Misc. Questions

1. Mar 29, 2006

Just a few things that I've been thinking about lately:

Are there any taylor/power series that converge to logarithmic functions (f(x)=log(x), etc.)? How would you do this?

Is there any series that will graph the traditional bell curve? How would you do this?

I remember the derivation of "e^[i(pi)] = -1" involved a series that converged to e. What series was that again?

Thanks.

2. Mar 29, 2006

HallsofIvy

Staff Emeritus
Yes, there are Taylor series that converge to log(x) etc. but they are not defined for all x. Since ln x is not defined for x<= 0, The Taylor series for ln x around x= a will only converge for 0< x< 2a.

The "traditional bell curve" is given by $y= e^{-x^2}$.
Take the Taylor's series for ex:
$$1+ x+ \frac{1}{2}x^2+ ...+ /frac{1}{n!}x^n+ ...$$
and replace x by -x2:
$$1- x^2+ \frac{1}{2}x^4- \frac{1}{3!}x^6+ ...+ \frac{(-1)^n}{n!}x^{2n}+...$$

The series that converges to ex (not just e) is the Taylor's series for e I just mentioned:
$$e^x= 1+ x+ \frac{1}{2}x^2+ ...+ \frac{1}{n!}x^n+...[/itex] If you replace x by ix you get [tex]e^{ix}= 1+ ix- \frac{1}{2}(ix)^2+ \frac{1}{3!}+ ...$$
$$= 1+ ix- \frac{1}{2}x^2- \frac{1}{3!}ix^3+ ...$$
Separating that into real and imaginary parts gives
$$e^{ix}= (1- \frac{1}{2}x^2+...)+ i(x- \frac{1}{3!}x^3+ ...)$$
which you can recognize as being the Taylor's series for cos(x) and sin(x):
eix= cos(x)+ i sin(x). Since $cos(\pi)= -1$ and $sin(\pi)= 0$, that gives $e^{i\pi}= -1$.

3. Apr 3, 2006