[VIEWED 6331
TIMES]
|
SAVE! for ease of future access.
|
|
|
guchcha chor
Please log in to subscribe to guchcha chor's postings.
Posted on 01-27-10 8:58
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Hello MatLab /C++ gurus out there, I tried to do it.. I couldn't. Please help me....
|
|
|
|
F22
Please log in to subscribe to F22's postings.
Posted on 01-28-10 12:15
AM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Its in no way complete and hasn't been verified but should be enough of a pointer to get you in the right track if using Matlab: clear clc N = 2; e = 100; e1 = 10^-5; piii(1)= 1/0.75; while e >= e1 pii=0; for k=1:N piii1= ((-1)^(k+1))/(0.5*k-0.25); pii=piii1+pii; end piii(N)=pii; e=abs(piii(N)-piii(N-1))/piii(N-1); N=N+1; end On a second note you could tell the Prof., " I can use pi function of Matlab to get the damn number, why worry about doing the homework"
|
|
|
santosh1984
Please log in to subscribe to santosh1984's postings.
Posted on 01-28-10 1:29
AM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Here is the code. It is perfectly running and does converge to 3.1416 for e=10^-8 with N= 63661979. Do the same for 10^-1 thru 10^-8. Hope you know how to make plots after this. Make sure the name of mat-file is pii.m and enjoy!!!!
function pii clc N=2; pii=((-1)^2)/(0.5*1-0.25); e=1; while e>=10^-8; pii_old=pii; k=N; pii1= ((-1)^(k+1))/(0.5*k-0.25); N=N+1; pii=pii_old+pii1; e=abs(pii-pii_old)/pii_old; end pii_old pii N e
|
|
|
guchcha chor
Please log in to subscribe to guchcha chor's postings.
Posted on 01-28-10 12:03
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0
?
|
|
Thank you very much F22 and Santosh1984. I was dying to get the right answer... Thank you
|
|