- #1

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Integal[{x^2},x]; and then have the output of x^3/3

preferably something easy to use like Derive. Any pointers are welcome.

- Mathematica
- Thread starter haki
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- #1

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Integal[{x^2},x]; and then have the output of x^3/3

preferably something easy to use like Derive. Any pointers are welcome.

- #2

chroot

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- Warren

- #3

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Check out this thread on www.mathlinks.ro .

- #4

HallsofIvy

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- #5

chroot

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Eh? Are you unaware of programs like Maxima and Octave? They are extremely good free replacements for the commercial packages Mathematica and MATLAB.heckof a lot of work, typically by a large crew of programmers. I doubt they are going to do all that work for free.

- Warren

- #6

uart

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To see how this software that clearly does require "one heck of a lot of work" to develop can become GPL freeware take look at the brief but interesting wikipedia article

http://en.wikipedia.org/wiki/Maxima" [Broken]

http://en.wikipedia.org/wiki/Maxima" [Broken]

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- #7

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Maxima rocks. I like it very much (commands and such, very intuitive) and the help system is excellent. For now it covers everything I need but there was one thing I couldn't find in maximas help.

I wanted to get the first pair of numbers (x,y) that are the solution to the following Diophantine equation 222x+255y=9 where y is the smallest possible positive intiger for the pair (x,y) to be the solution to the equation.

- #8

uart

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I don't know how to do that in Maxima, maybe someone else can help there, but that particular equation is very easy to solve by inspection. Soln :

Maxima rocks. I like it very much (commands and such, very intuitive) and the help system is excellent. For now it covers everything I need but there was one thing I couldn't find in maximas help.

I wanted to get the first pair of numbers (x,y) that are the solution to the following Diophantine equation 222x+255y=9 where y is the smallest possible positive intiger for the pair (x,y) to be the solution to the equation.

- #9

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I think that the pair (-8,7) that is x = -8, y = 7 is the first pair of solutions for which y is the smallest positive number. But anyway, thats a bit exotic problem to work on maxima.I don't know how to do that in Maxima, maybe someone else can help there, but that particular equation is very easy to solve by inspection. Soln :x=-3, y=3

- #10

uart

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Doh, I misread the equation as 2**5**2x+255y=9.

I need to start wearing my glasses more often.

I need to start wearing my glasses more often.

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