[xtremee]

Hi all,

i wanna to Know What is the BEST way to solve equation of order 6?

if you have equation : s^6+2S^5+S^4+8S^3+4S^2+15S+1562=0

What will be the MOST easiest way to find its roots!!

i know that you can use Bi-Section (Halfing Method)

Any Other Ideas..

N.B. I wanna to solve it manually without the use of PC or Calculator.



Any ideas will be excellent



Regards,

Xtremee

This post has been edited by xtremee: 08 November 2006 - 05:19 PM

[XP-is-a-CRAP]

Quote

if you have equation : s^6+2S^5+S^4+8S^3+4S^2+15S+1562=0

WOW ... trying to remember about algebra ...

Quote

What is the BEST way to solve equation of order 6?
What will be the MOST easiest way to find its roots!!
I wanna to solve it manually without the use of PC or Calculator.

SORRY but it definitely IS the (a smart one) calculator or PC.

Quote

Any Other Ideas..

Well, since the highest exponent is 6 and divisible by 2,
there is the risk that there is NO solution at all ...

The number of solutions may be 0 ... 6 .

There is a good method to solve equations of order 2 (possible result:
no solution exists).

There is a bad method for order 3 (exact solution for any equation)

But there are NO (exact and universal) methods for order >=4 !!!

What one can try is the brute-force method: test for solutions +1, -1, +2, -2, ...
and reduce through polynom division ... works for "artificial" "school" equations
with low integer solutions only

Otherwise, make a graph (preferably with a PC ), and then use an
inexact method (tangent method, ...) to approach the solutions.

BTW: If someone has the result (PC math prog) please post ...

This post has been edited by XP-is-a-CRAP: 03 November 2006 - 02:39 PM

[IcemanND]

the equation posted is not possible.

how about the constant being 62 instead of 1562?



But of course the answer to everything is 42.

[tain]

I'm no good with math but I can tell you that it is spelled Mathematical not Mathimatical.

[LLXX]

XP-is-a-CRAP, on Nov 3 2006, 03:37 PM, said:

There is a good method to solve equations of order 2 (possible result:
no solution exists).

There is a bad method for order 3 (exact solution for any equation)

But there are NO (exact and universal) methods for order >=4 !!!

Correction, a formula using conventional algebra exists for polynomials up to and including order 4 (quartic). It has been proved that no closed-form algebraic solution exists for polynomials of order 5 or higher.

http://en.wikipedia..../Cubic_equation
http://en.wikipedia....uartic_equation
http://en.wikipedia....uintic_equation

Quote

BTW: If someone has the result (PC math prog) please post ...

Here is Mathematica 4.1 attempting to solve your sextic equation. It cannot offer exact roots, as it does not have the capabilities to go beyond ordinary algebra. Below that is an approximation to each of the 6 roots (all complex numbers).

http://img145.images...72/equusuj6.png

(Yes, "equus" is Latin for "horse".)

[gamehead200]

Confirmed my my TI-89 and TI-83+ calculators. There are no solutions for that equation.

However:

d/dS(s^6+2S^5+S^4+8S^3+4S^2+15S+1562)
= 6S^5+10S^4+4S^3+24S^2+8S+15

6S^5+10S^4+4S^3+24S^2+8S+15 = 0
S = -2.18274

Don't know if that will help...

[LLXX]

gamehead200, on Nov 3 2006, 08:00 PM, said:

d/dS(s^6+2S^5+S^4+8S^3+4S^2+15S+1562)
= 6S^5+10S^4+4S^3+24S^2+8S+15

6S^5+10S^4+4S^3+24S^2+8S+15 = 0
S = -2.18274

Fixed.

[ripken204]

ya, just plug it into a calc and u should get the roots, although i havnt tried it yet. just graph and see where x=0

[Zxian]

@gamehead - that'll give you where the original equation is flat (i.e. a maxima or minima). It unfortunately won't give you the zeros of the equation.

[gamehead200]

I know. I just wanted to get in on the conversation!

[EchoNoise]

My head hurts

[RJM]

But S=2.9881167 gives you an error of only 0.000015

[XP-is-a-CRAP]

Quote

@gamehead - that'll give you where the original equation is flat (i.e. a maxima or minima). It unfortunately won't give you the zeros of the equation.

BUT I can check the "flat" points for minima, and if all of them are >0 then the fact is proven that
no usable (real numbers) solution exists.

BTW: A flat point does NOT necessarily bring an extreme ... check "y=x^3" function.

Anyone can generate and post the graph of the function on left side of the original equation, so we can see what we are trolling about ?

[LLXX]

The graph is very misleading. At different magnifications it looks very different.

http://img218.images...13/plot1bu7.png
http://img140.images...62/plot2tg3.png

[xtremee]

@ all
1st thanks for trying to help me
2nd sorry for late in my reply
i wanna an easy and quickly method that i can use to solve equation of order 6, 5 and 4

and i don't wanna the solution for 6 order equation that i post in the head of the topic. this equation for example only

@gamehead200,
Why you Diff. (d/ds) HOw it can help?

Regards,

Xtremee

This post has been edited by xtremee: 08 November 2006 - 05:20 PM

[azagahl]

I think the fundamental theorem of algebra says any N-degree polynomial (N > 0) can be factored using N roots. Finding them is hard though...

For 3rd/4th degree equations, there are exact algebraic solutions. look at http://mathworld.wolfram.com/ for Quartic Equations or Cubic equations.

Otherwise, I don't think there is a good method in general. For polynomials with degree 5 and above, you can also get exact solutions but they depend on nasty functions and polynomial transformations. Newton's method is easy and fast and is probably the best bet here, however it doesn't work very well for complex roots or coefficients. E.g. for x*x+1 = 0, Newton's method will probably oscillate or diverge stupidly because there is no real root. The above site is a great resource for this stuff. Here is another method, with its own limitations; I think it uses both first and second derivatives: http://mathworld.wol...dersMethod.html
There are actually buttloads of these, which shows how hard this problem is: http://mathworld.wol...gAlgorithm.html

For odd-degree equations (e.g. quintic) with real coefficients, they go toward negative infinity on one side, and positive infinity on the other. In this case you can just start far out on both sides and use a binary search to find a zero. This isn't very fast, but is very easy method to implement. If you use this plus the quartic (degree 4) formula you can easily solve any quintic (degree 5) equation. This is because, when you manage to find a root, you can factor (x - root) out of the polynomial (it should divide with no remainder) to reduce its degree by 1.

Hope this helps.

This post has been edited by azagahl: 08 November 2006 - 09:33 PM