Mathematical Rules of Thumb, illustrated reader · Chapter 4

04Calculus

Turn Local Change into Global Control

5 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 4 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-calculus from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 5

Check a tangent estimate against a remainder

Why is the tangent estimate above the square-root curve?

Linearization is useful when its error is small enough for the decision. The derivative bound supplies a certificate.

100+h≈10+h20,|R|≤h28min⁡(100,100+h)3/2 \sqrt{100+h}\approx10+\frac{h}{20},\quad |R|\leq\frac{h^2}{8\min(100,100+h)^{3/2}}

Change h from 100. Positive square-root domain and a second-derivative bound on the entire interval.

Predict first. Why is the tangent estimate above the square-root curve?

Choose an example

Check a tangent estimate against a remainder. For x=100+10, the linear estimate differs by 0.0119115. Bounding the second derivative on the intervening interval certifies error at most 0.0125.
Change h from 100: 10
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Exact
10.4881
Linear estimate
10.5
Actual error
0.0119115
Remainder bound
0.0125

For x=100+10, the linear estimate differs by 0.0119115. Bounding the second derivative on the intervening interval certifies error at most 0.0125.

Use the idea

Use rule 4.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive square-root domain and a second-derivative bound on the entire interval.

Check your understanding: Why is the tangent estimate above the square-root curve?
Square root is concave on the positive domain, so its tangent is an upper bound.

Book source: Rule 4.1.1: Linearize near a point you already understand. Demonstration C04-D01. Worked illustration.

2Demonstration 2 of 5

Read a Newton step from its tangent

What happens if you start at x=0?

The tangent-line intercept is a proposed next root estimate. Its residual can be checked immediately.

xnext=x−x2−22x x_{next}=x-\frac{x^2-2}{2x}

Starting x. Equation x²−2=0 with nonzero starting x. A local picture alone is not a global convergence proof.

Predict first. What happens if you start at x=0?

Choose an example

Read a Newton step from its tangent. The tangent from x=1 crosses zero at 1.5. The residual shrank from 1 to 0.25, so this step helped.
Starting x: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Starting x
1
Next x
1.5
Next residual
0.25

The tangent from x=1 crosses zero at 1.5. The residual shrank from 1 to 0.25, so this step helped.

Use the idea

Use rule 4.1.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Equation x²−2=0 with nonzero starting x. A local picture alone is not a global convergence proof.

Check your understanding: What happens if you start at x=0?
The derivative vanishes and the Newton quotient is undefined. A different start or a bracketed method is needed.

Book source: Rule 4.1.2: Interpret a Newton step as the tangent-line root. Demonstration C04-D02. Worked illustration.

3Demonstration 3 of 5

Distinguish a finite sum from an infinite tail

Does a finite 100-term sum prove convergence?

The partial sums may look tame even when the infinite series diverges. The convergence condition comes first.

∑n>Nn−p≤∫N∞x−pdx=N1−pp−1,p>1 \sum_{n>N}n^{-p}\leq\int_N^\infty x^{-p}dx=\frac{N^{1-p}}{p-1},\quad p>1

Exponent p. Positive decreasing p-series. The displayed finite tail bound applies only for p>1.

Predict first. Does a finite 100-term sum prove convergence?

Choose an example

Distinguish a finite sum from an infinite tail. Adding n⁻ᵖ forever gives a finite total only when p>1. A finite partial sum does not establish convergence; this selected series diverges.
Exponent p: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Exponent p
1
100-term sum
5.18738
Integral upper bound on tail
No finite tail bound

Adding n⁻ᵖ forever gives a finite total only when p>1. A finite partial sum does not establish convergence; this selected series diverges.

Use the idea

Use rule 4.3.6 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive decreasing p-series. The displayed finite tail bound applies only for p>1.

Check your understanding: Does a finite 100-term sum prove convergence?
No. Every finite partial sum exists here. Infinite convergence depends on the tail; p≤1 diverges.

Book source: Rule 4.3.6: Bracket a decreasing series tail with integrals. Demonstration C04-D03. Worked illustration.

4Demonstration 4 of 5

Use the first omitted Taylor term as an error scale

How many terms of 1+1+1/2+1/6+... give e to three decimals?

Before trusting a truncated Taylor polynomial, look at the first term you dropped. It sets the size of the error.

e−∑j=0n1j!≈1(n+1)! e-\sum_{j=0}^{n}\frac{1}{j!}\approx\frac{1}{(n+1)!}

Taylor degree n. Rapidly shrinking terms, here eˣ at x=1. Slowly shrinking or alternating terms need a real remainder bound.

Predict first. How many terms of 1+1+1/2+1/6+... give e to three decimals?

Choose an example

Use the first omitted Taylor term as an error scale. Stopping at degree 3 leaves error 0.0516. The first omitted term 0.0417 predicts it within a factor 1.24, so you can read the error budget before computing e. At this low degree the later terms still add a noticeable share.
Taylor degree n: 3
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Degree n
3
Actual error
0.0516152
First omitted term
0.0416667
Actual / omitted
1.23876

Stopping at degree 3 leaves error 0.0516. The first omitted term 0.0417 predicts it within a factor 1.24, so you can read the error budget before computing e. At this low degree the later terms still add a noticeable share.

Use the idea

Use rule 4.3.10 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Rapidly shrinking terms, here eˣ at x=1. Slowly shrinking or alternating terms need a real remainder bound.

Check your understanding: How many terms of 1+1+1/2+1/6+... give e to three decimals?
Degree 6: the first omitted term 1/7! ≈ 0.000198 and the actual error is 0.000226, both below 0.0005.

Book source: Rule 4.3.10: Use the first omitted Taylor term as an error scale. Demonstration C04-D04. Worked illustration.

5Demonstration 5 of 5

Confirm 0/0 before using L'Hôpital

What is the limit when a=1?

Differentiating top and bottom always gives cos 0/1=1. That answer is right only when the original form is 0/0. Compare the curve with the dashed L'Hôpital line.

limx→0sin⁡x+ax \lim_{x\to0}\frac{\sin x+a}{x}

Numerator constant a. L'Hôpital requires 0/0 or ∞/∞ and differentiable functions near the point.

Predict first. What is the limit when a=1?

Choose an example

Confirm 0/0 before using L'Hôpital. With a=0.1 the numerator tends to 0.1 while the denominator tends to 0, so the ratio blows up (it is 11 at x=0.01) and has no finite limit. Differentiating anyway still returns 1, a wrong answer.
Numerator constant a: 0.1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Numerator at x=0
0.1
Form at x=0
nonzero/0
Ratio at x=0.01
11
Blind L'Hôpital answer
1

With a=0.1 the numerator tends to 0.1 while the denominator tends to 0, so the ratio blows up (it is 11 at x=0.01) and has no finite limit. Differentiating anyway still returns 1, a wrong answer.

Use the idea

Use rule 4.3.9 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

L'Hôpital requires 0/0 or ∞/∞ and differentiable functions near the point.

Check your understanding: What is the limit when a=1?
There is no finite limit. The numerator tends to 1 and the denominator to 0, so the ratio grows without bound (to +∞ from the right and −∞ from the left). L'Hôpital's 1 is wrong.

Book source: Rule 4.3.9: Use L'Hôpital only after confirming an indeterminate form. Demonstration C04-D05. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.