Back to the on-screen lesson ·
Reading $|x - c| < r$ as a distance, bounding a sum by its parts, and the two-stage estimate for a product.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to read a modulus inequality as an interval about a centre, bound a sum or a difference with the triangle inequality, derive and use its reverse form, and run a two-stage estimate in which a crude restriction bounds a loose factor before the fine restriction is chosen. These are the estimates every later proof in the course is assembled from.
The Archimedean property, which produces the whole number a threshold needs. What it does not supply is the inequality that gets a quantity down to that threshold, and that is what this lesson is. Everything after it — limits, continuity, integrals — is estimates, and nearly every estimate is one of two inequalities applied carefully.
$|x|$ is $x$ when $x \ge 0$ and $-x$ otherwise. The expression $|x - y|$ is read as the distance from $x$ to $y$, and that reading, rather than the case split, is what makes the rest of the course legible: $|a_n - L| < \varepsilon$ says the term is within $\varepsilon$ of the limit.
The triangle inequality. $|a + b| \le |a| + |b|$, with equality exactly when $a$ and $b$ point the same way. In the distance reading: the journey from $x$ to $z$ is never shorter via $y$.
It is used in one direction only. You have bounds on pieces; you want a bound on the whole; the pieces add.
The reverse triangle inequality. $\bigl||a| - |b|\bigr| \le |a - b|$. This bounds a difference of sizes by the distance. It is what makes $x \mapsto |x|$ continuous, and it is the right tool whenever the quantity you must control is itself a modulus.
Reading a modulus inequality as a set. $|x - c| < r$ is the open interval $(c - r, c + r)$: the points within $r$ of $c$. Two such conditions at once give the overlap. A non-strict inequality closes the ends.
The splitting move. To bound $|A - B|$, insert a convenient middle term: $$|A - B| = |(A - C) + (C - B)| \le |A - C| + |C - B|.$$ Choose $C$ so that both pieces are things you already control. This single line is the skeleton of the proof that limits are unique, that a convergent sequence is Cauchy, that a product of limits is the limit of the product, and of half a dozen theorems later in the course.
Another way: picture
Two arrows laid end to end reach no further than their lengths added, and exactly that far when they point the same way. The reverse form is the same picture read for the gap between the two lengths: two arrows of very different lengths must have a large gap between their tips.
Another way: steps
To bound a quantity $Q$:
Step 1 is the only one that takes thought, and it is where the middle term is invented.
A proof that a sum converges begins: let $\varepsilon > 0$; choose $N_1$ so that the first part is within $\varepsilon/2$, and $N_2$ so that the second is within $\varepsilon/2$.
That looks like sleight of hand the first few times. It is not. The triangle inequality says the two errors add, so to land under $\varepsilon$ each must be given a share, and halves are the obvious shares. Three pieces get thirds. A piece multiplied by a constant $K$ gets $\varepsilon/(2K)$.
The only rule is that the shares are chosen before the estimate is run, and that they do not depend on anything the estimate produces. A share that depended on $n$ would not be a share at all.
To bound $|xy - ab|$ given that $x$ is near $a$ and $y$ is near $b$, insert a middle term: $$|xy - ab| = |xy - ay + ay - ab| \le |y||x - a| + |a||y - b|.$$ The pieces are now controlled — except for $|y|$, which is not a constant. So the estimate has two stages: first restrict $y$ to within $1$ of $b$, which makes $|y| < |b| + 1$; then, with that bound in hand, choose the rest.
| Stage | What is fixed | What it buys |
|---|---|---|
| 1 | a crude restriction, say within $1$ | a numerical bound on the loose factor |
| 2 | the fine restriction, in terms of $\varepsilon$ | the target |
The delta finally used is the smaller of the two. Nearly every epsilon-delta proof for a non-linear function has this shape, and the item about $|x^2 - a^2|$ in this lesson is its smallest case.
Bounding every factor by the same restriction. From $|x - a| < 1$ it does not follow that $|x + a| < 1$. Each factor is bounded on its own, and a factor that is near $2a$ is not near zero.
Using the triangle inequality backwards. $|a + b| \ge |a| - |b|$ is true, but it bounds from below, and an estimate that needs an upper bound has gained nothing from it.
Letting the share depend on the index. Giving the $n$-th piece $\varepsilon/n$ and then summing does not produce $\varepsilon$. The shares are fixed in advance and there are finitely many of them.
Show that a sequence has at most one limit: suppose $a_n \to L$ and $a_n \to M$.
Two limits supposed, to be compared.
Insert the middle term $a_n$: $|L - M| \le |L - a_n| + |a_n - M|$.
The middle term is chosen, not given.
Both pieces can be made below $\varepsilon/2$ for large $n$, so $|L - M| < \varepsilon$ for every $\varepsilon > 0$, forcing $L = M$.
A non-negative number below every epsilon is zero.
Bound $|x^2 - 9|$ when $|x - 3| < \delta$ and $\delta \le 1$. Factorise: $|x^2 - 9| = |x - 3|\,|x + 3|$.
Separate what is controlled from what is not.
Stage one: $\delta \le 1$ gives $2 \le x \le 4$, so $|x + 3| \le 7$.
A crude restriction bounds the loose factor.
Stage two: $|x^2 - 9| \le 7|x - 3| < 7\delta$, so taking $\delta = \min(1, \varepsilon/7)$ brings it under $\varepsilon$.
The smaller of the two deltas is the one used.
Combine over a common denominator: the quantity is $\dfrac{|2 - x|}{2|x|}$.
Write it in terms of the thing you control.
Stage one: $\delta \le 1$ gives $x \ge 1$, so $2|x| \ge 2$ and the quantity is at most $|x - 2|/2$.
Bound the denominator away from zero.
Stage two: so $\delta = \min(1, 2\varepsilon)$ brings it under $\varepsilon$. The first stage is not optional here — without a bound keeping $x$ away from $0$ the quantity is unbounded, however small $\delta$ is made.
Give the set of real $x$ with $|x - 8| < 4$.
This task has no paper form; do it on a device.
Suppose $|x - 2| < 2$ and $|y - 5| < 3$. What is the best bound the triangle inequality gives for $|(x + y) - 7|$?
Answer:
Put the steps that derive the reverse triangle inequality from the ordinary one into order.
Number the steps in order (write the number in the box):
Match each question to the statement about absolute values that answers it.
| $|a + b| \le |a| + |b|$ | $\bigl||a| - |b|\bigr| \le |a - b|$ | the points within $8$ of $c$, ends excluded | $|ab| = |a|\,|b|$ | |
|---|---|---|---|---|
| How large can a sum be, given bounds on each part? | ||||
| How much can two numbers differ in size, given how far apart they are? | ||||
| Which points satisfy $|x - c| < 8$? |
Here is an attempt to bound $|x^2 - 8^2|$ when $|x - 8| < 1$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Give the set of real $x$ satisfying both $|x - 9| < 7$ and $|x - 12| < 7$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the set of real $x$ with $|x - 6| < 4$.
This task has no paper form; do it on a device.
You can turn a distance condition into an interval, bound a quantity by splitting it through a middle term, and say why a proof about a sum gives each part half of the room. Say in your own words why bounding one factor does not bound the other. Next: the definition the whole subject turns on.
10. Your turn: bound $\left|\dfrac{1}{x} - \dfrac{1}{2}\right|$ when $|x - 2| < \delta$ with $\delta \le 1$, step 3