Friday, 8 April 2016

complex analysis - What if Euler's formula was defined differently?




At the start of my first year of university in a Foundations module, my lecturer stated Euler's formula for complex numbers, $e^{i \theta} = \cos\theta + i \sin\theta$. In passing he mentioned this is just a definition which conveniently works when you plug the numbers into the Taylor expansions and that you could actually define it differently (a number other than $e$). What would be the consequences of this? Is there any other number which gives interesting results?


Answer



Suppose you want the following desirable properties:




  1. $\operatorname{exp}:\theta \mapsto e^{i\theta}$ is smooth


  2. $\exp(x+y)=\exp(x)\cdot\exp(y)$





the first requirement is reasonable since we want the function to be nice, and the second is desirable, because it says most essentially that if we add angles,
we get



$(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))$



which after working out with some trigonometric identities reduces to
$\cos(x+y)+\sin(x+iy)$



so that adding angles does what we would expect on the unit circle.




For any function with these two properties: $f(0)=1$ and differentiating with respect to $x$ and letting $x=0$ gives
$$f^{\prime}(y) =f(y)\cdot f^{\prime}(0)$$



which solving returns an exponential function $Ce^{a \theta}$ where $a=f^{\prime}(0)$, although the requirement that $f(0)=1$ forces that $C=1$.



In this case, one could in principle define exponential functions with different bases, but this seems to me inconvenient.



The special thing about base $e$ is that the derivative at $0$ is precisely $1$, which is why we get the clean formulas.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...