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Fascinating. You care to show the math or are you just going to tell us the fact and leave it at that? [[User:Jhunsber|His Awesomeness, Josh Hunsberger]] | Fascinating. You care to show the math or are you just going to tell us the fact and leave it at that? [[User:Jhunsber|His Awesomeness, Josh Hunsberger]] | ||
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| + | It all started as me trying to swallow imaginary numbers. At first, I took <math>i^x</math> 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 | ||
| + | |||
| + | <math> i^i = .207879574...</math> | ||
| + | |||
| + | <math> i^{-i} = 4.810477381... </math> | ||
| + | |||
| + | 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 | ||
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| + | <math> e^{\pi/2} = i^{-i} </math> | ||
| + | |||
| + | then | ||
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| + | <math> \frac{\pi}{2} = -i \ln i </math> | ||
| + | |||
| + | 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. | ||
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| + | <math> e^{ix} = \cos x + i \sin x </math> | ||
| + | |||
| + | And where is cosine (the real portion) equal to 0? | ||
| + | |||
| + | <math> x = \frac{\pi}{2}</math> | ||
| + | |||
| + | <math> e^{i\frac{\pi}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} </math> | ||
| + | |||
| + | <math> e^{i\frac{\pi}{2}} = i </math> | ||
| + | |||
| + | <math> i\frac{\pi}{2} = \ln i </math> | ||
| + | |||
| + | Ah-ha! The natural log of i. A little more manipulation: | ||
| + | |||
| + | <math> \frac{\pi}{2} = \frac{\ln i}{i} = \frac{\ln i}{i} \frac{i}{i} = - i \ln i </math> | ||
| + | |||
| + | And there it is. You can have even more fun finding the imaginary portions of hyperbolic and trigonometric functions. --[[User:Jmason|John Mason]] | ||
Revision as of 14:59, 26 October 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... $
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
