Numerical computing fixed point method using matlab?
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
920579
use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
914601
http://mathworld.wolfram.com/PoincareConjecture.html
429670
In matlab, type\n>> syms x\n>> solve('x^3+4*x^2-5=0')\n \nans =\n \n[                1]\n[ -5/2+1/2*5^(1/2)]\n[ -5/2-1/2*5^(1/2)]
815728
first this is not a PDE but ODE\n\nsimple integration would do
215087
false position is simply a way of estimating the zeroes of a function. I don't know about matlab, but you can find code for C/C++, java, and others at www.planetsourcecode.com\n\nHowever, if you're trying to solve a problem rather than just code false position, I would recomend secent method or newton's method as they provide a higher order of convergence
676950
A few that come to mind at Euler, Maxima and Octave. Unfortunately, I don't think any of the free packages come close to Mathematica or Matlab, but maybe one of them will still meet your needs.
268125
just investigate for example ODE y"=-y and then add to the left part increasingly small non-linearity +c*y^2 and find solution of the first order of c - you will see how different modes interact with each other.
141538
Read the references at the end of the article or talk to an industrial engineer.\n\nhttp://en.wikipedia.org/wiki/Linear_programming
85418
dx/dt+ax=b :P\n\n\nmathematica had some online find-the-derivitive program, last I remember
601386
Through the fourier map generator.
437960
why not try Danube's Enigma, it is easy and fulfills the requirements of all matlab projects!\n:)
131534
http://www.google.com/search?hl=en&q=matlab+ocr&btnG=Google+Search
354245
the answer is 1.442250\n\nthe solution will be rigorous if tried to solve mathematically. The better way is trial and error and the following program does that pretty well. \n\n#include<stdio.h>\n#include<math.h>\n\nint main()\n{\nfloat error = .000001;\nfloat x = 1.0;\nfor (;;)\n{\nfloat y = pow (x,3);\nif ( pow (x,y) < 3 )\n{\nx = x+error;\n}\nelse break;\n}\nprintf ("%f",x);\nreturn 0;\n}
12763
There is a review called "ACM Transactions on Mathematical Software (TOMS)". Check out Volume 20, Issue 3 (September 1994)\nPages: 373 - 392   \n(ISSN:0098-3500)
902349
Hynton has it.  Just remember that you'll be restricted to x > 0 because of the original form of the problem.
742919
Free source codes for several numerical methods problems are available at http://www.sourcecodesworld.com/. please visit and download subjected to copyrights of distribution.