Help - Derivation of formula for Pi (2 Viewers)

Sy123 · Dec 28, 2012

Help - Real Analysis

So I read somewhere that:

$\sum_{k=1}^{\infty} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

However they derived it some weird way that I know nothing about (this is STEP Advanced Problems #43)

Now I set out to derive this, I first though of integrating the sum of $\cos k \theta$
And I knew how to derive a closed formula for this sum, via:

$z+z^2+z^3+...+z^{n} = \frac{z(1-z^n)}{1-z}$

And substituting z= cis theta in there, and equating real parts, then simplification. All that was easily done, then I had the task of integrating the beast.

I arrived at the result:

$\sum_{k=1}^{n} \cos k\theta = \frac{\cos \theta - \cos (n+1)\theta + \cos n\theta -1}{4\cos^2 \frac{\theta}{2} }$

I do realise I need to take a limit for n to infinity, but I will hope to do that later, but I got stuck, I was able to get up to:

$\sum_{k=1}^{n} k^{-1}\sin k\theta = \frac{\theta}{2} - \tan \frac{\theta}{2} + \int \frac{\cos n\theta}{4\cos^2 \theta/2} \ d\theta - \int \frac{\cos (n+1) \theta}{4\cos^2 \theta/2} \ d\theta + C$

I plugged in those integrals into Wolfram Alpha and it spat out something horrible so I am assuming I cannot find the exact definite integral for that.
But all I needed to do, was to find:

$\lim_{n\to \infty} (I_n - I_{n+1})$

Where, $I_n= \int \frac{\cos n\theta}{cos^2 \theta /2}$

Question 1: The limit is all I need to find in order to complete this yes?

Now, I do know the IBP formula and I have done it just a couple of times, so I barely know anything from it, because I haven't done 4U Integration at school yet. But I have yet to be able to complete this problem, the final goal of it is to be able to show that:

$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$

By utilising a substitution

Question 2: Can anyone give me guidance on how to do this using HSC techniques?

Shadowdude · Dec 28, 2012

"And substituting z= cis theta in there, and equating real parts,"

Just skimming...

Hmm, this might be a problem. What contour are you integrating around?

Sy123 · Dec 28, 2012

Shadowdude said:
"And substituting z= cis theta in there, and equating real parts,"

Just skimming...

Hmm, this might be a problem. What contour are you integrating around?

Not sure what you mean by that, but I only used complex numbers to get to the initial sum of cos identity to then integrate.
I need help evaluating the limit/or another approach using integration.

Shadowdude · Dec 28, 2012

ah okay, my bad

that seems fine but

$\sum_{k=1}^{\infty} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

If theta is zero, you get that sum equalling pi/2.

I'm not sure if that's right then...?

For your first question, if that above is true - then I think that would be fine. The limit converges and... whatnot.

sorry if i'm a bit rambly, i caught a cold and i'm not 100% and i feel bad not responding to this thread

SpiralFlex · Dec 28, 2012

Sy123 said:
So I read somewhere that:

$\sum_{k=1}^{\infty} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

However they derived it some weird way that I know nothing about (this is STEP Advanced Problems #43)

Now I set out to derive this, I first though of integrating the sum of $\cos k \theta$
And I knew how to derive a closed formula for this sum, via:

$z+z^2+z^3+...+z^{n} = \frac{z(1-z^n)}{1-z}$

And substituting z= cis theta in there, and equating real parts, then simplification. All that was easily done, then I had the task of integrating the beast.

I arrived at the result:

$\sum_{k=1}^{n} \cos k\theta = \frac{\cos \theta - \cos (n+1)\theta + \cos n\theta -1}{4\cos^2 \frac{\theta}{2} }$

I do realise I need to take a limit for n to infinity, but I will hope to do that later, but I got stuck, I was able to get up to:

$\sum_{k=1}^{n} k^{-1}\sin k\theta = \frac{\theta}{2} - \tan \frac{\theta}{2} + \int \frac{\cos n\theta}{4\cos^2 \theta/2} \ d\theta - \int \frac{\cos (n+1) \theta}{4\cos^2 \theta/2} \ d\theta + C$

I plugged in those integrals into Wolfram Alpha and it spat out something horrible so I am assuming I cannot find the exact definite integral for that.
But all I needed to do, was to find:

$\lim_{n\to \infty} (I_n - I_{n+1})$

Where, $I_n= \int \frac{\cos n\theta}{cos^2 \theta /2}$

Question 1: The limit is all I need to find in order to complete this yes?

Now, I do know the IBP formula and I have done it just a couple of times, so I barely know anything from it, because I haven't done 4U Integration at school yet. But I have yet to be able to complete this problem, the final goal of it is to be able to show that:

$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + ...$

By utilising a substitution

Question 2: Can anyone give me guidance on how to do this using HSC techniques?

This all seems HSCy techniques.

Sy123 · Dec 28, 2012

Shadowdude said:
ah okay, my bad

that seems fine but

$\sum_{k=1}^{\infty} \frac{\sin k \theta}{k} = \frac{\pi}{2} - \frac{\theta}{2}$

If theta is zero, you get that sum equalling pi/2.

I'm not sure if that's right then...?

For your first question, if that above is true - then I think that would be fine. The limit converges and... whatnot.

sorry if i'm a bit rambly, i caught a cold and i'm not 100% and i feel bad not responding to this thread

If theta is zero, then z = 1
And we cannot have z = 1 due to using the geometric series formula, and the denominator being 1-z

SpiralFlex said:
This all seems HSCy techniques.

Yes, but how would I use HSC techniques to complete the problem (this is the point of the thread)

SpiralFlex · Dec 28, 2012

Wait, which step do you need us to start from?

Sy123 · Dec 28, 2012

SpiralFlex said:
Wait, which step do you need us to start from?

Integration from when I have acquired:

$\sum_{k=1}^{n} \cos k\theta$

SpiralFlex · Dec 28, 2012

Wut? Integrate the sum cos blah blah cos theta-cos(n+1)\theta +cos n theta -1 over blah. Or the one with sine?

Sy123 · Dec 28, 2012

SpiralFlex said:
Wut? Integrate the sum cos blah blah cos theta-cos(n+1)\theta +cos n theta -1 over blah. Or the one with sine?

Integrate the load of cosine functions basically

Or if there is another method to acquiring this sum I would like to hear it.

SpiralFlex · Dec 28, 2012

Okay, I finally see what you mean now.

SpiralFlex · Dec 28, 2012

$P=\sum_{1}^{n} \frac{\sin k \theta}{k}$

$P'=\sum_{1}^{n}\cos k\theta$

Notice,

$Re(\cis k\theta)=\cos k \theta$

$\therefore P'=\sum_{k=1}^{n}\cos k\theta = \sum_{k=1}^{n} Re(cis k \theta)$

$P'=Re(cis \theta (\frac{1-cis n\theta}{1-cis theta}))$

$=Re(\frac{\sin \frac{1}{2} n \theta}{\sin \frac{1}{2}\theta}*cis \frac{n+1}{2}\theta)$

$=\frac{\sin (n+1)/2 \theta}{2\sin \frac{1}{2} \theta}-\frac{1}{2}$

$\int_{k_{1}}^{k_{2}}P' \;d\theta=\int_{k_{1}}^{k_{2}}[\frac{\sin (n+1)/2 \theta}{2\sin \frac{1}{2} \theta}-\frac{1}{2}]d\theta$

I will do the integration tomorrow. Let's skippp

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Arrive at this somehow

$\sum_{1}^{\infty} \frac{\sin k \theta}{k}=\frac{\pi}{2}-\frac{\theta}{2}$

Let $\theta = \frac{\pi}{2}$

$\frac{\pi}{4}=\sum_{1}^{\infty} \frac{\sin k \frac{\pi}{2}}{k}$

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+....$

seanieg89 · Dec 29, 2012

My method was the same as spiralflex's with k1=pi k2=theta in (0,2pi) (my original identity was in terms of phi to avoid confusion).

Regarding the integrand:

The -1/2 pops out to give you what you want, you can show the remaining term tends to zero as n->inf by using IBP once. If no-one posts a complete solution within a couple of hours I will finish spirals.

seanieg89 · Dec 29, 2012

$Okay, so let $S_n(\phi):=\sum_{k=1}^n \cos(k\phi)$ for $\phi\in(0,2\pi)$. We have by working similar to Spiral's that:\\$S_n(\phi)=-\frac{1}{2}+\frac{\sin(n+1/2)\theta}{2\sin(\theta/2)}.$\\ Now integrate both sides of this expression from $\pi$ to $\theta\in(0,2\pi)$. We obtain:\\$\sum_{k=1}^n \frac{\sin (k\theta)}{k}=\frac{1}{2}(\pi-\theta)+\int_\pi^\theta \frac{\sin(n+1/2)\phi}{2\sin(\phi/2)}\, d\phi$\\ so all that remains to be shown is that the final term $I$ tends to zero as $n\rightarrow\infty.$\\Integrating by parts yields: \\$I=\int_\pi^\theta \sin((n+1/2)\phi)\cdot\csc(\phi/2)\,d\phi=\left[\frac{-\cos((n+1/2)\phi)\csc(\phi/2)}{n+1/2}\right]^\theta_\pi-\frac{1}{2(n+1/2)}\int_\pi^\theta\frac{\cos((n+1/2)\phi)\cos(\phi/2)}{\sin^2(\phi/2)}\,d\phi.$\\ \\The first term clearly tends to zero and by the triangle inequality (*), the absolute value of the second is bounded above by:\\ $\frac{1}{2(n+1/2)}\int_\pi^\theta\frac{|\cos(\phi/2)|}{|\sin^2(\phi/2)|}\,d\phi\rightarrow 0.$$

(*) The triangle inequality for convergent integrals/infinite series is a relatively straightforward consequence of the regular triangle inequality, as the integral is constructed from a limiting process involving finite sums, and we already know the triangle inequality for finite sums.

Sy123 · Dec 29, 2012

seanieg89 said:
$Okay, so let $S_n(\phi):=\sum_{k=1}^n \cos(k\phi)$ for $\phi\in(0,2\pi)$. We have by working similar to Spiral's that:\\$S_n(\phi)=-\frac{1}{2}+\frac{\sin(n+1/2)\theta}{2\sin(\theta/2)}.$\\ Now integrate both sides of this expression from $\pi$ to $\theta\in(0,2\pi)$. We obtain:\\$\sum_{k=1}^n \frac{\sin (k\theta)}{k}=\frac{1}{2}(\pi-\theta)+\int_\pi^\theta \frac{\sin(n+1/2)\phi}{2\sin(\phi/2)}\, d\phi$\\ so all that remains to be shown is that the final term $I$ tends to zero as $n\rightarrow\infty.$\\Integrating by parts yields: \\$I=\int_\pi^\theta \sin((n+1/2)\phi)\cdot\csc(\phi/2)\,d\phi=\left[\frac{-\cos((n+1/2)\phi)\csc(\phi/2)}{n+1/2}\right]^\theta_\pi-\frac{1}{2(n+1/2)}\int_\pi^\theta\frac{\cos((n+1/2)\phi)\cos(\phi/2)}{\sin^2(\phi/2)}\,d\phi.$\\ \\The first term clearly tends to zero and by the triangle inequality (*), the absolute value of the second is bounded above by:\\ $\frac{1}{2(n+1/2)}\int_\pi^\theta\frac{|\cos(\phi/2)|}{|\sin^2(\phi/2)|}\,d\phi\rightarrow 0.$$

(*) The triangle inequality for convergent integrals/infinite series is a relatively straightforward consequence of the regular triangle inequality, as the integral is constructed from a limiting process involving finite sums, and we already know the triangle inequality for finite sums.

Thank you for taking the time to latex it all up, I appreciate it

But, in your explanation to why the second term tends to zero, what do you mean it is bounded above by...?

seanieg89 · Dec 29, 2012

Less than or equal to. As the second term's absolute value is less than some positive thing that tends to zero, the second term must itself tend to zero as n-> inf.

Carrotsticks · Dec 29, 2012

Sy123 said:
Thank you for taking the time to latex it all up, I appreciate it

But, in your explanation to why the second term tends to zero, what do you mean it is bounded above by...?

http://en.wikipedia.org/wiki/Limit_comparison_test

Sy123 · Dec 29, 2012

seanieg89 said:
Less than or equal to. As the second term's absolute value is less than some positive thing that tends to zero, the second term must itself tend to zero as n-> inf.

Carrotsticks said:
http://en.wikipedia.org/wiki/Limit_comparison_test

Oo, I see thank you.

EDIT: Is there a way of doing something similar for my integral?

Carrotsticks · Dec 29, 2012

Sy123 said:
Oo, I see thank you.

EDIT: Is there a way of doing something similar for my integral?

What was the integral?

seanieg89 · Dec 29, 2012

Sy123 said:
Oo, I see thank you.

EDIT: Is there a way of doing something similar for my integral?

You want to be taking definite integrals for these sorts of questions. The output of an indefinite integral is an equivalence class of functions which differ by a constant, the output of a definite integral is a number. We want to show that our integral is small, this notion is well defined for the latter but not for the former. (A notion of size could possible be given for the set of such equivalence classes of functions but it would be beyond mx2 and less straightforward).

Help - Derivation of formula for Pi (2 Viewers)

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