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The product of 0 and anything is $0$, and seems like it would be reasonable to assume that $0 Show that ∇· (∇ x f) = 0 for any vector field [duplicate] ask question asked 9 years, 5 months ago modified 9 years, 5 months ago I'm perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell).
Ah yeah my favorite language is 0|\\/|G |-|4xx0|2 !!!!111 : softwaregore
I heartily disagree with your first sentence How would you explain the There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also gadi's answer)
For all this, $0^0=1$ is extremely convenient, and i wouldn't know how to do without it
In my lectures, i always tell my students that whatever their teachers said in school about $0^0$ being undefined, we. It is possible to interpret such expressions in many ways that can make sense The question is, what properties do we want such an interpretation to have $0^i = 0$ is a good choice, and maybe the only choice that makes concrete sense, since it follows the convention $0^x = 0$
Is a constant raised to the power of infinity indeterminate Say, for instance, is $0^\\infty$ indeterminate Or is it only 1 raised to the infinity that is? Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number
It seems as though formerly $0$ was considered i.
The intention is if you have a number whose magnitude is so small it underflows the exponent, you have no choice but to call the magnitude zero, but you can still salvage the. In the context of limits, $0/0$ is an indeterminate form (limit could be anything) while $1/0$ is not (limit either doesn't exist or is $\pm\infty$) This is a pretty reasonable way to think about why it is that $0/0$ is indeterminate and $1/0$ is not However, as algebraic expressions, neither is defined
Division requires multiplying by a multiplicative inverse, and $0$ doesn't have one. So there is a sense in which the only really geometrically meaningful integral of $0$ is $0$ itself But your friend is still wrong, since the term integral in this context means antiderivative and not area function. My daughter is stuck on the concept that $$2^0 = 1,$$ having the intuitive expectation that it be equal to zero
