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# 偶函數一題 高斯平面兩題 (高中基礎數學請教)

偶函數一題 高斯平面兩題 (高中基礎數學請教)

001

請問第五選項如何快速判斷為偶函數?

在下使用-x帶入法

解了將近十分鐘

不知有無更快之方案

圖片參考：http://farm6.static.flickr.com/5290/5375231855_53d...

http://farm6.static.flickr.com/5290/5375231855_53d...

002 003

用極式好像做不出來

還請列位賜鑑

### 1 Answer

- 教書的Lv 610 years agoFavorite Answer
001

combining 1st and 3rd terms = x[1/(10^x-1)+1/2]=x[(10^x+1)/(10^x-1)]

a. show that [(10^x+1)/(10^x-1)] is odd ---not hard;

b. thus x[(10^x+1)/(10^x-1)] is even; [odd by odd --> even]

c. thus f(x) is even [even + even --> even]

002 & 003

w= (az+b)/(cz+d) , for a,b,c,d complex numbers satisfying ad-bc not 0 is the bilinear transformation from complex z-plane to complex w-plane. This type of transformation has the following properties:

a. It maps circles to circles, where "circles" include the straight lines as its limiting case [ radius of a line is infinity];

b. It maps 3 distinct points to 3 distinct points.

Both a. and b. are good exercises in complex analysis..

With these two properties in mind, we know the answers to 002&003 must be circle or straight line on w-plane. To find it out the equation for w, we may use the fact that 3 distinct points uniquely determine a circle (or a line).

In particular, in 002 if we select z1=1, z2=i, and z3=-i, which generates w1=0, w2=1+i, and w3=(1+i)/2? -----> It must be the line theta=pi/4 in polar form or arg z=pi/4 in complex form.

In 003, if we select z1=2-->w1=(2-i)/5, z2=-2-->w2=(6+3i)/5, z3=2i-->w3=(2+i)/3.---> a real circle in w-plane. I leave the job of finding radius and center to you.