Post your ideas for new math programs here, or build on ideas posted by other visitors.

I have a TI-83+ silver edition calculator and i was wondering if anyone knew of a simple factoring program that could just do something like x^2-9 = (x+3) (x-3)
If anyone knows of one, please let me know.
Lem419@aol.com

For a equ like you stated that would be:
mx^2+q=(x+a)(x-a)
where a = Sqrt(q/m) so in a program that is
Prompt M,Q
Disp (Q/M)^.5

OK...i made a great program that will do that but how would you do a program that would solve for a poly greater than 2nd degree?

my answer is to long so ill post it in two or more messages:

You question can be rewritten to find the zeropoints for a polynomal since a polynomal written in the form
(x-a)(x-b) is a polynomal with zeropoints a and b. So rewriting just means find the zeropoints. I believe CAS systems like yacas, mathematica etc usually do the rewriting through algebra but i dont think the TI-83 basic program code is powerfull enough, and i dont have any idea how such thing should be done. So i stick with a nummerical method:

Newton gives a method for finding zeropoints for any formula. The only problem that arrises is you do not know how much solutions exist, and how to get to what solution.

Using the formula we are certain we can get to one solution using newton. But with repeating this method of newton we can get to the other solutions. I'ill show you how:

f(x) is the formula that is to be rewritten to:
(x-a)(x-b)(x-c)... etc
First find using newton with a random starting point (i think its important to give some thought about the starting point though) the value of a. So that is:
a(n+1) = a(n) - f(a(n))/f'(a(n))
the deriative of f is easy to find, since you get the formula in an easy-to-derive formula from the user (prompt for the single terms and just lower the exponent with one (exp = exp - 1) and multiply the exponent with the factor)
now just repeat the line untill you notice that a(n+1)=a(n). The value of a has been determined.

to get b define a new function g as follows:
g(x) = f(x) / (x - a)
use the function g and newton to determine b
so b(n+1) = b(n) - g(b(n))/g'(b(n))
repeat till b(n+1)=b(n)
For c and the rest of the solutions you should keep repeating these steps. You are done when a newly defined function (like g(x) was defined as f(x) / (x -a)) appears to equal one. Note that it is possible that no solutions in the real plane exist (in other words, the solution is complex). To enable the program to find these solutions in the complex plane you should use numbers with a value not equal to zero for the imaginary unit of the starting point.
I hope this helped you
greets,

arno

Hello everyone, I'm brand spankin new to these forums and i have been programming for about 2 years. I have a program for anyone that is doing algebra, algebra 2, and trig. this is what it has:
Foiling(factoring)(x+3)(2x-5) that kind of stuff
Slope formula
Distance and midpoint formula
Synthetic Division
Heron Area
and Properties

Its pretty small but really useful (1696)
Im also working on a dictionary and an RPG for my TI83+

I have done some research on this and it appears possible to solve all polynomals up to degree four through algebra, though for polynomals up from degree five it has been proven to be impossible. Excluding special cases where a polynomal of higher degree can be rewritten to a product of two 'low' degree polynomals. For example:

f(x) = 3x^7+2x^6+7x^5+8x^4+11x^3+10x^2+4x = (3x^3+2x^2+4x) * (x^4+x^2+2x+1)

Solutions of a(x) = (3x^3+2x^2+4x) and b(x) (x^4+x^2+2x+1) are possible to find. Since f(x) is the product a(x) and b(x), f(x) is zero when either a(x) is zero or b(x) is zero.

uhhh, im a little confused. but i think what you need are formulas, like the quadratic formula-it finds the zeros and with those u put em in the (x-a)(x-b) form. use the quadratic for 2nd degree and general cubic for 3rd ect. but your right about 5th degree and above, there is no possible general formula. (they have to be solved with computer algorithms) hope that helped a bit-if not, im sorry.

What kind of calculator do you have?
I know that for the TI-83+/SE, you can download PolySmlt, at education.ti.com, to solve for polynomials up to the 30th degree.

go to the TI-83+ BASIC Math and look for mathstuff.zip (it has A LOT of other stuff) it has a foiling program for quadratics, but you have to look under the "quadratics" section of the main menu. look for it on 11/20/03

Hello everyone
I'm a programmer very lousy but getting better day by day, but somehow while in my free time, i wanted to make a program that chains others. I've got them all finished, but the chain is the problem. I need an algorithm good enough in basic so my program asks the ecuation and the user just input it in the line. I'm specific, i'm not asking a quadratic, i mean ANY FUNCTION like: e^Cos(x*Ln(pi-x/(x+1))) got it?? well, i did found a solution doing: "define f(x)= here your ecuation"" but that's the problem cause u have to write that whole line so my program work BEFORE entering to the program itself.
SO:
HOW THE HECK DO I PUT AN ALGORITHM THAT ASKS THE FUNCTION AND DEFINES IT AS F(X)????
Answer me please.

I GUESS YOU ARE PROGRAMING IN TI BASIC, WHY DONT YOU ASK THE FUNCTION WHEN YOU CALL THE PROGRAM, LIKE THIS MAIN \ YOUR_PROGRAM_NAME ( e^Cos(x*Ln(pi-x/(x+1))),x )

What you are saying is somewhat confusing, but if I understand, the program would be something like this: (Using the variable x, not the coordinate)
:Input "f(x)=",Str1
:Lbl 1
:Prompt x
:Str->equation~as the function~(Str1,Y1~as the variable~)
:Disp Y1
:Goto 1~if you want to reapeat~

I'm writing a very simple pascals triangle program on my ti-83 plus. It lists all the numbers in whatever row of pascals trianle you put in. I was wondering if anyone knew how to set it up so you could scroll through a long verticle set of numbers after they are displayed.

try this:
:Disp "#"
:Pause
:ClrHome
:Disp "#"
etc...
If you want to put it in a certain area, use output(5,7,"#") instead of Disp.
The 5 and 7 are your coordinates on a graph. I hope this helps you EEE55.

if you do this
Pause #
you can scroll it

or even better:
Try this...
Prgm
{1}->L1
For(x,1,#of rows to display)
Pause L1
augment(L1,{0})->L2
augment({0},L1)->L3
L2+L3->L1
EndFor
EndPrgm

This is for the TI-89...if you need it in another calcs format, feel free to email me at jrock7286@aol.com... Hope this helped!

Determine the number of displays algebraicly, then pause and ClrHome after each one. I wrote a sample program that has a number of errors. Tell me if you want to see it.

Or... determine the length of each number, output them in str1, and display it. The last line should be:
:Str1

Or... place the row into L1 and the last line of the program should be:
:L1

Or... place a good amount of Pascal's Triangle into a matrix. When you need a row, call up that row of the matrix.

I hope this helps.

Does anyone know of a good CAS(Computer Algebra System) for any ti calc?

Yes- The OS.

Hahahahaha.

Haha! Let's see you try to integrate e^-t*t^z on a ti 86!

or just Integral(e^(x^2), x)