Rigorous exponential logarithmic and trigonometric functions

A familiar graph is not a mathematical definition. We construct the elementary functions from convergent series, prove their algebraic and differential laws, and define their inverse branches only after proving the necessary ranges. This closes a different gap from applying a Taylor formula to a function whose existence and properties were already assumed.

1 Constructing the exponential

Define $E(x)=\sum_{n=0}^\infty x^n/n!$ for every real $x$. The ratio test gives absolute convergence. On each bounded interval, factorial majorants give uniform convergence of the series and its derivative series. The uniform derivative theorem therefore gives $E'=E$ and $E(0)=1$.

The array $x^my^n/(m!n!)$ is absolutely summable, because its absolute total is $E(|x|)E(|y|)<\infty$. The double-series Fubini theorem allows grouping by $m+n=k$. The binomial theorem then gives

$$E(x)E(y)=\sum_{k=0}^\infty\frac{(x+y)^k}{k!}=E(x+y).$$

In particular $E(x)E(-x)=1$, so $E$ never vanishes. Also $E(x)=E(x/2)^2>0$, hence $E'=E>0$ and the mean value theorem makes it strictly increasing. For $x\ge0$, the nonnegative series gives $E(x)\ge1+x$, so $E(x)\to\infty$ as $x\to\infty$. The reciprocal identity gives $E(x)\to0$ as $x\to-\infty$. Continuity and the intermediate value theorem show its range is exactly $(0,\infty)$.

We now write $\exp x=E(x)$ and $e=E(1)$. The usual real exponential laws and derivative have been proved from the definition, not assumed from a graph or a power notation.

2 Defining logarithms and real powers

Define $\log:(0,\infty)\to\mathbb R$ as the inverse of the continuous strictly increasing $E$. Its derivative is $1/x$. For a direct justification, apply the mean value theorem to $E$ between $\log x$ and $\log(x+h)$:

$$h=E(\xi_h)[\log(x+h)-\log x].$$

Continuity of the inverse makes $\xi_h\to\log x$, so the difference quotient tends to $1/E(\log x)=1/x$. The inverse is continuous because a continuous strictly monotone map onto an interval has a continuous inverse; this follows by bracketing the inverse value between images of two nearby domain points.

Since $\log1=0$, the fundamental theorem gives

$$\log x=\int_1^x\frac{dt}{t}\quad(x>0).$$

The exponential product law and injectivity prove $\log(xy)=\log x+\log y$ for positive $x,y$. For any real $a$, define $x^a=E(a\log x)$ on $x>0$; the chain rule gives derivative $ax^{a-1}$. This construction does not define arbitrary irrational powers of negative numbers in the real field.

For example, the limit $\log x/x\to0$ as $x\to\infty$ follows without an unproved logarithmic rule. Put $x=E(t)$ with $t\to\infty$. The exponential series gives $E(t)\ge t^2/2$ for $t>0$, hence $0\le t/E(t)\le2/t\to0$.

3 Defining sine and cosine by series

Define

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

As with $E$, factorial majorants control every derivative series on bounded intervals. Thus $S'=C$, $C'=-S$, $S(0)=0$, $C(0)=1$, with $S$ odd and $C$ even. Differentiating $S^2+C^2$ gives zero; its value at zero is 1, so

$$S(x)^2+C(x)^2=1.$$

To derive addition laws, note an elementary uniqueness fact: if $u'=-v$, $v'=u$ and both initial values vanish, then $(u^2+v^2)'=0$, forcing both functions to vanish. For fixed $y$, the shifted pair $(C(x+y),S(x+y))$ and the pair

$$(C(x)C(y)-S(x)S(y),\ S(x)C(y)+C(x)S(y))$$

satisfy the same differential system and agree at zero. Apply the uniqueness argument to their difference. The familiar addition formulas are therefore consequences of the series and derivatives, not prior trigonometric assumptions.

4 Defining pi and proving a fundamental period

The series at 2 satisfies $C(2)<1-2+2/3=-1/3$: after these terms, the alternating tail starts negative and has strictly decreasing magnitudes. Continuity therefore gives a positive zero below 2, while $C(0)=1$ gives an initial neighborhood with no zero. Let $\rho$ be the infimum of the positive zeros. A sequence of zeros approaching it and continuity prove $C(\rho)=0$ and $\rho>0$.

Before $\rho$, cosine is positive; a nonpositive value would produce an earlier zero by continuity. Therefore $S'=C>0$ there, so $S(\rho)>0$ and the unit-circle identity gives $S(\rho)=1$. Define $\pi=2\rho$. The addition laws yield

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

Four such shifts return the original pair, giving common period $4\rho=2\pi$. Its minimality needs a separate argument. On the four successive open quarter intervals, the signs of $(C,S)$ are $(+,+),(-,+),(-,-),(+,-)$; at intervening endpoints the values are $(0,1),(-1,0),(0,-1)$. Hence $(1,0)$ is not attained again for $0<x<4\rho$. Any smaller common period would force that return at zero. A period of $S$ is also a period of $C=S'$ by differentiating the shifted identity, and similarly for a period of $C$. Thus $2\pi$ is each function's least positive period.

Write $\sin=S$ and $\cos=C$. The parameterized unit-circle curve $(C(x),S(x))$ has speed 1. Strict first-quarter monotonicity and the quarter-turn identities make it cover the circle once during $[0,2\pi)$. Its parameter therefore agrees with oriented arc length on the unit circle, identifying the analytic parameter with geometric radian measure only after the construction.

For clarity, the arc-length identification uses an elementary C1-curve argument. If $\gamma:[a,b]\to\mathbb R^2$ is C1, define its length as the supremum of polygonal chord sums over finite partitions. The fundamental theorem and the integral norm inequality give each chord sum at most $\int_a^b\|\gamma'(t)\|dt$. Conversely, uniform continuity of $\gamma'$ gives a modulus $\omega(\eta)\to0$. On an interval of length $\Delta t$ in a partition of mesh $\eta$, the difference between its chord and $\gamma'$ at the left endpoint times $\Delta t$ has norm at most $\omega(\eta)\Delta t$. The difference of their lengths has the same bound. Summing shows the chord sums approach the Riemann sums for the speed integral, with error at most $(b-a)\omega(\eta)$. Thus the supremum equals that integral. For the constructed unit-circle curve the speed is one, so its positively oriented parameter measures exactly the arc length; the quarter-turn construction above gives one traversal before the least period.

5 Inverse branches and endpoint limits

On $[-\pi/2,\pi/2]$, sine is strictly increasing onto $[-1,1]$; on $[0,\pi]$, cosine is strictly decreasing onto that interval. These restrictions define $\arcsin$ and $\arccos$. The quotient $\tan=S/C$ has derivative $1/C^2>0$ on $(-\pi/2,\pi/2)$ and endpoint limits of opposite infinities, defining $\arctan$ on the whole real line.

Inverse differentiation, including the signs of the chosen branches, gives

$$\frac d{du}\arcsin u=\frac1{\sqrt{1-u^2}},\quad \frac d{du}\arccos u=-\frac1{\sqrt{1-u^2}}\quad(|u|<1),$$

and $(\arctan u)'=1/(1+u^2)$ for all real $u$. The first two formulas do not assert finite derivatives at endpoints. On the arcsine branch cosine is nonnegative and positive in the interior, fixing the positive square-root sign; this is not a freely chosen algebraic sign.

Exercises with complete solutions

E1 Construction. Explain why this double-series proof of the exponential product identity needs absolute double-series control, not just pointwise convergence of each factor's partial sums.

S1. Expanding the product introduces all pairs of term indices and grouping them by total degree. Absolute summability makes these row, column and diagonal groupings equivalent. Pointwise convergence alone does not justify such rearrangement; the double-series counterexample with different row and column sums shows the missing principle.

E2 Domain and derivative. Derive the derivative of $x^{\sqrt2}$ from its definition and state its real domain here.

S2. On $x>0$, define $x^{\sqrt2}=E(\sqrt2\log x)$. The chain rule gives $E(\sqrt2\log x)\sqrt2/x=\sqrt2 x^{\sqrt2-1}$. The construction supplies no real-valued definition for negative inputs.

E3 Uniqueness. Prove that $u''+u=0$, $u(0)=A$, $u'(0)=B$ implies $u=AC+BS$ without invoking a general ODE uniqueness theorem.

S3. Subtract $AC+BS$ to get $v''=-v$ with $v(0)=v'(0)=0$. The derivative of $v^2+(v')^2$ is zero, so it is identically its initial value zero. Both $v$ and $v'$ vanish, proving the formula. Direct differentiation verifies existence of the displayed solution.

E4 Inverse branch. Calculate $\arcsin(\sin\pi)$ and explain why unrestricted sine cannot have a global inverse.

S4. The quarter-turn identities give $\sin\pi=0$, and the inverse branch gives $\arcsin0=0$, not $\pi$. Periodicity sends many real inputs to each sine value; only a restricted one-to-one branch has an inverse.

E5 Integral identity. Prove $\int_a^b dt/t=\log b-\log a$ for $0<a<b$ from the constructed logarithm.

S5. The constructed logarithm has derivative $1/t$ on the positive interval, and that derivative is continuous on $[a,b]$. FTC part two therefore gives the endpoint difference. Neither zero nor a negative endpoint can be inserted without changing the real-domain problem.

Summary

Absolute series convergence constructs the exponential and its product law. Positivity and monotonicity make the logarithm a genuine inverse. Sine and cosine arise from their own convergent series; energy uniqueness yields addition, and a first-zero argument yields pi and a least period. Inverse formulas require branch restrictions and open-range derivative conditions.