The product of 0 and anything is $ 0 $, and seems like it would be reasonable to assume that $ 0 ! 之所以现在普遍用 0=\emptyset，主要原因是，0和空集似乎有天然的关系。 空集，不就是不包含任何元素的集合，或者说，有0个元素的集合吗。 Then it depends on what you count as first principles. · 四舍五入小数点 round 函数将数字四舍五入到指定的位数。 语法：=round (number, num_digits) number必需参数。 要四舍五入的数字。 num_digits必需参数。 要进行四 … If were dealing with the natural numbers, this follows from the peano axiom that the successor of a natural number is not 0 (1 being defined as the successor. · 0!等于几？说的简单一点是认为规定的，但它是有道理的，你想过没有，为什么不规定0！=0呢？因为阶乘是一个递推定义，n!=n* (n-1)!,那么必然有一个初值需要人为规定。我 … Even under ieee-754. 知乎，中文互联网高质量的问答社区和创作者聚集的原创内容平台，于 2011 年 1 月正式上线，以「让人们更好的分享知识、经验和见解，找到自己的解答」为品牌使命。知乎凭借认真、专业 … F (x,y)=e^ {ylnx} 当然这里仅考虑了x>0的情形，当x<0时可类似考察。 那么我们的问题：0的0次方是什么东西就转化成了这个二元函数在原点（0,0）的连续性问题，如果我们取点p趋向于原 … In my lectures, i always tell my students that whatever their teachers said in school about $ 0 ^ 0 $ being undefined, we. · is a constant raised to the power of infinity indeterminate? · you can also prove it by moving the space: 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. Say, for instance, is $ 0 ^\\infty$ indeterminate? @swivel but 0 does equal - 0. The derivative of the complex-valued sine function is everywhere well-defined. · in the context of natural numbers and finite combinatorics it is generally safe to adopt a convention that $ 0 ^ 0 =1$. = 1 $\leftrightarrow$ 0 != 1, which is computer notation for 0 $\neq$ 1 :-). Or is it only 1 raised to the infinity that is ? I am just curious. Your title says something else than infinity times zero. Im perplexed as to why i have to account for this condition in my factorial function (trying to learn haskell). It is also an indefinite form because $$\infty^ 0 = \exp ( 0 \log \infty) $$ but $\log\infty=\infty$, so the argument of the exponential is the indeterminate form zero times infinity discussed at the beginning. 在没有学习0和负 整数 之前，我们都知道： 1，3，5，7，9…叫 奇数. However, as algebraic expressions, neither is defined. Theres the binomial theorem (which you find too weak), and theres power series and polynomials (see also gadis answer). The only reason ieee-754 makes a distinction between + 0 and - 0 at all is because of underflow, and for +/- ∞, overflow. For all this, $ 0 ^ 0 =1$ is extremely convenient, and i wouldnt know how to do without it. Extending this to a complex arithmetic context is fraught with risks, as is the ambition to justify limits of this form generally by analogy to the value of a particular limit of this form. This is a pretty reasonable way to think about why it is that $ 0 / 0 $ is indeterminate and $1/ 0 $ is not. · in the context of limits, $ 0 / 0 $ is an indeterminate form (limit could be anything) while $1/ 0 $ is not (limit either doesnt exist or is $\pm\infty$). The peano axioms for natural numbers take $ 0 $ to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take $ 0 $ to be a natural number. It says infinity to the zeroth power. Division requires multiplying by a multiplicative inverse, and $ 0 $ doesnt have one. This definition of the 0 -norm isnt very useful because (1) it doesnt satisfy the properties of a norm and (2) $ 0 ^ { 0 }$ is conventionally defined to be 1. I heartily disagree with your first sentence. Inclusion of $ 0 $ in the natural numbers is a definition for them that first occurred in the 19th century.
