Prove the irrationality of pi by contradiction

[Delta2]

How $I_n$ will be an integer? By the summation equation I think $I_n$ will be a rational number.

Seems to me you are right on that, that part was also a bit unclear to me.

 

[etotheipi]

We have that $$f(x) = \frac{q^n x^n (\pi - x)^n}{n!}$$ and also $$I_n = \sum_{j=0}^n (-1)^j \left(f^{(2j)}(\pi) + f^{(2j)}(0)\right)$$I believe all the derivatives evaluated at $\pi$ or $0$ for which $j < n$ will be zero (since they will still contain a multiplicative $x$ or $(\pi - x)$ term).

So the sum will I think reduce to a constant term $(-1)^n \left(f^{(2n)}(\pi) + f^{(2n)}(0)\right)$. But after doing $2n$ derivatives, the $n!$ on the denominator should be canceled since we will have multiplied by all of $n$, $(n-1)$, etc.

I don't know if this is right, and it's certainly not a "proof" of any sort, it's just how I pictured it.

 

[Adesh]

I believe all the derivatives evaluated at ππ\pi or 000 for which j<nj<nj < n will be zero (since they will still contain a multiplicative xxx or (π−x)(π−x)(\pi - x) term).

No, that's not correct. Try evaluating the second derivative.

 

[etotheipi]

No, that's not correct. Try evaluating the second derivative.

If we let $g(x) = x^n (\pi - x)^n$, then

$g'(x) = nx^{n-1}(\pi - x)^n - nx^n(\pi - x)^{n-1}$

$g''(x) = n(n-1)x^{n-2}(\pi - x)^{n} - n^2 x^{n-1}(\pi - x)^{n-1} + nx^n (n-1)(\pi - x)^{n-2} - n^2(\pi - x)^{n-1}x^{n-1}$

So $g''(0) = g''(\pi) = 0$

And every subsequent derivative down the chain apart from the last one (the constant term) will evaluate to zero.

 

[Delta2]

I think @etotheipi cleared this out at least for me, I don't know if you @Adesh have any more questions.

 

[etotheipi]

I think I've confused myself too , since Wikipedia gives a slightly different rundown of this part of the proof. Ultimately the summation still becomes an integer, but they seem to say there are some more constant terms:

I'll need to read over this again, because it's giving me a headache . I don't think we're too far off but there is perhaps a little bit left to this part of the proof.

 

[etotheipi]

I tried evaluating the 3rd derivative of $x^4(\pi - x)^4$ at $x=0$ and $x=\pi$ and I get zero on both counts, so I struggle to see what Wikipedia is trying to push here!