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I'm trying to get through the proof of transcendence of $e$ (the base of the natural logarithm) already for a couple of days, but now I got seriously stuck.
Proof is in most sources roughly the same. I followed this version.

We prove the transcendence of $e$ by contradiction - suppose that $e$ is an algebraic number. That means there is a nonzero polynomial $P \in \mathbb{Z}[x]$ such that $P(e) = 0$.
Let $P(x) = \sum_{i=0}^{n}w_{i}x^{i}.$

Lemma
Let $f\in \mathbb{R}[x]$ be a polynomial, $t\in \mathbb{R}^{+}$. Then the following equality holds: $$ \int_{0}^{t} e^{-x}f(x)\text{d}x = \sum_{i=0}^{\infty}f^{(i)}(0) - e^{-t}\sum_{i=0}^{\infty}f^{(i)}(t)$$

Lemma can be proved by integration per partes. For clarity let's define $F(x) = \sum_{i=0}^{\infty}f^{(i)}(x)$.
Now the lemma looks like this:

$$ \int_{0}^{t} e^{-x}f(x)\text{d}x = F(0) - e^{-t}F(t)$$

$$ \int_{0}^{t} e^{t-x}f(x)\text{d}x = e^{t}F(0) - F(t)$$

Because $e$ is (by contradiction) an algebraic number, we have $\sum_{i=0}^{n}w_{i}e^{i} = 0$ for some $w_{i}\in \mathbb{Z}$. Now a rather complicated step. Write the last integral identity for $t = 0$, $1$, ..., $n$, multiply each of them by $w_{t}$ and add the results.
You should obtain: $$ \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x = F(0)\sum_{i=0}^{n}w_{i}e^{i} - \sum_{i=0}^{n}w_{i}F(i) $$ See why we did this? Using the property of an algebraic number mentioned above we get: $$ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x $$ Now the idea of the proof is to choose the polynomial $f$ so wisely that the left side is a non-zero integer and the right side is small (say, less than $\frac{1}{10}$ in absolute value), which gives the contradiction. The wisely chosen polynomial is $f(x) = \frac{1}{(p-1)!}x^{p-1}\prod_{i=1}^{n}(x-i)^p$, where $p\in \mathbb{P}$ is some prime that will be specified later.

And here comes my digging. I am brave, I don't need such a giant polynomial to conclude the contradiction. My chosen polynomial will be $f(x)=x$. Now my equality looks like this: $\sum_{i=0}^{n}w_{i}(i+1) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}x\text{d}x$. Uhh, no contradiction. :( Maybe wrong polynomial. Okay, let's take a general polynomial f(x). Now let's compute that integral and find out whether there's a contradiction or not: $$ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}\int_{0}^{i} e^{i-x}f(x)\text{d}x $$ $$ \sum_{i=0}^{n}w_{i}F(i) = - \sum_{i=0}^{n}w_{i}(e^{i}F(0) - F(i)) $$ $$ 0 = 0 $$ Damn! Neither for $f(x)=x$ nor for $f(x) = x^{2}$ there's no contradiction. Okay, that's understandable - the wisely chosen polynomial wouldn't be so giant if it didn't have to be. But now I showed that when I try to substitute any possible polynomial and compute the integral, it doesn't lead to contradiction! Can you help me? What am I missing here?

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You haven't. Note that when passing to $0=0$, you used the assumption $\sum_i w_ie^i=0$ that you want to demonstrate false, so you just effectively cancelled your initial step (suppose that...) when doing so. You need to keep it and not try to use it again (or, at least, use it more wisely than that). A simple example of what you've done would be "Assume 1=2". Take any $n$. Then $n=1+(n-1)=2+(n-1)=n+1$. Then, since $1=2$, we have $n-1=(n+1)-2$, i.e., $n-1=n-1$. Now I see nothing wrong with that! –  fedja Nov 5 '12 at 0:12
    
@fedja "you just effectively cancelled your initial step" Ah, now I see that. It all started with those easy polynomials. I guess that the trick is to do whatever but not to evaluate that integral, which always leads back. That might be the reason why the proof continues with establishing the upper bound for the integral instead of evaluating it. –  Jeyekomon Nov 5 '12 at 0:41
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