OpenEuphoria

Re: Suggestion for 2.5

• Posted by Martin Stachon <martin.stachon at worldonline.cz> Feb 21, 2003
• 417 views

Christian CUVIER wrote:
> > Some other indeterminate forms are :
> > 0^0 (although MS Windows Calculator says 0^0=1, my Casio says "Math error"
> > 
> 
> 1 is correct, because epsilon^0 is always 1.

Some problems arise when defining 0^0=1 :

    log(0^0) = 0*log(0) -- not defined

    (0^1)/(0^1) = 0^(1-1) = 0^0 = 1 -- but this is 0/0, indeterminate 

    natural n-th power of a is defined as:
        a^n = a*a^(n-1)
    This way,
        0^0=0*0^(-1)=0*inf -- indeterminate

There is also a proof using the fact that :
    lim_(x->0) 0^x = 0
    lim_(x->0) x^0 = 1

But I don't understand it (we haven't been taught using limits yet).


   Martin