Zhidong Leong

Posted on Thursday, 06 February, 2003 - 08:55 am:

Is (-1) ^ Pi defined? Prove it.

Hauke Worpel

Posted on Thursday, 06 February, 2003 - 12:42 pm:

- 1 = e^{π i}

therefore $(- 1)^{π} = (e^{π i})^{π} = e^{π^{2} i}$

$= \cos (π^{2}) + i \sin (π^{2})$

which comes out to about $- 0.903 - 0.430 i$

David Loeffler

Posted on Thursday, 06 February, 2003 - 02:38 pm:

Hauke, this expression is multi-valued; its values are

$\cos ((2 n + 1) π^{2}) + i \sin ((2 n + 1) π^{2})$ for any $n \in Z$ .

David

Hauke Worpel

Posted on Friday, 07 February, 2003 - 01:27 am:

And since

π^{2}

is not an integer multiple of

π

, 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.

David Loeffler

Posted on Friday, 07 February, 2003 - 09:42 am:

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

Hauke Worpel

Posted on Friday, 07 February, 2003 - 11:19 pm:

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

n

s 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 '' $- 1^{π}$ '' is equal to any number of modulus 1 you can think of.''

Demetres Christofides

Posted on Saturday, 08 February, 2003 - 11:07 am:

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

(2 n + 1) π^{2}

never equals

2 m π

for any integers

m

n

Demetres

Hauke Worpel

Posted on Sunday, 09 February, 2003 - 12:59 am:

Moral of the story: If you're raising a number to a transcendental power, stick with the principal value.

Zhidong Leong

Posted on Monday, 10 February, 2003 - 01:50 pm:

Oh, I am totally lost. How did you get your 1st step in the 1st reply?

Marcos

Posted on Monday, 10 February, 2003 - 02:41 pm:

Zhidong, are you familiar with complex numbers?

Do you know Euler's formula $e^{i x} = \cos x + i \sin x$ ?

If so, put $x = π$ ...

This gives us $e^{i π} = - 1$

(In fact, putting $x = (2 n + 1) π$ for $x$ still gives $- 1$ which isbasically what the rest of the thread is about...)

Marcos