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liza
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Joined: May 30th, 2003, 6:40 am

tri-diagonal system

November 10th, 2003, 2:22 pm

Can somebody advise me a numerical method for solution tri-diagonal system of equations? I am trying to implement implicit method for european call option.thanks.
 
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MattF
Posts: 6
Joined: March 14th, 2003, 7:15 pm

tri-diagonal system

November 10th, 2003, 2:27 pm

If you are willing to buy a book then Numerical Recipes in C/C++ has what you want, called tridiag() I think.
 
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robertral
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Joined: March 6th, 2003, 7:12 am

tri-diagonal system

November 10th, 2003, 2:53 pm

QuoteOriginally posted by: MattFIf you are willing to buy a book then Numerical Recipes in C/C++ has what you want, called tridiag() I think.Yeah it is Tridiag. I used it about 4 months ago.
 
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hazerider
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Joined: July 24th, 2003, 3:45 pm

tri-diagonal system

November 10th, 2003, 6:22 pm

QuoteOriginally posted by: MattFIf you are willing to buy a book then Numerical Recipes in C/C++ has what you want, called tridiag() I think.No need to buy the book, just go to http://www.nr.com.
 
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liza
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Joined: May 30th, 2003, 6:40 am

tri-diagonal system

November 11th, 2003, 7:26 am

thanks a lot guys, I've found this book.
 
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liza
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Joined: May 30th, 2003, 6:40 am

tri-diagonal system

November 11th, 2003, 9:42 am

hi guys,sorry for asking stupid questions, but can somebody explaine tridiag or advise some articles where it is explained:| b1 c1 0 .................| |u1| |r1| | a2 b2 c2 ...............| |u2|= |r2||..............................| | | | ||..................0 aN bN| |uN| |rN]solution:u1=r1/(bet=b1)for (j=2; j<=n; j++) { gam(j)=c(j-1)/bet; bet=b(j)-a(j)*gam(j); u(j)=(r(j)-a(j)*u(j-1))/bet}for (j=(n-1);j>=1;j--) u(j)-=gam(j+1)*u(j+1);why u(1)=r(1)/b(1) ?
 
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Graeme
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Joined: April 25th, 2003, 5:47 pm

tri-diagonal system

November 11th, 2003, 4:49 pm

The explanation in Burden and Faires is not bad at all.