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The Exponential Function: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The log-free product limit $(1-2/n)^n\to\exp(-2)$

Example

Define a sequence $(a_n)_{n\in\mathbb N}$ by $a_0:=0,\qquad a_n:=\left(1-\frac2{\iota(n)}\right)^n\quad(n\ge1).$ Then $a_n\longrightarrow e^{-2}$ . The separate value at $n=0$ avoids division by $\iota(0)=0$ ; the finitely many remaining initial indices with nonpositive base do not affect the limit.

Facts & Assumptions

Given: The sequence $(a_n)$ defined above.

[L1]

The product-limit theorem holds for every real input once $n>|x|$ (For every real $x$ , $(1+x/n)^n\to\exp x$ ).

[L2]

Since $e=\exp(1)$ , the addition and reciprocal formulas and the definition of negative integer powers give $\exp(-2)=1/\exp(2)=1/(\exp(1)\exp(1))=1/e^2=e^{-2}>0$ (The real exponential function and the number $e$ by a power series, The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$ , The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$ , Integer powers $a^m$ ).

[L3]

A sequence and any one of its tails have the same limit (Convergence depends only on the tail).

Verification

technique · direct

1.1

For every $n>2$ , the base is positive and $a_n=(1-2/\iota(n))^n$ , so [L1] applies at $x=-2$ and gives $a_n\to\exp(-2)$ on that tail.

2.1

By [L3], the whole sequence has the same limit as the tail in step 1.1, and [L2] identifies that limit as $e^{-2}>0$ .

step 1.1L2L3∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-01Open item page →

The convergence $(1+x/n)^n\to\exp x$ is not uniform on $\mathbb{R}$

Statement refuted

The pointwise convergence $(1+x/\iota(n))^n\to\exp(x)$ is uniform on all of $\mathbb R$ .

Facts & Assumptions

Given: $h_n(x)=(1+x/\iota(n))^n$ .

[L1]

Pointwise convergence is For every real $x$ , $(1+x/n)^n\to\exp x$ , while uniform convergence is Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions.

[L2]

$e>2$ (The elementary numerical bound $2<e<3$ ) and $\exp(n)=e^n$ by The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$ .

Counterexample

technique · direct

1.1

At the moving point $x=\iota(n)$ , $h_n(x)=2^n$ , whereas $\exp(x)=e^n$ .

givenL2

2.1

Since $e>2$ , the difference $e^n-2^n$ is at least $e-2>0$ and in fact grows; therefore $\sup_x|h_n(x)-\exp(x)|\not\to0$ .

step 1.1L2algebra

3.1

Hence the pointwise convergence is not uniform on $\mathbb R$ .

step 2.1L1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-01Open item page →

The one-sided flat function is $C^\infty$ with identically zero Taylor series

Example

Define $\phi(x)=\begin{cases}\exp(-1/x),&x>0,\\0,&x\le0.\end{cases}$ Then $\phi\in C^\infty(\mathbb R)$ and $\phi^{(m)}(0)=0$ for every $m\in\mathbb N$ , although $\phi(x)>0$ for $x>0$ .

Facts & Assumptions

Given: The displayed function.

[L1]

Exponential dominates every polynomial (The exponential dominates every fixed nonnegative integer power at $+\infty$ ).

[L2]

Exponential is smooth, with positive values (The exponential function is smooth and $(\exp)'=\exp$ , A power-series sum is infinitely differentiable inside its radius and satisfies $a_n=f^{(n)}(c)/\iota(n!)$ at its centre, The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$ ).

[L3]

Products and composites are differentiable by Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ and The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)'(c) = f'(g(c))\,g'(c)$ .

Verification

technique · induction

1.1

On $x>0$ , repeated product and chain rules give $\phi^{(m)}(x)=P_m(1/x)\exp(-1/x)$ for a polynomial $P_m$ ; on $x<0$ , every derivative is $0$ .

baseL2L3given

2.1

If the formula holds at order $m$ , differentiating produces another polynomial $P_{m+1}(1/x)$ times $\exp(-1/x)$ . By [L1], this tends to $0$ as $x\to0^+$ .

ihstep 1.1L1L2L3

3.1

The difference quotient for the $m$ -th derivative at $0$ is again a polynomial in $1/x$ times $\exp(-1/x)$ , so it tends to $0$ . Induction therefore extends every derivative continuously across $0$ , with value $0$ .

step 2.1L1givendischarge-induction∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-01Open item page →

A nonzero smooth compactly supported bump

Example

Let $\phi$ be The one-sided flat function is $C^\infty$ with identically zero Taylor series and define $b(x)=\phi(1-x^2)$ . Then $b$ is smooth, positive on $(-1,1)$ , and zero outside $[-1,1]$ ; its support is $[-1,1]$ .

Facts & Assumptions

Given: The displayed composite.

[L1]

$\phi$ is smooth and flat at $0$ (The one-sided flat function is $C^\infty$ with identically zero Taylor series).

[L2]

Products and composites preserve smoothness (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)'(c) = f'(g(c))\,g'(c)$ , Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ , A power-series sum is infinitely differentiable inside its radius and satisfies $a_n=f^{(n)}(c)/\iota(n!)$ at its centre).

[L3]

Closure, boundary, and interior are Interior, closure, boundary, limit point, isolated point and dense subset of a metric space.

Verification

technique · direct

1.1

On $|x|<1$ , $1-x^2>0$ , so $b(x)>0$ ; on $|x|\ge1$ , $1-x^2\le0$ , so $b(x)=0$ .

givenL1

1.2

Away from $\pm1$ , smoothness follows from [L2]. At $\pm1$ , every one-sided derivative from inside is a finite sum of derivatives of $\phi$ at $0$ , all zero by [L1], matching the zero function outside.

L1L2

2.1

Thus $b$ is smooth and its nonzero set is $(-1,1)$ , whose closure is $[-1,1]$ ; this is its support.

step 1.1step 1.2L3∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-01Open item page →

A discontinuous positive solution of $F(x+y)=F(x)F(y)$

Statement refuted

Every positive $F:\mathbb R\to(0,\infty)$ satisfying $F(x+y)=F(x)F(y)$ is continuous and equals an ordinary exponential.

Facts & Assumptions

Given: An irrational real $b$ , so $\{1,b\}$ is linearly independent over $\mathbb Q$ , and the Axiom of Choice.

[L1]

Every independent set extends to a basis (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $\operatorname{span}(S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ , Vector space over a field, The Axiom of Choice).

[L2]

A $\mathbb Q$ -linear map is additive (Cauchy's functional equation $f(x+y) = f(x) + f(y)$ , and the additive functions $\mathbb{R} \to \mathbb{R}$ ).

[L3]

The ordinary exponential is positive, injective, and multiplicative (The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$ , The exponential function is strictly increasing, The exponential addition formula $\exp(x+y)=\exp(x)\exp(y)$ ).

Counterexample

technique · constructive

1.1

Extend $\{1,b\}$ to a Hamel basis. Define the $\mathbb Q$ -linear map $A$ by $A(1)=1$ , $A(b)=0$ , and $A(v)=v$ on the remaining chosen basis elements. Then $A$ is additive but not the identity.

L1L2construct

2.1

Put $F(x)=\exp(A(x))$ . Positivity and additivity give $F(x+y)=F(x)F(y)$ , and $F(1)=e$ .

step 1.1L3

3.1

If $F$ were continuous, Regular normalized multiplicative Cauchy equations characterize the exponential would give $F=\exp$ ; injectivity of $\exp$ would then give $A(x)=x$ , contradicting $A(b)=0\ne b$ .

step 1.1step 2.1L3given

4.1

Thus $F$ is a discontinuous positive multiplicative solution. The construction uses Choice exactly in the basis extension.

step 3.1givendischarge-construct∎

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)Open item page →

The exponential is not uniformly continuous on $\mathbb{R}$

Statement refuted

The exponential function is uniformly continuous on $\mathbb R$ .

Facts & Assumptions

Given: $\exp'=\exp$ .

[L1]

Uniform continuity is Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ .

[L2]

The exponential is strictly increasing (The exponential function is strictly increasing), the mean value theorem is The mean value theorem, as the case $g(x) = x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a < b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$ , and exponential dominates polynomials (The exponential dominates every fixed nonnegative integer power at $+\infty$ ).

[L3]

The reciprocal sequence $1/\iota(n)$ tends to $0$ when started at $n\ge1$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

Counterexample

technique · direct

1.1

For $n\ge1$ , let $x_n=\iota(n)$ and $y_n=\iota(n)+1/\iota(n)$ . Then $|y_n-x_n|=1/\iota(n)\to0$ .

1.2

By the mean value theorem, $\exp(y_n)-\exp(x_n)=\exp(c_n)/\iota(n)$ for some $c_n\in(x_n,y_n)$ . Since exponential is increasing, this is at least $\exp(\iota(n))/\iota(n)$ , which tends to $+\infty$ by [L2].

L2given

2.1

Thus arbitrarily close pairs have image distances bounded away from $0$ , contradicting the uniform-continuity condition [L1].

step 1.1step 1.2L1∎

Remarks

On every compact interval, $\exp$ is uniformly continuous by Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness; the failure is global.

Sources

Standard references

Recommended treatments; not extraction sources.

The Exponential Function: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The log-free product limit (1−2/n)n→exp⁡(−2)(1-2/n)^n\to\exp(-2)(1−2/n)n→exp(−2)

Example

Facts & Assumptions

Verification

The convergence (1+x/n)n→exp⁡x(1+x/n)^n\to\exp x(1+x/n)n→expx is not uniform on R\mathbb{R}R

Statement refuted

Facts & Assumptions

Counterexample

The one-sided flat function is C∞C^\inftyC∞ with identically zero Taylor series

Example

Facts & Assumptions

Verification

A nonzero smooth compactly supported bump

Example

Facts & Assumptions

Verification

A discontinuous positive solution of F(x+y)=F(x)F(y)F(x+y)=F(x)F(y)F(x+y)=F(x)F(y)

Statement refuted

Facts & Assumptions

Counterexample

The exponential is not uniformly continuous on R\mathbb{R}R

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

Standard references

The log-free product limit $(1-2/n)^n\to\exp(-2)$

The convergence $(1+x/n)^n\to\exp x$ is not uniform on $\mathbb{R}$

The one-sided flat function is $C^\infty$ with identically zero Taylor series

A discontinuous positive solution of $F(x+y)=F(x)F(y)$

The exponential is not uniformly continuous on $\mathbb{R}$