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Locate and bisect a circular-equation root

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Bisection for a circular equation

Form a continuous difference function and check the endpoints for an exact root. Otherwise locate opposite endpoint signs and test the midpoint. Stop if its value is zero; if not, keep the sign-changing half and repeat

\[F(x)=f(x)-P(x),\quad M=\frac{L+R}{2}\]
  1. 1Form a difference function Write the circular side minus the algebraic side so that an intersection is a root of F \(\displaystyle F(x)=f(x)-P(x),\quad F(x)=0\)
  2. 2Check the endpoint signs Check that F is continuous throughout the interval. If an endpoint value is zero, that endpoint is an exact root; otherwise use non-zero endpoint values with opposite signs \(\displaystyle F(L)<0<F(R)\quad\text{or}\quad F(R)<0<F(L)\)
  3. 3Calculate the midpoint Average the current left and right endpoints to obtain the next test value \(\displaystyle M=\frac{L+R}{2}\)
  4. 4Keep the sign-changing half If F(M) = 0, stop: M is an exact root. Otherwise retain the half whose endpoints have opposite signs and repeat \(\displaystyle F(L)F(M)<0\quad\Rightarrow\quad [L,M]\)
Year 11, VCE Units 1 & 2, Chapter 9, 4 difficulty levels, Calculator allowedBuild a worksheet

Bisection for a circular equation

Form a continuous difference function and check the endpoints for an exact root. Otherwise locate opposite endpoint signs and test the midpoint. Stop if its value is zero; if not, keep the sign-changing half and repeat

\[F(x)=f(x)-P(x),\quad M=\frac{L+R}{2}\]
  1. Form a difference function Write the circular side minus the algebraic side so that an intersection is a root of F \(\displaystyle F(x)=f(x)-P(x),\quad F(x)=0\)
  2. Check the endpoint signs Check that F is continuous throughout the interval. If an endpoint value is zero, that endpoint is an exact root; otherwise use non-zero endpoint values with opposite signs \(\displaystyle F(L)<0<F(R)\quad\text{or}\quad F(R)<0<F(L)\)
  3. Calculate the midpoint Average the current left and right endpoints to obtain the next test value \(\displaystyle M=\frac{L+R}{2}\)
  4. Keep the sign-changing half If F(M) = 0, stop: M is an exact root. Otherwise retain the half whose endpoints have opposite signs and repeat \(\displaystyle F(L)F(M)<0\quad\Rightarrow\quad [L,M]\)

Worked example

Use two bisection steps to refine a root of the equation below on the stated interval

\[\sin x=\frac{1}{2},\quad 0\le x\le\frac{\pi}{2}\]
  1. \(\displaystyle F(x)=\sin x-\frac{1}{2},\quad F(0)=-\frac{1}{2}<0,\quad F\left(\frac{\pi}{2}\right)=\frac{1}{2}>0\)The endpoint signs are opposite, so the interval brackets a root
  2. \(\displaystyle M_1=\frac{0+\frac{\pi}{2}}{2}=\frac{\pi}{4},\quad F\left(\frac{\pi}{4}\right)\approx0.207>0\quad\Rightarrow\quad \left[0,\frac{\pi}{4}\right]\)The first midpoint is positive, so it is paired with the negative left endpoint
  3. \(\displaystyle M_2=\frac{0+\frac{\pi}{4}}{2}=\frac{\pi}{8},\quad F\left(\frac{\pi}{8}\right)\approx-0.117<0\quad\Rightarrow\quad \left[\frac{\pi}{8},\frac{\pi}{4}\right]\)The second midpoint is negative, so it is paired with the positive right endpoint
  4. \(\displaystyle x_{\rm estimate}=\frac{\frac{\pi}{8}+\frac{\pi}{4}}{2}=\frac{3\pi}{16}\)Average the endpoints of the final retained interval to report the midpoint estimate

Common mistakes

Keeping the same-sign half
✕\(\displaystyle F(0)<0,\quad F\left(\frac{\pi}{4}\right)>0\quad\Rightarrow\quad \text{retain }\left[\frac{\pi}{4},\frac{\pi}{2}\right]\)✓\(\displaystyle F(0)<0,\quad F\left(\frac{\pi}{4}\right)>0\quad\Rightarrow\quad \text{retain }\left[0,\frac{\pi}{4}\right]\)
Bisection retains the half whose endpoint signs are opposite because that is the half that brackets the root
Rounding a function value before its sign
✕\(\displaystyle F(M)=0.004\approx0.00\quad\Rightarrow\quad F(M)=0\)✓\(\displaystyle F(M)=0.004>0\)
The sign decision must use the unrounded function value because rounding can turn a small positive or negative value into zero
Using circular x-intercepts
✕\(\displaystyle \sin x=0\)✓\(\displaystyle \sin x=P(x)\)
An equation between two functions is solved where their graphs intersect, so the circular function must be compared with the algebraic function

Difficulty levels

introChoose the difference function, root-count conclusion and supplied sign-changing bracket
standardComplete a short bisection table after choosing the starting interval
challengeBisect a mixed circular and algebraic equation for more iterations and report a midpoint estimate
examChoose the justified root statement and bracket, then record the retained interval

Questions about this topic

Why must the endpoint signs be opposite?+

A continuous difference function with opposite endpoint signs crosses zero somewhere between the endpoints, so the interval brackets a root

Why should midpoint function values not be rounded before choosing a half?+

A small non-zero value can round to zero. Bisection needs the sign of the unrounded value

What happens to the interval width after each bisection step?+

If no exact root has been found, each completed halving keeps one half of the current interval. After n such halvings its width is the initial width divided by two raised to the power n

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 9: Circular functions

In this chapter

Convert angle measure and use the radian arc relationLocate extended angles and generate coterminal anglesDeduce exact trigonometric ratios from one ratioUse unit-circle coordinates to evaluate circular functionsEvaluate exact circular-function values by symmetry and periodicityKey points and features of the basic sine and cosine graphsLocate and bisect a circular-equation rootUse local polynomial approximations for circular functions