How can i find the roots for this equation ?

[SphinX]

mathcad newton

Hi,

What is the best program\method to find the roots of this functions

x^6-5x^3+7=0

another one

x^7+3x+5=0

I don't know any of the roots .

Thanks

 

[gorkin]

You can use MatLab...
See more about MatLab in www.matlab.com , install demo version and in
"prompt", use: help root and You can learn more about Your necessites.
gorkin

 

[bastos4321]

use matematica. Its much more powerfull than matlab for that type of stuff. You can use the symbolic evaluation to check if your equation has an analitic solution, however for that order of polinom mith not have.

Bastos

 

[SphinX]

Hi,

I don't have a matlab or mathmatica now.
Is there any small soft for calculate the roots direct ?


Thanks for you help

 

[mayyan]

look here at a small program ( about 2 M ) called "Functions". it finds the roots draw and more. by the way it uses the Newton method to sove the equtions.

https://www.numericalmathematics.com/functions_and_equations.htm

p.s
this is a trail version

 

[E-design]

Hi,

What is the best program\method to find the roots of this functions

x^6-5x^3+7=0

another one

x^7+3x+5=0

I don't know any of the roots .

Thanks

Most graphic calculators will be able to calculate this. Another cheap program is Mathcad - student edition which I have seen advertised for as little as $15 on the web.

As far as the above equations, my HP calculator gets

an extremum at 1.36 for the first and a root at -1.08 for the 2nd

E

 

[flatulent]

solutions

Here are the solutions

the one with X^6 as the first term

-0.8201 +-j 1.114
-0.5544 +-j 1.267
1.375 +-j 0.1534

This could have been solved by hand by first substituting y=x^3 into the equation, solving for y and then for x

The one with x^7 as the first term

1.211 +-j 0.6299
0.1790 +-j 1.302
-0.8486 +-j 0.8447
-1.083

 

[Al Farouk]

Why you did not use one of the numeric techniques and translat it into C program for example, istead of searching for a tool.

 

[yoyo]

Mathematically, it was proven by a genius (Galois) more than 200 years ago that, "in general", there exists no solution for any polynomial equation of degree 5 or more. Sound strange? It's just that we can not separate the unknown x = (some means of algebraic manipulation. e.g +, -, *, /, root..). Too bad that this genius died too young in a tragedy.

On the other hand, it was also proven that there always exists a zero (real or complex) for any polynomial. Not remember who proved this, Gauss?

The two statements above seem confict each other. But no. Solution is there, but we can not get there.

So if you want an exact solution. Good luck! By chance, you may find it or you will waste your time.

If you want approximation, there are many means to do the job....
yoyo

 

[ynhe]

Hi
USUALLY I USE sigmaplot2000(www.sigmaplot.com) SOFTWARE TO PLOT THE FUNCTION AND SEARCH THE DOT CROSS THE X AXIS.
You could find the approximation solution.


ynhe

 

[jimjim2k]

Hi

You can use from the following approaches too:

1. use scientific workplace and its maple/mupad engine for online documentation plus equation soloving tools. It is very easy to use. Just like writing a paper draft.

2. Use maple stand alone program.


tnx

 

[BigBoss]

Evaluate that..

>> p=[1 0 0 -5 0 0 7]

p =

1 0 0 -5 0 0 7

>> roots (p)

ans =

-0.8201 + 1.1137i
-0.8201 - 1.1137i
-0.5544 + 1.2671i
-0.5544 - 1.2671i
1.3746 + 0.1534i
1.3746 - 0.1534i

>>

 

[nebisman]

well you can use MATLAB as our frien told you. You can get a powerfull free software called scilab which is a french MATLAB clone.

If your need is to develop a routine for polynomial roots calculation there are several methods. One of them is laguerre method you can deep on it if you see numerical recipes on C or numerical recipes on Pascal.

see:


http://www.unilim.fr/pages_perso/jean.debord/tpmath/tpmath.htm
**broken link removed**




Hope this helps. Mos algorithms use a newton search and an iterative polynomal division for finding a root, one a time. Dididing the polynomial by the foounded root and serch a root again.