Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Following from the book "An Introduction to the Theory of Numbers" - Hardy & Wright I am having trouble with this proof. The book uses a familiar proof for the irrationality of e and continues into some generalizations that lose me.

In the following statement where is the series coming from or how is the statement derived?

$ f = f(x) = \frac{x^n(1 - x)^n}{n!} = \frac{1}{n!} \displaystyle\sum\limits_{m=n}^{2n} c_mx^m $

I understand that given $ 0 < x < 1 $ results in

$ 0 < f(x) < \frac{1}{n!} $

but I become confused on . . .

Again $f(0)=0$ and $f^{(m)}(0)=0$ if $m < n$ or $m > 2n.$ But, if $n \leq m \leq 2n $,

$ f^{(m)}(0)=\frac{m!}{n!}c_m $

an integer. Hence $f(x)$ and all its derivatives take integral values at $x=0.$ Since $f(1-x)=f(x),$ the same is true at $x=1.$

All wording kept intact!

The proof that follows actually makes sense when I take for granted the above. I can't however take it for granted as these are, for me, the more important details. So . . .

share|cite|improve this question
up vote 4 down vote accepted

I see 8 assertions in the highlighted region - are all 8 worrying you?

  1. $f(0)=0$. I hope that one is no problem.

  2. $f^{(m)}(0)=0$ for $m\lt n$. We have $f(x)=x^ng(x)$ where $g$ is just a polynomial, so $f'(x)$ has $x^{n-1}$ as a factor, $f''(x)$ has $x^{n-2}$ as a factor, dotdotdot, $f^{(m)}(x)$ has $x^{n-m}$ as a factor, and is thus zero at $x=0$. OK?

  3. $f^{(m)}(0)=0$ for $m\gt2n$. Well, $f$ is a polynomial of degree $2n$, so any derivative beyond $2n$ is identically zero.

  4. If $n\le m\le2n$ then $f^{(m)}(0)=m!c_m/n!$. If you differentiate $\sum_{j=n}^{2n}c_jx^j$ $m$ times, the value at zero is the constant term, and the constant term is the one that comes from differentiating the $c_mx^m$ term, which becomes $m!c_m$. OK?

  5. $m!c_m/n!$ is an integer. Well, $c_m$ is a coefficient in $(1-x)^n$, so it is certainly an integer, and $m\ge n$, so $m!/n!$ is certainly an integer. OK?

  6. $f$ and its derivatives take integer values at zero. Well, that's what we've just established, right?

  7. "Since $f(x-1)=f(x)$, .... This should be $f(1-x)=f(x)$, which you can verify by just replacing $x$ in the formula for $f$ with $1-x$ and seeing what happens.

  8. "The same is true at $x=1$." Apply $d/dx$ any number of times to $f(1-x)=f(x)$ and then put in $x=0$ and you'll see the relation between the values and derivatives at $x=0$ and at $x=1$.

share|cite|improve this answer
x-1 was my typo, thanks! – Xittenn Sep 1 '11 at 4:51

For your first question, this follows from the binomial theorem

$$x^n(1-x)^n=x^n\sum_{m=0}^{n}{n\choose m}(-1)^mx^m=\sum_{m=0}^{n}{n\choose m}(-1)^{m}x^{m+n}=\sum_{m=n}^{2n}{n\choose m-n}(-1)^{m-n}x^m$$

where the last equality is from reindexing the sum. Then let $c_m={n\choose m-n}(-1)^m$, which is notably an integer. I'm not quite clear what your next question is.

share|cite|improve this answer
Thank you for pointing that out! – Xittenn Sep 1 '11 at 4:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.