一道数学题，需要高中数学知识

新用户

iⁱ =?

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Ein

1

新用户

不好意思，我的数学很糟糕，尽管有点喜欢。不过，这里倒是有一些数学高手，慢慢的时间长了，你就会发现的。   该题目的i 是复树里面的那个i，就是i2 = -1 你给的答案，能给出解题过程吗？

Ein

我是新手。不知道能否用MS Word的Equetion打出所有方程后贴上。 www.ddhw.com

新用户

哈，这个我没有试过。但是凭感觉我认为应该不能，特别一些复杂的数学表达式。   要不怎么我们在其他网站见到的所有的数学公式都是用图形表示。   我觉得最简单的办法就是，你在WORD输入好以后，把它转换成图形文件（.jpg或者.gif 文件）。然后发贴的时候，上传该图形文件就可以了。上传的链接在编辑器的上方： 免费上传图片、音频、视频等文件，请点击这里

有空想想

0.207879576 www.ddhw.com

新用户

答案正确，能否给出解法？ www.ddhw.com

Ein

答案是1和-1。 pdf文件无法上传（哪位高手能告诉我如何上传pdf文件？)。只能勉强重新打出。   First, we should assume that the real field (the set of all real numbers with the common addition and multiplication satisfying some axioms) has been extended to be the complex field, that is, assume that all common operations for real numbers are also valid for complex numbers, and 0 and 1 are still the zero element and the unit element respectively. Thus, ii is well defined.  www.ddhw.com Let ii = x. Then x2 =(ii)2 = (i2)i = (-1)i. Taking ith power in the both sides, we have x2i =((-1)i)i =(-1)-1 = 1/(-1) = -1. Thus, taking the square root in both sides, we obtain xi = (-1)1/2 = i and, therefore, (xi)i = ii = x. Noting that its left side is just x-1, from x-1 = x, we get x2 = 1. So, x = 1 or -1. www.ddhw.com

新用户

这里好像不能上传pdf文件，只能图形文件，所以你可以把公式的pdf文件转换成.gif or .jpg文件，就可以上传。   你的推倒：www.ddhw.com Let ii = x. Then x2 =(ii)2 = (i2)i = (-1)i. Taking ith power in the both sides, we have x2i =((-1)i)i =(-1)-1 = 1/(-1) = -1. Thus, taking the square root in both sides, we obtain xi = (-1)1/2 = i and, therefore, (xi)i = ii = x. Noting that its left side is just x-1, from x-1 = x, we get x2 = 1. So, x = 1 or -1.www.ddhw.com  www.ddhw.com 应该是 xi = (-1)1/2 = i    or    xi = (-1)1/2 = -i 也就是说，最终有两条path....一条是对的，一条是错的。www.ddhw.com  www.ddhw.com 假设 xi = (-1)1/2 = -i ，then  -xi = i , therefore, (-xi)i  = ii  => (-1) i x-1  = ii www.ddhw.com 根据你的第一个推倒： x2 =(ii)2 = (i2)i = (-1)iwww.ddhw.com 我可以得出：(-1) i x-1  = ii  => x2 x-1 = ii   www.ddhw.com 所以，x = ii  www.ddhw.com 很显然，回到了原来的表达式，所以，我们选取  -i 是正确的（xi = (-1)1/2 = -i）www.ddhw.com 所以，你所选取的 i 是不正确的 （xi = (-1)1/2 = i）   www.ddhw.com       www.ddhw.com 原贴： 文章来源: Ein® 于 2006-6-12 11:22:43 标题：回复：一道数学题，需要高中数学知识www.ddhw.com 答案是1和-1。 pdf文件无法上传（哪位高手能告诉我如何上传pdf文件？)。只能勉强重新打出。  www.ddhw.com First, we should assume that the real field (the set of all real numbers with the common addition and multiplication satisfying some axioms) has been extended to be the complex field, that is, assume that all common operations for real numbers are also valid for complex numbers, and 0 and 1 are still the zero element and the unit element respectively. Thus, ii is well defined.  www.ddhw.com Let ii = x. Then x2 =(ii)2 = (i2)i = (-1)i. Taking ith power in the both sides, we have x2i =((-1)i)i =(-1)-1 = 1/(-1) = -1. Thus, taking the square root in both sides, we obtain xi = (-1)1/2 = i and, therefore, (xi)i = ii = x. Noting that its left side is just x-1, from x-1 = x, we get x2 = 1. So, x = 1 or -1. www.ddhw.com   www.ddhw.com

Ein

有道理。我用了乘方和开方，很可能会产生增根。新新看看下面的求解过程怎么样?   Let x = ii. Then xx = (ii)^ (ii) = ii(i)^i = ii^(i+1) = xi+1. Taking logarithm in both sides, we have xlogx = (i+1)logx. So, either x = i+1, or logx = 0, i.e. x = 1. www.ddhw.com   以前从未涉足虚数的研究。没经验。多指教。 www.ddhw.com

Ein

Let x = ii. Then xx = (ii)^(ii) = ii(i^i) = ii^(i+1) = xi+1. Taking logarithm on both sides, we have xlogx = (i+1)logx. So, either x = i+1, or logx = 0, i.e., x = 1. www.ddhw.com

huxlnn

cos(PI/2)+isin(PI/2)=i e^(PI/2*i)=i PI/2*i=ln(i) -PI/2=i*ln(i)=ln(i^i) i^i=e^(-PI/2)=0.2078

有空想想

e^(i*a) = cos(a) + i*sin(a) 是学“数学物理方法”时学的了，没敢用。 www.ddhw.com

新用户

但是Ein用不同的方法，结果也不同，如果错，也不知道错在何处？   对不起。没发好。现重发。 Ein® www.ddhw.com