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I understand the definition of $e^x$ by limit. But I do not know how to come up with:

$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$

without using Taylor series. more explicitly without using calculus. how do we know if a function can be expressed as series or not ?

Yirmidokuz
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  • Are you familiar with Taylor series? Series solutions of differential equations at regular points? From what foundation/background are you approaching this problem? – Eric Towers May 01 '14 at 16:51
  • @EricTowers In fact, I am familiar with Taylor/Mac Laurin series. But is it possible to express the summation definition of $e^x$, without using them ? Since, I am regenerating my math knowledge I want to go step by step to calculus, differential equations and so forth. – Yirmidokuz May 01 '14 at 16:55
  • might be useful: http://math.stackexchange.com/a/763121/59234 – Aloizio Macedo May 01 '14 at 17:16

2 Answers2

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One definition of the exponential function is the limit $$ \lim_{n\to\infty} \Big(1 + \frac{x}{n}\Big)^n=e^x. $$ Let $P_n(x)$ denote the polynomial $(1+x/n)^n$, so that $e^x=\lim_{n\to\infty} P_n(x)$; I will show that $$ \lim_{n\to\infty} P_n(x) = \sum_{n=0}^\infty \frac{x^n}{n!}. $$

If you expand out $P_n(x)$ using the binomial theorem, you'll find that $$ P_n(x) = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}. $$ Now \begin{align} \frac{1}{n^k}\binom{n}{k} &= \frac{1}{n^k}\cdot\frac{n}{k}\cdot\frac{n-1}{k-1}\cdots\frac{n-k+1}{1} \\ &= \frac{1}{k}\cdot\frac{1-1/n}{k-1}\cdots\frac{1-(k-1)/n}{1}. \end{align} Since $k$ is fixed, $\lim_{n\to\infty}\binom{n}{k}/n^k=\frac{1}{k!}$. It follows that $$ \lim_{n\to\infty} P_n(x) = \sum_{k=0}^\infty\lim_{n\to\infty}\binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^\infty\frac{x^k}{k!}. $$

You should look at the Wikipedia article about characterizations of the exponential function; it has five.

As for your question about whether a function can be expressed as a series or not, to answer it I believe you need to say something about calculus. What I mean is that if a "nice" function $f(x)$ has a series representation at a point $a$ then the series is given by $$ f(x)=\sum_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k, $$ where $f^{(k)}(a)$ is the $k$th derivative of $f$ at $a$.

user134824
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If you take the fact that $(e^x)'=e^x$ then by taylor series, we can expand around $0$, taking into account that each $n$-th derivative at $0$ is $1$.

Then $e^x=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{1}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{n!}$

Lets try this without Taylor series. A function that can be expressed by a real power series is called real analytic. All that is needed is that all derivatives are greater than or equal to $0$. Clearly this holds for $e^x$.

So let us write $e^x=\sum_{n=0}^\infty a_nx^n$ taking derivatives we get:

$\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty a_nx^n$

Comparing coefficients of each power of $x$ we get:

$a_{n+1}=\frac{1}{n+1}a_n$ and we know $a_0=1$ (because $e^0=1$) this gives as $a_n=\frac{1}{n!}$

Thus $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

Ellya
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  • Thank you for the answer! I forgot the mention that I would like to express the defition without applying Taylor series, because it uses calculus, which that I do not want to consider at the moment. – Yirmidokuz May 01 '14 at 17:06
  • Where would you need the exponential function before calculus? Your argument does not really stand... – b00n heT May 01 '14 at 17:10
  • there are some convergence tests for series, but this is still the beginings of calculus and analysis: http://en.wikipedia.org/wiki/Convergence_tests – user136920 May 01 '14 at 17:31
  • Then maybe I should ask how do we know when to use Taylor Series to define a function ? – Yirmidokuz May 01 '14 at 17:41
  • @Yirmidokuz I've put in an edit that does not use Taylor's series. – Ellya May 01 '14 at 18:11