}.\cr
}
$$
The graphs of $y=e^x$ and the Taylor Polynomials $P_0$, $P_1$, $P_2$, and $P_3$ are shown below:
 
Note that the approximations are very good, for $x$ close to 0, once $n\ge2$.  Also, the graphs of the Taylor Polynomials seem to be converging to the graph of $f$.  However, once $x$ becomes big, the approximation for a fixed $P_n$ becomes bad. 
What if we kept going? That is, what if we formed the infinite degree polynomial?

The power series you have is obtained by taking the Taylor polynomial for $f(x)=e^x$ of "infinite degree".  

That this can actually be done, and that the resulting series represents $f(x)=e^x$ takes a good deal of machinery. One needs the following results and definitions (whose proofs can be found in any Calculus text worthy of the name):
Fact 1: Suppose that $f$ is $n$-times continuously  differentiable on the interval $[0,x]$ (or $[x,0]$) and that  $f^{(n+1)}$ exists on $(0,x)$ (or $(x,0)$).  Then
$$
f(x)=P_n(x) + R_n(x),$$
where $P_n(x)$ is as in (1) and
$$
\tag{2}R_n(x)= {f^{(n+1)}(c)\over (n+1)! } x^{n+1}
$$
for some number $c$ between 0 and $x$.
Definition:
The Taylor Series of $f$ is:
$$
P(x)= f(0)+{f'(0)\over 1!} x +{f''(0)\over 2!} x^2+{f'''(0)\over 3!} x^3+\cdots.
$$
Note that the Taylor Series of the function $f$ is an infinite series whose first $(n+1)$-terms give the Taylor Polynomial of degree $n$ of $f$.
Fact 2: The Taylor Series $P(x)$ of the function $f$ converges to the function value $f(x)$ if and only if the remainder term $R_n(x)$ for the Taylor Polynomial $P_n(x)$ of $f$ converges to zero. That is
$$
f(x) = \sum_{n=0}^\infty {f^{(n)}(0)\over n!} x^n,
$$
If and only if 
$$
\lim_{n\rightarrow\infty} R_n(x) =0,
$$
where $R_n(x)$ is defined as in (2).

Now, the Taylor series for $f(x)=e^x$ can be found as in the previous Example. In fact, we have:
$$
P(x)=\sum_{n=0}^\infty {x^n\over n!}
$$

Using Fact 2, one can show that the Taylor Series for $f$ indeed converges to $e^x$ for any $x$:
$$
e^x=\sum_{n=0}^\infty {x^n\over n!}.
$$
(You just show that for fixed $x$, $R_n(x)$ has limit 0. Here, note that with $x$ fixed, one can bound the terms $|f^{(n+1)(c)|}$ above.)
A: $e$ is defined to be the number such that $\frac{d}{dx}e^x = e^x$. It is easy to show that $\forall x \in \mathbb{R}$, the series $\sum_{i = 0}^\infty \frac{x^i}{i!}$ converges. It's also easy to show that $\frac{d}{dx}\sum_{i = 0}^\infty \frac{x^i}{i!} = \sum_{i = 0}^\infty\frac{x^i}{i!}$ but $e^x$ is the only function that is its own derivative and assigns 1 to 0. Therefore, $\forall x \in \mathbb{R}e^x = \sum_{i = 0}^\infty \frac{x^i}{i!}$
