1

I am trying to solve this problem:

Use Newton's method to find the intersection points of the two circles defined by

$$x^2 + y^2 = 2, \quad (x-1)^2 + (y+1.5)^2 = 1.$$

I used this code in Maple:

Newton(4+(-y^2+2)*(2*y+3)^2-4*(-y^2+2), y = .5, output = sequence);

and this gave me:

.5, -1.062500000, -.7532718336, -1.001899321, -.5138663374,
-.7942990696, -1.069866630, -.7693943666, -1.025772773,
-.6418803790, -.8854387837

From this I do not understand what the solution is, and also how can I choose my initial value more accurately?

Community
  • 1

1 Answers1

2

First I couldn't understand why you used that particular formula. the formula should be:

$$(x-1)^2 + (y + \frac 32)^2 = 1$$ $$x^2 - 2x + 1 + y^2 + 3y + \frac 94 = 1$$ $$(x^2 + y^2) - 2x + 3y + \frac 94 = 0$$ $$2 - 2x + 3y + \frac94 = 0$$ $$3y - 2x + \frac{17}{4} = 0$$ $$y = \frac{2x - \frac{17}{4}}{3}$$

Now substituting we have:

$$x^2 + \left(\frac{2x - \frac{17}{4}}{3}\right)^2 - 2 = 0$$ $$x^2 + \frac{4x^2 - 17x + \frac{289}{16}}{9} - 2 = 0$$ $$9x^2 + 4x^2 - 17x + \frac{289}{16} - 18 = 0$$ $$144x^2 + 64x^2 - 272x + 289 - 288 = 0$$ $$208x^2 - 272x + 1 = 0$$

Now it should be easy to find the solutions using Newton's method.

Also Newton's method may fail for numbers of reasons. First is "bad" starting point. For some starting points the value of x doesn't converges to the right value, but diverges

For more info you can read here

Stefan4024
  • 36,357
  • I made a mistake in simplifying the second circle. I thought (x - 1)^2 was (x^-1)^2. –  Sep 23 '13 at 22:13
  • I've edited the soluton. I think it's as good as it gets. – Stefan4024 Sep 23 '13 at 22:31
  • The roots of this polynomial are $x_1 \approx 0.00368687$ and $x_2 \approx 1.30401$. Depending on the initial guess you'll get one soluton. – Stefan4024 Sep 23 '13 at 22:35
  • For circle2:

    Newton((2x-17/4)(1/3), x = .5, output = sequence); print(output redirected...); # input placeholder 0.5, 2.125000001, 2.125000001 For circle1: Newton(x^2+((2x-17/4)(1/3))^2-2, x = .5, output = sequence); print(output redirected...); # input placeholder 0.5, -0.796874994, -0.2172030317, -0.0243209366, 0.00310851556,

    0.003686608518, 0.003686865674

    So how can I figure out their intersection point from this?

     –  Sep 23 '13 at 22:36
  • I don't how to use Maple, because I don't use it ussaualy, but as far as I know the code should be:

    Newton(x^2 + ((2x - 17/4)/3)^2 - 2, x = .5, output = sequence)...

     – Stefan4024 Sep 23 '13 at 22:39
  • Your code gives the same output of circle1. –  Sep 23 '13 at 22:41
  • $y = \frac{2x - \frac{17}{4}}{3}$ won't give solution. We just use it to express y in terms of x and then substitute in the first circle equation. If you want to find the intersection point it doesn't matter which circle you'll use because at the intersection point (x,y) for the first circle are the same as (x,y) for the second – Stefan4024 Sep 23 '13 at 22:44
  • If you see more carefully you'll see that the sequence converges to 0.00368687... which means that's the solution so the one intersection point have x-value of 0.00368687. Now just substitute back in the $$y= \frac{2x - \frac{17}{3}}{3}$$ and you get the y-value and the coordinates of the point of intersection.

    If you initial guess is 2, then you shoul obtain the second point of intersection.

     – Stefan4024 Sep 23 '13 at 22:46
  • Hey I got it! Thanks. –  Sep 23 '13 at 22:49
  • Just for confirmation the intersection points are (0.00368687, -1.4164) and (1.30401, -0.547). –  Sep 23 '13 at 22:54
  • let us continue this discussion in chat – Stefan4024 Sep 23 '13 at 22:56