Posted by: Dave Richeson | September 28, 2010

The transcendence of e (part 2)

This is the second part in a 3-part blog post in which we prove that {e} is transcendental.

Three-step proof that e is transcendental
Step 1
Step 2
Step 3

Recall that in step 1 we proved the following lemma.

Lemma 1. Suppose {e} is a root of the polynomial {\varphi(x)=c_{0}+c_{1}x+c_{2}x^{2}+\cdots+c_{n}x^{n}}. Let {f} be a polynomial and {F(x)=\sum_{i=0}^{\infty}f^{(i)}(x)}. Then there exist {\alpha_{1},\ldots,\alpha_{n}\in(0,1)} such that {c_{0}F(0)+\cdots+c_{n}F(n)=c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}}, where {\beta_{k}=-ke^{k(1-\alpha_{k})}f(k\alpha_{k})}.

Step 2.

Now we are ready pick a specific polynomial {f} for Lemma 1. Since there are infinitely many primes, we can find a prime number {p} greater than both {n} and {c_{0}}. Let

\displaystyle f(x)=\frac{x^{p-1}(1-x)^{p}(2-x)^{p}\cdots(n-x)^{p}}{(p-1)!}.

Our goal for this step is to show that {c_{0}F(0)+\cdots+c_{n}F(n)} is a nonzero integer. In fact we will show is that it is an integer not divisible by {p}, which implies that it is nonzero.

We will accomplish this by proving that for {m=1,\ldots n}, {F(m)} (and hence {c_{m}F(m)}) is an integer divisible by {p}, but that {c_{0}F(0)} is an integer that is not divisible by {p}.

First we state but do not prove the following lemma. (This is a homework problem in Herstein.)

Lemma 2. If {g} is a polynomial with integer coefficients and {h(x)=g(x)/(p-1)!}, then for {i\ge p}, {h^{(i)}} is a polynomial with integer coefficients, each of which is divisible by {p}.

Applying Lemma 2 to our function {f} we see that when {i\ge p}, {f^{(i)}(x)} is a polynomial with integer coefficients, each of which is divisible by {p}. In particular, when {i\ge p} and {m} is any integer, {f^{(i)}(m)} is an integer that is divisible by {p}.

Now let us restrict our attention to {m\in\{1,\ldots,n\}}. From the definition of {f} we see that {m} is a root of {f} with multiplicity {p}. The following lemma tells us that {f(m)=f^{(1)}(m)=\cdots=f^{(p-1)}(m)=0}. (We leave the proof of this lemma as an exercise for the reader.)

Lemma 3. Suppose {x_{0}} is a root of a polynomial {g} with multiplicity {m}. Then {g^{(k)}(x_{0})=0} for {k=0,\ldots,m-1}.

So,

F(m)=f(m)+f^{(1)}(m)+\cdots+f^{(r)}(m)
=f^{(p)}(m)+\cdots+f^{(r)}(m).

We know from above that each of the terms {f^{(p)}(m),\cdots,f^{(r)}(m)} is an integer divisible by {p}. Thus {F(m)} is an integer divisible by {p}.

Now let us consider {m=0}. By the definition of {f}, {0} is a root with multiplicity {p-1}. By Lemma 3, {f(0)=f^{(1)}(0)=\cdots=f^{(p-2)}(0)=0}.

Thus,

F(0)=f(0)+f^{(1)}(0)+\cdots+f^{(r)}(0)
=f^{(p-1)}(0)+f^{(p)}(0)+\cdots+f^{(r)}(0).

From the result above, we know that {f^{(p)}(0),\cdots,f^{(r)}(0)} are all divisible by {p}. What about {f^{(p-1)}(0)}?

If we multiply out the terms in {f}, we obtain

\displaystyle f(x)=\frac{(n!)^{p}}{(p-1)!}x^{p-1}+\{\text{higher order terms in }x\}

So,

\displaystyle f^{(p-1)}(x)=(n!)^{p}+\{\text{higher order terms in }x\},

and hence {f^{(p-1)}(0)=(n!)^{p}}. Since {p>n}, {p} does not divide {f^{(p-1)}(0)}. Therefore {p} does not divide {F(0)}. Moreover, {p>c_{0}}, so {p} does not divide {c_{0}}. We conclude that {p} does not divide {c_{0}F(0)}.

Finally this implies that {c_{0}F(0)+\cdots+c_{n}F(n)} is an integer not divisible by {p}. In particular, it is a nonzero integer.

We are almost finished with the proof. In the next step we will show that {|c_{1}\beta_{1}+\cdots+c_{n}\beta_{n}|<1}, thus it cannot be a nonzero integer and we will obtain our sought-after contradiction.

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Responses

  1. […] The transcendence of e (part 2) […]

  2. […] proof that is transcendental Step 1 Step 2 Step […]

  3. Thanks for posting this, it’s very helpful. I have one question though, I know I’m being stupid but I can’t seem to prove lemma 2. I’d really appreciate any help on this. Thank you.

    • For simplicity, suppose g(x)=x^n. Then \displaystyle h^{(i)}(x)=\frac{n(n-1)\cdots(n-i+1)}{(p-1)!}x^{n-i}. Rewritten, the coefficient is \displaystyle \frac{n!}{(p-1)!(n-i)!}=(i(i-1)\cdots p)\frac{n!}{i!(n-i)}=(i(i-1)\cdots p){{n}\choose{i}}. Seen in this way, it is clearly an integer divisible by p.

  4. Thanks a lot for your help, I really should have been able to work that out myself!

  5. Sorry to bother you again, but I’ve been trying to prove that $e^{m/n}$ is transcendental, where $m>0$ and $n$ are integers (it’s an exercise in Herstein on p.178), but I’ve had no luck. Any help would be much appreciated. Thanks.


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