Power Series & Taylor Series

A power series centered at $c$ is an expression of the form

$\sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots$

where $a_0, a_1, a_2, \ldots$ are real constants called the coefficients and $c$ is the center. The special case $c = 0$ gives $\sum a_n x^n$.

$\sum_{n=0}^{\infty} x^n$ has $a_n = 1$ for all $n$ and center $c = 0$. From Series Convergence & Tests, this converges to $\frac{1}{1-x}$ for $|x| < 1$ and diverges for $|x| \geq 1$. This is the prototype: convergence on an open interval, divergence outside it.

$\sum_{n=0}^{\infty} \frac{x^n}{n!}$ has $a_n = 1/n!$ and converges for all $x \in \mathbb{R}$. From Mean Value Theorem & Taylor Expansion, we know this equals $e^x$. The ratio of consecutive terms is $|x|/(n+1) \to 0$ for every fixed $x$, so the ratio test gives convergence everywhere.

$\sum_{n=0}^{\infty} n! \, x^n$ diverges for every $x \neq 0$. The ratio $|a_{n+1} x^{n+1}/(a_n x^n)| = (n+1)|x| \to \infty$ for any fixed $x \neq 0$. The “radius of convergence” is $0$ — this series is useless as a function of $x$.

A polynomial of degree $N$ is a power series with $a_n = 0$ for $n > N$. It converges everywhere ($R = \infty$). Power series extend this to “infinite-degree polynomials,” but the trade-off is that convergence is no longer automatic.

For any power series $\sum a_n (x - c)^n$, exactly one of the following holds:

(i) The series converges only at $x = c$ ($R = 0$).

(ii) The series converges for all $x \in \mathbb{R}$ ($R = \infty$).

(iii) There exists $R > 0$ such that the series converges absolutely for $|x - c| < R$ and diverges for $|x - c| > R$.

Suppose $\sum a_n (x_0 - c)^n$ converges at some $x_0 \neq c$. Then $a_n(x_0 - c)^n \to 0$ (by the divergence test from Series Convergence & Tests), so $|a_n(x_0 - c)^n| \leq M$ for some bound $M$. For any $x$ with $|x - c| < |x_0 - c|$, set $r = |x - c|/|x_0 - c| < 1$. Then

$|a_n(x-c)^n| = |a_n(x_0-c)^n| \cdot r^n \leq M r^n.$

Since $\sum M r^n$ converges (geometric series with $r < 1$), the comparison test gives absolute convergence at $x$.

Now define $R = \sup\{|x_0 - c| : \sum a_n(x_0-c)^n \text{ converges}\}$. If the series converges only at $c$, then $R = 0$ (case i). If the supremum is infinite, then $R = \infty$ (case ii). Otherwise, $R$ is a positive real number (case iii): the argument above shows convergence for $|x - c| < R$, and divergence for $|x - c| > R$ follows because if the series converged at some $x_1$ with $|x_1 - c| > R$, it would also converge at all $x$ with $|x - c| < |x_1 - c|$, contradicting $R = \sup$.

The number $R$ from Theorem 1 is the radius of convergence of $\sum a_n(x-c)^n$. We allow $R = 0$ and $R = \infty$. The open interval $(c-R, c+R)$ is the open interval of convergence.

$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}$

with the convention $1/0 = \infty$ and $1/\infty = 0$.

Apply the root test from Series Convergence & Tests to $\sum |a_n(x-c)^n|$:

$\limsup_{n \to \infty} |a_n(x-c)^n|^{1/n} = |x-c| \cdot \limsup_{n \to \infty} |a_n|^{1/n}.$

The root test gives convergence when this is $< 1$, i.e., $|x-c| < 1/\limsup |a_n|^{1/n} = R$, and divergence when $> 1$, i.e., $|x-c| > R$.

If $\lim_{n \to \infty} |a_{n+1}/a_n|$ exists (possibly $= 0$ or $= \infty$), then

$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|.$

The Cauchy-Hadamard formula always works (it uses limsup). The ratio test requires the limit to exist. When both apply, they give the same $R$. From Series Convergence & Tests, the root test is strictly stronger — there exist series where the ratio test is inconclusive but the root test determines $R$.

(a) $\sum x^n/n!$: ratio gives $R = \lim_{n \to \infty} (n+1) = \infty$.

(b) $\sum n! \, x^n$: ratio gives $R = \lim_{n \to \infty} 1/(n+1) = 0$.

(c) $\sum x^n/n$: ratio gives $R = \lim_{n \to \infty} n/(n+1) = 1$.

(d) $\sum x^n/n^2$: ratio gives $R = \lim_{n \to \infty} n^2/(n+1)^2 = 1$.

(e) $\sum n^n x^n$: Cauchy-Hadamard gives $1/R = \limsup n = \infty$, so $R = 0$.

The interval of convergence of $\sum a_n(x-c)^n$ is the set of all $x$ where the series converges. It always includes the open interval $(c-R, c+R)$ and may or may not include either endpoint.

All three of the following have $R = 1$ centered at $c = 0$:

(a) $\sum x^n$: diverges at both endpoints. At $x = 1$: $\sum 1$ diverges. At $x = -1$: $\sum (-1)^n$ diverges. Interval: $(-1, 1)$.

(b) $\sum x^n/n^2$: converges at both endpoints. At $x = 1$: $\sum 1/n^2$ converges ($p$-series, $p = 2$). At $x = -1$: $\sum (-1)^n/n^2$ converges absolutely. Interval: $[-1, 1]$.

(c) $\sum x^n/n$: mixed. At $x = -1$: $\sum (-1)^n/n$ converges by the alternating series (Leibniz) test. At $x = 1$: $\sum 1/n$ diverges (harmonic). Interval: $[-1, 1)$.

With two endpoints and two possible verdicts (converge/diverge) at each, there are four combinations: $(c-R, c+R)$, $[c-R, c+R]$, $[c-R, c+R)$, $(c-R, c+R]$. All four occur in practice. The algorithm is always the same: (1) compute $R$ via ratio or Cauchy-Hadamard; (2) substitute $x = c \pm R$ and apply a convergence test from Series Convergence & Tests.

For $\sum \frac{(-1)^n x^n}{\sqrt{n+1}}$, the ratio test gives $R = 1$. At $x = 1$: $\sum (-1)^n/\sqrt{n+1}$ converges by the Leibniz test (alternating, terms decrease to $0$). At $x = -1$: $\sum 1/\sqrt{n+1}$ diverges by comparison with $\sum 1/\sqrt{n}$ ($p$-series with $p = 1/2 < 1$). Interval: $(-1, 1]$.

If $\sum a_n(x-c)^n$ has radius of convergence $R > 0$, then the series converges uniformly on every closed interval $[c-r, c+r]$ for $0 < r < R$.

Fix $r$ with $0 < r < R$ and let $M_n = |a_n| r^n$. For $|x - c| \leq r$, we have

$|a_n(x-c)^n| \leq |a_n| r^n = M_n.$

Since $r < R$, the series $\sum M_n = \sum |a_n| r^n$ converges (by the definition of $R$ — the power series converges absolutely for $|x - c| < R$, and $r < R$). By the Weierstrass M-test from Uniform Convergence, the series $\sum a_n(x-c)^n$ converges uniformly on $[c-r, c+r]$.

The power series $\sum x^n = 1/(1-x)$ converges pointwise on $(-1, 1)$ but does not converge uniformly on all of $(-1, 1)$. The partial sums $S_n(x) = (1-x^{n+1})/(1-x)$ satisfy

$\sup_{x \in (-1,1)} |S_n(x) - 1/(1-x)| = \sup_{x \in (-1,1)} \frac{|x|^{n+1}}{|1-x|} = \infty$

for every $n$ (the supremum blows up as $x \to 1^-$). The uniform convergence theorem only guarantees uniformity on $[-r, r]$ for $r < 1$ — compact subsets strictly inside the interval of convergence. This is the same pointwise-vs-uniform distinction from Uniform Convergence, now appearing in a concrete power-series context.

If $f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n$ has radius of convergence $R > 0$, then $f$ is differentiable on $(c-R, c+R)$ and

$f'(x) = \sum_{n=1}^{\infty} n a_n (x-c)^{n-1}.$

The differentiated series has the same radius of convergence $R$.

Fix $x_0 \in (c-R, c+R)$ and choose $r$ with $|x_0 - c| < r < R$. On $[c-r, c+r]$, the series converges uniformly (Theorem 4). The partial sums $S_N(x) = \sum_{n=0}^{N} a_n(x-c)^n$ are polynomials, hence differentiable. Each $S_N'(x) = \sum_{n=1}^{N} n a_n(x-c)^{n-1}$.

The differentiated series $\sum n a_n(x-c)^{n-1}$ has

$\limsup_{n \to \infty} |n a_n|^{1/n} = \limsup_{n \to \infty} \bigl(n^{1/n} |a_n|^{1/n}\bigr) = 1 \cdot \limsup_{n \to \infty} |a_n|^{1/n} = \frac{1}{R}$

since $n^{1/n} \to 1$. So the differentiated series has the same radius $R$ and converges uniformly on $[c-r, c+r]$.

By the interchange theorem for differentiation from Uniform Convergence, $f'(x_0) = \lim_{N \to \infty} S_N'(x_0) = \sum_{n=1}^{\infty} n a_n(x_0-c)^{n-1}$.

Applying Theorem 5 repeatedly, $f^{(k)}(x) = \sum_{n=k}^{\infty} \frac{n!}{(n-k)!} a_n (x-c)^{n-k}$ for all $k \geq 0$, each with radius $R$. A power series is infinitely differentiable inside its radius of convergence.

If $f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n$ has radius $R > 0$, then

$\int_c^x f(t)\,dt = \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}$

for $|x - c| < R$. The integrated series has radius of convergence $R$.

Differentiating $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ term by term gives

$\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=0}^{\infty} (n+1) x^n \quad \text{for } |x| < 1.$

Integrate $\frac{1}{1+x} = \sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n$ term by term from $0$ to $x$:

$\ln(1+x) = \int_0^x \frac{dt}{1+t} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1} x^{n+1} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n \quad \text{for } |x| < 1.$

The series also converges at $x = 1$ by the alternating series test (Leibniz) from Series Convergence & Tests, giving $\ln 2 = 1 - 1/2 + 1/3 - 1/4 + \cdots$.

Integrate $\frac{1}{1+t^2} = \sum_{n=0}^{\infty} (-1)^n t^{2n}$ to get

$\arctan(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} \quad \text{for } |x| \leq 1.$

Setting $x = 1$ gives the Leibniz formula $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$.

If $f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n$ on some interval $(c-R, c+R)$ with $R > 0$, then

$a_n = \frac{f^{(n)}(c)}{n!}$

for all $n \geq 0$. A power series representation of a function is necessarily its Taylor series.

By Corollary 1, $f^{(k)}(x) = \sum_{n=k}^{\infty} \frac{n!}{(n-k)!} a_n (x-c)^{n-k}$. Setting $x = c$, all terms with $n > k$ vanish (they contain a factor $(c-c)^{n-k} = 0$), leaving $f^{(k)}(c) = k! \, a_k$. Solving gives $a_k = f^{(k)}(c)/k!$.

If two power series $\sum a_n x^n$ and $\sum b_n x^n$ are equal on any interval containing $0$, then $a_n = b_n$ for all $n$. You cannot have two different power series representations of the same function centered at the same point. This makes power series representations canonical.

The six essential Taylor series at $c = 0$ (Maclaurin series):

1. $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, $R = \infty$

2. $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$, $R = \infty$

3. $\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}$, $R = \infty$

4. $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} x^n$, $R = 1$, also converges at $x = 1$

5. $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$, $R = 1$

6. $(1+x)^\alpha = \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n$ (binomial series), $R = 1$ for non-integer $\alpha$

A function $f$ is real analytic at $c$ if its Taylor series at $c$ converges to $f(x)$ in some neighborhood of $c$:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n \quad \text{for } |x - c| < R, \; R > 0.$

A function is analytic on an interval if it is analytic at every point of that interval. The class of analytic functions is denoted $C^\omega$.

If $f$ is $C^\infty$ on an interval $I$ containing $c$ and there exist constants $M > 0$ and $K > 0$ such that

$|f^{(n)}(x)| \leq M \cdot K^n \cdot n! \quad \text{for all } n \text{ and all } x \in I,$

then $f$ is analytic at $c$ with $R \geq 1/K$.

The Lagrange remainder from Mean Value Theorem & Taylor Expansion satisfies

$|R_n(x)| \leq \frac{|f^{(n+1)}(\xi)|}{(n+1)!} |x-c|^{n+1} \leq \frac{M K^{n+1} (n+1)!}{(n+1)!} |x-c|^{n+1} = M (K|x-c|)^{n+1}.$

For $|x-c| < 1/K$, this is a geometric sequence converging to $0$, so $R_n(x) \to 0$ and the Taylor series converges to $f$.

For $e^x$: on any interval $[-B, B]$, $|f^{(n)}(x)| = e^x \leq e^B$. Setting $M = e^B$ and $K = 1$ gives the bound $|f^{(n)}(x)| \leq e^B \cdot 1^n \cdot n!$ — but we need $M K^n n!$, and the actual derivatives satisfy the much tighter bound $|f^{(n)}(x)| \leq e^B$ (without the $n!$ factor). This means $R_n(x) \leq e^B |x|^{n+1}/(n+1)! \to 0$ for every $x$, so $R = \infty$.

The same argument applies to $\sin x$ and $\cos x$, where all derivatives are bounded by $1$.

This extends the discussion from Mean Value Theorem & Taylor Expansion, Example 8. Define

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

All derivatives at $0$ are $0$ (the proof by induction uses L’Hôpital’s Rule repeatedly: each $f^{(n)}(x)$ is a polynomial in $1/x$ times $e^{-1/x^2}$, and $e^{-1/x^2}$ decays faster than any polynomial as $x \to 0$). So the Taylor series at $0$ is $T(x) = 0$. But $f(x) > 0$ for $x \neq 0$. The Taylor series converges — but to the wrong function.

The derivative bound $|f^{(n)}(0)| \leq M K^n n!$ fails for any finite $K$: the derivatives at points near $0$ (but not at $0$) grow faster than any geometric rate.

In a precise sense (Baire category), “most” smooth functions are not analytic. Yet in practice — in calculus, physics, and ML — almost every function we encounter is analytic (or piecewise analytic). The standard function zoo ($e^x$, $\sin$, $\cos$, $\ln$, polynomials, rational functions, compositions thereof) is closed under the operations that preserve analyticity. The smooth-but-not-analytic examples are constructed to violate the derivative growth condition.