approximate the root of perturbed polynomial

Question

Approximate the root of $$f(x)=(x-1)(x-1)(x-3)(x-4)-10^{-6}x^6$$ near $r=4$.

Do I have to use iterative method of finding the root, such as Bisection, Secant, etc? Is there other way?

Answer 1

We are given the function:

$$f(x) = (x-1) (x-2) (x-3) (x-4)-\dfrac{x^6}{1000000}$$

A local plot of the function shows:

We clearly see there is a root $r \approx 4$ as well as other roots.

One method we can use is called the Newton-Raphson Method. The iteration is given by:

$$\tag 1 x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{(x_n-1) (x_n-2) (x_n-3) (x_n-4)-\dfrac{x_n^6}{1000000}}{-\dfrac{3 x_n^5}{500000}+4 x_n^3-30 x_n^2+70 x_n-50}$$

We will start the iteration with an initial guess of $x_0 = 5$ and iterate using $(1)$.

• $x_0 = 5.0000000000000000000000000000000000000000000000000$
• $x_1 = 4.520132549706139802425909716143553832687257721646$
• $x_2 = 4.213728406139413811621689461664147945319554657302$
• Continuing this with $50-$ digits of accuracy, we converge after $9$ iterations to:
• $x_9 = 4.0006825115317206360517121593821239383700711643283$

Note that you mentioned the Secant Method and it also converged to the root, but the Bisection Method did not.

It is worth noting that this is a sixth order function and we can find all of the roots using this process (even imaginary ones) as:

• $x = -1004.9801730979548412226307120082711312441645974604$
• $x = 0.99999983333355092551054620226892671527779215577814$
• $x = 2.0000320035846145257264041034190706369224905276042$
• $x = 2.9996356990888364529650498845326637064948826629689$
• $x = 4.0006825115317206360517121593821239383700711643283$
• $x = 994.97982305041611868237699965866834624709936094976$

Lastly, there are methods that converge even faster like the fourth or seventh order method, for example, see How to develop fourth and seventh order iterative methods?, but many other methods are available. See, for example Root Finding Algorithms

Answer 2

If $f(x) =(x-1)(x-2)(x-3)(x-4)-10^{-6}x^6 $, if $c$ is small,

$\begin{array}\\ f(4+c) &=(3+c)(2+c)(1+c)(c)-10^{-6}(4+c)^6\\ &=6c(1+c/2)(1+c/3)(1+c)-10^{-6}4^6(1+c/4)^6\\ &\approx 6c(1+c/2+c/3+c)-(2/5)^6(1+6c/4)\\ &=6c(1+11c/6)-(2/5)^6(1+3c/2)\\ &=c(6+(3/2)(2/5)^6) +11c^2-(2/5)^6\\ &\approx 6c -(2/5)^6\\ \end{array} $

If this is zero, $c = \frac{(2/5)^6}{6} \approx 0.0006826 $ so the root is about $4.0006826$, which agrees very nicely with Moo's much more accurate answer.

Answer 3

If you want an approximation, the implicit function theorem is a useful tool.

Let $\phi(x,\epsilon) = (x-1)(x-2)(x-3)(x-4)-\epsilon x^6$. It is not too difficult to compute ${\partial \phi(4,0) \over \partial x} = 6$, ${\partial \phi(4,0) \over \partial \epsilon} = -4^6$. Hence there is a function $\xi$ defined in a neighbourhood of $\epsilon=0$ such that $\phi(\xi(\epsilon), \epsilon) = 0$, and ${\partial \xi(0) \over \partial \epsilon} = - ({\partial \phi(4,0) \over \partial x})^{-1} {\partial \phi(4,0) \over \partial \epsilon} = - {-4^6 \over 6} = {4^6 \over 6}$.

Hence we expect $\xi({1 \over 10^6}) \approx \xi(0) + {\partial \xi(0) \over \partial \epsilon} {1 \over 10^6}=4 +{4^6 \over 6} {1 \over 10^6} \approx 4.000683$.