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A smooth function not equal to its Maclaurin series

Statement refuted

If a smooth real function has a Maclaurin series that converges everywhere, then the function equals the sum of that series everywhere.

Counterexample

Define

$ψ (x) = {\begin{cases} e^{- 1 / x^{2}}, & x \neq 0, \\ 0, & x = 0. \end{cases}$

Then $ψ \in C^{\infty} (R)$ and $ψ^{(n)} (0) = 0$ for every $n \geq 0$ . Consequently its Maclaurin series is the zero series, which converges for every real $x$ , while $ψ (x) > 0$ whenever $x \neq 0$ .

Facts & Assumptions

Given: The function $ψ$ displayed above.

[C1]

Let $ϕ (u) = {\begin{cases} e^{- 1 / u}, & u > 0, \\ 0, & u \leq 0, \end{cases} q (x) = x^{2},$ so that $ψ = ϕ \circ q$ .

[L1]

The function $ϕ$ belongs to $C^{\infty} (R)$ , $ϕ^{(j)} (0) = 0$ for every $j \geq 0$ , and $ϕ (u) > 0$ for $u > 0$ (The one-sided flat function is $C^{\infty}$ with identically zero Taylor series).

[L2]

If $g$ is differentiable at $x$ and $h$ is differentiable at $g (x)$ , then $(h \circ g)^{'} (x) = h^{'} (g (x)) g^{'} (x)$ (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L3]

Sums and products of differentiable functions are differentiable, with the sum and product rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L4]

The Maclaurin series of a smooth function $f$ is $\sum_{n \geq 0} f^{(n)} (0) x^{n} / n!$ ; its definition alone asserts neither convergence nor equality with $f$ (Taylor and Maclaurin series).

[L5]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable at every real with derivative $n x^{n - 1}$ , and for $n = 0$ it is the constant $1$ , with derivative $0$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L6]

A function differentiable at a limit point $c$ of its domain is continuous at $c$ ; hence a function differentiable on a set is continuous at every point of that set (A function differentiable at $c$ is continuous at $c$ ).

[L7]

A function is of class $C^{k}$ on an interval when $f^{(j)}$ exists there for every $j \leq k$ and each such $f^{(j)}$ is continuous there, and it is smooth, or $C^{\infty}$ , when it is $C^{k}$ for every $k \in N$ (Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ ).

Proof

technique · direct

1.1

The function $q (x) = x^{2}$ is a polynomial, so it is differentiable at every real with $q^{'} (x) = 2 x$ , and $ψ (x) = ϕ (q (x))$ for every $x \in R$ .

C1L5algebra

1.2

The derivative of any finite sum of functions of the form $p (x) ϕ^{(j)} (q (x))$ , with $p$ a polynomial, is again a finite sum of this form: each $p$ is differentiable with polynomial derivative $p^{'}$ by [L5], each $ϕ^{(j)}$ is differentiable by [L1], and $q$ is differentiable with $q^{'} (x) = 2 x$ by [L5], so [L2] gives $(ϕ^{(j)} \circ q)^{'} (x) = 2 x ϕ^{(j + 1)} (q (x))$ and then [L3] gives $(p \cdot (ϕ^{(j)} \circ q))^{'} (x) = p^{'} (x) ϕ^{(j)} (q (x)) + 2 x p (x) ϕ^{(j + 1)} (q (x)),$ in which $p^{'}$ and $2 x p$ are again polynomials; [L3] then adds the finitely many summands.

L1L2L3L5algebra

1.3

If $x \neq 0$ , then $q (x) = x^{2} > 0$ , and therefore $ψ (x) = ϕ (q (x)) > 0$ .

C1L1algebra

2.1

Starting from $ψ = ϕ \circ q$ and applying step 1.2 repeatedly shows that every derivative of $ψ$ exists; moreover, for each $m \geq 0$ , $ψ^{(m)}$ is a finite sum of functions $p (x) ϕ^{(j)} (q (x))$ .

step 1.1step 1.2

3.1

For every $m \geq 0$ , step 2.1 makes $ψ^{(m)}$ differentiable at every real, because $ψ^{(m + 1)}$ exists there, and [L6] then makes $ψ^{(m)}$ continuous on $R$ . So every derivative of $ψ$ exists on $R$ and is continuous there, which by [L7] is exactly $ψ \in C^{\infty} (R)$ .

step 2.1L6L7

3.2

At $x = 0$ , every summand in step 2.1 vanishes because $q (0) = 0$ and $ϕ^{(j)} (0) = 0$ . Hence $ψ^{(m)} (0) = 0$ for every $m \geq 0$ .

L1step 2.1algebra

4.1

By the definition of the Maclaurin series, every coefficient of the Maclaurin series of $ψ$ is zero, so the series converges everywhere to $0$ .

L4step 3.2algebra

5.1

Thus the everywhere-convergent Maclaurin series agrees with $ψ$ at $x = 0$ but disagrees with it at every $x \neq 0$ , refuting the stated claim.

step 1.3step 4.1∎

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Sources

J. K. Hunter, An Introduction to Real Analysis, Example 10.31 and Corollary 10.30 (standard reference, not scraped)
W. F. Trench, Introduction to Real Analysis, §4.5 (standard reference, not scraped)