More with Newton's method and introduction to Mean Value Theorem (Calc 1; Lecture 3-3; Fall 26)

Overview

The lecture develops Newton’s method as a way to approximate roots by repeatedly using tangent-line intersections, then applies it to finding √2, a curve intersection, and a root of x − cos(x). It introduces Rolle’s theorem and the Mean Value Theorem, explains their relationship through a transformed function, and verifies the theorem for h(t) = 2t/(1 + t²) on [0, 1].

Key takeaways

Newton’s method updates a guess using xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ), geometrically following the tangent line to its x-intercept.
A Newton recurrence can reveal its target root by rewriting it in the form xₙ − f(xₙ)/f′(xₙ); the recurrence 4xₙ/5 + 3/xₙ⁴ corresponds to f(x) = x⁵ − 15.
For f(x) = x² − 2, starting Newton’s method at 1 produces 3/2 and then 17/12, an estimate of √2 accurate to about four decimal places after two updates.
Newton’s method can turn an intersection problem into root-finding: x³ = 5 + 5x becomes x³ − 5x − 5 = 0, with two updates from x₀ = 2 producing 59/22.
Rolle’s theorem is the equal-endpoint special case of the Mean Value Theorem: continuity and differentiability guarantee an interior derivative equal to the interval’s average rate of change.
For h(t) = 2t/(1 + t²) on [0, 1], the Mean Value Theorem’s matching point is c = √(√5 − 2), found by setting h′(c) equal to the average slope of 1.

Chapters

0:00 Quickfire Review: Algebra and Derivative Notation

The opening review checks exponent arithmetic and combining 3t⁴ − t/5 over the common denominator 5t⁴.
The class recalls that finding the second derivative of arctan(x) begins with its first derivative, 1/(1 + x²).

2:07 Newton’s Method: Interpreting a Recurrence as Root-Finding

Newton’s method seeks a function’s x-intercept by improving an initial guess through repeated tangent-line estimates.
A recurrence’s limit represents the root the iterations approach, not necessarily its stated starting value of 11.

5:13 Recovering the Function Behind the Newton Recurrence

The recurrence 4xₙ/5 + 3/xₙ⁴ is rewritten as xₙ − (xₙ⁵ − 15)/(5xₙ⁴).
Matching the numerator and denominator to f(xₙ) and f′(xₙ) identifies f(x) = x⁵ − 15 and f′(x) = 5x⁴.
The target root is x = 15^(1/5), so the recurrence is expected to converge to that value.

13:00 From Newton’s Method to the Mean Value Theorem

The lecture distinguishes the Mean Value Theorem’s use of “mean” as average from the everyday meaning of unkind.
Its central comparison is between average rate of change, [f(b) − f(a)]/(b − a), and instantaneous rate of change, f′(c).
Newton’s method remains the first topic, followed by an introduction to the Mean Value Theorem.

15:48 Approximating √2 with Newton’s Method

To find √2, choose f(x) = x² − 2, whose root is √2, and use Newton’s update xₙ₊₁ = xₙ − (xₙ² − 2)/(2xₙ).
Starting at x₀ = 1 gives x₁ = 3/2 and x₂ = 17/12, approximately 1.4167.
The example shows how repeated iterations quickly improve an estimate; ancient Mesopotamian calculations used related iterative ideas to approximate square roots.

24:43 Finding a Curve Intersection with Newton’s Method

The intersection of y = x³ and y = 5 + 5x becomes the root-finding problem f(x) = x³ − 5x − 5 = 0.
With f′(x) = 3x² − 5 and x₀ = 2, the first update is x₁ = 3.
The second update is x₂ = 3 − 7/22 = 59/22, approximately 2.682; the in-class arithmetic is corrected to this value.

31:45 Newton Iteration for the Nonpolynomial Function x − cos(x)

For f(x) = x − cos(x), the derivative is 1 + sin(x), giving xₙ₊₁ = xₙ − (xₙ − cos(xₙ))/(1 + sin(xₙ)).
Starting from x₀ = 0, computer iterations converge to the root in only a few steps.
Repeatedly applying x ↦ cos(x) also converges to the same value, but the lecture contrasts its much slower convergence, taking about 50 iterations for comparable displayed precision.

35:23 Rolle’s Theorem from the Extreme Value Theorem

For a continuous function on a closed, bounded interval, the Extreme Value Theorem guarantees an absolute maximum and minimum.
Rolle’s theorem adds equal endpoint values and differentiability, concluding that f′(c) = 0 for at least one interior point.
If both extrema occur at the endpoints while the endpoint values match, the function must be constant on the interval; otherwise an interior extremum has derivative zero.

40:21 Mean Value Theorem: A Tangent Parallel to the Secant

The Mean Value Theorem keeps Rolle’s continuity and differentiability conditions but drops the requirement that f(a) = f(b).
It guarantees a point c where f′(c) = [f(b) − f(a)]/(b − a), meaning a tangent line is parallel to the endpoint secant.
A speeding example applies the result to average speed: if a car averages 75 miles per hour over an interval, its instantaneous speed must reach at least 75 miles per hour during that interval.

44:36 Proving the Mean Value Theorem and Verifying h(t)

The proof defines g(x) = f(x) − [f(b) − f(a)](x − a)/(b − a), which has equal endpoint values and inherits continuity and differentiability from f.
Applying Rolle’s theorem gives g′(c) = 0, which rearranges to the Mean Value Theorem equation for f′(c).
For h(t) = 2t/(1 + t²) on [0, 1], the average rate of change is 1; solving h′(c) = 1 yields c = √(√5 − 2), the solution inside the interval.

Transcript excerpt

0:00 We'll start as we often do. We start with our quickfire. Okay. So our first problem, if you have 2 5th subtract 6^ 2 add 4 to ^ 1, what do we get? >> Zero. >> Zero. Yes. Good. All right. And zero is a good friend of ours. Some a little bit of an algebra combine. So putting it as to one term 3 t 4th subtract t over 5 what will this become? get a common denominator.

0:53 If you can't see anything that the denominators currently have in common, it's just oh, the common denominator is the multiplying the terms together. So the common denominator would be the t 4th * 5 or 5 t 4th. So the first term you multiply by 5 over 5. That gives you your 15 upstairs. Second term t 4th over t 4th and that gives you your t 5th. And don't forget the minus. [snorts] All right. Well, an oldie but a goodie. We have to make sure we always keep track of our arc tangent. What's the second d? That's our notation. d^2 dx^2 of arc tangent of x. The second dative of arc tangent of x -2x over 1 + x^2. Good. And of course we

1:43 should keep in mind what's an intermediate step to finding the second d of arc tangent. >> Find the first der of arc tangent. And that we should just know. We love the first d of arc tangent. It's such a nice answer. 1 over 1 + x^2. And uh then you can go from there to get your second derivative. Now, at the very end of last time, we just barely had a chance to say hello to this idea of Newton's method. And so, this is it's going to be a

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