I was going through some basic recap of complex numbers and in the book (M. Boas. Mathematical Methods in the Physical Sciences) she says we define $e^{ix}$ by the Taylor series with $x$ replaced by $ix$ and we define $\sin(z)$ by the exponentials $\displaystyle\left(\frac1{2i}\right)(e^{iz}-e^{-iz})$. Why does she say we define the functions to be that, that makes it sound like we are choosing that statement to be true rather than it being logically true. Can we not just replace the real number inputs with the complex inputs and see what happens? Sorry this is quite vague and ambiguous but could someone help clear my view on this? Sorry if this has been asked already this is my first post and I hope I am following any rules.

Edit: the example here is supposed to just illustrate the point of what it means to define something in Mathematics.

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    $\begingroup$ Can we not just replace the real number inputs with the complex inputs and see what happens? How would you do that? $\endgroup$ – Daniel R Aug 14 '14 at 15:15
  • $\begingroup$ I see we can't just replace x with ix in the exponential without using the series otherwise we couldn't make sense of it, is this why we define it? $\endgroup$ – Matt Gaskell Aug 14 '14 at 15:22

In high school one begins by defining $\sin$ for angles in the interval $\bigl[0^\circ, 90^\circ\bigr]$, by considering certain ratios in right triangles. Then one introduces a new angle measure by replacing degrees by corresponding arc lengths on the unit circle $S^1$. Now one has $\sin$ on the interval $\bigl[0,{\pi\over2}\bigr]$. But one readily recognizes that looking at the $y$-coordinates of the points on all of $S^1$ would allow to define a periodic function $\sin:\>{\mathbb R}\to[-1,1]$ with lots of desirable properties.

This is all well and fine, but does not answer the question how to compute the value $\sin t$ for given $t\in{\mathbb R}$. This is where calculus comes in. There are several approaches; and one of them is the following: Consider the map $${\rm cis}:\quad {\mathbb R}\to{\mathbb C}, \qquad t\mapsto e^{it}:=\sum_{k=0}^\infty{(it)^k\over k!}\ .$$ From basic properties of the exponential function (defined by its power series) it follows that ${\rm cis}$ winds ${\mathbb R}$ onto $S^1\subset{\mathbb C}$ in a locally isometric way, i.e., small intervals $[t_1,t_2]$ are mapped onto small arcs on $S^1$ having length $t_2-t_1$. But this means that $${\rm Re}\bigl(e^{it}\bigr)=\cos t, \quad {\rm Im}\bigl(e^{it}\bigr)=\sin t\ ,$$ where $\cos$ and $\sin$ have there original "geometric" meaning. This fact not only allows to compute $\cos t$ and $\sin t$ using a well convergent series, but in the first place gives a universal analytic description of these functions. It is only natural that we now define once and for all $$\cos t:={\rm Re}(e^{it}),\qquad \sin t:={\rm Im}(e^{it})\qquad (t\in{\mathbb R})\ ,$$ so that we from now on can work with these functions without referring to a cloudy notion of "angle", and get for free and without case distinctions their well known properties (addition theorems, etc.): These properties can immediately be derived from corresponding properties of the exponential function.


Taylor series for $e^x$ is: $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\infty$$ For $\sin x$ and $\cos x$ is: $$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\infty$$ $$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\infty$$ Putting $x=i\theta$ $$e^{i\theta}=1+\frac{i\theta}{1!}+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}\cdots\infty$$ $$e^{i\theta}=\left(1+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^4}{4!}+\cdots\infty\right)+\left((i\theta)+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^5}{5!}+\cdots\infty\right)$$ $$e^{i\theta}=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}+\cdots\infty\right)+i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}+\cdots\infty\right)$$ $$e^{i\theta}=\cos\theta+i\sin\theta$$ So $$e^{i(-\theta)}=\cos(-\theta)+i\sin(-\theta)$$ Now $$\begin{cases}e^{i\theta}=\cos\theta+i\sin\theta\\e^{-i\theta}=\cos\theta-i\sin\theta\end{cases}$$ So, $$\begin{cases}\sin\theta=\frac1{2i}(e^{i\theta}-e^{-i\theta})\\\cos\theta=\frac12(e^{i\theta}+e^{-i\theta})\end{cases}$$

  • $\begingroup$ I understand how to derive it that is simple just why do we use the term definition? $\endgroup$ – Matt Gaskell Aug 14 '14 at 15:29
  • $\begingroup$ @MattGaskell IMO there's no such definition, depends on context of book $\endgroup$ – RE60K Aug 14 '14 at 15:33
  • $\begingroup$ @MattGaskell I think the author meant "let ... equals ..." $\endgroup$ – RE60K Aug 14 '14 at 15:33

You can define the exponential function in many ways, one of those ways is a by a Taylor series. The other way you could define it is as a function that is equal to its own derivative, i.e. let such a function be $f(x)$, then it must satisfy the differential equation with inittial condition $f(0) = 1$.

$$ f'(x) = f(x) $$

From this equation you can construct the exponential function by construting $f(x)$ and $f'(x)$ simultaneously. First set $f(x) = 1$ (to satisfy the initial condition), but then to satisfy the diff. eqn. $f'(x)$ must be $1$, so you set $f(x) = 1 + x$, now to satisfy the de, $f'(x)$ must be $1 + x$, so then $f(x) = 1 + x + \frac{x^2}{2} $ ... If you continue with this process you arive for the infinite series representation of $f(x) = e^x$, because no finite series can be equal to its own derivative.

You can do the same for the sine function, its differential equation is

$$ f''(x) = -f(x), f(0) = 0, f'(0) = 1 $$

You can construct a function from the properties that you want the function to have, and you state the properties via a differential equation.

  • $\begingroup$ As a side note, is the exponential function the only function that is its own differential? $\endgroup$ – Matt Gaskell Aug 14 '14 at 16:16
  • $\begingroup$ if we use the differential equation and set the function f to be a power series, then the definition comes as a consequence. On the other hand, if you define the exponential function directly by the complex power series, the fact that it is the derivative of itself comes as the consequence. It depends on where you start. The power series is motivated from real analysis series. The ODE gives the definition of an analytic function that is the derivative of itself, which is nothing more the exponential function as a consequence! $\endgroup$ – user144410 Aug 14 '14 at 16:30
  • $\begingroup$ @MattGaskell f(x) = 0 is a trivial solution to the differential equation (if the initial condition was f(0)=0). Also you could change the initial condition, i.e. f(0) = C, that would shift the function by a constant. But up to a constant and not including 0, $e^x$ is the only such function. $\endgroup$ – vladimirm Aug 14 '14 at 17:33

since you are changing the domain of the function, you are creating a new object that you have to define (a relations on $\mathbb{C} \times\mathbb{C}$ that must satisfy some rule) such that your definition makes sense (i.e well defined).

If we are to approach the topic from the point of view of real numbers, we should not expect any relationship between the exponential function and the trigonometric functions $sin(x)$ and $cos(x)$. Indeed the Taylor expansions are quite similar and if one uses imaginary arguments it will be possible to get the Euler formula. This is not rigorous though because it doesn't prove that the power series make sense when we allow complex variables instead of reals! However, we can define them in the complex analysis setting. One way is the following "motivation": define the exponential function as the solution of the differential equation \begin{equation} f'(z) = f(z) \end{equation} with the initial value $f(0)=1$. And set $f(z) = a_0 + a_1z + \dots + a_nz^n +\dots$ To satisfy the differential equation we should have $f'(z) = a_1 + 2a_2z+ \dots + na_nz^{n-1} +\dots$, therefore \begin{equation} a_{n-1} = na_n \end{equation} and $a_0 = 1$. It then follows by induction that \begin{equation} a_n = \frac{1}{n!} \end{equation}

We get the solution \begin{equation} e(z) = 1 + \frac{z}{2!} + \frac{z^2}{2!} + \dots + \frac{z^n}{n!} + \dots \end{equation}

It is then follow the step of showing that this definition makes sense for each complex number $z$ and that this series converges in $\mathbb{C}$. The series converges absolutely for every $z$ and converges uniformly on every bounded subset of $\mathbb{C}$. This shows that it is continuous.

We can now define the trigonometric functions and get Euler's formula.


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