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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?

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Do you know Taylor series? See here: en.wikipedia.org/wiki/Taylor_series The power series you wrote down is the Taylor series of $e^x$ at $a=0$. –  Paul Dec 24 '11 at 13:32
    
tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx (in example 1, the Taylor Series for $f(x)=e^x$ is derived) –  David Mitra Dec 24 '11 at 13:33
    
Let me go through it.Hang on! –  alok Dec 24 '11 at 13:34

4 Answers 4

up vote 11 down vote accepted

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: enter image description here

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.)

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There is a mistake which i tried to fix, but forum says thant my fix needs to be at least 6 characters long. I can't fix $a_0 = f'(0)$ to get the proper form $a_0 = f(0)$. Please could YOU fix this because forum won't let me. –  71GA Feb 8 '13 at 10:36

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 $$e^x=\sum_{k=0}^\infty\frac{1}{k!}x^k=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$

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Use "Blockquote", that is the icon $``$ which you can find when you type your question. Try it! –  Paul Dec 24 '11 at 13:50

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.

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Yes it does.What tool do i use to get that grey rectangular box which you use to highlight math statements? –  alok Dec 24 '11 at 13:49
    
@alok: One can add the symbol > before a line to make it appear in a block quote (the grey box). For example, to see how I added it to your question: math.stackexchange.com/revisions/… –  Zev Chonoles Dec 24 '11 at 13:50
    
@alok: See this page for more information about how to format posts. –  Zev Chonoles Dec 24 '11 at 13:53

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.

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