wu :: forums
« wu :: forums - Series for pi »

Welcome, Guest. Please Login or Register.
Oct 31st, 2024, 4:10pm

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   general problem-solving / chatting / whatever
(Moderators: towr, Eigenray, william wu, Icarus, ThudnBlunder, SMQ, Grimbal)
   Series for pi
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Series for pi  (Read 873 times)
Sir Col
Uberpuzzler
*****




impudens simia et macrologus profundus fabulae

   
WWW

Gender: male
Posts: 1825
Series for pi  
« on: May 9th, 2007, 1:40am »
Quote Quote Modify Modify

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.
« Last Edit: May 9th, 2007, 1:48am by Sir Col » IP Logged

mathschallenge.net / projecteuler.net
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Series for pi  
« Reply #1 on: May 9th, 2007, 7:54am »
Quote Quote Modify Modify

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
IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler
*****



Boldly going where even angels fear to tread.

   


Gender: male
Posts: 4863
Re: Series for pi  
« Reply #2 on: May 9th, 2007, 3:22pm »
Quote Quote Modify Modify

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.
IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
Sir Col
Uberpuzzler
*****




impudens simia et macrologus profundus fabulae

   
WWW

Gender: male
Posts: 1825
Re: Series for pi  
« Reply #3 on: May 9th, 2007, 3:37pm »
Quote Quote Modify Modify

That was a real "eureka" moment; I'd never properly grasped the significance of that result until now. Thanks guys.
IP Logged

mathschallenge.net / projecteuler.net
Icarus
wu::riddles Moderator
Uberpuzzler
*****



Boldly going where even angels fear to tread.

   


Gender: male
Posts: 4863
Re: Series for pi  
« Reply #4 on: May 9th, 2007, 5:04pm »
Quote Quote Modify Modify

Of course, since it is at the very verge of divergence, you can be sure that any such series will converge slowly. Thus
 
/4 = 1 - 1/3 + 1/5 - 1/7 + ...
 
is a very poor choice for calculating .
IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
SWF
Uberpuzzler
*****





   


Posts: 879
Re: Series for pi  
« Reply #5 on: May 9th, 2007, 9:11pm »
Quote Quote Modify Modify

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.
IP Logged
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: Series for pi  
« Reply #6 on: May 10th, 2007, 4:34am »
Quote Quote Modify Modify

on May 9th, 2007, 9:11pm, SWF wrote:
Although getting the right answer is tough to argue with...

Can you argue with this?   Tongue
 
ex = ex
 
ex(1 - ) = 0
 
ex = (1 - )-1 * 0
 
= ( 1 + + 2 + 3 + 4 + .......) * 0  
 
= (1 + + + + + .......) * 0
 
And
0 = 1
1 = x
x = x2/2!
x2/2! = x3/3!
etc.
 
Hence ex = 0 + 1 + x + x2/2! + x3/3!  + .......  
 
« Last Edit: May 11th, 2007, 2:35pm by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
rmsgrey
Uberpuzzler
*****





134688278 134688278   rmsgrey   rmsgrey


Gender: male
Posts: 2873
Re: Series for pi  
« Reply #7 on: May 10th, 2007, 6:59am »
Quote Quote Modify Modify

16/64
 
cancel the sixes...
IP Logged
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Series for pi  
« Reply #8 on: May 10th, 2007, 11:01am »
Quote Quote Modify Modify

on May 9th, 2007, 9:11pm, 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] ?  
IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler
*****



Boldly going where even angels fear to tread.

   


Gender: male
Posts: 4863
Re: Series for pi  
« Reply #9 on: May 10th, 2007, 7:56pm »
Quote Quote Modify Modify

on May 10th, 2007, 4:34am, ThudanBlunder wrote:

Can you argue with this?   Tongue
 
ex = ex
 
(1 - )ex = 0
 
ex = [1/(1 - )] * 0
 
= ( 1 + + 2 + 3 + 4 + .......) * 0  
 
= (1 + + + + + .......) * 0
 
And
0 = 1
1 = x
x = x2/2
x2 = x3/3
etc.
 
Hence ex = 0 + 1 + x + x2/2 + x3/3  + .......  
 

 
With the corrections that you need  
 
ex = (1 - )-1 * 0
 
and
 
x2/2 = 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.
 
« Last Edit: May 10th, 2007, 8:34pm by Icarus » IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Series for pi  
« Reply #10 on: May 10th, 2007, 11:07pm »
Quote Quote Modify Modify

on May 10th, 2007, 4:34am, ThudanBlunder wrote:

Can you argue with this?   Tongue

 
It would make more sense to do the following:  Define a linear operator T by
 
(Tf)(x) = 0x 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 C(0,1), or uniformly on compacta, or some such.  In other words, letting A be the operator I-T, and fn = Un1, then
 
Afn 1 = A(ex),
 
and we want to conclude fn ex.  That is, is A-1 continuous?
 
Note that, in L1 norm, Tf < f for all f 0, so A is one-to-one.  But in fact Tn = 1 for all n, so Un does not converge in operator norm, although Un1 does converge.  For which f does Unf converge?
« Last Edit: May 10th, 2007, 11:14pm by Eigenray » IP Logged
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: Series for pi  
« Reply #11 on: May 11th, 2007, 2:04pm »
Quote Quote Modify Modify

on May 10th, 2007, 7:56pm, Icarus wrote:

 
With the corrections that you need  
 
ex = (1 - )-1 * 0
 
and
 
x2/2 = 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.)   Embarassed
 
Also, I remember from when I studied elementary differential equations that differential operators can also be manipulated as though they are linear operators.
 
« Last Edit: May 11th, 2007, 2:39pm by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: Series for pi  
« Reply #12 on: May 11th, 2007, 3:34pm »
Quote Quote Modify Modify

on May 10th, 2007, 11:07pm, Eigenray wrote:

 
It would make more sense to do the following:  Define a linear operator T by
 
(Tf)(x) = 0x 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 C(0,1), or uniformly on compacta, or some such.  In other words, letting A be the operator I-T, and fn = Un1, then
 
Afn 1 = A(ex),
 
and we want to conclude fn ex.  That is, is A-1 continuous?
 
Note that, in L1 norm, Tf < f for all f 0, so A is one-to-one.  But in fact Tn = 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?
 
IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Series for pi  
« Reply #13 on: May 11th, 2007, 6:18pm »
Quote Quote Modify Modify

on May 11th, 2007, 2:04pm, 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 unbounded, and not everywhere defined.
 
on May 11th, 2007, 3:34pm, ThudanBlunder wrote:

Are these Banach spaces?

L1(0,1) and C(0,1) are Banach, but the normed space C(0,1) isn't (a uniform limit of smooth functions need not be smooth).  And the topology of compact convergence doesn't even come from a norm.
IP Logged
Icarus
wu::riddles Moderator
Uberpuzzler
*****



Boldly going where even angels fear to tread.

   


Gender: male
Posts: 4863
Re: Series for pi  
« Reply #14 on: May 11th, 2007, 6:53pm »
Quote Quote Modify Modify

You can make C a Banach space too, but you have to use a Sobolev norm.
 
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.
« Last Edit: May 11th, 2007, 6:56pm by Icarus » IP Logged

"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

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