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The real function 1/(1+x^2) is smooth on the real line but its Maclaurin series has radius one

Example

For real $x$ with $∣ x ∣ < 1$ , $\frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} (- 1)^{n} x^{2 n} .$ The function on the left is smooth on all of $R$ , but the displayed Maclaurin series has radius $1$ .

Facts & Assumptions

Given: The real rational function $f (x) = 1 / (1 + x^{2})$ .

[L1]

If $L = {lim sup}_{k \to \infty} ∣ c_{k + 1} ∣^{1 / (k + 1)}$ , Cauchy–Hadamard gives radius $+ \infty$ for $L = 0$ , radius $1 / L$ for $0 < L < + \infty$ , and radius $0$ for $L = + \infty$ (Cauchy-Hadamard for complex power series, including zero and infinite radius).

[L2]

Let $A \subseteq R$ , let $c \in A$ be a limit point of $A$ , and let $f, g : A \to R$ be differentiable at $c$ . Then $f + g$ , $α f$ and $f g$ are differentiable at $c$ with the usual formulas, and if $g (c) \neq 0$ the quotient $f / g$ is differentiable at $c$ with the quotient rule. Differentiability of the inputs is a hypothesis, not a conclusion (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 $g (c) \neq 0$ ).

[L3]

Sums, products, and quotients with nonzero denominator of continuous real functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L4]

Smooth means having continuous derivatives of every order (Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ ).

[L5]

For $n \in N$ the power function $p_{n} (x) = x^{n}$ is differentiable at every $c \in R$ ; in particular $p_{0}$ is the constant function $1$ with $p_{0}^{'} (c) = 0$ , and $p_{1}$ is the identity with $p_{1}^{'} (c) = 1$ (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).

Verification

technique · direct

1.1

The finite geometric identity with ratio $- x^{2}$ gives the displayed series for $∣ x ∣ < 1$ ; its coefficients at even indices have modulus $1$ , so [L1] gives radius $1$ .

L1algebra

1.2

The hypothesis of [L2] is that the inputs are already differentiable, so the induction needs a base: by [L5] the constant and identity functions are differentiable everywhere, and [L2] applied to sums and products makes every real polynomial differentiable everywhere, $1 + x^{2}$ among them. Since $1 + x^{2} > 0$ for every real $x$ , the quotient clause of [L2] applies at every point, and an induction on the order — each step differentiating a quotient of polynomials with denominator a positive power of $1 + x^{2}$ , which [L5] and [L2] make differentiable — expresses every derivative of $f$ as such a quotient. [L3] then makes each derivative continuous.

L2L3L5algebra

2.1

Therefore $f$ is smooth by [L4] while its Maclaurin series has finite radius.

step 1.1step 1.2L4∎

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