by **Ben G** » Wed May 18, 2011 10:04 pm

Astrogirl wrote:

>

K^2 wrote:shining2k1 wrote:e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole

> numbers n

> e^(-2i*Pi*n) = 1 for all integer n, but what you actually get is only 1 for n=0.

> That doesn't go away on its own, and has to be part of definition of imaginary exponents.

> How does this have to be part of the definition of imaginary exponents? How would

> you decide which n you use? Don't say 0, I could just shift the index.

>

Okay, so a clarification of the ambiguity involved (for anyone who cares):

Euler's formula is that e^(ix) = cosx + isinx (e is Euler's constant, that is 2.71282..., i is the square root of -1 and pi is the ratio of circumference to diameter, that is 3.14159...). This means that e^(2pi*i) = cos(2pi) + isin(2pi) = 1. In fact, we can equivalently state that 1^n = [e^(2*pi*i)]^n = e^(2*pi*i*n).

So, now we have infinitely many ways of expressing any number by multiplying or dividing by 1 as many times as one pleases.

So, for example, "i" can be expressed as e^(pi*i/2), but it can also be expressed as e^(pi*i/2) * 1^n = e^(pi*i/2 + 2*pi*i*n), where n is any integer.

So far so good?

While the exponent of the expression for i has infinitely many potential values (corresponding to the choice of "n") there is only 1 value this expression represents, since we've been multiplying it by powers of 1. However, things change when we take the "i"th power of this expression:

i^i = [e^(pi*i/2 + 2*pi*i*n)] = e^(-pi/2 - 2*pi*n); n is any integer

this expression has infinitely many (non-equal) values, each of which is equally an answer to "what is i^i?" In fact, this is generally true when you take the complex or irrational power of a complex number.

If we were to stick with this convention, however, we would be stuck with a multivalued function, which in the context of practical computation can be unwieldy. So, often a single member from this solution set is chosen, and the one chosen is usually the one for which "n=0," which can mean different things in different contexts. In this context, we state that i^i = e^(-pi/2) is a representative, useful, selection from our solution set.

Astrogirl wrote:

> [quote="K^2"][quote="shining2k1"]e^(Pi/2-Pi/2 - 2*Pi*n)=e^(-2Pi*n)=1 for all whole

> numbers n[/quote]

> e^(-2i*Pi*n) = 1 for all integer n, but what you actually get is only 1 for n=0.

> That doesn't go away on its own, and has to be part of definition of imaginary exponents.[/quote]

> How does this have to be part of the definition of imaginary exponents? How would

> you decide which n you use? Don't say 0, I could just shift the index.

>

Okay, so a clarification of the ambiguity involved (for anyone who cares):

Euler's formula is that e^(ix) = cosx + isinx (e is Euler's constant, that is 2.71282..., i is the square root of -1 and pi is the ratio of circumference to diameter, that is 3.14159...). This means that e^(2pi*i) = cos(2pi) + isin(2pi) = 1. In fact, we can equivalently state that 1^n = [e^(2*pi*i)]^n = e^(2*pi*i*n).

So, now we have infinitely many ways of expressing any number by multiplying or dividing by 1 as many times as one pleases.

So, for example, "i" can be expressed as e^(pi*i/2), but it can also be expressed as e^(pi*i/2) * 1^n = e^(pi*i/2 + 2*pi*i*n), where n is any integer.

So far so good?

While the exponent of the expression for i has infinitely many potential values (corresponding to the choice of "n") there is only 1 value this expression represents, since we've been multiplying it by powers of 1. However, things change when we take the "i"th power of this expression:

i^i = [e^(pi*i/2 + 2*pi*i*n)] = e^(-pi/2 - 2*pi*n); n is any integer

this expression has infinitely many (non-equal) values, each of which is equally an answer to "what is i^i?" In fact, this is generally true when you take the complex or irrational power of a complex number.

If we were to stick with this convention, however, we would be stuck with a multivalued function, which in the context of practical computation can be unwieldy. So, often a single member from this solution set is chosen, and the one chosen is usually the one for which "n=0," which can mean different things in different contexts. In this context, we state that i^i = e^(-pi/2) is a representative, useful, selection from our solution set.