Prove the irrationality of pi by contradiction

I think? I'm not really sure what I'm doing with this next step.In summary, the conversation discusses the process of solving a reduction formula type question using the method of Integration by Parts (IBP). The original integral is transformed into a summation of derivatives of a function, and the conversation focuses on finding the sum of these derivatives. It is also mentioned that the sum is related to the value of the original integral. The conversation then moves on to discussing the second part of the problem, which involves showing that the integral is less than or equal to a specific value. Finally, the conversation mentions the ultimate goal of using this solution to prove the irrationality of pi.
  • #1
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!
 
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  • #2
Not sure this works:

$$I_n = \int_0^\pi f(x) \sin(x)dx \leq \int_0^\pi \frac{q^n x^n (\pi-x)^n}{n!}dx = \dots$$
 
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  • #3
What do you want us to show?
The sum equals to what?
 
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  • #4
MathematicalPhysicist said:
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!
 
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  • #5
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.
 
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  • #6
Delta2 said:
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.
 
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  • #7
etotheipi said:
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)##
 
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  • #8
Delta2 said:
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!
 
  • #9
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##
 
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  • #10
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...
 
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  • #11
I sort of forgot about this... does anyone have any ideas?
 
  • #12
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.
 
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  • #13
Isn't there an official reference for this proof?

Just asking.
 
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  • #14
MathematicalPhysicist said:
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!
 
  • #15
etotheipi said:
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.
 
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  • #16
haruspex said:
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!

haruspex said:
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.
 
  • #17
etotheipi said:
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?
 
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  • #18
Delta2 said:
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.
Delta2 said:
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! :wideeyed:
 
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  • #21
etotheipi said:

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.
 
  • #22
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}##
 
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  • #23
Delta2 said:
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 $$\int_{0}^{\pi} f^{2n+1} (x) \cos x dx = 0$$ ?
 
  • #24
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## .
 
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  • #25
Delta2 said:
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)## ?
 
  • #26
No I mean the highest term in ##f(x)##.
 
  • #27
Delta2 said:
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.
 
  • #28
Adesh said:
##\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)##?
 
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  • #29
Delta2 said:
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.
 
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  • #30
Delta2 said:
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.
 
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  • #31
Adesh said:
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.
 
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  • #32
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.
 
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  • #33
etotheipi said:
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.
 
  • #34
Adesh said:
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.
 
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  • #35
I think @etotheipi cleared this out at least for me, I don't know if you @Adesh have any more questions.
 
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<h2>1. What is the definition of irrational numbers?</h2><p>Irrational numbers are numbers that cannot be expressed as a ratio of two integers. They cannot be written as a simple fraction and their decimal representation never repeats or terminates.</p><h2>2. How can we prove the irrationality of pi?</h2><p>We can prove the irrationality of pi by contradiction, which means assuming the opposite of what we want to prove and showing that it leads to a contradiction or an impossibility.</p><h2>3. What is the contradiction in the proof of the irrationality of pi?</h2><p>The contradiction in the proof of the irrationality of pi is that if we assume that pi is rational, then it can be expressed as a ratio of two integers. However, this leads to a contradiction because we know that pi is a non-terminating and non-repeating decimal, which means it cannot be expressed as a ratio of two integers.</p><h2>4. Is the proof of the irrationality of pi universally accepted?</h2><p>Yes, the proof of the irrationality of pi by contradiction is universally accepted by mathematicians and is considered a fundamental proof in mathematics.</p><h2>5. Can we use the same proof technique to prove the irrationality of other numbers?</h2><p>Yes, the proof by contradiction can be used to prove the irrationality of other numbers as well, as long as they satisfy the definition of irrational numbers. However, the specific steps and details of the proof may vary depending on the number being proven.</p>

1. What is the definition of irrational numbers?

Irrational numbers are numbers that cannot be expressed as a ratio of two integers. They cannot be written as a simple fraction and their decimal representation never repeats or terminates.

2. How can we prove the irrationality of pi?

We can prove the irrationality of pi by contradiction, which means assuming the opposite of what we want to prove and showing that it leads to a contradiction or an impossibility.

3. What is the contradiction in the proof of the irrationality of pi?

The contradiction in the proof of the irrationality of pi is that if we assume that pi is rational, then it can be expressed as a ratio of two integers. However, this leads to a contradiction because we know that pi is a non-terminating and non-repeating decimal, which means it cannot be expressed as a ratio of two integers.

4. Is the proof of the irrationality of pi universally accepted?

Yes, the proof of the irrationality of pi by contradiction is universally accepted by mathematicians and is considered a fundamental proof in mathematics.

5. Can we use the same proof technique to prove the irrationality of other numbers?

Yes, the proof by contradiction can be used to prove the irrationality of other numbers as well, as long as they satisfy the definition of irrational numbers. However, the specific steps and details of the proof may vary depending on the number being proven.

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