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We have $P(x)=x^3-2x^2-5$. I know the formula of Newthon's method. That is given here.

The problem is, how do I find the approximations to within $10^{-4}$ to all the real zeros of the following polynomials using Newton's method?

graydad
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user3543012
  • 139

3 Answers3

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The Newton-Raphson iteration is given by:

$$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{x_n^3-2 x_n^2-5}{3x_n^2-4 x_n}$$

If we plot the function, we have:

enter image description here

From this image, we can see there is a real root between $x \in (2,3)$, so lets choose $x_0 = 2.0$ and iterate until we get an error, $E_n \le 10^{-4}$ between consecutive iterates.

  • $x_0 = 2.0$
  • $x_1 = 3.25$
  • $x_2 = 2.8110367893$
  • $x_3 = 2.69798950247$
  • $x_4 = 2.69067715286$
  • $x_5 = 2.69064744852$ (Bingo, we have liftoff since consecutive iterates agree to 4 places!)

So, we converged in five steps. The actual root is given by:

$$x^* = 2.690647448028613750350789$$

Note: If we had started at $x_0 = 3.0$, we would have converged in only four steps. Try it.

Update: Complex Roots

The other two roots are complex and you can use Newton's Method to find those too.

  • $x_0 = ~~~1.0000000+2.0000000 i$
  • $x_1 = ~~~0.51351351+1.08108108 i$
  • $x_2 = -0.57219283+0.75601472 i$
  • $x_3 = -0.06472804+1.27707430 i$
  • $x_4 = -0.37438100+1.27691783 i$
  • $x_5 = -0.34383959+1.31845095 i$
  • $x_6 = -0.34532472+1.31872594 i$
  • $x_7 = -0.34532372+1.31872678 i$

The actual root is given by:

$$x^* = -0.3453237240143068751753944+1.3187267795713238829519526 i$$

Note that since roots come in complex conjugate pairs, this provides both of them (just negate the imaginary part).

Also note, you could have used the first real root and written $(x-a)(x^2 + b x + c)$ and then just used the quadratic equation to find the two imaginary roots.

Amzoti
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By Descartes' rule of signs, there is only one real root and it is positive. Applying your iteration formula, $$ x_{n+1} = x_n - \frac{x_n^3-2x_n^2-5}{3x_n^2-4x_n} $$ Since $P(x=2) = -5$ and $P(x=3) = 4$, starting with $x_0 = 3$ gives successively, $3, 2.7333, 2.6916, 2.6906, 2.6906, \ldots$.

Victor Liu
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Apply Newton's method until two successive values are within $10^{-4}$. Do this for a number of initial values chosen depending on the roots of $P'(x)$.

marty cohen
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  • What Po in this case? – user3543012 Feb 20 '15 at 20:28
  • @user3543012 Initial conditions with Newton's method are in general more art than science. In this case one thing you can note is that the root must be positive (why?), and it can't be "huge". More specifically, a monic polynomial can't have a root which is larger than the sum of the absolute values of the non-leading coefficients. (This is not very hard to prove with the triangle inequality.) So your root must be somewhere in $[0,7]$. So a decent, if naive, guess is to "run bisection" for a single iteration, obtaining $3.5$, and then use $3.5$ as your initial condition. – Ian Feb 21 '15 at 00:08
  • @user3543012 (Cont.) I didn't check whether this works, but unless there winds up being a root of $P'$ between $3.5$ and the root of $P$, this should work. – Ian Feb 21 '15 at 00:09