wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> general problem-solving / chatting / whatever >> Series for pi
        
(Message started by: Sir Col on May 9th, 2007, 1:40am)
Title: Series for pi
Post by Sir Col on May 9th, 2007, 1:40am
I am probably showing my ignorance here, but I was wondering if someone could shed some light on a query I have relating to the following well known infinite series for pi?

pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ...

It is usually derived using the following infinite geometric series:

S = 1 - x2 + x4 - x6 + ...
x2S = x2 - x4 + x6 - ...
S(x2+1) = 1
S = 1/(x2+1)

That is, 1/(x2+1) = 1 - x2 + x4 - x6 + ...

Integrating both sides from 0 to x gives:

arctan(x) = x - x3/3 + x5/5 - ...

And if x = 1 then we get the desired result, arctan(1) = pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

However, my query relates to the validity of allowing x to be equal to 1. Surely the infinite geometric series, from which the arctan(x) series is derived, is only valid for |x| < 1? I appreciate that this relates to the limit as x tends towards one, but it seems somewhat naļve to assume that the limit of arctan(x) and its series necessarily tend towards the desired result as x tends towards one. Is there a formal demonstration of this, or am I missing something really obvious? After all the geometric series tends towards an undefined state as x tends towards one.
Title: Re: Series for pi
Post by Aryabhatta on May 9th, 2007, 7:54am
It follows from Abel's limit theorem:

if f(x) = sum anxn (n = 1 to infinity) for |x| < r

and if sum anrn is convergent to L, then

Limit x->r- f(x) = L
Title: Re: Series for pi
Post by Icarus on May 9th, 2007, 3:22pm
Yes. The radius of convergence of a power series guarantees convergence in the interior, and divergence on the exterior. On the boundary the series may converge or diverge.

However, the function defined by the power series is continuous wherever the power series converges, both in the interior, and on the boundary. Since this power series is equal to arctan(x) on the interior, continuity requires they have the same value at all boundary points for which the series converges.
Title: Re: Series for pi
Post by Sir Col on May 9th, 2007, 3:37pm
That was a real "eureka" moment; I'd never properly grasped the significance of that result until now. Thanks guys.
Title: Re: Series for pi
Post by Icarus on May 9th, 2007, 5:04pm
Of course, since it is at the very verge of divergence, you can be sure that any such series will converge slowly. Thus

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4 = 1 - 1/3 + 1/5 - 1/7 + ...

is a very poor choice for calculating http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif.
Title: Re: Series for pi
Post by SWF on May 9th, 2007, 9:11pm
Although getting the right answer is tough to argue with, the ends don't justify the means. You also switched the order of integration and summation of an infinte series when it was not uniformly convergent.

You can get the series for atan(x) by expanding in a Taylor series about x=0.  For a faster converging series that is not difficult to find pi, arcsin(1/2) is pretty good, but not nearly as good as some of the other methods.
Title: Re: Series for pi
Post by ThudanBlunder on May 10th, 2007, 4:34am

on 05/09/07 at 21:11:10, SWF wrote:
Although getting the right answer is tough to argue with...

Can you argue with this?   :P

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifex = ex

ex(1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif) = 0

ex = (1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif)-1 * 0

= ( 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif3 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif4 + .......) * 0

= (1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + .......) * 0

And
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0 = 1
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1 = x
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx = x2/2!
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx2/2! = x3/3!
etc.

Hence ex = 0 + 1 + x + x2/2! + x3/3!  + .......

Title: Re: Series for pi
Post by rmsgrey on May 10th, 2007, 6:59am
16/64

cancel the sixes...
Title: Re: Series for pi
Post by Aryabhatta on May 10th, 2007, 11:01am

on 05/09/07 at 21:11:10, SWF wrote:
Although getting the right answer is tough to argue with, the ends don't justify the means. You also switched the order of integration and summation of an infinte series when it was not uniformly convergent.

You can get the series for atan(x) by expanding in a Taylor series about x=0.  For a faster converging series that is not difficult to find pi, arcsin(1/2) is pretty good, but not nearly as good as some of the other methods.


It is not uniformly convergent in (-1,1), but isn't it the case that for any 0 < r < 1, the series is uniformly convergent in [-r,r] ?
Title: Re: Series for pi
Post by Icarus on May 10th, 2007, 7:56pm

on 05/10/07 at 04:34:36, ThudanBlunder wrote:
Can you argue with this?   :P

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifex = ex

(1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif)ex = 0

ex = [1/(1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif)] * 0

= ( 1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif2 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif3 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif4 + .......) * 0

= (1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif + .......) * 0

And
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0 = 1
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif1 = x
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx = x2/2
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx2 = x3/3
etc.

Hence ex = 0 + 1 + x + x2/2 + x3/3  + .......


With the corrections that you need

ex = (1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif)-1 * 0

and

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx2/2 = x3/3!
etc.

and have conclusion ex = 0 + 1 + x + x2/2! + x3/3!  + .......,

it can all be justified - if you define it just right.
Title: Re: Series for pi
Post by Eigenray on May 10th, 2007, 11:07pm

on 05/10/07 at 04:34:36, ThudanBlunder wrote:
Can you argue with this?   :P


It would make more sense to do the following:  Define a linear operator T by

(Tf)(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0x f(t)dt,

and let

Un = I + T + T2 + ... + Tn.

If 1 is the function given by 1(x) = 1 for all x, then

(I - T)Un1 = (I - Tn+1)1 = 1 - xn+1/(n+1)!,

which converges to 1, say in L1(0,1), or Chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif(0,1), or uniformly on compacta, or some such.  In other words, letting A be the operator I-T, and fn = Un1, then

Afn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 1 = A(ex),

and we want to conclude fn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif ex.  That is, is A-1 continuous?

Note that, in L1 norm, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gifTfhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.giffhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif for all f http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0, so A is one-to-one.  But in fact http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gifTnhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif = 1 for all n, so Un does not converge in operator norm, although Un1 does converge.  For which f does Unf converge?
Title: Re: Series for pi
Post by ThudanBlunder on May 11th, 2007, 2:04pm

on 05/10/07 at 19:56:32, Icarus wrote:
With the corrections that you need

ex = (1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif)-1 * 0

and

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifx2/2 = x3/3!
etc.

and have conclusion ex = 0 + 1 + x + x2/2! + x3/3!  + .......,

it can all be justified - if you define it just right.

Yes, how careless of me! Now corrected. (I was a little drunk at the time.)   :-[

Also, I remember from when I studied elementary differential equations that differential operators can also be manipulated as though they are linear operators.
Title: Re: Series for pi
Post by ThudanBlunder on May 11th, 2007, 3:34pm

on 05/10/07 at 23:07:11, Eigenray wrote:
It would make more sense to do the following:  Define a linear operator T by

(Tf)(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif0x f(t)dt,

and let

Un = I + T + T2 + ... + Tn.

If 1 is the function given by 1(x) = 1 for all x, then

(I - T)Un1 = (I - Tn+1)1 = 1 - xn+1/(n+1)!,

which converges to 1, say in L1(0,1), or Chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif(0,1), or uniformly on compacta, or some such.  In other words, letting A be the operator I-T, and fn = Un1, then

Afn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 1 = A(ex),

and we want to conclude fn http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif ex.  That is, is A-1 continuous?

Note that, in L1 norm, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gifTfhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif < http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.giffhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif for all f http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0, so A is one-to-one.  But in fact http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gifTnhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/parallel.gif = 1 for all n, so Un does not converge in operator norm, although Un1 does converge.  For which f does Unf converge?

Are these Banach spaces?
Title: Re: Series for pi
Post by Eigenray on May 11th, 2007, 6:18pm

on 05/11/07 at 14:04:43, ThudanBlunder wrote:
Also, I remember from when I studied elementary differential equations that differential operators can also be manipulated as though they are linear operators.

They are linear operators.  Unfortunately they tend to be [link=http://en.wikipedia.org/wiki/Unbounded_operator]unbounded[/link], and not everywhere defined.


on 05/11/07 at 15:34:17, ThudanBlunder wrote:
Are these Banach spaces?

L1(0,1) and C(0,1) are Banach, but the normed space Chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif(0,1) isn't (a uniform limit of smooth functions need not be smooth).  And the topology of [link=http://en.wikipedia.org/wiki/Compact_convergence]compact convergence[/link] doesn't even come from a norm.
Title: Re: Series for pi
Post by Icarus on May 11th, 2007, 6:53pm
You can make Chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supinfty.gif a Banach space too, but you have to use a Sobolev norm (http://en.wikipedia.org/wiki/Sobolev_space).

And while the topology of compact convergence doesn't come from a norm, it still makes C0 a complete topological vector space - so the essentials of a Banach space are still there.


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board