# A question about complex exponential

The famous power series $1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\ldots$ converges normally in the whole complex plane $\mathbb C$ and mathematicians chose to call $$\exp(z):=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$

now my question is:

Why, they where sure that the above limit wasn't a rational function? In a nutshell, who does ensure that there aren't two polynomials $p$ and $q$ such that $\frac{p}{q}=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ ?

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You may be interested in my question: math.stackexchange.com/questions/170373/… –  Argon Jul 28 '12 at 18:25
I think I may have posted the simplest answer. But it relies on knowing certain basic facts about the function defined by the powers series, which my answer doesn't prove. –  Michael Hardy Jul 29 '12 at 1:31

There are several possible ways to eliminate the possibility of a rational function, for example:

It is not hard to prove that $\exp(z)$ tends to infinity faster than any power of $x$.

Also, looking at the function in the complex plane, it is periodic, and it is not hard to prove that a rational function can never be periodic.

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A rational function has a limit as $z\to\infty$ (either a complex number or $\infty$). On the other hand, $\lim_{z\to\infty}e^z$ does not exist.

A different reasoning. Since $e^z$ has no singularities other than $\infty$, if it were rational, it should be a (non-constant) polynomial. But a nonconstant polynomial vanishes at some point, while $e^z\ne0$ for any complex number $z$.

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+1 for the second argument. This is a new thought for me... –  Gottfried Helms Jul 28 '12 at 18:29
Most rational functions do not have finite limits as $z\to\infty$! –  Michael Hardy Jul 29 '12 at 1:28
@MichaelHardy How do you define "most"? If $R(z)$ has an infinite limit at $\infty$, then $1/R(z)$ has a finite limit. So it seems hat here as many rational functions with a finite limit as with an $\infty$ limit. –  Julián Aguirre Jul 29 '12 at 9:31
@JuliánAguirre : OK, let's say "most" means "half". –  Michael Hardy Jul 29 '12 at 19:42

$$\exp(z):=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$

Let's assume that we have 2 polynomials and we express $e^x=\frac{P(x)}{Q(x)}$

The polynomials are $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$$

$$Q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$$

$(e^x)'=e^x=\left(\frac{P(x)}{Q(x)}\right)'$

We assumed $e^x=\frac{P(x)}{Q(x)}$ Thus,

$$e^x=\frac{P(x)}{Q(x)}=\left(\frac{P(x)}{Q(x)}\right)'$$

$$\frac{P(x)}{Q(x)}=\left(\frac{P(x)}{Q(x)}\right)'=\left(\frac{P'(x)Q(x)- P(x)Q'(x)}{Q^2(x)}\right)$$

$$\frac{P(x)}{Q(x)}=\frac{P'(x)Q(x)- P(x)Q'(x)}{Q^2(x)}$$

$$P(x)Q(x)=P'(x)Q(x)- P(x)Q'(x)$$

$$P(x)(Q(x)+Q'(x))=P'(x)Q(x)$$

$$P(x)(Q(x)+Q'(x))=(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0)(b_mx^m+(mb_m+b_{m-1})x^{m-1}+\cdots)=a_nb_mx^{n+m}+\cdots$$

$$P'(x)Q(x)=(na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1)(b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0)=na_nb_mx^{n+m-1}+\cdots$$

As we see $P(x)(Q(x)+Q'(x))=P'(x)Q(x)$ cannot be equal because their degrees are not equal. The degree of polynomial $P(x)(Q(x)+Q'(x))$ is $m+n$, The degree of other polynomial $P'(x)Q(x)$ is $m+n-1$ . Thus it is impossible to find $e^x=\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials.

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Very nice! Only as a matter of taste I would leave the derivation at the "symmetric" stage $P(x)Q(x) = P'(x)Q(x) - P(x)Q'(x)$ and would then focus on the highest exponents of $x$, which is on the lhs $n+m$ and on the rhs $\max((n-1)+m,n+(m-1))=n+m-1$ which I would see immediately without need of resorting to the explicite polynomial representation. –  Gottfried Helms Jul 28 '12 at 21:19

Another argument that can easily be checked with power series: $$\left(\sum_{n=0}^\infty\frac{z^n}{n!}\right)'=\sum_{n=0}^\infty\frac{z^n}{n!}$$ which cannot be true for no non-zero rational functions.

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If a rational function of $z$ has a limit as $z\to-\infty$, then it has the same limit as $z\to\infty$. If you can show that this function approaches $0$ as $z\to-\infty$ and approaches $\infty$ as $z\to\infty$, then you've shown it doesn't behave at all like a rational function.

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