Animation · Bisection - COMP 521: Introduction to Scientific Computing and Data Science

For $f(x)=x^2-2$ , start with $[1,2]$ . Each midpoint evaluation identifies a half-interval that retains opposite endpoint signs. The film follows three decisions.

Before watching¶

At the first midpoint, $m=1.5$ . Calculate the sign of $f(m)$ and predict which half-interval will remain.

For f(x)=x^2-2, start with [1,2]. Each midpoint evaluation identifies a half-interval that retains opposite endpoint signs. The film follows three decisions.

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Mathematical connection¶

If $f$ is continuous on $[a,b]$ and the endpoints have opposite signs, at least one root lies inside. Check endpoint roots before requiring opposite signs.

After $k$ halvings of an initial width $W_0$ , the retained width and its midpoint error satisfy

W_k=\frac{W_0}{2^k},\qquad |m_k-r|\leq\frac{W_0}{2^{k+1}}.

(1)

Here $m_k$ is the midpoint of the new retained interval and $r$ is a root inside it. This bound does not describe the old midpoint discarded or retained as an endpoint. Opposite signs across a discontinuity do not establish the theorem; a sign change also does not establish uniqueness.

Explain what you saw¶

Separate the sign test from the width bound. Which justifies retaining a half-interval, and which supports a claim about root location? Explain why a small residual is a different statement.

Use the player controls to pause, replay, or view fullscreen. The animation has no narration; its labels and the explanation above carry the content.

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