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How does one find the interval that encloses the root for large valued function like without graphing:

$$x^5-x^2+x=20000$$

I know that graphing with a software will give you the interval easily, but how do I find the two consecutive integers that enclose the root?

Loo Soo Yong
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  • You can guess some values - when $x = 7$, the left hand side is $16765$, and when $x = 8$ he left hand side is $32712$, so one real root is between $7$ and $8$. – Bilbottom May 04 '18 at 13:39
  • This is easy because you have a polynomial with odd degree, just choose the intervall as $[-x,x]$ where $x$ is a large positive value. – gammatester May 04 '18 at 13:39
  • Are you looking for an interval that encloses the root? or an optimal interval that encloses the root? The latter is a point of research for many functions. – Dando18 May 04 '18 at 13:47
  • @BillWallis But how does one guess the values, as the problem is the root may lie in a far interval which can be tricky unless a graphical method is used. – Loo Soo Yong May 06 '18 at 10:38
  • @Loo Soo Yong I noticed when evaluated at $x = 0$ the left hand side is zero, and when evaluated at $x = 10$ the left hand side is $100000 - 100 + 10 = 99910$ which is easy to work out in your head, and gives an easy yet crude interval. Working with $x = 0$ and some power of 10 is usually an easy combination of integers to determine a crude bound in your head. This can then be refined by any method of your choosing. – Bilbottom May 06 '18 at 11:18

3 Answers3

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I'm not sure why you're looking for a bracketing interval. That doesn't necessarily help for Newton's method. Consider, for example, $$f(x) = -8 x^3+12 x^2-2.$$ It's easy to show by direct computation that $x=1/2$ is a simple root of $f$ and that $[0,1]$ is a bracketing interval. Unfortunately, $0$ and $1$ are both poor starting points for Newton's method here, since $f'(0)=f'(1)=0$.

Newton's method requires a single starting point. For this particular example, I took a look and thought - maybe $x_0$ should be around $$\sqrt[4]{20000} = 10\times \sqrt[4]{2},$$ which is probably just a little bit bigger than $10$. So I tried $x_0=12$ and it worked.

Application of Newton's method to arbitrarily contrived examples like this will almost certainly require ad hoc methods. When working with some particular family of functions that arise from an actual application, however, it's quite common that you can find reasonable starting values for that application. For example, this answer describes how Newton's method can be used to solve an equation that arises in the context of fractal dimension.

Mark McClure
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  • By you second method one would look for roots for the first example polynomial close to the non-zero root of the first two terms $-8x^3+12x^2=0\implies x=\frac32$ and the second two terms $0=12x^2-2\implies x=\pm\sqrt{\frac16}\approx\pm 0.4$. – Lutz Lehmann May 04 '18 at 17:05
  • @LooSooYong Refining a bracket around a root is the corner stone of any robust root finder. If the fast solver leaves the current bracket, then that step is discarded and the bracket is refined using bisection. MATLABs root finder fzero uses a combination of bisection, the secant method and inverse quadratic interpolation to create a solver which is robust, accurate and fast. – Carl Christian May 05 '18 at 14:58
  • @CarlChristian Agreed - but the question concerns Newton's method. – Mark McClure May 05 '18 at 15:02
  • @MarkMcClure You are right. I do not know how, but I missed the title completely. My apologies. – Carl Christian May 05 '18 at 15:10
  • @CarlChristian No biggie - I appreciate the input; it's definitely worth mentioning that bracketing roots is generally a big deal! It does bother me sometimes that crucial details are left to titles without making it clear in the title. :) – Mark McClure May 05 '18 at 15:12
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A bound (probably also named after Cauchy) for positive roots tells us to count the number $m$ of negative coefficients in $0=p(x)=x^n+a_{n-1}x^{n-1}+..+a_1x+a_0$ and then compute $$ R = \max\left\{\sqrt[n-k]{m|a_k|}\middle|0\le k<n\land a_k<0\right\}. $$ This is an upper bound for the positive roots. Because for $x>R$ one has $$ p(x)\ge x^n-\sum_{a_k<0}|a_k|x^k > x^n-\sum_{a_k<0}\frac{R^{n-k}}mx^k >x^n-\frac{m}mx^n=0$$

In this case, $$0=p(x)=x^5-x^2+x-20000,$$ we get $m=2$ and thus $$ R=\sqrt[5]{4\cdot 10^4}=10\cdot \sqrt[5]{\tfrac25}<8.5. $$

As $-p(-x)=x^5+x^2+x+20000$, there will be no negative roots.

Lutz Lehmann
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Hint.- Since the degree is odd, necessarily there is at least a real root $x_0$. What you have to do is to locate two consecutive integers $n$ and $n+1$ such that $f(n)$ and $f(n+1)$ have distinct signes. This procedure can fails when $x_0$ is a double roots in which case $x_0$ should be zero of the derivative $4x^4-2x+1$ which is not the case in this example.

Ataulfo
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