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Bisection, Newton & Error

Useful preparation: Derivatives, Chain Rule & Implicit Change

For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

Finding a root numerically means producing an approximation together with evidence about its accuracy. Bisection buys certainty by repeatedly shrinking an interval known to contain a root. Newton uses local slope information to move much faster near a suitable root, but that speed comes without a general guarantee from an arbitrary starting value.

Root and residual
A root r satisfies f (r)=0\displaystyle f\, \left(r\right) = 0. The residual at an approximation x is f(x); it measures failure to satisfy the equation, not directly the distance from a root.
Bisection bracket
A bracket [a,b] has a continuous function with opposite nonzero signs at the endpoints. The intermediate value theorem guarantees at least one root inside.
Newton update
Newton replaces the curve near x by its tangent line and uses that line’s zero as the next approximation. The derivative at x must be nonzero.

Conditions before calculation

  • Bisection requires continuity throughout the bracket, not just evaluations with opposite signs. If an endpoint already has value zero, the root is known without iteration.
  • Local quadratic convergence of Newton is guaranteed near a simple root under suitable smoothness, such as a twice continuously differentiable function with nonzero derivative at the root and a sufficiently close starting point.
  • A small residual implies a small position error only with additional conditioning information. For example, a derivative magnitude bounded below on the interval connecting the approximation and root gives an error bound.

Why the method works

1. Keep the invariant

After evaluating the midpoint, retain a half whose endpoints have opposite signs. The root remains bracketed; the interval width halves each time.

2. State the error convention

After k halvings the bracket width is its initial width divided by 2 to the k. Reporting the midpoint of that final bracket guarantees half that width as an error bound.

∣mk−r∣≤b0−a02k+1|m_k-r|\le\frac{b_0-a_0}{2^{k+1}}

3. Derive and monitor Newton

Setting the tangent approximation to zero yields the update. Monitor residual, step size, and domain validity; safeguards can return to a known bracket.

0≈f(xn)+f′(xn)(xn+1−xn),xn+1=xn−f(xn)f′(xn)0\approx f(x_n)+f'(x_n)(x_{n+1}-x_n),\quad x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

Worked examples

EXAMPLE 1

Bracket the positive root of x squared-5, beginning with [2,3], using three bisection halvings.

  1. Endpoint values are -1 and 4. The midpoint 2.5 gives 1.25, so keep [2,2.5].

  2. The midpoint 2.25 gives 0.0625, so keep [2,2.25].

  3. The midpoint 2.125 gives -0.484375, so the final bracket is [2.125,2.25]. Its midpoint is 2.1875.

    ∣2.1875−5∣≤0.1252=0.0625|2.1875-\sqrt5|\le\frac{0.125}{2}=0.0625

Conclusion: The final midpoint has a certified error at most 0.0625, even without knowing the exact root numerically.

EXAMPLE 2

Take two Newton steps toward the positive root of x squared-10 from x 0=3\displaystyle x\, 0 = 3.

  1. Differentiate f; the update averages x with 10x\displaystyle \frac{10}{x}.

    f′(x)=2x,xn+1=12(xn+10xn)f'(x)=2x,\qquad x_{n+1}=\frac12\left(x_n+\frac{10}{x_n}\right)

  2. The first step is 196\displaystyle \frac{19}{6}, using a nonzero derivative at 3.

    x1=3−−16=196x_1=3-\frac{-1}{6}=\frac{19}{6}

  3. The next step simplifies exactly, allowing a residual check.

    x2=721228≈3.162280702,f(x2)=151984x_2=\frac{721}{228}\approx3.162280702,\quad f(x_2)=\frac1{51984}

Conclusion: The second approximation is close to the positive root. Its residual is small; the slope near the root is safely away from zero.

Pitfalls & counterexamples

Tempting claim: Opposite signs always guarantee a root between two points.

Correction: Continuity on the entire interval is essential.

Test the claim: The function 1x\displaystyle \frac{1}{x} has opposite signs at -1 and 1 but is undefined at zero and has no root.

Check your understanding

Does Newton necessarily converge from every starting point?

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Compare your reasoning

No.

For f (x)=x\displaystyle f\, \left(x\right) = x cubed-2 x+2\displaystyle 2\, x + 2, the starting value 0 alternates between 0 and 1 instead of converging.

Connect this idea

Conditioning describes sensitivity of the mathematical answer; convergence describes the algorithm. Floating-point rounding, poor scaling, and derivative errors can affect both practical stopping decisions and reported digits.

Practise this topic →Review concept flashcards

Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

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