is it posible to solve a non-liner system...and how!!!

for example

$\displaystyle x^2 +y^3 =10 $

$\displaystyle x^3 -y^4 =-6 $

if any one has any thoughts on how you do this please let me know

~dan

Printable View

- Sep 4th 2006, 06:33 PMdanhmmm... just wondering
is it posible to solve a non-liner system...and how!!!

for example

$\displaystyle x^2 +y^3 =10 $

$\displaystyle x^3 -y^4 =-6 $

if any one has any thoughts on how you do this please let me know

~dan - Sep 4th 2006, 06:36 PMThePerfectHackerQuote:

Originally Posted by**dan**

$\displaystyle x^3=y^4-6$

$\displaystyle x=\sqrt[3]{y^4-6}$

Substitute,

$\displaystyle (y^4-6)^{2/3}+y^3=10$

Thus,

$\displaystyle (y^4-6)^{2/3}=(10+y^3)$

Cube both sides,

$\displaystyle (y^2-6)^2=(10+y^3)^3$

Expand and solve for the nasty nontic.

Note, the solutions not necessarily can be expressed in algebraic terms. - Sep 5th 2006, 06:34 AMdanQuote:

Originally Posted by**ThePerfectHacker**

i hate to ask...but how do you slolve for y in that equation.

it seems i'v spent hours working on probelms of that type before....

would i use the "binomal therem"?

dan - Sep 5th 2006, 07:17 AMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

$\displaystyle x=\pm \sqrt{10-y^3}$

and substituting into

$\displaystyle

x^3 -y^4 =-6

$

will give two octics - a much easier problem than a single nonic. :D?

RonL - Sep 5th 2006, 07:20 AMCaptainBlackQuote:

Originally Posted by**dan**

all or some or the roots. Otherwise numerically.

RonL - Sep 5th 2006, 07:24 AMCaptainBlackQuote:

Originally Posted by**dan**

These two equations represent curves in the x-y plane. Sketch them

and your first approximation to the soution/s will be the point/s of

intersection.

Then they can be refined with an appropriate numerical procedure (say

linearising about the approximate roots, or something like that).

RonL - Sep 5th 2006, 07:46 AMThePerfectHackerQuote:

Originally Posted by**dan**

Yes, you do expand with binomial theorem then work with polynomial equation.

(Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!") - Sep 5th 2006, 09:05 AMdanQuote:

Originally Posted by**ThePerfectHacker**

- Sep 5th 2006, 09:12 AMdan
so if i had a polynomal. could i solve with newton??

i dont so how you get a nontic though $\displaystyle (10 +y^3)^3$ just because 3*3 is nine???

some how i got from $\displaystyle (y^2 -6)^2 =(10 +y^3)^3 $

to $\displaystyle 0= y^5 + 59y^4 +90y^3 + 12y^2 + 964 $

is that right???

dan - Sep 5th 2006, 09:15 AMdanQuote:

Originally Posted by**CaptainBlack**

mabye i'll try that - Sep 5th 2006, 09:37 AMThePerfectHackerQuote:

Originally Posted by**dan**

Thus,

$\displaystyle y^9+30y^6-y^4+300y^3+12y^2+9964=0$ - Sep 5th 2006, 09:48 AMdanQuote:

Originally Posted by**ThePerfectHacker**

- Sep 5th 2006, 09:58 AMdan
so if

$\displaystyle f(y)=y^9 + 30y^6 - y^4 +300y^3 + 12y^2 +9964$

then

$\displaystyle f'(y)=y(9y^7 +180y^4 -4y^2 +900y +24) $

so could i use newton for here to approximate [tex] y [tex] - Sep 5th 2006, 10:03 AMdan
by my calc $\displaystyle y \approx -3.13425651413923 $

look right?? - Sep 5th 2006, 01:28 PMCaptainBlackQuote:

Originally Posted by**dan**

Is there a x which with this as a value of y solves the equations?

RonL

(when I do this I get no consistent solution for x, but I could be doing the

arithmetic wrongly :( )