PG5.4 The series for eˣ converges for every x

Simulate

Check the claim with numbers

Lesson 2.7.2 wrote $e^x$ as a series and said it works for every $x$ . Its partial sums are the Taylor polynomials

p_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!}

Far from 0 the terms first grow. The cell prints the term $\frac{x^k}{k!}$ for $x = 10$ , then the ratio of the next term to it, $\frac{x}{k+1}$ .

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The terms climb to 2755.7319 at $k = 9$ and $k = 10$ , then fall: 41.1032 at $k = 20$ , 0.0038 at $k = 30$ . They turn round once the ratio $\frac{x}{k+1}$ drops below 1, and the ratio keeps falling towards 0.

Now compare $p_n(x)$ with $e^x$ . The cell prints the error $|e^x - p_n(x)|$ next to the remainder bound of lesson 2.7.2, which the next section writes out.

import math from math import factorial def p(x, n): return sum(x**k / factorial(k) for k in range(n + 1)) for x in [3, -5]: for n in [5, 10, 20]: error = abs(math.exp(x) - p(x, n)) bound = math.exp(abs(x)) * abs(x)**(n + 1) / factorial(n + 1) print(x, n, f"{error:.3e}", f"{bound:.3e}")

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Every error is below its bound, and both shrink fast. At $x = -5$ the bound is loose, $1.385 \times 10^{-3}$ against a true error of $7.593 \times 10^{-6}$ for $n = 20$ , but it still heads to 0.

Conjecture

x

Two values of $x$ , three degrees each. The claim is about every real $x$ , and the proof must also handle the terms that grow before they shrink.

The plan: a bound, then a limit

Every derivative of $e^t$ is $e^t$ (lesson 2.3.4). Between 0 and $x$ , $t$ is at most $|x|$ , so $e^t \le e^{|x|}$ . The remainder bound of lesson 2.7.2, with $M = e^{|x|}$ , becomes:

Rule: Remainder bound for the exponential

\big| e^x - p_n(x) \big| \le e^{|x|}\, \frac{|x|^{n+1}}{(n+1)!}

For every real $x$ and every $n \ge 0$ . The first factor stays fixed as $n$ grows, so the proof only needs the second to tend to 0.

This lesson takes the remainder bound from lesson 2.7.2 as given. Its proof needs the mean value theorem, which this stage does not cover. Everything else is proved here.

Format 1

Reorder the steps

Here is the proof of the claim, with its lines shuffled. Put them in order with the arrows, or drag a line by its handle.

The series $\sum_{k \ge 0} \frac{|x|^k}{k!}$ has positive terms, and the ratio of each term to the one before it is $\frac{|x|}{k+1}$ , which tends to 0.
So $p_n(x) \to e^x$ : the series for $e^x$ converges to $e^x$ , for every real $x$ .
By the term test (lesson PG5.1), its terms tend to 0. In particular $\frac{|x|^{n+1}}{(n+1)!} \to 0$ as $n \to \infty$ .
By the ratio test (lesson PG5.2), with $L = 0 < 1$ , the series $\sum_{k \ge 0} \frac{|x|^k}{k!}$ converges.
Let $\varepsilon > 0$ , and choose $N$ with $\frac{|x|^{n+1}}{(n+1)!} < \varepsilon\, e^{-|x|}$ for every $n \ge N$ . Then $|e^x - p_n(x)| < \varepsilon$ for every $n \ge N$ .
Fix $x \ne 0$ ; at $x = 0$ every $p_n(0) = 1 = e^0$ . By the remainder bound, $|e^x - p_n(x)| \le e^{|x|} \frac{|x|^{n+1}}{(n+1)!}$ for every $n$ .

An $N$ you can compute

The ratio test shows that $\frac{|x|^n}{n!} \to 0$ , but it does not say from where the terms are small. A direct proof names $N$ . It waits until every new factor $\frac{|x|}{j}$ is below $\frac{1}{2}$ , then halves.

x = 10

import math from math import factorial C = 10**20 / factorial(20) print(round(C, 4), round(math.log2(C / 1e-6), 2)) print(next(n for n in range(20, 200) if all(10**k / factorial(k) < 1e-6 for k in range(n, 200))))

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Format 2

Fill the gap

The direct proof that $\frac{|x|^n}{n!} \to 0$ , with its key steps left out. Type each missing piece as you would a quick answer, such as n+m or 1/3, then check.

x

How to type answers

Powers with ^, brackets round a power that is a sum: r^k, r^(k+1).
Fractions with / and brackets: (1-r)/(1+r). Multiply with * or side by side: 2k+1.
Press Enter in any box to check them all.

The terms really drop below $10^{-6}$ for good from $n = 37$ , so 46 is not the smallest place. A proof needs an $N$ that works, not the best one.

Simulate

One N for every x?

The proof finds an $N$ for each $x$ . Does one $N$ serve every $x$ at once?

For $x > 0$ every term left out is positive, so the error only shrinks as $n$ grows. The cell finds the first $n$ with error below $10^{-6}$ , using 60-digit decimals so rounding cannot fake an answer.

from decimal import Decimal, getcontext from math import factorial getcontext().prec = 60 def error(x, n): x = Decimal(x) return x.exp() - sum(x**k / factorial(k) for k in range(n + 1)) def first_n(x, eps=Decimal("1e-6")): n = 0 while error(x, n) >= eps: n += 1 return n for x in [1, 5, 10, 20, 40]: print(x, first_n(x)) for x in [10, 20, 40]: print(x, f"{error(x, 20):.4e}")

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The degree needed grows with $x$ : $n = 9$ at $x = 1$ , 36 at $x = 10$ and 119 at $x = 40$ . With $n = 20$ fixed, the error is 34.984 at $x = 10$ and $2.353 \times 10^{17}$ at $x = 40$ .

Conjecture

\varepsilon > 0

Five values of $x$ and a pattern. The next proof claims the opposite, and gets one line wrong.

Format 3

Find the error

A proof that one $N$ serves every real $x$ at once, which the simulation contradicts. One line does not follow. Select it.

Uniform convergence, lightly

On a bounded interval the trouble disappears. Take $R > 0$ . For every $x$ in $[-R, R]$ , the remainder bound gives $|e^x - p_n(x)| \le e^R \frac{R^{n+1}}{(n+1)!}$ , a number that does not depend on $x$ .

That number tends to 0, by the proof above with $x = R$ . So one $N$ makes it below $\varepsilon$ , and that $N$ works for every $x$ in $[-R, R]$ at once.

Rule: Uniform convergence

\begin{gathered} p_n \to f \text{ uniformly on a set } I: \\ \text{for every } \varepsilon > 0 \text{ there is one } N \\ \text{with } |f(x) - p_n(x)| < \varepsilon \\ \text{for every } n \ge N \\ \text{and every } x \in I \end{gathered}

The series for $e^x$ converges uniformly on every interval $[-R, R]$ , but not on the whole real line.

R = 5

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Uniform convergence is what lets you integrate or differentiate a series term by term, a topic for a full analysis course. Lebl's Basic Analysis, in the further reading below, treats it in chapter 6.

|e^x - p_n(x)| \le e^{|x|} \frac{|x|^{n+1}}{(n+1)!}

The series for eˣ converges for every x

Check the claim with numbers

The plan: a bound, then a limit

Rule: Remainder bound for the exponential

Reorder the steps

An $N$ you can compute

Fill the gap

One N for every x?

Find the error

Uniform convergence, lightly

Rule: Uniform convergence

Summary card

Free further reading

Check the claim with numbers

The plan: a bound, then a limit

Rule: Remainder bound for the exponential

Reorder the steps

An NNN you can compute

Fill the gap

One N for every x?

Find the error

Uniform convergence, lightly

Rule: Uniform convergence

Summary card

Free further reading

An $N$ you can compute