Hwhm

#1
I need some help on this problem.

Find the half width at half maximum (HWHM) for the density

f(x)=1/2 exp (-abs(x))

I understand that f(x_1/2)=1/(sigma(sqrt(2pi)))exp(-(x_1/2-m)^2/2sigma^2 and that abs(x_1/2-m)=sigma(sqrt(2log2)^1/2).

However, I don't understand what I need to do with the given f(x) to find the HWHM. Do I need to find m for f(x) and then plug in? I'm very confused.

Any help on this is greatly appreciated.

Thanks.
 

Dragan

Super Moderator
#2
I need some help on this problem.

Find the half width at half maximum (HWHM) for the density

f(x)=1/2 exp (-abs(x))

However, I don't understand what I need to do with the given f(x) to find the HWHM. Do I need to find m for f(x) and then plug in? I'm very confused.

The max of the Double Exponential distriubtion (fx) is 1/2 given the way it's scaled.

Half of this max is 1/4.

Solve fx - 1/4 = 0. Use these solutions (+ - Ln[2]).