| Zhidong
Leong |
Is (-1) ^ Pi defined? Prove it. |
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| Hauke
Worpel |
-1=epi therefore (-1)p=(epi)p=ep2 i =cos(p2)+i sin(p2) which comes out to about -0.903-0.430 i |
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| David
Loeffler |
Hauke, this expression is multi-valued; its values are cos((2n+1)p2)+i sin((2n+1)p2) for any n Î Z. David |
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| Hauke
Worpel |
And since p2 is not an integer multiple of p, the sines and cosines can take any value at all, which means that every number of modulus 1 technically fulfils the equation. I have given the principal value, which I always find to be the most useful one. |
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| David
Loeffler |
Hauke, this is not true. Clearly the expression I gave can take only countably many values. These are dense in the unit circle, but they certainly do not constitute all of it. David |
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| Hauke
Worpel |
You might be right. I assumed that if you pick any point on the unit circle, you can also choose a value of n in your formula that will get me there, since I have (countably) infinite ns to choose from. I still think that's true, but I did not take into account for the uncountably many numbers that exist but cannot be described. I should have said ''-1p'' is equal to any number of modulus 1 you can think of.'' |
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| Demetres
Christofides |
David is right. You can get 'as close as you want' to any number of modulus 1 but you cannot get all of them. Not even all those you can think of. For example the above expression never equals 1 since (2n+1)p2 never equals 2mp for any integers m, n! Demetres |
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| Hauke
Worpel |
Moral of the story: If you're raising a number to a transcendental power, stick with the principal value. |
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| Zhidong
Leong |
Oh, I am totally lost. How did you get your 1st step in the 1st reply? |
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| Marcos |
Zhidong, are you familiar with complex numbers? Do you know Euler's formula ei x=cos x+i sin x? If so, put x=p... This gives us eip=-1 (In fact, putting x=(2n+1)p for x still gives -1 which isbasically what the rest of the thread is about...) Marcos |