some math problems

Tue Nov 10, 2009 9:01 pm, by wxydx00

[ Currently: some which i though recently ]

1.a_{1},a_{2},...,a_{n} > 0,prove:\lim_{x\longrightarrow 0}(\frac {\sum a_{n}^{x}}{n})^\frac {1}{x}

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311091


2.find \lim_{n\longrightarrow \infty}\prod_{k = 1}^n\cos\dfrac{x}{2^k}(We can prove viete theorem from it )

3,n堆球,每堆若干个,2人取,每次取一堆中的任意多个,取到最后一球胜,问,谁有必胜策略.
3.n piles of balls, each pile has some balls which we don't know the exact number, A,B are two persons,they begin carry the balls from the piles, but they can only carry from one pile each time. the game says that the person we take away the last ball is the winner. Question: who is the winner if two persons are all quite clever.
(i have solved the situation when n=2 and 3 yesterday...)

4.Find a function defined on ( - \infty, + \infty), and is continuous iff x=1.

5.find or prove there exist/not exist f(x),s.t.
f(f(x)) = x^2 + 1

6.x_1,x_2,...,x_n \in \mathbb{N}^ +,\sum \frac {1}{x_n^2} = 1,find:
\sum\frac {1}{x_n^n}

7.f,g are continuous function,f(0) = 0,g(0) = 1
f(x + y) = f(x)g(y) + f(y)g(x)
g(x + y) = g(x)g(y) - f(x)f(y)
prove
f(x) = \sin ax or- \sin ax
g(x) = \cos ax
(a \in \mathbb{R})

the proof of integrabel

Wed Nov 04, 2009 4:05 am, by wxydx00

Problem:
If f(x) is a continous function at [a,b],then:
exist F(x), such that:
\dfrac{dF}{dx} = f(x).

Proof:
反证:假设在[a,b]上.f(x)的原函数不存在.
即:\all F(x),\exists x_0\in[a,b],\exists \epsilon > 0,\all \delta > 0,\exists x\in U(x_0,\delta)\backslash\{x_0\}, such that:
|\frac {F(x) - f(x_0)}{x - x_0} - f(x_0)|\ge \epsilon
令a_0 = a,b_0 = b,(a < b)
考虑[a_0,\frac {a_0 + b_0}{2}],[\frac {a_0 + b_0}{2}], 先证两区间至少有一者原函数不存在。
反证，设两区间对于f(x)的原函数均可求，那么不妨设：
两段区间所对应的函数分别为:F_1(x),F_2(x).
为今后方便起见，我们把F_1(x),F_2(x)的定义域以如下方式拓展到( - \infty, + \infty)
在( - \infty,a_0),有F_1(x) - F_1(a_0) = f(a_0)(x - a)
在(\frac {a_0 + b_0}{2}, + \infty),有F_1(x) - F_1(\frac {a_0 + b_0}{2}) = f(\frac {a_0 + b_0}{2})(x - \frac {a_0 + b_0}{2})
在( - \infty,\frac {a_0 + b_0}{2}),有F_2(x) - F_2(\frac {a_0 + b_0}{2}) = f(\frac {a_0 + b_0}{2})(x - \frac {a_0 + b_0}{2})
在(b_0, + \infty),有F_2(x) - F_2(b_0) = f(b_0)(x - b_0)
这样，我们就把F_1(x)与F_2(x)同时扩展定义到了\mathbb{R}上.
此时考虑
F(x) = F_1(x)..................................................................................(x\le \frac {a_0 + b_0}{2})
F(x) = F_2(x) + F_1(\frac {a_0 + b_0}{2}) - F_2(\frac {a_0 + b_0}{2})...........(x\ge \frac {a_0 + b_0}{2})
下证:F(x)在\mathbb{R}上连续且可导.
由于我们知F_1(x),F_2(x)在\mathbb{R}上连续且可导.
故F(x)在\mathbb{R}\backslash\{\frac {a_0 + b_0}{2}\}上连续且可导.
而在x = \frac {a_0 + b_0}{2}:
\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ + }F(x) = \lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ + }F_2(x) + F_1(\frac {a_0 + ...
\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ - }F(x) = \lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ - }F_1(x) = F_1(\frac {a_0 + ...
\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ + }F(x) = \lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ - }F(x) = F(\frac {a_0 + b_0}...
即F(x)在x = \frac {a_0 + b_0}{2}连续.
\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ + }\frac {F(x) - F(\frac {a_0 + b_0}{2})}{x - \frac {a_0 + b_0}{2}} \\ = \lim_{x\r...

\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ - }\frac {F(x) - F(\frac {a_0 + b_0}{2})}{x - \frac {a_0 + b_0}{2}} \\ = \lim_{x\r...
由f(x)连续知f(\frac {a_0 + b_0}{2})有限.
故\lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ - }F'(x) = \lim_{x\rightarrow (\frac {a_0 + b_0}{2})^ +} F'(x)
即F(x)在x = \frac {a_0 + b_0}{2}处亦可导.
于是构造出一函数F(x)使得假设不成立。
于是两个区间至少有一个区间的原函数不可求。
记该区间为[a_1,b_1]
同理，把[a_1,b_1]二等分,亦可找到[a_2,b_2]
....
于是我们构造出一系列的闭区间套I_1\supset I_2 \supset...\supset I_n \supset...，其中I_n = [a_n,b_n]
由闭区间套定理，有且仅有一点\xi = \cap_{n = 1}^\infty I_n
此时，依据假设，该点性质如下:
即f(x)在x = \xi连续，但对于\all F(x),\frac {dF}{dx} = f(x)在x = \xi均不成立。
而上句在F(x) = f(\xi)x时显然不成立。
说明假设矛盾，不成立。
证明完毕

Triangle inequality from Liu SuYuan

Fri Jan 30, 2009 2:31 am, by wxydx00

To any p,q \in N^{ + },Prove that:
\frac {\sum_{i = 0}^{p - 1}sin\frac {i\pi}{2p}}{p} \le \frac {\sum_{i = 1}^{q}sin\frac {i\pi}{2q}}{q}

Notice that e^{i\frac {\pi}{2p}} = cos\frac {\pi}{2p} + isin\frac {\pi}{2p},hence we obtain:

\sum_{i = 0}^{p - 1}sin\frac {i\pi}{2p}
= Im(\sum_{i = 0}^{p - 1}e^{i\frac {i\pi}{2p}})
= Im(\frac {1 - (e^{i\frac {\pi}{2p}})^p}{1 - e^{i\frac {\pi}{2p}}})
= Im(\frac {1 - i}{1 - cos\frac {\pi}{2p} - isin\frac {\pi}{2p}})
= Im(\frac {(1 - i)(1 - cos\frac {\pi}{2p} + isin\frac {\pi}{2p})}{(1 - cos\frac {\pi}{2p})^2 + (sin\frac {\pi}{2p})^2})
= \frac {cos\frac {\pi}{2p} + sin\frac {\pi}{2p} - 1}{2 - 2cos\frac {\pi}{2p}}

In the same way, we get:

\sum_{i = 1}^{q}sin\frac {i\pi}{2q}} = \frac {1 + sin\frac {\pi}{2q} - cos\frac {\pi}{2q}}{2 - 2cos\frac {\pi}{2q}}

Since, it is enough to prove the following inequality:
\dfrac{\dfrac{sin\frac {\pi}{2p}}{1 - cos\frac {\pi}{2p}} - 1}{p}\le \dfrac{\dfrac{sin\frac {\pi}{2q}}{1 - cos\frac {\pi}{2q}...

Denote:
F(x) = \dfrac{\dfrac{sin\frac {\pi}{2x}}{1 - cos\frac {\pi}{2x}} - 1}{x}

G(x) = \dfrac{\dfrac{sin\frac {\pi}{2x}}{1 - cos\frac {\pi}{2x}} + 1}{x}


Find that F(x) is increaseing in [1,\infty),and G(x) is decreasing in [1,\infty).

So it is equivalent to prove that:
F(x) - G(x) < 0 which x \in N^{ + }
What's more,we prove:
F(x) - G(x) < 0 (x \in [1,\infty) )
which is obvious

limit

Mon Apr 28, 2008 3:30 am, by wxydx00

Find \lim_{n - > + \infty}\prod(1 + \frac {1}{9^n})
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181443
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1112927#1112927

inequlity

Sun Apr 27, 2008 7:56 am, by wxydx00

[ Currently: unsolved ]

\frac {1}{a(1 + b)} + \frac {1}{b(1 + c)} + \frac {1}{c(1 + a)}\ge \frac {3}{1 + abc}

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202187

Algebra Problem

Sun Apr 27, 2008 5:32 am, by wxydx00

Click to reveal hidden content (点击此处可展开或折叠)

Some method

Thu Apr 24, 2008 7:56 am, by wxydx00

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1064949#1064949

An inequality to be finished

Tue Apr 22, 2008 3:04 am, by wxydx00

[ Currently: to practise using SOS method and do not finished ]

Click to reveal hidden content (点击此处可展开或折叠)
we have abc = 1, to prove:
\sum\dfrac{1}{a^2 + 2b^2 + 1} \le \frac {1}{2}

solution:
use the substitution:a^2 = \frac {x}{y},b^2 = \frac {y}{z},c^2 = \frac {z}{x}
it is equal to prove:\sum(\frac {1}{6} - \frac {yz}{xz + 2y^2 + 3yz})\ge 0
it is equivalent to \sum\frac {xz + 2y^2 - 3yz}{xz + 2y^2 + 3yz}\ge 0
LHS = \\ \sum\frac {(y - z)(2y - z) - y(z - x)}{xz + 2y^2 + 3yz} \\ \sum(\frac {2y - z}{xz + 2y^2 + 3yz} - \frac {x}{yz + 2x^...
it is easy that \sum(\frac {3x^2 + 4xy + yz}{(xz + 2y^2 + 3yz)(yz + 2x^2 + 3xy)}(y - z)^{2} \ge 0
so that we have to prove \sum y(x + z)(x - z)(xy + 2z^2 + 3xz) \ge 0

I think the inequality can be solved by using schur inequality,but I cannot finish.
maybe what arqady said may solve it.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98484

An inequality

Tue Apr 22, 2008 3:03 am, by wxydx00

Click to reveal hidden content (点击此处可展开或折叠)

An inequality

Sun Apr 20, 2008 3:48 am, by wxydx00

[ Currently: SOS ]

Click to reveal hidden content (点击此处可展开或折叠)

Some inequalities with perfect solution

Sat Apr 19, 2008 6:52 am, by wxydx00

Click to reveal hidden content (点击此处可展开或折叠)

已知a,b,c > 0且abc = 1
求证：\frac {1}{1 + a + b} + \frac {1}{1 + b + c} + \frac {1}{1 + c + a}\le 1


证明：令a = x^3,b = y^3,c = z^3,则xyz = 1,且有
1 + a + b = 1 + x^3 + y^3 \ge xyz + x^2y + xy^2 = xy(x + y + z) = \frac {x + y + z}{z},
同理 1 + b + c \ge \frac {x + y + z}{x},1 + c + a \ge\frac {x + y + z}{y},
因此 \frac{1}{1 + a + b} + \frac {1}{1 + b + c} + \frac {1}{1 + c + a} \le \frac{z}{x + y + z} + \frac{x}{x + y + z}+\frac{y}{x + ...

[06 China TST]x,y,z\ge 0,x + y + z = 1
\sum\frac {xy}{\sqrt {xy + yz}}\le \frac {\sqrt {2}}{2}
proof:
with Cauchy-Schwartz inequality,
it is equavalent to:
\sum xy\sum\frac {xy}{xy + yz}\le \frac{1}{2}
\rightleftarrow \sum yz\sum\frac {x}{x + z} = [xz + y(x + z)]\sum\frac {x}{x + z} = \sum\frac {x^2z}{x + z} + \sumxy\le \frac...
\rightleftarrow 2\sum\frac {x^z}{x + z}\le \sum x^2
that:
2\sum\frac {x^2z}{x + z}\le\sum\frac {x(x + z)}{2}\le\sum x^2

四次多项式非负性判准

Sat Apr 19, 2008 6:35 am, by wxydx00

[ Currently: 转自-中国不等式小组论坛 ]

Click to reveal hidden content (点击此处可展开或折叠)
四次多项式 x^4 + p*x^3 + q*x^2 + r*x + s（设 s\ne 0）对于一切 x > 0 保持非负的充分必要条件是下面的公式满足：

s > 0\wedge((p\geq 0\wedge q\geq 0\wedge r\geq0)\vee

(d8 > 0\wedge(d6\leq0\vee d4\leq 0))\vee

(d8 < 0\wedge d7\geq 0\wedge(p\geq 0\vee d5 < 0))\vee

(d8 < 0\wedge d7 < 0\wedge p > 0\wedge d5 > 0)\vee

(d8 = 0\wedge d6 < 0)\vee

(d8 = 0\wedge d6 > 0\wedge d7 > 0\wedge (p\geq 0\vee d5 < 0))\vee

(d8 = 0\wedge d6 = 0\wedge (d4\leq0\vee E1 = 0))\ )

其中

d4 = - 8*q + 3*p^2,

d5 = 3*r*p + q*p^2 - 4*q^2,

d6 = 14*q*r*p - 4*q^3 + 16*s*q - 3*p^3*r + p^2*q^2 - 6*p^2*s - 18*r^2,

d7 = 7*r*p^2*s - 18*q*p*r^2 - 3*q*p^3*s - q^2*p^2*r + 16*s^2*p + 4*r^2*p^3 + 12*q^2*p*s + 4*r*q^3

- 48*r*s*q + 27*r^3,

d8 = p^2*q^2*r^2 + 144*q*s*r^2 - 192*r*s^2*p + 144*q*s^2*p^2 - 4*p^2*q^3*s + 18*q*r^3*p

- 6*p^2*s*r^2 - 80*r*p*s*q^2 + 18*p^3*r*s*q - 4*q^3*r^2 + 16*q^4*s - 128*s^2*q^2

- 4*p^3*r^3 - 27*p^4*s^2 - 27*r^4 + 256*s^3,

E1 = 8*r - 4*p*q + p^3;

符号∧表示“并且”，符号∨表示“或者”。

其中若干表达式可直接用于Maple中计算.

一道解析不等式

Sat Apr 19, 2008 5:37 am, by wxydx00

[ Currently: from yxz ]

Click to reveal hidden content (点击此处可展开或折叠)
x,y,z \ge 0
\sum\frac {a^3}{2a^2 - ab + 2b^2} > = \frac {a + b + c}{3}

Here is a solution from Chenji.
\sum\frac {x^3}{2x^2 - xy + 2y^2} - \frac {x + y + z}{3} \\ = \frac {8F(x,y,z)}{3\prod(2x^2 - xy + 2y^2)}
assume x = min(x,y,z),F(x,y,z) = F(x,x + s,x + t)
Thus,
LHS = 18(s^2 - st + t^2)x^5 + 9(3s^3 + 5s^2t - 4st^2 + 3t^3)x^4 + 3(7s^4 + 22s^3t + 9s^2t^2 - 14st^4 + 7t^4)x^3 + 3(2s^5 + 13...
because:
2s^5 + 13s^4t + 17s^3t^2 - 7s^2t^3 - 2st^4 + 2t^5 = s^3t^2 + (2s + t)(s^4 + 6s^3t + 5s^2t^2 - 6st^3 + 2t^4)\ge 0
2s^4 + 12s^3t - 2s^2t^2 - 3st^3 + 2t^4 = s^4 + 8s^3t + (s + t)^2(s - t)^2 + t(s + t)(2s - t)^2\ge 0

Who can post some other method which donot use mechanical method.

Page 1 of 1 [13 Entries]

write down something

wxydx00

Calendar

Select a timeframe

Shoutbox

Shouts

  • Hey, is that you in your avatar? Somehow you look like that starcraft pro called jaedong maybe

    By Nerd_of_the_Ages, on Tue Apr 22, 2008 6:00 pm

  • First shout! Mr. Green

    By shalex, on Thu Apr 10, 2008 10:23 pm

Contact owner

About owner

  • Joined: 22 Mar 2007
  • Location: Beijing China
  • Occupation: A student without a smart brain, but love math and try the best to find more happiness from learning

Blog details

  • Blog started: 05 Apr 2008
  • Total entries: 13
  • Total visits: 3580

RSS Feed

RSS Feed