Mathematical Rules of Thumb, illustrated reader · Chapter 1

01Algebra

See the Structure Before You Solve

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 1 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-algebra 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

Watch a small change become a large response

Does a 10% input increase imply exactly a 50% response increase?

Compare the curved exact response with the straight first-order estimate. Look at the numerical error before accepting the shortcut.

(1+x)5≈1+5x (1+x)^5\approx1+5x

Fractional input increase. Positive base, fifth-power response. The approximation needs a tolerance and a sufficiently small change.

Predict first. Does a 10% input increase imply exactly a 50% response increase?

Choose an example

Watch a small change become a large response. An input increase of 10% gives 1.61051, versus 1.5 from linearization. Judge the 0.11051 error against your tolerance.
Fractional input increase: 0.1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Exact multiplier
1.61051
Linear estimate
1.5
Absolute error
0.11051

An input increase of 10% gives 1.61051, versus 1.5 from linearization. Judge the 0.11051 error against your tolerance.

Use the idea

Use rule 1.1.3 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive base, fifth-power response. The approximation needs a tolerance and a sufficiently small change.

Check your understanding: Does a 10% input increase imply exactly a 50% response increase?
No. The exact multiplier is 1.1⁵=1.61051, a 61.051% increase; 50% is only the linear estimate.

Book source: Rule 1.1.3: Linearize a small binomial perturbation. Demonstration C01-D01. Worked illustration.

2Demonstration 2 of 5

Price the omitted geometric tail

Is the first omitted term a reliable tail estimate when r=.9?

Compare the first omitted term with the whole omitted tail. Their ratio stays constant for a fixed r.

∑j=k∞rj=rk1−r,|r|<1 \sum_{j=k}^{\infty}r^j=\frac{r^k}{1-r},\quad |r|<1

Geometric ratio r. Displayed positive ratios below one. The infinite tail formula fails at r=1.

Predict first. Is the first omitted term a reliable tail estimate when r=.9?

Choose an example

Price the omitted geometric tail. The tail starting at r^5 equals r^5/(1-r). At r=0.5, its multiplier over the first omitted term is 2; a ratio near one makes the tail much larger.
Geometric ratio r: 0.5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Ratio r
0.5
Tail from r^5 onward
0.0625
Tail / first term
2

The tail starting at r^5 equals r^5/(1-r). At r=0.5, its multiplier over the first omitted term is 2; a ratio near one makes the tail much larger.

Use the idea

Use rule 1.2.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Displayed positive ratios below one. The infinite tail formula fails at r=1.

Check your understanding: Is the first omitted term a reliable tail estimate when r=.9?
It supplies the scale, but the entire positive tail is ten times that term because 1/(1−.9)=10.

Book source: Rule 1.2.2: Estimate a geometric tail from its first omitted term. Demonstration C01-D02. Worked illustration.

3Demonstration 3 of 5

See roots appear at a discriminant boundary

What changes at b=2?

The parabola crosses, touches, or misses the horizontal axis as its discriminant changes sign.

x2+bx+1=0,Δ=b2−4 x^2+bx+1=0,\qquad\Delta=b^2-4

Linear coefficient b. Real coefficients; the count refers to distinct real roots, with the repeated root identified separately.

Predict first. What changes at b=2?

Choose an example

See roots appear at a discriminant boundary. The discriminant b²-4 is 0. This gives one repeated real root.
Linear coefficient b: 2
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Discriminant
0
Real root count
1
Vertex x
-1

The discriminant b²-4 is 0. This gives one repeated real root.

Use the idea

Use rule 1.3.4 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Real coefficients; the count refers to distinct real roots, with the repeated root identified separately.

Check your understanding: What changes at b=2?
The discriminant is zero and x²+2x+1=(x+1)². Two distinct real roots merge into one repeated root.

Book source: Rule 1.3.4: Use the discriminant to triage quadratic roots. Demonstration C01-D03. Worked illustration.

4Demonstration 4 of 5

Check the rule of 70 for doubling time

At 7% per year, does money double in exactly 10 years?

The rule of 70 turns a growth rate into a doubling time with one division. The chart shows how its error drifts as the rate grows.

T2=ln⁡2ln⁡(1+r)≈70100r T_2=\frac{\ln 2}{\ln(1+r)}\approx\frac{70}{100r}

Growth rate per period. Constant compound growth once per period. 70 comes from 100 ln 2 for continuous growth.

Predict first. At 7% per year, does money double in exactly 10 years?

Choose an example

Check the rule of 70 for doubling time. At 7% per period the exact doubling time is 10.24 periods; the rule of 70 says 10. Good enough for mental arithmetic.
Growth rate per period: 0.07
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Rate per period (%)
7
Exact doubling time (periods)
10.2448
Rule-of-70 estimate (periods)
10
Estimate error (periods)
-0.244768

At 7% per period the exact doubling time is 10.24 periods; the rule of 70 says 10. Good enough for mental arithmetic.

Use the idea

Use rule 1.1.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Constant compound growth once per period. 70 comes from 100 ln 2 for continuous growth.

Check your understanding: At 7% per year, does money double in exactly 10 years?
Not quite. The exact time is ln 2/ln 1.07 = 10.24 years. The rule is about 2.4% short, fine for a quick estimate.

Book source: Rule 1.1.2: Estimate doubling time from the exponential rate. Demonstration C01-D04. Worked illustration.

5Demonstration 5 of 5

Check the denominator sign before cross-multiplying

Is x=1 a solution of (x+1)/(x−3)≤0?

Cross-multiplying by x−3 without a sign check gives x+1≤0, that is x≤−1. The sign chart gives the true answer. Pick a test point and compare the two verdicts.

x+1x−3≤0⇔−1≤x<3 \frac{x+1}{x-3}\le0\iff -1\le x<3

Test point x. Real x with x≠3. Multiplying an inequality by a negative quantity reverses it.

Predict first. Is x=1 a solution of (x+1)/(x−3)≤0?

Choose an example

Check the denominator sign before cross-multiplying. At x=1 the fraction equals -1, so the inequality is true. Cross-multiplying without a sign check gives x+1 ≤ 0, which says false. They disagree because x<3 makes the denominator negative, which reverses the inequality.
Test point x: 1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Test x
1
(x+1)/(x-3)
-1
Truly satisfies
yes
Naive x ≤ -1 says
no

At x=1 the fraction equals -1, so the inequality is true. Cross-multiplying without a sign check gives x+1 ≤ 0, which says false. They disagree because x<3 makes the denominator negative, which reverses the inequality.

Use the idea

Use rule 1.3.7 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Real x with x≠3. Multiplying an inequality by a negative quantity reverses it.

Check your understanding: Is x=1 a solution of (x+1)/(x−3)≤0?
Yes. The fraction equals 2/(−2)=−1≤0. The naive answer x≤−1 rejects it because it ignored that x−3<0 there.

Book source: Rule 1.3.7: Cross-multiply inequalities only after checking signs. Demonstration C01-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.