**Infinite series**Hi:
We have the following identity:
\sum_{m=1}^{infinity} (-1)^m/((2m-3)^2*(2m-1)*(2m+1)^2)=-Pi/32.
When we type the command,
In[1]:=Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,Infinity}]
we get
2 1 1
-16 Pi + 2 Pi - HurwitzZeta[2, -(-)] - Zeta[2, -]
4 4
Out[1]= --------------------------------------------------
512
The command Simplify[%] does not simplify it further.
I am sure the above expression must be equal to -Pi/32, ...

**Infinite series**Is anyone out there familiar with using MATLAB to manipulate infinite series
calculations and where do I go to find?...cmw.
...

**Re: Infinite series**On 12/15/09 at 7:27 AM, jogc@mecheng.iisc.ernet.in (Dr. C. S. Jog)
wrote:
>We have the following identity:
>\sum_{m=1}^{infinity} (-1)^m/((2m-3)^2*(2m-1)*(2m+1)^2)=-Pi/32.
>When we type the command,
>In[1]:=Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,Infinity}]
>The command Simplify[%] does not simplify it further.
>I am sure the above expression must be equal to -Pi/32, but a user
>would prefer this answer than the above one.
There are a variety of reasons Simplify often does not achieve
what you are looking for. But the obvious thing to try when...

**Mathematica and infinite series**Hi,
I am about to embark on a project that operates heavily in infinite
series, so I started figuring out Mathematica's basis capabilities. I
found them very impressive, but I came across this:
f[x_] := Sum[Log[n]/(n^2 Factorial[n]) x^n, {n, 1, Infinity}]
Assuming[n > 0, SeriesCoefficient[f[x], {x, 0, 4}]]
Answer:
SeriesCoefficient[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]
\*FractionBox[\(
\*SuperscriptBox[\(x\), \(n\)]\ Log[n]\), \(
\*SuperscriptBox[\(n\), \(2\)]\ \(n!\)\)]\), {x, 0, 4}]
Why doesn't Mathematica produce Log[n]/(n...

**How to solve this infinite series...**The cosine function can be evaluated by the following
infinite series:
cos x = 1 – [x^2 / 21] + [x^4 / 41] – [x^6 / 61] ……
Write a Matlab function to implement this formula so that it
computes and prints out the values of cos(x) as each term in
the series is added. In other words, compute and print in
sequence the values for:
cos x = 1
cos x = 1 - [x^2 / 21]
cos x = 1 - [x^2 / 21] + [x^4 / 41]
In article <fnlmg7$9g7$1@fred.mathworks.com>,
Melvin <melvin1974@mathwoks.com> ...

**infinite series #2**New to MATLAB. Trying to write code to determine the covergence value
of an infinite series. I also need to display to 5 sig. fig. Any help
would be appreciated.
Thanks, Danny
Danny Little wrote:
>
>
> New to MATLAB. Trying to write code to determine the covergence
> value
> of an infinite series. I also need to display to 5 sig. fig. Any
> help
> would be appreciated.
>
> Thanks, Danny
Take a look at the function symsum, from the Symbolic Math toolbox.
Nilton Quoirin wrote:
>
>
> Danny Little wrote:
>>
>>
>> New to MATLAB. Trying to w...

**How to fit with an infinite serie function ?**I would like to fit non-linear data points I have collected from FRAP experiments. The function to be fitted is an infinite serie :
f = Somme (((-b)^n/factorial(n))/(1+n*(1+2*t/tauD))) with n=1 to 100
I need to find the best value for my parameters b and tauD.
I have tried to use the curve fitting tool but the function was not accepet as custom equation.
I'm totally novice in matlab so I would be pleased if someone could help me.
"Aurelie Chabaud" <aurelie.billon@univ-nantes.fr> wrote in message <i16rvl$hmf$1@fred.mathworks.com>...
> I would like to f...

**How to solve this infinite series... #2**The cosine function can be evaluated by thefollowing
infinite series:
cos x = 1 – [x^2 / 21] + [x^4 / 41] – [x^6 / 61] ……
Write a Matlab function to implement this formula so that it
computes and prints out the values of cos(x) as each term in
the series is added. In other words, compute and print in
sequence the values for:
cos x = 1
cos x = 1 - [x^2 / 21]
cos x = 1 - [x^2 / 21] + [x^4 / 41]
In article <fnqlf2$lmp$1@fred.mathworks.com>,
Melvin <melvin1974@mathwoks.com> w...

**A model problem for infinite series**Hi,
As a follow up to my previous post and to give a little bit more
information, I would like to briefly describe a mathematically
nonsensical problem which has some of the elements that I need.
To solve Laplace's equation for u(r, alpha) on the unit circle subject
to Dirichlet boundary conditions U(alpha), one needs to decompose U as a
Fourier series, multiply each term by r^|n| and add them back up.
The model problem is this. Starting with a boundary condition U0, solve
for u and let U1(alpha) = du/dr(evaluated at r=1)*f[alpha], where
f[alpha] is relatively simple a...

**Can Mathematica do this (infinite series)?**Hi,
I'm working on a project involving infinite series and I don't know how
to do it or even ask a sensible question about it. So I cooked up a
question the answer to which might give me ideas.
f[x_]:=Sum[c[n]x^n, {n, 1, Infinity}]
What's the infinite series for
f[x]^2 + Sin[x]f[x]
in terms of c[n]?
What's the simplest way that Mathematica can answer this question for
general c[n]?
The pipe dream is this:
f[c_][x_] := Sum[c[n] x^n, {n, 1, Infinity}]
g[c_][x_] := f[c][x]^2 + Sin[x] f[c][x]
d[c_][n_] := SeriesCoefficient[g[c][x], {x, 0, n}]
...

**A Series of Convolution of Infinite Sequences**I have a set of $M$ infinite data sequences, and I hope to perform a serie
of convolutions such that
SEQ_result = SEQ_1 * SEQ_2 * SEQ_3 * ... SEQ_M
I am only interested at the first few terms of SEQ_result.
Right now, I can only think of one method, which is to use fas
convolution (truncation all to finite sequences, FFT, pointwise multiply
iFFT). Is there other efficient way to implement such system?
Thanks for input.
CJ
On Nov 28, 11:21 am, "cjlam" <jethro...@gmail.com> wrote:
> I have a set of $M$ infinite data sequences, and I hope to perform a series
> of ...

**Infinite Series Error**Thanks to everyone who responded (not sure why I cannot see the
responses at the newsgroup).
Here is something peculiar:
Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
gives an error, but
(Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, 1}] /. {q ->
0.1}) + (Sum[
q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 2, Infinity}] /. {q -> 0.1})
gives the correct answer
-21.1012
...

**Infinite Series Error**Can anyone explain why evaluating the following expression gives an
error?
Sum[q^(-6 + 4*n)/(1 - q^(-5 + 4*n)), {n, 0, Infinity}] /. {q -> 0.1}
None of the denominators are 0 (would only happen if -5 + 4*n=0, which
is impossible for an integer n, yet I get an error message "Infinite
expression 1/0 encountered" (several times in fact).
Any help on a work around greatly appreciated.
(Please reply to my personal e-mail also).
Thanks,
Jimmy
Hello Jimmy,
Thank you for reporting the error with the infinite
sum considered by you.
The unexpected Power::infy...

**problems with hypergeom and infinite series**Dear all,
I have encountered the following problems with MatLab R2010b.
1. The following instruction
>> hypergeom(62,[64,26],-400
cannot get the answer and I dont know why.
2. The following m-file cannot give the plot. In fact it stops at one point and I found the mistake maybe the hypergeom instruction, but I have no idea how to fix it.
Could anyone help me in overcoming this problem?
Many thanks in advance!
Kindest regards,
Dale Guo
clc
clear all
close all
r=3.;
lam=5.32e-4;
k=2*pi/lam;
n=25;
f=500;
alpha=9.9;
rho=linspace(0.,1,100);
len=length(rho)...

**Infinite loop calculating a series value**I'm trying to calculate the value of this expresion:
ABS(\sum(K=1, \infty, ((1+i)/2)^k))
the result should be 1 but my calculator enter in a infinite loop. How
can I fix it?
Thank you, bye.
On Mon, 10 Mar 2008 09:26:58 -0500, mambro wrote:
> I'm trying to calculate the value of this expression:
> ABS(sum(K=3D1,infinity,((1+i)/2)^k))
> the result should be 1
> my calculator enters an infinite loop.
An explicit sum of an infinite number of terms will tend to do that.
You could replace "infinity" by 100, say, or else use a limit:
First do command CASCFG, the...

**polynomial division which gives a infinite series**hi
is there any code or built in function in matlab to perform polynomial division betwee two finite polynomials to get an infinite series ,,, pls dont suggest me to use deconv function this gives zero as quotient for division of lower order polynomial with the higher order one,,, please post in the answer as soon as possible
thanx in advance..................
"sree vidyanikethan engineering college Morusu" <madan_morusu@gmail.com> wrote in message <gn917l$l0v$1@fred.mathworks.com>...
> hi
>
> is there any code or built in function in matlab to perfo...

**Print value of p from the infinite series**Hi, new here.
I'm trying to print a table that shows how many terms needed to
accomplished 3.14, 3.141, 3.1415, 3.14159.
But, dead end with weird logic-error (at least for me, right now).
Here is how, suppose the output should something like:
<code>
full: p: flag: p in int: counter:
4.000000 4.000 true 4000 1
2.666667 2.667 false 2666 2
3.466667 3.467 true 3466 3
....
<snipped>
....
3.142186 3.142 true 3142 1685
3.141000 3.141 f...

**calculate the value of pi from the infinite series**does anyone know how to calculate the value of pi from the infinite series
pi=4-4/3+4/5-4/7+4/9-4/11+...
then how to shows the value of pi approximated by one term of this series, by two terms, by three terms,etc. How many terms of this series do you have to use before you first get a)3.14 b)3.141 c)3.1415 d)3.14159
"atiqa eiqa" <atiqa_eiqa@yahoo.com.my> wrote in message <gr2me5$2h8$1@fred.mathworks.com>...
> does anyone know how to calculate the value of pi from the infinite series
>
> pi=4-4/3+4/5-4/7+4/9-4/11+...
>
> then how to shows the value of...

**Re: Mathematica and infinite series #2**No, this sort of thing certainly won't work. Unless Mathematica can find a
closed form of the sum it will not be able to compute a general series coefficient from an infinite series. However, it will work if you truncate the series than you can get particular series coefficients, like this:
f == Sum[BesselJZero[0, n^2] Sin[n^2] Log[
Sin[Cos[n]]] Log[n]/(n^2 Factorial[n]) x^n, {n, 1, 10}] +
O[x]^11;
SeriesCoefficient[f, 4]
and so on.
Andrzej
On 17 Oct 2010, at 21:44, Sam Takoy wrote:
> Hi,
>
> Thanks for the response!
>
> I don't have Ma...

**this is my solution of calculate pi using infinite series**%include <iostream>
%include <cmath>
%include <iomanip>
using namespace std;
int main()
[
// Declare and initalise variables
int decPlaces;
double fraction=0, pi=0;
cout << "Program to calculate pi using the sum of an infinite series" << endl;
cout << "Please input the number of decimal places to display: " << endl;
cin >> decPlaces;
cout << end1;
// calculate the value for pi using a 100 step loop
for ( int n = 1; n <= 100; n++ )
[
fraction = (4*pow(-1.0, n+1))/(2*n...

**compute cos(x) using taylor series**I need to compute using the taylor series but i must also name the function, i have looked around and from what i can understand this is how i thought it needs to be done:
%x- the argument
%N – the initial number of terms in the expansion
function[x N]=ty_cos(x,N)
for x=1:N
y=feval(taylor(cos(x)))
end
end
either my understanding of this is very wrong or my coding is wrong seeign as it doesnt work, all i need to be able to do, is create the taylor series expansion of cos(x) for the amount of terms N
Dear Kevin!
> I need to compute using the taylor series but i must...

**this is my solution of calculate pi using infinite series #2**%include <iostream>
%include <cmath>
%include <iomanip>
using namespace std;
int main()
[
// Declare and initalise variables
int decPlaces;
double fraction=0, pi=0;
cout << "Program to calculate pi using the sum of an infinite series" << endl;
cout << "Please input the number of decimal places to display: " << endl;
cin >> decPlaces;
cout << end1;
// calculate the value for pi using a 100 step loop
for ( int n = 1; n <= 100; n++ )
[
fraction = (4*pow(-1.0, n+1))/(2*n...

**Command stand for Series(Sigma from n=1 to infinite**Which command can be used as inf
=A1=C6(function)
n=3D1
...

**Which Command stands for Series(Sigma from n=1 to infinite)??? ThankU...**
Which command can be used as inf
=A1=C6(function)
n=3D1
...