Originally posted by: lcoy
Someone told me 0^0 was undefined but I always thought that any number to the zeroth power was 1. So what does 0^0 equal?
It is most convenient to define it to be 1. You can write polynomials nicely that way.Originally posted by: lcoy
Someone told me 0^0 was undefined but I always thought that any number to the zeroth power was 1. So what does 0^0 equal?
d/dx (x ln x)=lnx+1<0 for small x, so x ln x is decreasing for small x, so tends to l in (-inf, 0] as x->0.Originally posted by: pcy
Lt {x -> 0, for positive x} x*x = 1 (but I can't prove this)
Originally posted by: CSMR
d/dx (x ln x)=lnx+1<0 for small x, so x ln x is decreasing for small x, so tends to l in (-inf, 0] as x->0.Originally posted by: pcy
Lt {x -> 0, for positive x} x*x = 1 (but I can't prove this)
Consider a sequence x_(n+1)=x_n/e tending to 0 with y_n=x_n ln x_n. y_n tends to l also.
y_(n+1)=x_n ((ln x_n) -1)/e=(y_n-x_n)/e
(x_n,y_n)->(0,l)
(x_(n+1),y_(n+1))=f(x_n,y_y), f given above, f continuous.
f(0,l)=(0,l)
So l=(l-0)/e, so l=0.
So e^(x ln x)=x^x tends to e^0=1.
Originally posted by: inspire
It doesn't seem like it can't be proven, so evaluating the arguements for defining 0^0=1 as you would a proof doesn't really make a whole lot of sense. What is probably a better point to argue is whether the extension would be appropriate in certain applied settings. Caclulus Limits generally make pretty decent approximations in practice. I wouldn't use it in algebra, but in analysis, it shouldn't make a difference. Like using deleted residuals to get t-statistics.
Originally posted by: bwanaaa
In simple words, The expression 0^0 is an example of an unusual kind of mathematics. The value, x^y can be examined and watching what happens to the value as each variable is changed independently. Keeping the (X) fixed at 0 and shrinking y from a real number, the value of the expression should remain 0. The square root of 0 is 0, the cube root, fourth root are all 0.
Likewise taking any x^0 is 1. Shrinking x to an infinitesimal value should continue to yield the number 1.
So, 0^0 is an expression whose value depends on how you calculate it.
Rather like the expression, 4 x 3 +2. Whether multiplication or addition is performed first, the results are different. By convention, we agree that multiplication precedes addition so we are really saying (4 x 3) +2.
Likewise, we have to agree first on the variable which takes precedence in calculating an exponent.
Does this make sense?
What axiom is 0^0=1 inconsitent with?Originally posted by: inspire
Said mathematically, 0^0 is an expression whose value is undefined. We can't define it mathematically in any one way because any definition would be inconsistent with the current axioms.
Originally posted by: TheoPetro
umm my TI-89 says its 1 and it never lies.