Prove the irrationality of pi by contradiction

[etotheipi]

Homework Statement

Suppose $\pi = \frac{p}{q}$ and let $$I_n = \int_0^{\pi} f(x) \sin{x} dx$$ where $f(x) = \frac{q^n x^n (\pi - x)^n}{n!}$. Show that $$I_n = \sum_{j=0}^n (-1)^j \left(f^{(2j)}(\pi) + f^{(2j)}(0)\right)$$

Relevant Equations

N/A

Normally you can do reduction formula type questions with I.B.P., but here that just results in something like $$I_n = \left[-f(x) \cos{x}\right]_0^{\pi} + \frac{q^n n}{n!} \int_0^{\pi} \cos{(x)} x^{n-1} (\pi - x)^{n-1} (\pi - 2x) dx$$I can't seem to get anywhere from here though. Substitutions like $u = \pi - x$ are useless since it just gets you back to the original thing! I did I.B.P. again but that just made it worse...

I'm not too sure either how we get a sum from it; I suspect a series expansion might be involved but I tried expanding $(1 - \frac{x}{\pi})^n$ binomially as well as substituting in the Maclaurin for $\sin$ but that gets me nowhere either.

Once you do obtain this form you're then supposed to show also that $$I_n \leq \frac{\pi}{n!} \left(\frac{q\pi^2}{4}\right)^n$$ and that $I_n \rightarrow 0$ as $n \rightarrow \infty$ to finally deduce $\pi$ is irrational.

Though at the moment I'm still struggling to get the first part out! I wondered if anyone could provide any hints... thanks!

 

[member 587159]

Not sure this works:

$$I_n = \int_0^\pi f(x) \sin(x)dx \leq \int_0^\pi \frac{q^n x^n (\pi-x)^n}{n!}dx = \dots$$

 

[MathematicalPhysicist]

What do you want us to show?
The sum equals to what?

 

[etotheipi]

What do you want us to show?
The sum equals to what?

Ah sorry, yes it's supposed to be shown to be equal to $I_n$. I've fixed the problem statement!

 

[Delta2]

I think you must do IBP once more cause its the only way to relate $I_n$ with $I_{n-1}$ along some other integrals. But the way you wrote it after the first IBP doesn't help, I believe it is better to write it down as
$$I_n=[-f(x)cosx]_0^{\pi}+\frac{q^nn}{n!}(\int_0^{\pi}x^{n-1}(\pi-x)^{n}\cos xdx-\int_0^{\pi}x^{n}(\pi-x)^{n-1}\cos x dx)$$.
Now each of the above two integral terms will give a contribution to a $I_{n-1}$ term along with other integrals (which other integrals will cancel out, not so sure about that) when you do IBP once more.

 

[etotheipi]

I think you must do IBP once more cause its the only way to relate $I_n$ with $I_{n-1}$ along some other integrals. But the way you wrote it after the first IBP doesn't help, I believe it is better to write it down as
$$I_n=[-f(x)cosx]_0^{\pi}+\frac{q^nn}{n!}(\int_0^{\pi}x^{n-1}(\pi-x)^{n}\cos xdx-\int_0^{\pi}x^{n}(\pi-x)^{n-1}\cos x dx)$$.
Now each of the above two integral terms will give a contribution to a $I_{n-1}$ term along with other integrals (which other integrals will cancel out, not so sure about that) when you do IBP once more.

If I do it again then I get
$$I_n=\frac{q^nn}{n!}\int_0^{\pi} \sin{(x)} x^{n-2}(\pi - x)^{n-2}\left(2xn(\pi - x) - (n-1)[x^2 + (\pi -x)^2]\right)$$Now we do get something similar looking to $I_{n-2}$ but it's got those extra terms multiplied on the end so I can't just substitute to get rid of the integral.

 

[Delta2]

If I do it again then I get
$$I_n=\frac{q^nn}{n!}\int_0^{\pi} \sin{(x)} x^{n-2}(\pi - x)^{n-2}\left(2xn(\pi - x) - (n-1)[x^2 + (\pi -x)^2]\right)$$Now we do get something similar looking to $I_{n-2}$ but it's got those extra terms multiplied on the end so I can't just substitute to get rid of the integral.

This looks fine to me, just notice that the first term is
$$2\int_0^{\pi}\sin x x^{n-1}(\pi-x)^{n-1}dx$$ which is essentially $I_{n-1}$

I think the rest two terms add to zero. To see this what do you get from the integral
$$\int_0^{\pi}\sin x x^n(\pi-x)^{n-2}dx$$ if you do the substitution $y=\pi-x$ and knowing that $\sin(\pi-y)=\sin(y)$

 

[etotheipi]

I think the rest two terms add to zero. To see this what do you get from the integral
$$\int_0^{\pi}\sin x x^n(\pi-x)^{n-2}dx$$ if you do the substitution $y=\pi-x$ and knowing that $\sin(\pi-y)=\sin(y)$

Yes that works, you get the same integral back except negated which results in $2I=0\implies I=0$. That gives us $I_n = \frac{2q^n n^2}{n!} I_{n-1}$. I'll see if I can do anything with that!

 

[Delta2]

You must have some "constants" $g(n,q)$ in front which come from IBP an also it must be $I_n=g(n,q)+2qnI_{n-1}$ because $I_{n-1}=\int_0^{\pi}\frac{q^{n-1}x^{n-1}(\pi-x)^{n-1}}{(n-1)!}\sin x dx$

 

[etotheipi]

Sorry for the delay! I've been working on it for a little while! I think I've made some progress; I had a go just leaving $f(x)$ as itself without substituting in the full expression because we're aiming to get a sum of derivatives of sum sort. I get, for subsequent applications, \begin{align*}
I_n &= [-f(x) \cos{x}]_0^{\pi} + \int_{0}^{\pi} \cos{x} f'(x) dx\\
I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} - \int_{0}^{\pi} \sin{x} f''(x) dx\\
I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} + [f''(x) \cos{x}]_0^{\pi} - \int_{0}^{\pi} \cos{x} f^{(3)}(x) dx\\
I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} + [f''(x) \cos{x}]_0^{\pi} + [-f^{(3)}(x) \sin{x}]_0^{\pi} + \int_{0}^{\pi} \sin{x} f^{(4)}(x) dx
\end{align*}
So each term is of the form $[f^{k}(x) \frac{d^k}{dx^k}(-\cos{x})]_0^{\pi}$. Also, considering that the highest possible power in $f(x)$ is $x^{2n}$, the $(2n+1)^{th}$ derivative of $f(x)$ is going to be zero and the integral at the very end disappears.

So I tried evaluating these and I end up getting $$I_n = (f(\pi) + f(0)) + 0 + (-f''(\pi)-f''(0)) + 0 + \cdots$$ and since only the terms containing even derivatives (with $\cos$) remain, it does indeed turn out to be
$$I_n = \sum_{j=0}^n (-1)^j \left(f^{(2j)}(\pi) + f^{(2j)}(0)\right)$$ Now I've got to try and do the second part, showing that $I_n \leq \frac{\pi}{n!} \left(\frac{q\pi^2}{4}\right)^n$. You can say $$I_n \leq \int_{0}^{\pi} f(x) dx = \frac{q^n}{n!}\int_0^{\pi} x^n (\pi - x)^n dx$$ so we really just need to show that $$\int_0^{\pi} (x (\pi - x))^n dx \leq \frac{\pi^{2n+1}}{4^n}$$I said that the equation $y = x\pi - x^2$ is just an inverted parabola with vertex at $y = \frac{\pi^2}{4}$. So it then follows that $$\int_0^{\pi} (x (\pi - x))^n dx \leq \int_0^{\pi} (\frac{\pi^2}{4})^n dx = \frac{\pi^{2n+1}}{4^n}$$ so the statement $I_n \leq \frac{\pi}{n!} \left(\frac{q\pi^2}{4}\right)^n$ is indeed also true. As $n \rightarrow \infty$ then $I_n \rightarrow 0$.

Now I just need to try and find out why all of that shows $\pi$ is irrational...

 

[etotheipi]

I sort of forgot about this... does anyone have any ideas?

 

[Delta2]

An idea that might work, if $\pi=\frac{p}{q}$ then work out $I_n$ as
$$I_n=\int_0^{\frac{1}{q}}f(x)\sin x dx+\int_{\frac{1}{q}}^{\frac{2}{q}} f(x) \sin x dx+...+\int_{\frac{p-1}{q}}^{\frac{p}{q}} f(x) \sin x dx$$

and then try to find a lower bound for each term of the above and finally conclude that $I_n$ cannot converge to zero.

 

[MathematicalPhysicist]

Isn't there an official reference for this proof?

Just asking.

 

[etotheipi]

Isn't there an official reference for this proof?

Just asking.

I'm not sure, I just found the question on a problem sheet I came across. I would also be interested to see if anyone can find a reference!

 

[haruspex]

let $$I_n = \int_0^{\pi} f(x) \sin{x} dx$$ where $f(x) = \frac{q^n x^n (\pi - x)^n}{n!}$. Show that ... $$I_n \leq \frac{\pi}{n!} \left(\frac{q\pi^2}{4}\right)^n$$

There seems to be a far easier way for that step.
$x(\pi-x)\le\frac{\pi^2}4$, and $\sin(x)\le 1$, so $f(x)\sin(x)\le\frac{1}{n!}(\frac{q\pi^2}4)^{^n}$.
Am I missing something?

Take a look at https://en.m.wikipedia.org/wiki/Proof_that_π_is_irrational#Niven's_proof and the Bourbaki group proof that follows it.

 

[etotheipi]

Take a look at https://en.m.wikipedia.org/wiki/Proof_that_π_is_irrational#Niven's_proof and the Bourbaki group proof that follows it.

Ah okay, so the reason is that $I_n$ is an integer from the first part, but approaches zero from the second part, and that is the contradiction. Thanks for the link!

There seems to be a far easier way for that step.
$x(\pi-x)\le\frac{\pi^2}4$, and $\sin(x)\le 1$, so $f(x)\sin(x)\le\frac{1}{n!}(\frac{q\pi^2}4)^{^n}$.
Am I missing something?

Whoops, you're quite right! That's definitely an easier route.

 

[Delta2]

Ah okay, so the reason is that $I_n$ is an integer from the first part, but approaches zero from the second part, and that is the contradiction. Thanks for the link!

I understand that $I_n$ is an integer due to the summation equation but why it is a positive integer, couldn't it be zero?

Whoops, you're quite right! That's definitely an easier route.

I don't understand you and @haruspex here , didn't you essentially do the same thing??or not?

 

[etotheipi]

I don't understand you and @haruspex here , didn't you essentially do the same thing??or not?

I think it's essentially the same idea, it's just his is slightly more direct.

I understand that $I_n$ is an integer due to the summation equation but why it is a positive integer, couldn't it be zero?

I had just presumed that since both $f(x)$ and $\sin{x}$ are greater than zero (since $q>0$) in $(0, \pi)$, $I_n$ is also going to be strictly greater than zero. There's probably a better way of saying that, but I imagined that the graph between $(0,\pi)$ is going to bound a non-zero area no matter how large you make $n$.

This could be wrong, in which case I'm also not too sure!

 

[Infrared]

This looks like Niven's proof. You can find a full argument here: https://en.wikipedia.org/wiki/Proof_that_π_is_irrational#Niven's_proof

 

[haruspex]

This looks like Niven's proof. You can find a full argument here: https://en.wikipedia.org/wiki/Proof_that_π_is_irrational#Niven's_proof

See post #15.

 

[Adesh]

​

In​

=[−f(x)cosx]π0+∫π0cosxf′(x)dx

In​

=[−f(x)cosx]π0+[f′(x)sinx]π0−∫π0sinxf′′(x)dx

In​

=[−f(x)cosx]π0+[f′(x)sinx]π0+[f′′(x)cosx]π0−∫π0cosxf(3)(x)dx

In​

=[−f(x)cosx]π0+[f′(x)sinx]π0+[f′′(x)cosx]π0+[−f(3)(x)sinx]π0+∫π0sinxf(4)(x)dxIn=[−f(x)cos⁡x]0π+∫0πcos⁡xf′(x)dxIn=[−f(x)cos⁡x]0π+[f′(x)sin⁡x]0π−∫0πsin⁡xf″(x)dxIn=[−f(x)cos⁡x]0π+[f′(x)sin⁡x]0π+[f″(x)cos⁡x]0π−∫0πcos⁡xf(3)(x)dxIn=[−f(x)cos⁡x]0π+[f′(x)sin⁡x]0π+[f″(x)cos⁡x]0π+[−f(3)(x)sin⁡x]0π+∫0πsin⁡xf(4)(x)dx​

\begin{align*} I_n &= [-f(x) \cos{x}]_0^{\pi} + \int_{0}^{\pi} \cos{x} f'(x) dx\\ I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} - \int_{0}^{\pi} \sin{x} f''(x) dx\\ I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} + [f''(x) \cos{x}]_0^{\pi} - \int_{0}^{\pi} \cos{x} f^{(3)}(x) dx\\ I_n &= [-f(x) \cos{x}]_0^{\pi} + [f'(x) \sin{x}]_0^{\pi} + [f''(x) \cos{x}]_0^{\pi} + [-f^{(3)}(x) \sin{x}]_0^{\pi} + \int_{0}^{\pi} \sin{x} f^{(4)}(x) dx \end{align*}

How can we ignore the integral which are appearing at last? Although, I have arrived at the same result but still I want to know what argument we should use to ignore the integral that are appearing after each step.

 

[Delta2]

We don't ignore that integral for the first $2n$ steps, we do integration by parts of that integral to reach the next step and so on till we reach the $(2n+1)$ step where the integral will be $\int \cos x f^{(2n+1)}(x)dx$ which we can prove it will be zero because $f^{(2n+1)}$ is zero because the highest power of $f$ is $x^{2n}$

 

[Adesh]

We don't ignore that integral for the first $2n$ steps, we do integration by parts of that integral to reach the next step and so on till we reach the $(2n+1)$ step where the integral will be $\int \cos x f^{(2n+1)}(x)dx$ which we can prove it will be zero because $f^{(2n+1)}$ is zero because the highest power of $f$ is $x^{2n}$

Sir I didn’t get how $$\int_{0}^{\pi} f^{2n+1} (x) \cos x dx = 0$$ ?

 

[Delta2]

First of all it is $\int_0^{\pi} f^{(2n+1)}(x) \cos x dx=0$ ( we have the (2n+1) derivative of f and not f raised to the (2n+1) power)

Second, can you see that from the definition of f we can conclude that the highest power term is $x^{2n}$? You might have to use binomial expansion of the $(\pi-x)^n$ .

 

[Adesh]

First of all it is $\int_0^{\pi} f^{(2n+1)}(x) \cos x dx=0$ ( we have the (2n+1) derivative of f and not f raised to the (2n+1) power)

Second, can you see that from the definition of f we can conclude that the highest power term is $x^{2n}$? You might have to use binomial expansion of the $(\pi-x)^n$ .

Do you mean the highest term in $f^{(2n+1)} (x)$ ?

 

[Delta2]

No I mean the highest term in $f(x)$.

 

[Adesh]

No I mean the highest term in $f(x)$.

$\left (\pi + (-x) \right)^n = \pi^n + ... + (-x)^{n}$ when we multiply this by $x^n$ we would have the highest term $x^{2n}$. Everything is clear till here.

 

[Delta2]

$\left (\pi + (-x) \right)^n = \pi^n + ... + (-x)^{n}$ when we multiply this by $x^n$ we would have the highest term $x^{2n}$. Everything is clear till here.

Fine so now if you take the 1st derivative what will be the highest power of $f'(x)$?

 

[Adesh]

Fine so now if you take the 1st derivative what will be the highest power of $f'(x)$?

Got you! Thank you so much! After $2n$ the differentiation $f(x)$ will become a constant and therefore $f^{(2n+1)}(x)$ is just zero.

 

[Adesh]

I understand that InInI_n is an integer due to the summation equation

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