标注i=1/12表示坡度，是百分比坡度表示法，是表示坡度最为常用的方法，是两点的高程差与其水平距离的百分比，计算公式是：坡度i ＝ (高程差/水平距离)x100%

2019年8月11日 · I am new to complex numbers and wondered, what is $i^{(1/2)}$ and how do I actually create a complex number, so that $i$ has a different power than $1$ in tuple form, like $(a, b) = a + b*i^2$? Is it just $(a, bi)$ ?

2009年6月27日 · 当n=1时，求和公式为1+2+3+...+n，这被称为高斯求和，其结果是n*(n+1)/2。 对于平方和，即n次幂为2的情况， 求和 公式 为n*(n+1)*(2n+1)/6。 立方和，即n次幂为3的情况， 求和 公式 是n^2*(n+1)^2/4。

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2008年4月29日 · 1、i++先赋值后自加，i--同理先赋值后自减 int a1= 2; //先赋值后自加 b1=a1; a1=a1+1; int b1 = a1++; //b1=2 a1=3 System.out.println(b1); System.out.println(a1); 2、++i先自加后赋值，--i同理先自减后赋值 ...

An imaginary number is the product of a real number and the imaginary unit i, [note 1] which is defined by its property i2 = −1. [1][2] The square of an imaginary number bi is −b2. For example, 5i is an imaginary number, and its square is −25. The number zero is considered to be both real and imaginary. [3]

2014年10月20日 · If we call (0, 1) "i", we are saying that [itex]i^2= (i)(i)= -1[/itex]. Then (a, b)= (a, 0)+ (0, b)= (a, 0) + b(0, 1) so a+ bi. Last edited by a moderator: Oct 20, 2014

I believe the proper definition of $i$ is $i^2 = -1$ and not $i = \sqrt{-1}$ as is commonly stated. It is useful in many deep areas of mathematics, but as a starting point, it serves as a solution to $x^2 + 1 = 0$. I recommend looking deeper into the theory of complex numbers, particularly the polar form. To help you understand further.

2013年1月15日 · The main reason that it takes so long to get to $1+1=2$ is that Principia Mathematica starts from almost nothing, and works its way up in very tiny, incremental steps. The work of G. Peano shows that it's not hard to produce a useful set of axioms that can prove 1+1=2 much more easily than Whitehead and Russell do.

2019年3月20日 · 感觉网上用立方差证明的思路太low了，今天偶然间在Miskcoo大佬的博客中看到了一种脑洞清奇通俗易懂的证明方法. 我们要求的是 Sn = ∑n i=1i2 S n = ∑ i = 1 n i 2，现在我们对 Cn = ∑n i=1i3 C n = ∑ i = 1 n i 3 来进行"扰动"。 首先列一个恒等式. n+1 ∑ i=1i3 = Cn+(n +1)3 ∑ i = 1 n + 1 i 3 = C n + (n + 1) 3. 这里有个骚操作是把前面的转化一下. n ∑ i=0(i +1)3 =Cn +(n+1)3 ∑ i = 0 n (i + 1) 3 = C n + (n + 1) 3. 然后就可以推柿子啦。