Math: two exponent questions

Factual Questions
21

That’s because I cheated. I plugged j^j into matlab, and backed out the e^(-pi/2) bit. Here’s a derivation.

C = j^j
ln© = ln(j^j) = j * ln(j)
but
ln(j) = j*pi/2
so
ln© = -pi/2
C = exp(-pi/2)

panamajack, when I first derived it myself, I got the same answer as you. Maybe it’s a branch cut issue?

22

Take special case of Euler’s formula:

e[sup]i*pi[/sup] = -1

Take square root:

e[sup]i*pi/2[/sup] = i

Raise to power of i:

esup*i[/sup] = i[sup]i[/sup]

So e[sup]-pi/2[/sup] = i[sup]i[/sup]

That is, the imaginary unit i, raised to the power of i is a real number.

23

I wrote :

Why yes, I am, one more j to be precise. I figured the angle for ln j but failed to use Euler’s Formula correctly.

so ln j = j* pi/2 which yields the correct answer,
exp(-pi/2) for j[sup]j[/sup] for the method I showed. Though jcgmoi did it cleanly (and right the first time).

Dr. Matrix, Euler’s formula is :

e[sup]j*theta[/sup] = cos(theta) + j sin(theta)

It actually is a pretty simple formula if you understand what’s behind it, though I probably shouldn’t try and explain it right now considering how I mishandled it earlier.

24

What is it with you engineers using j for the square root of -1? As all good mathematicians and physicists know, the correct notation is i.

Now off to Great Debates to reside with the Java/C Curly Bracket Position War. :slight_smile:

Rick