Constructing sine, cosine, radians and inverse branches

Proof

Prerequisite proofs used below

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

$$s_N=\sup\{u_n:n\ge N\},\qquad L=\inf_N s_N=\limsup_{n\to\infty}u_n,$$

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

$$|a_{N+k}|\le |a_N|q^k\qquad(k\ge0).$$

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.

Theorem

Uniform limit preserves continuity

If each $f_n$ is continuous at $a\in A$ relative to $A$ and $f_n\to f$ uniformly on $A$, then $f$ is continuous at $a$.

Proof

The three-term estimate

Given $\varepsilon>0$, choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x\in A$. By continuity of the fixed function $f_N$ at $a$, choose $\delta$ so $|f_N(x)-f_N(a)|<\varepsilon/3$ whenever $|x-a|<\delta$. Then

$$|f(x)-f(a)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\varepsilon.$$

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

$$|f_n(x)-f_m(x)|\le|f_n(x_0)-f_m(x_0)|+(b-a)\sup_{(a,b)}|f_n'-f_m'|.$$

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 $x+h$:

$$\left|\frac{f_n(x+h)-f_n(x)}h-\frac{f_N(x+h)-f_N(x)}h\right|\le\varepsilon.$$

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.

Theorem

Intermediate value theorem

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, there is $c\in[a,b]$ with $f(c)=y$.

Proof

Completeness locates a crossing

If $y=f(a)$ or $y=f(b)$, take the corresponding endpoint. Otherwise, assume $f(a)<y<f(b)$; the reverse order follows by replacing both $f$ by $-f$ and $y$ by $-y$. The set $S=\{x\in[a,b]:f(x)\le y\}$ is nonempty and bounded above. Let $c=\sup S$. If $f(c)<y$, continuity gives nearby points to the right with value below $y$; since $c<b$, this contradicts the upper-bound property. If $f(c)>y$, continuity gives a neighbourhood of $c$ with value above $y$; the supremum approximation property supplies an element of $S$ in that neighbourhood, a contradiction. Thus $f(c)=y$.

Learning goals and proof dependencies

Construct sine and cosine analytically, prove their identities and a genuine fundamental period, and obtain the inverse-function domains and derivatives. We use factorial series convergence, uniform differentiation on compact intervals, the mean value theorem, the intermediate value theorem and real completeness. Familiar geometric sine/cosine identities are not assumptions in this construction.

Define for every real $x$,

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

On $|x|\le M$, factorial majorants bound both series and their derivative series by convergent numerical series: successive factorial denominators eventually make each ratio at most $1/2$. The uniform derivative theorem for polynomial partial sums applies. Reindexing the differentiated series gives

$$S'=C,\qquad C'=-S,\qquad S(0)=0,\quad C(0)=1.$$

Repeating proves both functions are infinitely differentiable. Their definitions also give $S(-x)=-S(x)$ and $C(-x)=C(x)$, since their powers are respectively odd and even.

1. The unit-circle identity is a theorem

Differentiate $S^2+C^2$. Its derivative is $2SC-2CS=0$. The mean value theorem makes it constant on the real line, and its value at zero is one. Therefore $S(x)^2+C(x)^2=1$ for all real $x$. In particular both functions have absolute value at most one. No previously defined geometric circle parametrization was used to prove this identity.

We will also use a uniqueness observation. If differentiable $U,V$ satisfy $U'=-V,V'=U$ and both vanish at zero, then $(U^2+V^2)'=0$ and its initial value is zero. Both functions must therefore vanish everywhere. This elementary energy argument proves uniqueness for this two-equation system, without invoking an unproved general ODE theorem.

2. Addition formulas follow from uniqueness

Fix real $y$ and define

$$U(x)=C(x+y)-[C(x)C(y)-S(x)S(y)],$$ $$V(x)=S(x+y)-[S(x)C(y)+C(x)S(y)].$$

The derivative identities give $U'=-V,V'=U$, and both initial values are zero. Uniqueness therefore gives

$$C(x+y)=C(x)C(y)-S(x)S(y),\qquad S(x+y)=S(x)C(y)+C(x)S(y).$$

They are valid for every pair of real inputs. Familiar double-angle formulas are now consequences obtained by setting the two inputs equal, not prior identities used in a circular definition.

3. Prove a first zero exists before defining pi

It is not enough to draw an oscillating curve and announce a period. The cosine series at two has first terms $1-2+2/3$. Starting with its magnitude-two term, the absolute terms strictly decrease, because the subsequent magnitude ratio is $4/[(2n+1)(2n+2)]<1$ for $n\ge1$. The remaining tail after $2/3$ starts negative and its alternating pair sums are negative. Thus

$$C(2)<1-2+\frac23=-\frac13<0.$$

Since $C(0)=1$ and $C$ is continuous, it has a zero in $(0,2)$. Continuity also makes it positive on some interval around zero. Let $\alpha$ be the infimum of its positive zeros. A sequence of zeros approaching that infimum and continuity give $C(\alpha)=0$; the initial positive interval ensures $\alpha>0$. For $0\le x<\alpha$, $C(x)>0$: any nonpositive value would give an earlier zero by continuity.

Therefore $S'=C>0$ before $\alpha$, and $S(\alpha)>0$. The unit-circle identity forces $S(\alpha)=1$. Also $C'=-S<0$ on $(0,\alpha)$, so the first quarter has strictly decreasing $C$ and increasing $S$.

Define the positive constant $\pi=2\alpha$. This is an analytic definition of the usual constant, not an assumption about where a previously known sine vanishes.

4. Quarter turns prove the period and its minimality

Insert $(C(\alpha),S(\alpha))=(0,1)$ into the addition identities:

$$C(x+\alpha)=-S(x),\qquad S(x+\alpha)=C(x).$$

Applying twice changes both signs; applying four times returns both values. Hence

$$C(x+2\pi)=C(x),\quad S(x+2\pi)=S(x),\qquad C(x+\pi)=-C(x),\quad S(x+\pi)=-S(x).$$

To prove $2\pi=4\alpha$ is the smallest positive period, use the first-quarter signs and these quarter-turn relations. On successive open quarters the signs of $(C,S)$ are $(+,+),(-,+),(-,-),(+,-)$. At the intermediate endpoints the pairs are $(0,1),(-1,0),(0,-1)$. Thus the pair $(1,0)$ does not recur for $0<x<4\alpha$. Any common positive period smaller than $4\alpha$ would force such a return at zero, impossible. A period of $S$ alone would also be a period of its derivative $C$; a period of $C$ likewise forces a period of $S=-C'$. Thus each function has that same fundamental period.

We now name $\sin x=S(x)$ and $\cos x=C(x)$. The curve $x\mapsto(C(x),S(x))$ covers the unit circle once during $[0,2\pi)$: continuity and strict first-quarter monotonicity cover that quadrant exactly, and the quarter-turn formulas cover the rest. Its speed is $\sqrt{(-S)^2+C^2}=1$. Under the usual $C^1$ arc-length formula its parameter is therefore oriented arc length on the unit circle, which identifies this analytic parameter with geometric radian measure. The algebraic construction and period proof do not depend on assuming that geometric interpretation first.

Continuous monotone inverses and principal branches

Let $I$ be a real interval and let $f:I\to\mathbb R$ be continuous and strictly monotone. Its image $J=f(I)$ is an interval: every value between two image values is attained between their preimages by the intermediate value theorem. Strict monotonicity makes $f:I\to J$ bijective. Its inverse is continuous relative to $J$. Indeed fix $a\in I$ and $\varepsilon>0$. If $I$ has a point below $a$, choose $l\in I$ with $a-\varepsilon<l<a$; this is possible by the interval property. If there is no such point, every inverse value is already at least $a$. Similarly, when there are points above $a$, choose $a<r<a+\varepsilon$ in $I$. The positive numbers $|f(l)-f(a)|$ and $|f(r)-f(a)|$ for the sides that exist have a positive minimum $\delta$; if neither side exists, $I=\{a\}$ and continuity is immediate. For $y\in J$ with $|y-f(a)|<\delta$, monotonicity places $f^{-1}(y)$ between the chosen brackets, on each side that exists. Thus $|f^{-1}(y)-a|<\varepsilon$. This argument works for increasing and decreasing functions and for included endpoints.

If $a$ is interior to $I$, $f$ is differentiable at $a$ and $f'(a)\ne0$, continuity and strict monotonicity give an open output interval around $b=f(a)$. For $y\ne b$ in that interval put $x=f^{-1}(y)$. Injectivity gives $x\ne a$, inverse continuity gives $x\to a$, and

$$\frac{f^{-1}(y)-f^{-1}(b)}{y-b} =\left(\frac{f(x)-f(a)}{x-a}\right)^{-1}\longrightarrow\frac1{f'(a)}.$$

The nonzero limit justifies the reciprocal and proves existence of the inverse derivative. No derivative continuity is required once strict monotonicity on the interval has been established.

For $u<v$ in $[-\pi/2,\pi/2]$, the addition formulas give $\sin v-\sin u=2\cos((u+v)/2)\sin((v-u)/2)>0$: the midpoint lies strictly between $-\pi/2$ and $\pi/2$, and $0<(v-u)/2\le\pi/2$. Thus sine is strictly increasing on this closed interval; its endpoint values are $-1,1$. For $0\le u<v\le\pi$, $\cos v-\cos u=-2\sin((u+v)/2)\sin((v-u)/2)<0$, since both sine factors are positive. Cosine is therefore strictly decreasing on $[0,\pi]$, and its endpoint values are $1,-1$. Thus their continuous principal inverses have domains $[-1,1]$ and ranges $[-\pi/2,\pi/2]$ and $[0,\pi]$, respectively. Tangent is continuous on $(-\pi/2,\pi/2)$ and is strictly increasing there: for $u<v$ in that interval, $\tan v-\tan u=\sin(v-u)/(\cos u\cos v)>0$, because $0<v-u<\pi$ and both cosine factors are positive. As its argument tends to the endpoints from within, its sine tends to $-1$ or $1$ and its cosine tends to zero through positive values. Hence tangent tends to $-\infty$ and $+\infty$. IVT gives its range $\mathbb R$ and a continuous inverse into that open interval. The inverse difference quotient therefore applies to all three branches at every interior input. The positive square-root signs in the following formulas follow from cosine being positive on the arcsine interior range and sine being positive on the arccosine interior range.

5. Tangent and inverse branches follow

Define $\tan x=S(x)/C(x)$ where $C(x)\ne0$. The quotient rule gives $\tan' x=1/C(x)^2>0$. On $(-\alpha,\alpha)=(-\pi/2,\pi/2)$, cosine is positive, and the endpoint limits of tangent are $-\infty,+\infty$. It is a bijection there onto $\mathbb R$, defining $\arctan$.

Sine is a bijection from $[-\pi/2,\pi/2]$ onto $[-1,1]$, defining $\arcsin$. Cosine is a decreasing bijection from $[0,\pi]$ onto $[-1,1]$, defining $\arccos$. The inverse derivative theorem on their open ranges gives

$$(\arcsin u)'=\frac1{\sqrt{1-u^2}},\quad (\arccos u)'=-\frac1{\sqrt{1-u^2}}\quad(|u|<1), \qquad (\arctan u)'=\frac1{1+u^2}\quad(u\in\mathbb R).$$

For arcsine, the branch cosine is positive and equals $\sqrt{1-u^2}$. For arccosine the branch sine is positive, with a negative derivative of cosine. These branch signs are part of the proof, not optional conventions. Endpoint inverse derivatives at $u=\pm1$ are not finite conclusions of this theorem.

Continuous real functions on a closed interval are uniformly continuous

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.

Let $a\le b$ be finite real numbers and let $f:[a,b]\to\mathbb R$ be continuous relative to this interval. Uniform continuity means that for every $\varepsilon>0$ there is $\delta>0$ such that all $x,y\in[a,b]$ with $|x-y|<\delta$ satisfy $|f(x)-f(y)|<\varepsilon$. Fix $\varepsilon>0$. For each $c\in[a,b]$, relative continuity gives $r_c>0$ such that $|f(x)-f(c)|<\varepsilon/2$ for interval points with $|x-c|<r_c$. The intervals $(c-r_c/2,c+r_c/2)$ cover the compact interval. Choose a finite subcover with centres $c_1,\ldots,c_k$, and set $\delta=\min_i r_{c_i}/2>0$. The cover is nonempty since $[a,b]$ is nonempty. Given $x,y$ as above, choose $i$ with $|x-c_i|<r_{c_i}/2$. Then $|y-c_i|<\delta+r_{c_i}/2\le r_{c_i}$, so the triangle inequality gives $|f(x)-f(y)|<\varepsilon/2+\varepsilon/2=\varepsilon$. The same $\delta$ works for all points. This includes a singleton interval and uses the actual finite-cover compactness theorem, rather than citing an unproved scalar Heine–Cantor assertion.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Fundamental theorem of calculus

Theorem

Integral and derivative

If $f$ is continuous on $[a,b]$, the accumulation function $F(x)=\int_a^x f(t)dt$ is differentiable on $(a,b)$ with $F'(x)=f(x)$. If $G$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $G'(x)=f(x)$ for $x\in(a,b)$, then $\int_a^b f(x)dx=G(b)-G(a)$.

Proof

Average-value estimate

For small nonzero $h$ with $x+h\in[a,b]$,

$$\frac{F(x+h)-F(x)}h-f(x)=\frac1h\int_x^{x+h}[f(t)-f(x)]dt.$$

Continuity makes the absolute value of the right-hand side arbitrarily small, including for negative $h$ with oriented-integral convention. Thus $F'=f$. Both $G$ and $F$ are continuous on $[a,b]$, and $G-F$ has derivative zero on $(a,b)$, so it is constant by the mean value theorem; evaluating endpoints gives the second conclusion.

Worked example

Integrate a continuous function

For $f(x)=2x+1$ on $[0,2]$, $G(x)=x^2+x$ is an antiderivative. The theorem gives $\int_0^2(2x+1)dx=G(2)-G(0)=6$. The antiderivative shortcut rests on the theorem, not the definition of an integral.

Integration by parts under the actual continuous-derivative assumptions

Let $u,v$ be real $C^1$ functions on the closed interval with endpoints $a,b$. The functions $u'v$ and $uv'$ are continuous and therefore properly Riemann integrable. For nonzero real $h$ the product difference quotient is

$$\frac{u(t+h)v(t+h)-u(t)v(t)}h=u(t+h)\frac{v(t+h)-v(t)}h+v(t)\frac{u(t+h)-u(t)}h\longrightarrow u(t)v'(t)+v(t)u'(t).$$

Continuity of $u$ and existence of both derivatives justify this limit, proving $(uv)'=u'v+uv'$. The already proved continuous-function fundamental theorem, applied to $uv$, and linearity of the proper integral give

$$u(b)v(b)-u(a)v(a)=\int_a^b(u'v+uv')\,dt,$$

and hence

$$\int_a^b u(t)v'(t)\,dt=u(b)v(b)-u(a)v(a)-\int_a^b u'(t)v(t)\,dt.$$

For reversed endpoints use oriented integrals, and for equal endpoints all terms are zero. This is sufficient for the Taylor induction in U15: for fixed real $x$, take $u(t)=(x-t)^n/n!$ and $v(t)=f^{(n)}(t)$ when $f\in C^{n+1}$ on the segment. It does not claim integration by parts for merely differentiable functions with unproved integrability of their derivatives.

The order-$n$ Taylor polynomial at $a$ is $T_n(x)=\sum_{k=0}^nf^{(k)}(a)(x-a)^k/k!$, whose degree may be smaller than $n$ if leading derivatives vanish.

Theorem: Taylor's integral remainder

Let $n\ge0$ and let $f$ have continuous derivatives through order $n+1$ on the closed interval between $a$ and $x$. Then

$$f(x)=T_n(x)+\frac1{n!}\int_a^x f^{(n+1)}(t)(x-t)^n\,dt.$$

Consequently, if $|f^{(n+1)}(t)|\le M$ there, then $|f(x)-T_n(x)|\le M|x-a|^{n+1}/(n+1)!$.

Proof. At $n=0$, the identity is the fundamental theorem $f(x)-f(a)=\int_a^xf'(t)dt$. For $n\ge1$, integrate the remainder by parts, taking $u=(x-t)^n/n!$ and $dv=f^{(n+1)}(t)dt$. The boundary at $x$ is zero and at $a$ contributes $-f^{(n)}(a)(x-a)^n/n!$. The remaining integral is the order-$(n-1)$ remainder. Induction therefore yields the identity. This uses oriented integrals, so it also holds when $x<a$. Taking absolute values, reversing integration limits if needed, and integrating $|x-t|^n$ gives the stated bound. Continuity of the highest derivative makes every use of parts and the bound valid.

This theorem provides a sufficient regularity setting, not the weakest possible hypotheses. For a Taylor series to represent $f$ at a fixed $x$, the remainders must tend to zero as $n\to\infty$. A finite-order error bound alone does not prove that without control of its dependence on $n$.

Worked identification and error control

The differential system gives successive derivatives of sine as $S,C,-S,-C$ and of cosine as $C,-S,-C,S$. All have magnitude at most one. Taylor's theorem at zero therefore bounds any order-$N$ remainder by $|x|^{N+1}/(N+1)!$, which tends to zero for each fixed $x$ by eventual geometric ratio comparison. Their power-series definitions agree with their Taylor representations on all of $\mathbb R$. Here remainder control supplies the representation theorem rather than relying on the false general assertion that every smooth function equals its Taylor series.

Graded proof and domain tasks

1. Construction. Explain why differentiating the defining series is justified on every bounded interval. What extra information beyond pointwise convergence is used?
2. Uniqueness. If a twice-differentiable real function satisfies $u''+u=0$, $u(0)=A$, $u'(0)=B$, prove $u(x)=AC(x)+BS(x)$ without assuming a general ODE uniqueness theorem.
3. Period audit. Explain why proving a return at $2\pi$ alone would not prove it is the fundamental period. Supply the missing sign/initial-value argument.
4. Inverse boundary. Justify the arcsine derivative sign and show why no finite one-sided derivative exists at input one.
5. Taylor rigor. Prove $|x|^{N+1}/(N+1)!\to0$ for fixed $x$, then explain why the same factorial argument cannot automatically be applied to an arbitrary smooth function's derivative bounds.
6. Branch diagnosis. Explain why $\arcsin(\sin x)$ cannot equal $x$ for all real $x$, even though $\sin(\arcsin u)=u$ for every $u\in[-1,1]$.

Full solutions

1. On a fixed $[-M,M]$, the function series and their differentiated polynomial terms admit convergent factorial majorants. The M-test gives uniform convergence of the derivative series, and the partial sums converge at zero. The uniform derivative theorem therefore identifies the derivative of the limit. Pointwise convergence alone would not justify that interchange.
2. Let $v=u-AC-BS$. It has $v''=-v$ and zero initial value and initial derivative. Differentiate $v^2+(v')^2$ to obtain $2vv'+2v'v''=0$. Its constant value is zero, so $v=v'=0$ everywhere, proving the claimed expression. Existence is independently verified by differentiating that expression.
3. A function can return after several smaller periods. Here the quarter-turn sign table and endpoint values show $(C(t),S(t))\ne(1,0)$ for every $0<t<2\pi$. A smaller common period would force equality to that initial pair. A period of either function also shifts its derivative, so it is common to both. Hence there is no smaller individual period either.
4. On the chosen arcsine branch, $C>0$, so the inverse derivative is $1/C=1/\sqrt{1-u^2}$ for $|u|<1$. For $u<1$, the mean value theorem applied to arcsine on $[u,1]$ gives its secant slope as $1/\sqrt{1-c^2}$ for some $c\in(u,1)$. As $u\uparrow1$, every such $c\to1$, so the secant slopes tend to $+\infty$, not to a finite derivative.
5. For $x\ne0$, the ratio of consecutive positive bounds is $|x|/(N+2)$, eventually at most $1/2$, so they are bounded by a geometric sequence tending to zero. For $x=0$, every bound is zero. For an arbitrary smooth function, the bound for its $(N+1)$st derivative may itself grow with $N$ fast enough to defeat the factorial; no common magnitude-one bound was proved. Smoothness by itself is insufficient.
6. Arcsine always outputs a value in $[-\pi/2,\pi/2]$, so it cannot return arbitrary real inputs outside that interval. Its defining inverse identity holds in the other order because sine restricted to that interval is bijective onto $[-1,1]$. Periodicity already prevents a global inverse of unrestricted sine.

Summary and terms

An analytic construction defines functions by convergent expressions and proves their laws. An energy identity makes a squared difference constant and supplies uniqueness. A fundamental period is the least positive period, requiring a minimality proof. A principal inverse branch restricts a periodic function to a bijection. The construction establishes existence, identities, period and branch conditions in that order, without circular reliance on geometric trigonometric formulas.