Can Taylor Series Represent Logarithmic Functions and the Bell Curve?

[Quadratic]

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.

 

[HallsofIvy]

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

 

[Quadratic]

Thanks a bunch, Halls.