Understanding the Controversy: Simplifying 0^0 in Calculus

  • Thread starter Thread starter Null_
  • Start date Start date
Null_
Messages
227
Reaction score
0

Homework Statement



It's a word problem, but I just need to simplify 0^0.

Homework Equations



n/a

The Attempt at a Solution



0^n = 0.
n^0 = 1.




I'm just going through a calculus book on my own this summer. It's not even a calculus question, but I've never come across this before. Which is it? 0 or 1? Why?
 
Physics news on Phys.org


I believe it's 0 (or not even defined, such as in poles of a polynom), because from what I can remember n^0 = 1 is by definition only true for those n that are different from 0.
 
Last edited:


0^0 is undefined.
 


From Wolfram Mathworld http://mathworld.wolfram.com/ExponentLaws.html

The definition 0^0=1 is sometimes used to simplify formulas, but it should be kept in mind that this equality is a definition and not a fundamental mathematical truth (Knuth 1992; Knuth 1997, p. 56).
 


Okay, thanks guys. Wolfram is awesome!
 


Consider the binomial expansion:

1 = 1^2=(x+(1-x))^2 =\binom 2 0 x^0(1-x)^2+\binom 2 1 x^1(1-x) + \binom 2 2 x^2(1-x)^0

This doesn't work for x = 0 or 1 unless you define 00 = 1. That is an example of why it is usually defined as 1. Also why 0! = 1 in the same example.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top