# Exponential Functions

The general exponential function $b^x$ for base $b > 0$ and real number $x$ (*) is defined as the function that satisfies the conditions $b^x > 0 \\ b^x\cdot b^y=b^{x+y} \\ b^1=b$ It follows that: $\prod_{k=1}^{N}b^{x_k}=b^{\sum_{k=1}^{N}x_k} \\ b^0=1 \\ b^{-x}=1/b^x \\ (ab)^x=a^x b^x \\ b^{m/n}=\sqrt[n]{b^m}=\left ( \sqrt[n]{b} \right )^m \\ b^x=\underset{n \to \infty}{\lim}b^{\left \lfloor xn \right \rfloor/n}$
(*) We will extend this definition to complex $x$, for which, we will find, that $b^x>0$ may not hold. Moreover, there is some ambiguity for non-integer $x$, as, for example, $4^{1/2}$ may be $2$ or $-2$.

# Some Exponential Inequalities

Let $b>0$. By a simple argument we find: $0 \leq \left ( b^{(y-x)/2}-1 \right )^2 \\ b^{(y-x)/2} \leq \frac{b^{(y-x)}+1}{2} \\ b^xb^{(y-x)/2}\leq b^x\left (\frac{b^{y-x}+1}{2} \right ) \\ b^{(y+x)/2} \leq \tfrac{1}{2}b^y+\tfrac{1}{2}b^x$ Suppose that $0 \leq \alpha,\beta \leq 1$ and that $b^\alpha\leq\alpha b + (1-\alpha) \\ b^\beta\leq\beta b + (1-\beta)$ Then $b^{(\alpha+\beta)/2} \leq \tfrac{1}{2}b^\alpha+\tfrac{1}{2}b^\beta \\ b^{(\alpha+\beta)/2} \leq \tfrac{1}{2}(\alpha b + (1-\alpha))+\tfrac{1}{2} (\beta b + (1-\beta)) \\ b^{(\alpha+\beta)/2} \leq \tfrac{\alpha+\beta}{2}b+(1-\tfrac{\alpha+\beta}{2})$ As $b^0=1\leq 0 \cdot b + (1-0)=1$, and $b^1=b\leq 1 \cdot b + (1-1)=b$, it follows that, for all dyadic fractions of the form $x=M/2^N$ for some whole numbers M and N with $0 \leq M \leq 2^N$: $b^x \leq x b + (1-x)$ Moreover, as all real numbers $0 \leq x \leq 1$ can be written as the limit $x=\underset{N \to \infty}{\lim} \frac{\left \lfloor x \cdot 2^N \right \rfloor}{2^N}$ It follows that $b^x \leq x b + (1-x)$ Holds for all real x in the interval $[0,1]$ for all $b>0$, with equality holding only at the extremes. It follows that $2^x < 1+x$. Additionally, $(1/2)^x < 1-x/2$. We may then make the following argument: for $0 < x < 1$ $x^2 > 0 \\ 1-x^2=(1+x)(1-x) < 1 \\ 1+x < \frac{1}{1-x} \\ (1/2)^x < 1-x/2 \\ 2^x > \frac{1}{1-x/2} \\ 2^x >{1+x/2} \\ 4^x >(1+x/2)^2 > 1+x$ Thus, we have $2^x < 1+x < 4^x$.

# Derivatives and Derivatives of Exponentials

The definition of a derivative of a function is: $\frac{\mathrm{d} }{\mathrm{d} x}f(x)=f'(x) \triangleq \underset{h \to 0}{\lim}\frac{f(x+h)-f(x)}{h}$ Thus, for an exponential, the derivative would be given by: $\frac{\mathrm{d} }{\mathrm{d} x}b^x\triangleq \underset{h \to 0}{\lim}\frac{b^{x+h}-b^x}{h}=b^x\underset{h \to 0}{\lim}\frac{b^{h}-1}{h}=b^x L(b)$ Where $L(b)=\underset{h \to 0}{\lim}\frac{b^{h}-1}{h}$, provided this limit exists. This limit can be proven to exist as follows: for $0 < q < 1$, and $0 < x$, for $y = q x < x$ by the derived inequality $(b^x)^q = b^{qx} < (b^x -1) q +1 \\ \frac{b^{qx}-1}{qx} < \frac{(b^x -1)}{x} \\ \frac{b^{y}-1}{y} < \frac{(b^x -1)}{x}$ Thus, the limit is monotonically decreasing (from the right, increasing from the left). Moreover, the limit is bounded from below and above (for |x| < 1), as $1-\tfrac{1}{b} < \frac{(b^x -1)}{x} < b-1$ Thus, the limit exists, and so $b^x$ is everywhere differentiable. As exponentials with $b > 0$ are eveywhere differentiable and thus continuous, we may take the limit for $h > 0$. From the above inequalities, we have $L(2)=\underset{h \to 0}{\lim}\frac{2^{h}-1}{h} < \underset{h \to 0}{\lim}\frac{1+h-1}{h}=1 \\ L(4)=\underset{h \to 0}{\lim}\frac{4^{h}-1}{h} > \underset{h \to 0}{\lim}\frac{1+h-1}{h}=1$ As the limits are decreasing and both are bounded below ( $L(2) > 1/2, \; L(4) > 1$), it follows that both limits converge. Thus $L(2) < 1 < L(4)$. As L is clearly continous, by the intermedate value theorem, it follows that there is some real number $2 < e < 4$ such that $L(e)=1$. Let us define this number $e$ to be that number that satisfies $L(e)=\underset{h \to 0}{\lim}\frac{e^{h}-1}{h}=1$ This implies that $\frac{\mathrm{d} }{\mathrm{d} x}e^x=e^x$ This is a defining feature of the number $e$. We may also notice that, for any real x, if $h \to 0$ then $xh \to 0$. Thus $\underset{h \to 0}{\lim}\frac{e^{xh}-1}{xh}=1 \\ \underset{h \to 0}{\lim}\frac{e^{xh}-1}{h}=x$ Thus $L(e^x)=x$. This implies that, by definition, $L(x)=\log_e (x)=\ln (x)$. Moreover, given the chain rule $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$, we find $\frac{\mathrm{d} }{\mathrm{d} x} L(e^x)=L'(e^x)e^x=1$ And thus $L'(x)=1/x$. This is a very helpful result. For example, by rewriting and using the chain rule, we find: $\frac{\mathrm{d} }{\mathrm{d} x} x^a=\frac{\mathrm{d} }{\mathrm{d} x} e^{aL(x)}=e^{aL(x)} \frac{a}{x}=a x^{a-1}$ A result that is otherwise difficult to establish in the general case. We may write the limit derived above in an equivalent way as $\underset{n \to \infty}{\lim}n \cdot (e^{x/n}-1)=x$ Which directly implies that $e^x=\underset{n \to \infty}{\lim} \left ( 1+\frac{x}{n} \right )^n$ Let us expand the above expression using the binomial theorem: $e^x=\underset{n \to \infty}{\lim} \left ( 1+\frac{x}{n} \right )^n \\ e^x= \underset{n \to \infty}{\lim}\sum_{k=0}^{n}\binom{n}{k}\left ( \frac{x}{n} \right )^k \\ e^x= \underset{n \to \infty}{\lim}1+\sum_{k=1}^{n}\frac{x^k}{k!}\prod_{j=1}^{k}\left ( 1-\frac{j-1}{n} \right )$ Clearly, in the limit, all the factors in the products from 1 to k go to 1. Thus, we find: $e^x=1+\sum_{k=1}^{\infty}\frac{x^k}{k!}$ It can be checked that this series converges for all real x by the ratio test. This is an extremely useful formula, and can be taken to be a more robust and easy-to-work-with definition for $e^x=\exp(x)$. Note this formula directly implies that: $e=\sum_{k=0}^{n}\frac{1}{k!}$ Note that, as a verification, we can check that $e^0=1=1+\sum_{k=1}^{n}\frac{0^k}{k!}$ and $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x=1+\sum_{k=2}^{n}\frac{kx^{k-1}}{k!}=1+\sum_{k=2}^{n}\frac{x^{k-1}}{(k-1)!}=1+\sum_{k=1}^{n}\frac{x^{k}}{k!}$ Which verifies the differentiation formula.

# Trigonometric Functions and Inequalities

The definitions of the basic trigonometric functions are given by Figure 1. The curve between points C and D is the set of points equidistant from A between the line segments $AC$ and $AD$, i.e. a circular arc. Let us call the length of this curve $L$. Then the standad definition for the basic trigonometic functions is given by: $\theta=\frac{L}{\overline{AD}} \\ \\ \sin(\theta)\triangleq \frac{\overline{BD}}{\overline{AD}}, \; \;\; \cos(\theta)\triangleq\frac{\overline{AB}}{\overline{AD}}, \; \;\; \tan(\theta)\triangleq\frac{\overline{BD}}{\overline{AB}}$ Using these, let us look at figure 2. This figure will serve to evaluate bounds on the trigonometric functions for small angles ($0 < \theta < 1$) Let us denote the length of the curve $BE$, which is a circular arc, by $L$. It is clear that $\overline{BD} < L < \overline{BF}$ (An alternative way to demonstrate this is through areas, as triangle ABD is a strict subset of sector ABE which is a strict subset of triangle ABF.) Using the definitions above, and defining $\theta=L/\overline{AB}$, we have: $\frac{\overline{BD}}{\overline{AB}}=\sin(\theta) < \frac{L}{\overline{AB}}=\theta < \frac{\overline{BF}}{{\overline{AB}}}=\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$ And so it follows that $\theta\cdot\cos(\theta) < \sin(\theta) <\theta$ It follows from the Pythagorean theorem that $\sin(\theta)^2+\cos(\theta)^2=1$ From which we find: $\cos(\theta)^2 > 1-\theta^2 > (1-\theta^2)^2$ The last inequality following from the fact that $0 < \theta < 1$. We thus find $1-\theta^2 < \cos(\theta) < 1 \\ \theta-\theta^3 < \sin(\theta) <\theta$ Let us now find the summation formulas for sine and cosine. These are easily found using the construction in figure 3. $RB=QA \;\;\;\;\;\;\;\;\;\; RQ=BA$ $\frac{RQ}{PQ}=\frac{QA}{OQ}=\sin(\alpha) \;\;\;\;\;\;\;\; \frac{PR}{PQ}=\frac{OA}{OQ}=\cos(\alpha)$ $\frac{PQ}{OP}=\sin(\beta) \;\;\;\;\;\;\;\; \frac{OQ}{OP}=\cos(\beta)$ $\frac{PB}{OP}=\sin(\alpha+\beta) \;\;\;\;\;\;\;\; \frac{OB}{OP}=\cos(\alpha+\beta)$ $PB=PR+RB=\frac{OA}{OQ}PQ+QA$ $\frac{PB}{OP}=\frac{OA}{OQ}\frac{PQ}{OP}+\frac{QA}{OP}=\frac{OA}{OQ}\frac{PQ}{OP}+\frac{QA}{OQ}\frac{OQ}{OP}$ $\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)$ $OB=OA-BA=\frac{OA}{OQ}OQ-\frac{BA}{PQ}PQ$ $\frac{OB}{OP}=\frac{OA}{OQ}\frac{OQ}{OP}-\frac{BA}{PQ}\frac{PQ}{OP}$ $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$

# Complex Numbers

Complex numbers can be defined and used in the usual way, namely, as algebraic objects with the symbol $i$ having the property that $i^2=-1$. Additionally, we can define the norm of a complex number as $|a+bi|^2=a^2+b^2$. Some simple theorems we will make use of: $(a+bi)\cdot (c+di)=(ac-bd)+i(ad+bc) \\ |(a+bi)\cdot (c+di)|=|a+bi|\cdot|c+di| \\ \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}$ Let us define the function $\mathrm{cis}(x)=\cos(x)+i\sin(x)$ This function has the property that $\mathrm{cis}(\alpha)\cdot \mathrm{cis}(\beta)= \left (\cos(\alpha)+i\sin(\alpha) \right ) \cdot \left(\cos(\beta)+i\sin(\beta) \right ) \\ \mathrm{cis}(\alpha)\cdot \mathrm{cis}(\beta)= \left (\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) \right ) + i\left(\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha) \right ) \\ \mathrm{cis}(\alpha)\cdot \mathrm{cis}(\beta)= \cos(\alpha+\beta) + i\sin(\alpha+\beta)$ And thus $\mathrm{cis}(\alpha)\cdot \mathrm{cis}(\beta)=\mathrm{cis}(\alpha+\beta)$. It follows by induction and the definition of the exponential that, for any natural number $n$: $(\mathrm{cis}(x))^n=\mathrm{cis}(nx)$ And, thus $\mathrm{cis}(x)=(\mathrm{cis}(x/n))^n$ Importantly, this is true in the limit of large $n$. We can always pick $n$ large enough to make $x/n$ as small as needed. Thus, we can use the inequalities derived above, namely: $\mathrm{cis}\left ( \frac{x}{n} \right )=1+i\frac{x}{n}-\frac{x^2}{n^2}g(x)$ Let $g(x)=g_r(x)+i g_i(x)$ where $g_r, g_i$ are real. Then $0 < g_r(x) < 1$ and $0 < g_i(x) < \tfrac{x}{n}$. Clearly, then, for $n > |x|$, $|g(x)|^2 < 1+\frac{x^2}{n^2} < 2$ And so $|g(x)| < 2$. Also important to note is that a generic complex number can be written as $z=a+bi=r \cdot\mathrm{cis}(\theta)$ Where $r=|z|$ and $\theta$ satisfies $r \cos(\theta)=a, \;\;\; r \sin(\theta)=b$. From the above geometric argument, assuming $a,b > 0$ we have $\sin(\theta)=\frac{b}{|z|} < \theta < \tan(\theta)=\frac{b}{a}$ From the fact that $(\mathrm{cis}(x))^n=\mathrm{cis}(nx)$, we find that $z^n=(a+bi)^n=r^n \cdot\mathrm{cis}(n\theta)$

# A Lemma for a Family of Limits

From the above we have, for $0 < x < 1$: $2^{x} < 1+x < 4^{x}$ Let $x=B/n^2$, for $B>0$ and sufficiently large $n$. Then $2^{B/n^2} < 1+\frac{B}{n^2} < 4^{B/n^2} \\ 2^{B/n} < \left (1+\frac{B}{n^2} \right )^n < 4^{B/n}$ In the limit of large $n$, $B/n \to 0$. As $2^0=4^0=1$, we have $\underset{n \to \infty}{\lim} \left (1+\frac{B}{n^2} \right )^n=1$ A similar argument applies to the case that $B < 0$. In fact, suppose B is complex, then: $\underset{n \to \infty}{\lim} \left (1+\frac{B}{n^2} \right )^n =\underset{n \to \infty}{\lim} \left |1+\frac{B}{n^2} \right |^n \mathrm{cis}\left ( n\theta \right )$ Where $\frac{1+\frac{B}{n^2}}{\left | 1+\frac{B}{n^2} \right |}=\mathrm{cis}(\theta)$ For sufficiently large $n$, the real part is always positive. It's clear that $-|B|/n^2 \leq b \leq |B|/n^2$, and so $-\frac{|B|}{n^2} \leq \theta \leq \frac{|B|}{n^2}$. It clearly follows that $-\frac{|B|}{n} \leq n\theta \leq \frac{|B|}{n}$. Thus, i nthe limit of large n, $n\theta \to 0$, and so $\mathrm{cis}(n\theta)\to 1$. Therefore, for all complex $B$: $\underset{n \to \infty}{\lim} \left (1+\frac{B}{n^2} \right )^n=1$ Finally, let us note that $1+\frac{A}{n}+\frac{B}{n^2}=\left ( 1+\frac{A}{n} \right )\frac{1+\frac{A}{n}+\frac{B}{n^2}}{1+\frac{A}{n}}= \left ( 1+\frac{A}{n} \right )\left ( 1+\frac{1}{n^2}\frac{B}{1+\frac{A}{n}} \right )$ For sufficiently large $n$, we have, then $\left |\frac{B}{1+\frac{A}{n}} \right | < 2|B|$ It follows from the above that $\underset{n \to \infty}{\lim}\left (1+\frac{A}{n}+\frac{B}{n^2} \right )^n=\underset{n \to \infty}{\lim}\left ( 1+\frac{A}{n} \right )^n$ Clearly this applies to any $B(n)$ such that, there is some M such that, for $n>M$, $|B(n)| < K$ for some real $K>0$.

# Euler's Formula and Identity

We recall the following from a previous section: $\mathrm{cis}(x)=(\mathrm{cis}(x/n))^n$ And, for sufficiently large $n$: $\mathrm{cis}\left ( \frac{x}{n} \right )=1+i\frac{x}{n}-\frac{x^2}{n^2}g(x)$ Where $|g(x)| < 2$. Combining yields: $\mathrm{cis}(x)=\left ( 1+i\frac{x}{n}-\frac{x^2}{n^2}g(x) \right )^n$ Equality must hold in the limit of large $n$, and so, using the above lemma, we have: $\mathrm{cis}(x)=\underset{n \to \infty}{\lim}\left ( 1+i\frac{x}{n} \right )^n$ Using the limit definition of $e^x$, this yields, at last, Euler's celebrated formula: $e^{ix}=\cos(x)+i\sin(x)$ This has the special case, by the definition of $\pi$ and the trigonometric functions: $e^{i\pi}+1=0$ Using the power series expansion for the exponential function, and equating realand imaginary parts yields the two power series expansions: $\cos(x)=1+\sum_{k=1}^{\infty}\frac{(-x^2)^k}{(2k)!} \\ \sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^k x^{2k+1}}{(2k+1)!}$