x. This is, in fact, how e^{x} is traditionally defined for real numbers. The other properties can be proved from this one.
y=e^{x}, then y�=e^{x}. In general, if y=e^{mx}, then y�=me^{mx}.
e^{x+y}=e^{x}e^{y}, (e^{x})^{y}=e^{xy} and e^{-x}=1/e^{x}.
e^{x} is 1 + x + x^{2}/2 + x^{3}/6 + ... + x^{n}/n! + ....
e^{ax} is 1/(s-a).
e^{ix}.
## 2 Definition by IntegralWhen I want to work out
x, it's fairly clear what this means. After all, there's really only one way to move from 1 to x along the real number line. Well, we could vary our speed a bit. If u was a function u(t) of t, with u(t_{0})=1 and u(t_{1})=x, we could write the above integral as
u to t. It's drilled into us in first year calculus that this gives the same answer, and sure enough, it does.
So if we had a complex value of x, we could define a complex valued function u(t), with u(0)=1, u(1)=x, and work out the complex integral
u. Let's try one example, at least, to try to work out
u(t) I'll use will be u(t)=cos(pt/2)+isin(pt/2). If you check, you'll find that u(0)=1 and u(1)=i. If you differentiate it, you get u�(t) = (p/2)(-sin(pt/2)+icos(pt/2)). Then,
u�(t)/u(t) is a bit messy, until you notice a lot of the messy stuff cancels out. This is because
i×u(t) equals icos(pt/2)-sin(pt/2), so that u�(t) = (ip/2)u(t)!
So the integral we want is just
ip/2.
In fact, if you take u(t) = cos(xt)+isin(xt), for any x, you get
x)+isin(x))=ix? Or that that e^{ix} = cosx + isinx? Or does the integral depend on
the choice of u(t)?
Without going into further details about complex integrals here, let me say that
- The answer for ln
*x**almost*doesn't depend on the choice of*u*(*t*). It only depends on whether the path of*u*(*t*) passes the origin on the right or the left - or twists around it a few times. For example, choosing*u*(*t*) = (1-*t*) +*it*still gives ln*i*=*i*p/2, although the integral is a lot messier to solve. You get� �1 0(-1+ *i*)*dt*(1-*t*)+*it*= � �1 0(-1+ *i*)((1-*t*)-*it*)*dt*(1-*t*)^{2}+*t*^{2}= � �1 0((-1+ *i*)+2*t*)*dt*2*t*^{2}-2*t*+1 which is a messy, but doable, problem. Try it, and you'll get= 1 2� �1 0((-1+ *i*)+2*t*)*dt**t*^{2}-*t*+1/2, *i*p/2. Unless you make a mistake, of course, which (if you are anything like me) is highly likely. - Even if you zip around the origin a few times, the different values you can get for ln
*x*are all related to each other - if*a*and*b*are two different values of ln*x*calculated from two different paths, then*a*-*b*is an integer times 2p*i*, that is, either 0, or 2p*i*, or 4p*i*, or -2p*i*, or 1248p*i*, or whatever. Never anything else. - And for
*e*^{i}*x*? Well, it turns out that the only way to get ln*a*equal to*ix*is to make*a*=cos*x*+*i*sin*x*.
t from 1 to x, it seems that the "right" answer for e^{ix} is cosx + isinx. If we chose some other value, then it wouldn't fit the definition of e^{x} for real numbers x.
## 3 DerivativesWe know that the derivative ofe^{mx} is me^{mx}. So, the derivative of e^{ix} should be ie^{ix}, and the second derivative should be -e^{ix}. Therefore e^{ix} satisfies the
differential equation
x and sinx are also solutions to the very same differential equation. The general solution of a second order linear differential equation will be a linear combination of two independent solutions, so we could say the general solution of y"+y=0 is y=Acosx+Bsinx.
By "general" solution, we mean any solution can be written Acosx+ Bsinx. Now y=e^{ix} is a solution. Does this mean e^{ix}=Acosx + Bsinx for some A and B? Yes!
We can find the values of A and B by comparing the LHS and the RHS of e^{ix}=Acosx + Bsinx at particular values of x. Choosing x=0, for example, gives 1 = A + 0, so A=1. Differentiating both sides and then substituting x=0 gives ie^{0i} = -Asin0 + Bcos0, so i=B. Therefore, e^{ix} = cosx+isinx as before.
## 4 Exponent lawsIfe^{i}x=cosx+isinx, do the exponent laws still work? What is e^{i(x+y)}, for example?
According to the exponent law, we should have e^{i(x+y)}=e^{ix}e^{iy}. If we assume e^{ix}=cosx+isinx, we can use this formula to simplify both sides of the equation.
The left hand side becomes cos(x+y)+isin(x+y). Do you remember your trigonometry? If so, you'll remember that this is
x + isinx)(cosy + isiny). Multiplying this out, we again get
e^{ix}=cosx+isinx is consistent (at least this much) with the exponent law we've just tested.
If we tried (e^{ix})^{2}=e^{2ix} or (e^{ix})^{3}=e^{3ix} and so on, we'd get the same result.
## 5 Taylor seriesThe Taylor series fore^{x} is
x=1 gives the famous formula
x=1. We can substitute anything we like. Why not substitute ix?
i, we get
e^{ix} = cosx+isinx.
## 6 Laplace TransformsIf the Laplace Transform ofe^{ax} is 1/(s-a), doesn't that mean that the Laplace Transform of e^{ix} should be 1/(s-i)?
Just to refresh your memory, the Laplace Transform of f(x) is a function of s given by
f(x)=e^{ax}, this gives
e^{ix} is 1/(s-i), what about the Laplace transform of cosx + isinx?
Looking up a table of Laplace transforms tells me that the Laplace transforms of cosx and sinx respectively are
x+isinx is
s-i)? Well, we haven't simplified things yet. Remembering that s^{2}+1=(s-i)(s+i) means
that we can do some simplifying.
e^{ix} and cosx+isinx are equal, this means the functions are also equal.
## 7 SummaryIn each case, we took one property of the exponential function, extended it to the imaginary axis, and found the formulae^{ix} = cosx+isinx. In general, we say e^{u+iv} = e^{u} (cosv+isinv).
Since so many different ways of working out e^{ix} all give the same answer, it seems that maths is something real that mathematicians are busy discovering, rather than something artificial that mathematicians invent. Maybe this is related to the fact that maths is so useful for describing the real world?
As a post-scriptum, let me introduce the famous "Euler's formula"
e, p and i.
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