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(John, let me interject here that <math>i^i</math> has infinitely many possible values. Given a complex number <math>w</math>, there are infinitely many complex numbers <math>z</math> such that the complex exponential <math>e^z</math> is equal to <math>w</math>. Thus, the complex logarithm is infinitely valued. The symbol <math>i^i</math> represents <math>e^{i\log i}</math> for all possible values of <math>\log i</math>. Your puney little TI-84 merely told you the ''principal value'' of <math>i^i</math>. | (John, let me interject here that <math>i^i</math> has infinitely many possible values. Given a complex number <math>w</math>, there are infinitely many complex numbers <math>z</math> such that the complex exponential <math>e^z</math> is equal to <math>w</math>. Thus, the complex logarithm is infinitely valued. The symbol <math>i^i</math> represents <math>e^{i\log i}</math> for all possible values of <math>\log i</math>. Your puney little TI-84 merely told you the ''principal value'' of <math>i^i</math>. | ||
| − | The complex logarithm is given by <math>\log z = \ln|z|+i\theta</math> where <math>\theta</math> is any angle such that <math>z=|z|e^{i\theta}</math>. Using this, you can show that all the possible values of <math>\log i</math> are <math>i\frac{\pi}{2}+2\pi in</math> as <math>n</math> ranges over all the integers. OK, you may now continue. --[[User:Bell|Bell]] 00:18, 27 October 2008 (UTC) ) | + | The complex logarithm is given by <math>\log z = \ln|z|+i\theta</math> where <math>\theta</math> is any angle such that <math>z=|z|e^{i\theta}</math>. Using this, you can show that all the possible values of <math>\log i</math> are <math>i\frac{\pi}{2}+2\pi in</math> as <math>n</math> ranges over all the integers. OK, you may now continue. --[[User:Bell|Steve Bell]] 00:18, 27 October 2008 (UTC) ) |
Which was weird enough. I felt like I had seen that last number before, but I couldn't quite place it. I searched for the number itself and eventually found that it was equal to <math> e^{\pi/2} </math>, which obviously implicated Euler's. It then occurred to me that, if | Which was weird enough. I felt like I had seen that last number before, but I couldn't quite place it. I searched for the number itself and eventually found that it was equal to <math> e^{\pi/2} </math>, which obviously implicated Euler's. It then occurred to me that, if | ||
Revision as of 07:43, 9 November 2008
As hard as it is to believe or even interpret, the following is, in fact, true:
$ i^i e^\frac{\pi}{2} = 1 $
You can probably already guess that this has to do with Euler's Formula. The fact can be proven if you understand the natural log of $ i $, something that also follows from Euler's. --John Mason
Fascinating. You care to show the math or are you just going to tell us the fact and leave it at that? His Awesomeness, Josh Hunsberger
It all started as me trying to swallow imaginary numbers. At first, I took $ i^x $ and graphed it, which helped me understand Euler's formula. Then I had to wonder what an imaginary power would act like. As an imaginary power causes a real base to act like trig functions, an imaginary power should, possibly, cause an imaginary base to act like an exponential function. Curious to see what good 'ole TI-84 had to say, I found that
$ i^i = .207879574... $
$ i^{-i} = 4.810477381... $
(John, let me interject here that $ i^i $ has infinitely many possible values. Given a complex number $ w $, there are infinitely many complex numbers $ z $ such that the complex exponential $ e^z $ is equal to $ w $. Thus, the complex logarithm is infinitely valued. The symbol $ i^i $ represents $ e^{i\log i} $ for all possible values of $ \log i $. Your puney little TI-84 merely told you the principal value of $ i^i $.
The complex logarithm is given by $ \log z = \ln|z|+i\theta $ where $ \theta $ is any angle such that $ z=|z|e^{i\theta} $. Using this, you can show that all the possible values of $ \log i $ are $ i\frac{\pi}{2}+2\pi in $ as $ n $ ranges over all the integers. OK, you may now continue. --Steve Bell 00:18, 27 October 2008 (UTC) )
Which was weird enough. I felt like I had seen that last number before, but I couldn't quite place it. I searched for the number itself and eventually found that it was equal to $ e^{\pi/2} $, which obviously implicated Euler's. It then occurred to me that, if
$ e^{\pi/2} = i^{-i} $
then
$ \frac{\pi}{2} = -i \ln i $
Crazy! I had looked into the complex domain of natural log before, but apparently it's not a simple topic. However, I did, as a result, happen upon a fact that is always true about the natural logarithm. Now it was a matter of proving it. I could use Euler's, and take the log of both sides.
$ e^{ix} = \cos x + i \sin x $
And where is cosine (the real portion) equal to 0?
$ x = \frac{\pi}{2} $
$ e^{i\frac{\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} $
$ e^{i\frac{\pi}{2}} = i $
$ i\frac{\pi}{2} = \ln i $
Ah-ha! The natural log of i. A little more manipulation:
$ \frac{\pi}{2} = \frac{\ln i}{i} = \frac{\ln i}{i} \frac{i}{i} = - i \ln i $
And there it is. You can have even more fun finding the imaginary portions of hyperbolic and trigonometric functions. --John Mason
After today's lecture, I can readily see why $ i^i $ has infinitely many values, as there are an infinite number of ways to cause cosine to equal 0, all along that pattern. Weirder still, but starting to make sense. --John Mason
