3.3 Power series and Taylor expansions

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

For a sequence $u_n\ge0$, define

allowing the value $+\infty$. The tail suprema are nonincreasing, so their infimum is their extended limit. Two consequences are needed.

• If $L<q$, then $u_n<q$ eventually. Indeed some tail supremum is below $q$; otherwise their infimum would be at least $q$.
• If $q<L$, then $u_n>q$ for infinitely many indices. Otherwise a tail would have every term at most $q$, forcing its supremum and hence $L$ to be at most $q$.

These statements remain useful when the original sequence has no limit. A finite number of unusually large terms affects some early tail suprema, but not $L$.

Proof of the convergent ratio case. Suppose $a_n\ne0$ for all sufficiently large $n$ and $\limsup |a_{n+1}/a_n|<1$. Choose $0<q<1$ strictly larger than this limsup. For some $N$, every $n\ge N$ satisfies $|a_{n+1}/a_n|\le q$. Iterating gives

The geometric series $\sum_{k\ge0}|a_N|q^k$ converges, so comparison proves convergence of the absolute tail. A finite prefix does not affect convergence; hence $\sum a_n$ converges absolutely. A uniform conclusion for a family of series requires a bound valid uniformly in the input.

For real functions on a nonempty set $A$, uniform convergence means $\sup_{x\in A}|f_n(x)-f(x)|\to0$, with the errors eventually finite. Uniform Cauchy means every $\varepsilon>0$ has $N$ such that $|f_n(x)-f_m(x)|<\varepsilon$ for all $x\in A$ and all $m,n\ge N$. If $A=\varnothing$, the epsilon–N conditions are vacuously satisfied; no real supremum of the empty set is used.

Theorem: uniform completeness

A sequence of real-valued functions is uniformly convergent if and only if it is uniformly Cauchy.

Proof. Uniform convergence gives the Cauchy estimate by two errors of size $\varepsilon/2$. Conversely, uniform Cauchy implies that for each fixed $x$, $(f_n(x))$ is a real Cauchy sequence, so define $f(x)=\lim_n f_n(x)$. Fix $\varepsilon>0$ and use uniform Cauchy with tolerance $\varepsilon/2$. For every $x$ and $n\ge N$, pass $m\to\infty$ in the estimate to get $|f_n(x)-f(x)|\le\varepsilon/2<\varepsilon$. The same $N$ works everywhere, proving uniform convergence. Completeness of the target $\mathbb R$ supplied the pointwise limits.

Derivative interchange without $C^1$ hypotheses

Theorem

Let $a<b$. Suppose every $f_n$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f_n(x_0)$ converges at one fixed $x_0\in[a,b]$ and $f_n'$ converges uniformly on $(a,b)$ to $g$, then $f_n$ converges uniformly on $[a,b]$ to a continuous $f$, differentiable on $(a,b)$, and $f'=g$. Continuity of the individual derivatives is not required.

Proof of existence. Uniform convergence of derivatives makes them uniformly Cauchy. For $m,n$ sufficiently large, the mean value theorem applied to $f_n-f_m$ between $x_0$ and any $x$ gives

The anchor sequence is Cauchy, so the right side is uniformly small. Uniform completeness gives a uniform limit $f$, which is continuous as a uniform limit of continuous functions.

Proof of the derivative identity. Fix an interior point $x$ and $\varepsilon>0$. Choose $N$ with $|f_n'(t)-f_N'(t)|\le\varepsilon$ for all interior $t$ and all $n\ge N$, and also $|g(x)-f_N'(x)|\le\varepsilon$. For sufficiently small nonzero $h$, apply the mean value theorem to $f_n-f_N$ on the segment from $x$ to $x+h$:

Pass $n\to\infty$ at these two fixed inputs. The same inequality holds with $f$ replacing $f_n$. Since the fixed $f_N$ is differentiable at $x$, its difference quotient is within $\varepsilon$ of $f_N'(x)$ for small enough $h$. The triangle inequality bounds the difference quotient of $f$ from $g(x)$ by $3\varepsilon$. Arbitrary $\varepsilon$ proves differentiability and equality. This argument does not integrate $g$, so it does not silently assume $g$ continuous or Riemann integrable.

Define

For fixed $x\ne0$, the ratio of successive absolute terms is $|x|/(n+1)\to0$, so the series converges absolutely; at $x=0$ it equals one directly. On any fixed interval $|x|\le M$, both its series and derivative series have summable factorial majorants. The uniform derivative theorem therefore applies to polynomial partial sums and gives $E'(x)=E(x)$ everywhere, with $E(0)=1$.

Theorem: positivity and exponential law

For every real $x,y$, $E(x)>0$, $E(-x)=1/E(x)$, and $E(x+y)=E(x)E(y)$.

Proof. Differentiate $H(x)=E(x)E(-x)$ to obtain $H'(x)=E(x)E(-x)-E(x)E(-x)=0$. The mean value theorem makes $H$ constant; at zero it equals one. Thus $E$ never vanishes. Since it is continuous and $E(0)>0$, an intermediate-value argument prevents it from becoming negative anywhere, proving positivity and the reciprocal identity. For fixed $y$, differentiate $K(x)=E(x+y)E(-x)$; the same cancellation gives zero derivative. At zero $K(0)=E(y)$, so $E(x+y)E(-x)=E(y)$. Multiplication by $E(x)$ yields the addition law. No rearrangement of a double infinite series was needed.

Full flat-function differentiation at zero

Use the previously proved real exponential series, positivity, reciprocal law and derivative identity $\exp'=\exp$. Define $F(0)=0$ and $F(x)=\exp(-1/x^2)$ for $x\ne0$. For every real $r\ge0$, set $u=1/x^2$. Choose an integer $k>r/2$. The nonnegative exponential series gives $\exp u\ge u^k/k!$ for $u>0$, hence

At nonzero inputs, induction gives $F^{(n)}(x)=P_n(1/x)\exp(-1/x^2)$ for a polynomial $P_n$, starting with $P_0=1$ and using $P_{n+1}(z)=-z^2P_n'(z)+2z^3P_n(z)$. Suppose the order-$n$ derivative has this formula off zero and value zero at zero. Every polynomial has finite degree, so the estimate above first shows $F^{(n)}(x)\to0$ and then $F^{(n)}(h)/h\to0$. Thus $F^{(n)}$ is continuous at zero, its derivative at zero exists and equals zero, and the recurrence gives the next off-zero formula. Induction proves $F$ is smooth with every derivative zero at zero. Its Taylor series there is identically zero, whereas $F(x)>0$ for each $x\ne0$. This completes the claimed counterexample without treating smoothness as analyticity.

If the series converges at $x_0\ne a$, its terms are bounded: $|c_n(x_0-a)^n|\le M$. For any $|x-a|<|x_0-a|$, the absolute terms are bounded by $M(|x-a|/|x_0-a|)^n$, a convergent geometric series. Conversely, if the series diverges at $x_1$, it cannot converge at a point farther from $a$, since that would force convergence at $x_1$ by the preceding argument. These implications define a radius separating the two behaviors; endpoint behavior is not determined by them.

Fix $0<r<s<R$. Since $|c_n|s^n$ is bounded, the derivative terms on $|x-a|\le r$ are bounded by a constant times $n(r/s)^{n-1}$, whose series converges. The M-test gives uniform convergence of derivatives on the smaller interval, and the partial sums converge at the center $a$. The uniform derivative theorem gives termwise differentiation. Uniform convergence also permits termwise integration. The root-test formula or the same comparison argument shows multiplication or division of coefficients by $n$ does not change the radius.