Some math questions

1. Mar 7, 2006

eljose

Hello it has been more than three months since i don,t post now here are my questions...

a)the number $$i=\sqrt{-1}$$ is it bigger or smaller than 0?..

b)we know we can define the Bernoulli numbers as the Taylor expansion of the function $$\frac{x}{e^{x}-1}$$ my question is how would we define for example the Bernoulli number $$B_{1/2}$$ ?..

2. Mar 7, 2006

chingkui

You cannot define "order" for complex number the "usual" way you define that for real number. For real number, the order relation has some special properties that the product of two positive numbers is positive, the product of two negative numbers is also positive, the product of a positive number and a negative is negative, the sum of negative numbers is negative and the sum of positive numbers is positive. That's the "usual" properties we look at when ordering real numbers. Can you define two sets "positive complex" and "negative complex" numbers in a meaningful way and get similar relations?

3. Mar 7, 2006

AlphaNumeric

As my 1st year lecturer once commented sounded very much like a football (that's soccer if you're American) chant : 'There's only one complete ordered field', that being the Reals.

http://mathworld.wolfram.com/BernoulliNumber.html implies there is no notion of fractional Bernoulli numbers, and that you're also using an 'old' definition.

4. Mar 7, 2006

matt grime

I can choose to define (total) orderings on C such that i is greater than 0 or less than zero. Surely that tells you to reexamine your question? In anycase, tying this into the other other posts, there is no way of defining an order that is consistent with the field structure (ie so that a>b and c>0 implies ac>bc, in particular look at squaring i)

5. Mar 8, 2006

HallsofIvy

Staff Emeritus
What reason do you have for thinking that Bernoulli numbers are defined for fractional index?

6. Mar 21, 2006

eljose

But another question let be the identity:

$$( \sqrt(-2)+1)(\sqrt(-2)-1)=-3$$ that can be re-written as:

$$\sqrt(-2)-1=\frac{-3}{\sqrt(-2)-1+2}$$

so from this we could obtain a continued fraction approach to (-2)^{0.5}