NAU POTW-2011.09.20

I have found another POTW website, which is called northern Arizona university.

http://www.cefns.nau.edu/Academic/Math/departmentActivities/POTW/

The problem says that:


A very tiny bug starts at the origin in the (x, y) plane. He walks to the right one unit and ends up at (1, 0). Then he makes a left turn and moves forward half a unit, reaching (1, 1/2). Then he turns left again and goes half again as far, reaching (3/4, 1/2). If he keeps doing this (turning left and going half again as far) again and again, what is his limiting position?


It maybe not easy to consider two dimension sequence, and also have the direction changing all the time. But the bug always turn left. That exactly the complex multiple ( multiple a number by i means turn the vector anti-clock wise in 90 degree.).

So the iterative function is z(n)=z(n-1)+(i/2)^(n-1). z(0)=0. 
z(n)-z(n-1) =(i/2)^(n-1) ....(n)
z(n-1)-z(n-2)=(i/2)^(n-2)....(n-1)
...
z(1)-z(0)=(i/2)^0 ...(1)


Add all the n equations get z(n) - z(0) = [(i/2)^n-1]/(i/2-1) . So z(n) = [(i/2)^n-1]/(i/2-1)
z(n) -> 1/(1-i/2), when n-> Infinity.
In coordinates expression is {4/5, 2/5}.


I plot the graph with Mathematica

lis = {Re[#], Im[#]} & /@ Table[((I/2)^n - 1)/(I/2 - 1), {n, 0, 18}];
ListPlot[lis, Joined -> True, PlotRange -> {{-0.1, 1.1}, {-0.1, 0.6}}]

problem 3 fall 2011 of purdue POW

To solve problem 3, the pigeonhole principle should be knew first. The pigeonhole principle states that if n items put into m pigeonholes with n>m, then there is at least one pigeonhole must contain more than one item. In Chinese version is called 抽屜原理( drawer principle), which replace the pigeonholes with drawers. Beside this definition, there is another definition such n is greater than k times of m, then there is at least one pigeonhole must contain more than k items.

The problem 3 says: There are nine points in the interior of a cube of side 1. Show that at least two of the points are less than sqrt(3)/2. Can sqrt(3)/2 be replaced by a smaller number?

The idea of how to solve the problem is to set up the pigeonhole, then put the points in to the holes. Since the  side of cube equals to 1, half of the side is 1/2. Then we can cut the cube into 8 small cube with side of 1/2. The diagonal of small cube is just sqrt(3). sqrt((1/2)^2+(1/2)^2+(1/2)^2)=sqrt(3)/2. The distance of any two points in the interior of the small cube is less than sqrt(3). Then we set up 8 pigeonholes. But there are 9 points, so at least one small cube contain more than two point. Then the problem is solved.

The further question is easy, but it is hard to explain with english.

容易知道sqrt(3)/2不可以被更小的數代替，其實最極端的放置方法是：將8個點放在大正方形的四個角，而第九個點就放在大正方形的中心位置，那麼他們的距離就剛好都是sqrt(3)/2。但是全部點都放在正方形內部（interior），所以如果sqrt(3)/2換成比sqrt(3)/2的數d，至少有一種方法可以使全部點兩兩之間的距離小於d。

fall 2010 problem 2

fall 2010 problem 2
Show that sin(x) >= x-(x^2)/pi.  x belongs to [0,pi].
This problem seems very easy, but spend a lot of my time to finish it.  But I post this solution due to the deadline.
let f(x)=sin(x)-x+(x^2)/pi, and going to prove f(x)>=0, which is a necessary and sufficient condition of statement.


[ English is not my first language, so it may hard to read my prove. Then you be better guess what I was going to say ]

pf:
let f(x)=sin(x)-x+(x^2)/pi. then let x=t+pi/2, t belongs to [-pi/2, pi/2].
g(t)=f(t+pi/2)=sin(t+pi/2)+((t+pi/2-pi/2)^2)/pi-pi/4. 
so g(t) defines on (-pi/2, pi/2), which is a even function.
Simplify g(t)=cos(t)+((t+pi/2-pi/2)^2)/pi-pi/4.
Here going to prove g(t) >= 0 on [0,pi/2].
Consider g'(t)= -sin(t)+2*t/pi. 

Jordan's inequality said sin(x)>= 2x/pi, for 0< x < pi/2.

g'(t) is strictly decreasing on the interval of (0,pi/2). Because g(t) -> 0 as t->pi/2. That means g(t)>=0 on the interval of (0,pi/2). 
Since g(t) is a even function on (0,pi/2). So g(t)>=0 on (-pi/2,pi/2). Then f(x)>=0.  
 #


I also think about the f'(t)=cos(x)-1+2x/pi= -2(sin(x/2))^2+2x/pi. but it seems hard to separate the interval [0, pi] into [0,pi/2] and [pi/2, pi]. Then prove f'(x) is greater or equal to zero on [0,pi/2],  and less or equal to zero on [pi/2,pi/2]. I think it is easy to follow. But that will look a little complicate to show. If there is somebody know how to, please tell me.

Purdue Problem of the week, Fall 2011 Problem 4

The problem 4 of 2011 fall is:

 Show that if: u(x)= Sum { x^(3k)}/{(3k)!};  

                     v(x)= Sum {x^(3k+1)}/{(3k+1)!};

                     w(x)=Sum {x^(3k+2)}/{(3k+2)!}; where k=0,1,2,3,...

then u^3+v^3+w^3-2uvw=1.

Let f(x) = u^3+v^3+w^3-2uvw, then consider f'(x).

f'(x)=3u'u^2+3v'v^2+3w'w^2-3u'vw-3uv'w-3uvw'.

It is easy to calculate u'=w, v'=u, w'=v. Therefore f'(x)=0,  that means f(x) is constant.

Because f(0)=1, so f(x)=1 for all value of x.

Purdue的數學問題

Purdue大學的數學問題真的很有意思。這幾天也嘗試著去解一些。發現自己的水平確實比中學時有些退步，當然技術上有進步。不過我覺得它給出的解決方式有些問題（Fall 2011 problem1）。
http://www.math.purdue.edu/pow/
這個問題是：
Show that x^400+x^380+...+x^20+1 is divisible by x^20+x^19+...+x+1.

不過我想到的是，需要在複數域上面解決。因為只有在複數域上面才可以將多項式做完整的factorization，將多項式寫成product of lowest polynomials. 如果不是在複數域上面討論，它給出的解法就是不對的。
f(x)=x^400+x^380+...+x^20+1, g(x)=x^20+x^19+...+x+1.
f(x)=(x^420-1)/(x^20-1);  g(x)=(x^21-1)/(x-1);
Consider f(x)=0, x_n=cos(2n*Pi/420)+i sin(2n*Pi/420), where n=0,1,2,...,419.
              g(x')=0, x'_n=cos(2n*pi/21)+i sin(2n*Pi/21), where n=0,1,2,...,20.


since cos(2n*pi/21)+i sin(2n*Pi/21)= cos( 40n*Pi/420)+i sin(40n*Pi/420).


Therefore, g(x)|f(x).


因為，解題中沒有提到在複數域中解答，其實有些不好。