Mathematical Rules of Thumb, illustrated reader · Chapter 11

11Asymptotics

Finding What Matters at Scale

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

Use math-thumb-asymptotics 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

Improve factorial scale on the log axis

Why compute on the logarithmic scale?

Compare the leading Stirling formula with its first log correction against an exact log-factorial calculation.

log⁡(n!)≈nlog⁡n−n+12log⁡(2πn)+112n \log(n!)\approx n\log n-n+\tfrac12\log(2\pi n)+\frac1{12n}

Factorial argument n. Positive integer n. The plot checks finite cases; a general remainder statement needs an appropriate bound.

Predict first. Why compute on the logarithmic scale?

Choose an example

Improve factorial scale on the log axis. At n=10, the leading Stirling formula misses n! by 0.83%. Adding 1/(12n) to the log cuts the miss to about 1 part in 361,023. Working with log n! keeps huge factorials from overflowing. A few checked values do not prove the correction always works this well.
Factorial argument n: 10
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

n
10
Leading relative error
0.00829596
Corrected relative error
2.7699e-06

At n=10, the leading Stirling formula misses n! by 0.83%. Adding 1/(12n) to the log cuts the miss to about 1 part in 361,023. Working with log n! keeps huge factorials from overflowing. A few checked values do not prove the correction always works this well.

Use the idea

Use rule 11.2.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive integer n. The plot checks finite cases; a general remainder statement needs an appropriate bound.

Check your understanding: Why compute on the logarithmic scale?
Products and factorials grow rapidly. Their logarithms avoid overflow and make relative errors easier to compare.

Book source: Rule 11.2.1: Use Stirling's formula for factorial scale. Demonstration C11-D01. Worked illustration.

2Demonstration 2 of 5

Find the scale where competing terms exchange roles

What happens to the crossing if a grows by a factor of one hundred?

The crossing of two term scales marks a transition where dropping either term is unjustified.

x∼a/x⇒x∼a x\sim a/x\quad\Longrightarrow\quad x\sim\sqrt a

Parameter a. Positive x and a; this is a two-term balance, not a solved full physical model.

Predict first. What happens to the crossing if a grows by a factor of one hundred?

Choose an example

Find the scale where competing terms exchange roles. The competing terms x and a/x are equal at x=√a=100. For x below 100, a/x is the bigger term; above it, x is. Near the crossing neither term can be dropped.
Parameter a: 10000
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Parameter a
10000
Balance scale x
100

The competing terms x and a/x are equal at x=√a=100. For x below 100, a/x is the bigger term; above it, x is. Near the crossing neither term can be dropped.

Use the idea

Use rule 11.1.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive x and a; this is a two-term balance, not a solved full physical model.

Check your understanding: What happens to the crossing if a grows by a factor of one hundred?
The balance scale grows by √100=10.

Book source: Rule 11.1.2: Balance competing terms to find the transition scale. Demonstration C11-D02. Worked illustration.

3Demonstration 3 of 5

Preserve the term left after cancellation

Why is x−x=0 not the right asymptotic answer?

Rationalization recovers a small difference between two large terms. Subtracting their shared leading equivalent loses the answer.

x2+x−x=xx2+x+x→12 \sqrt{x^2+x}-x=\frac{x}{\sqrt{x^2+x}+x}\longrightarrow\frac12

Large x. Positive x tending to infinity. Exact algebra precedes asymptotic approximation.

Predict first. Why is x−x=0 not the right asymptotic answer?

Choose an example

Preserve the term left after cancellation. √(x²+x)−x is 0.4998750625 at x=1,000 and tends to 1/2. Replacing each large term by x and subtracting would incorrectly give zero.
Large x: 1000
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Exact difference
0.499875
Gap to limit 1/2
0.000124938
Naively subtract leading equivalents
0
Correct limiting value
0.5

√(x²+x)−x is 0.4998750625 at x=1,000 and tends to 1/2. Replacing each large term by x and subtracting would incorrectly give zero.

Use the idea

Use rule 11.3.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Positive x tending to infinity. Exact algebra precedes asymptotic approximation.

Check your understanding: Why is x−x=0 not the right asymptotic answer?
The next term survives the cancellation. The exact rationalized expression tends to one half.

Book source: Rule 11.3.1: Do not subtract asymptotic equivalents blindly. Demonstration C11-D03. Worked illustration.

4Demonstration 4 of 5

Watch an exponential overtake a power

Can a big enough power beat 2^n forever?

Compare log n, a power n^k and the exponential 2^n on a log axis. The dotted line marks where the exponential takes over for good.

nk≪2n as n→∞ n^k\ll2^n\text{ as }n\to\infty

Power k. Integer n from 2 to 80; base-2 exponential. The hierarchy is about eventual behavior, not small n.

Predict first. Can a big enough power beat 2^n forever?

Choose an example

Watch an exponential overtake a power. With k=5, the power n^5 matches or beats 2^n up to n=22; from n=23 on, 2^n wins and the gap grows without limit. A bigger power only delays the takeover; it never prevents it.
Power k: 5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Power k
5
Last n where n^k ≥ 2^n
22
First n where 2^n > n^k
23
n^k at n=100
1e+10
2^n at n=100
1.26765e+30

With k=5, the power n^5 matches or beats 2^n up to n=22; from n=23 on, 2^n wins and the gap grows without limit. A bigger power only delays the takeover; it never prevents it.

Use the idea

Use rule 11.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Integer n from 2 to 80; base-2 exponential. The hierarchy is about eventual behavior, not small n.

Check your understanding: Can a big enough power beat 2^n forever?
No. A larger k only moves the crossover later. For k=10 it comes at n=59.

Book source: Rule 11.1.1: Use the log-power-exponential growth hierarchy. Demonstration C11-D04. Worked illustration.

5Demonstration 5 of 5

Stop a divergent series at its smallest term

At x=.1, are 30 terms better than 10?

Plot the error of each partial sum against the exact integral. The error falls, bottoms out near k≈1/x, then grows without limit.

∫0∞e−t1+xtdt∼∑k≥0(−1)kk!xk \int_0^\infty\frac{e^{-t}}{1+xt}\,dt\sim\sum_{k\ge0}(-1)^k k!\,x^k

Small parameter x. x>0; exact value by 150-point Gauss-Laguerre quadrature. The series diverges for every x>0; it is asymptotic, not convergent.

Predict first. At x=.1, are 30 terms better than 10?

Choose an example

Stop a divergent series at its smallest term. The series 1 − 1!x + 2!x² − … for ∫e^(−t)/(1+xt)dt diverges for every x>0, yet at x=0.1 keeping 10 terms gives error 0.00018. Terms shrink until k is near 1/x=10, then grow. With 30 terms the error explodes to 66: adding terms past the smallest one makes things worse. Stop near the smallest term.
Small parameter x: 0.1
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Small parameter x
0.1
Exact integral
0.915633
Best number of terms
10
Smallest error
0.000177019
Error with 30 terms
65.8963

The series 1 − 1!x + 2!x² − … for ∫e^(−t)/(1+xt)dt diverges for every x>0, yet at x=0.1 keeping 10 terms gives error 0.00018. Terms shrink until k is near 1/x=10, then grow. With 30 terms the error explodes to 66: adding terms past the smallest one makes things worse. Stop near the smallest term.

Use the idea

Use rule 11.3.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

x>0; exact value by 150-point Gauss-Laguerre quadrature. The series diverges for every x>0; it is asymptotic, not convergent.

Check your understanding: At x=.1, are 30 terms better than 10?
No. Ten terms give error about 0.00018; thirty terms give error about 66. Past the smallest term, each extra term makes the sum worse.

Book source: Rule 11.3.2: Stop a divergent asymptotic series near its least term. Demonstration C11-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.