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 現有會員可[按此]登入。未成為會員可[按此]註冊。 [公司模式 - 關]  [懷舊模式 - 開]  [Youtube 預覽 - 關]  [大字型]  [小字型] 您現在聚腳在 學術台內。 將已解封的舊帳戶升級高級會員 我們上年12月曾解封部分只輕微觸犯版規或懷疑被駭的帳戶，但有會員反映一些於2009年前登記的舊帳戶在解封後未有自動升級為高級會員，現在我們已將該批帳戶升級，謝謝。

112 個回應
 Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.目前搵到一條:f(x)=x^3-2x^2+2xStep: f(0)=0^3f(1)-2(0)^4+2(0)^2=0f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2f(1)=0f(1)=(1)f(2)-2+2=f(2)hence x, (x-1), (x-2) are factors of f(x)This means degree of f(x) >=3Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*degree of L.H.S. of (*)= 2ndegree of R.H.S. of (*)= n+3L.H.S.=R.H.S. implies 2n=n+3, n=3Since x, (x-1), (x-2) are three factors of f(x) (proven),Let f(x)=Cx(x-1)(x-2)= C(x^3-3x^2+2x) where C is constantBy(*),C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2Hence -3C=-2-C and 2C=2 gives C=1,f(x)=x^3-3x^2+2x Q.E.D. 13/1/2020 16:28
 Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.目前搵到一條:f(x)=x^3-2x^2+2xStep: f(0)=0^3f(1)-2(0)^4+2(0)^2=0f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2f(1)=0f(1)=(1)f(2)-2+2=f(2)hence x, (x-1), (x-2) are factors of f(x)This means degree of f(x) >=3Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*degree of L.H.S. of (*)= 2ndegree of R.H.S. of (*)= n+3L.H.S.=R.H.S. implies 2n=n+3, n=3Since x, (x-1), (x-2) are three factors of f(x) (proven),Let f(x)=Cx(x-1)(x-2)= C(x^3-3x^2+2x) where C is constantBy(*),C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2Hence -3C=-2-C and 2C=2 gives C=1,f(x)=x^3-3x^2+2x Q.E.D.你邊度搵架 13/1/2020 16:29
 你define 個 hole 就點都要define f(0)=0 但而家唔俾define做1 就變左做constant function首先f(0)=k f(0*0)=k^2=k咁k=0 or 1但因為k=/=0k=11=f(0*x)=kf(x)=f(x) for all x我覺得拎走f(0)=/=0會有趣d 13/1/2020 16:59
 Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.f(x) = 1/xf(y) = 1/yf(xy) = 1/xy = f(x)f(y)搵到一個答案 應該唔得，因為0嘅時候係undefined不過如果幫0加exception嘅話，咁就無問題無錯, 因為f(0)應該undefined.quoting this 13/1/2020 17:04

define做1 就變左做constant function

f(0*0)=k^2=k

k=1
1=f(0*x)=kf(x)=f(x) for all x

13/1/2020 17:07
 Find all real numbers a, b, c, d such that:a^2+b^2+c^2+d^2=a(b+c+d) 13/1/2020 20:57
 Find all real numbers a, b, c, d such that:a^2+b^2+c^2+d^2=a(b+c+d)0 0 0 0易到 13/1/2020 22:10
Find all real numbers a, b, c, d such that:
a^2+b^2+c^2+d^2=a(b+c+d)

0 0 0 0

13/1/2020 22:15
 Find all real numbers a, b, c, d such that:a^2+b^2+c^2+d^2=a(b+c+d)0 0 0 0易到Step 呢你係咪google出黎Uniqueness都冇你玩野? 13/1/2020 22:16
 Find all real numbers a, b, c, d such that:a^2+b^2+c^2+d^2=a(b+c+d)0 0 0 0易到Step 呢你係咪google出黎心算都得啦駛乜google 13/1/2020 22:58
Find all real numbers a, b, c, d such that:
a^2+b^2+c^2+d^2=a(b+c+d)

Assuming b, c, d are not all 0, otherwise (a, b, c, d) = (0, 0, 0, 0) is the trivial solution.
x^2-(b+c+d)x+(b^2+c^2+d^2)=0 .... (*)
If (*) has real roots, then for fixed (b, c, d), there will be at least one real value of a satisfying the stated equation.
We want to find the ranges of (b, c, d) for which this happens.

The discriminant of (*) is given by
delta = (b+c+d)^2 - 4(1)(b^2+c^2+d^2)

Using the Cauchy-Schwarz inequality, we have
(b+c+d)^2 <= 3(b^2+c^2+d^2)

thus, delta = (b+c+d)^2 - 4(b^2+c^2+d^2) <= 3(b^2+c^2+d^2) - 4(b^2+c^2+d^2)
which is in turn <= -(b^2+c^2+d^2) < 0 (as b, c, d not all 0)

Thus (*) will always have no real solution for x for any value b, c, d

Hence (a, b, c, d) = (0, 0, 0, 0) is the only solution.

BTW, do you have a degree related to math? Or just interested in math?

13/1/2020 23:50
 x^5+x^4+x^3+x^2+x^1+1=0Let x^5+x^4+x^3+x^2+x^1=-1-1+1=0z^6-1-z^6+1=0z^6-1-(z^6-1)=0(z^6-1)(1-1)=0z^6-1=0z^6=1z=1 or -1Since LHS will be same as RHS if sub z=-1 into x^5+x^4+x^3+x^2+x^1+1=0. So z=1 is rejected.For the equation z^6=1, the OP did not specify whether z must be a real number(given the notation, it is more likely there is no such restriction)The solutions of z^6=1 is just the roots of unity (there are totally 6 roots (real or complex), recalling a polynomial of degree 6 must have 6 roots)The general solution is given by z=cos(2k*pi/6)+i sin(2k*pi/6), where k=0, 1, 2, .., 5Coincidentally, when k is 3, z is just -1. 14/1/2020 00:17
 Find all real numbers a, b, c, d such that:a^2+b^2+c^2+d^2=a(b+c+d)Assuming b, c, d are not all 0, otherwise (a, b, c, d) = (0, 0, 0, 0) is the trivial solution.Consider the quadratic equationx^2-(b+c+d)x+(b^2+c^2+d^2)=0 .... (*)If (*) has real roots, then for fixed (b, c, d), there will be at least one real value of a satisfying the stated equation.We want to find the ranges of (b, c, d) for which this happens.The discriminant of (*) is given bydelta = (b+c+d)^2 - 4(1)(b^2+c^2+d^2)Using the Cauchy-Schwarz inequality, we have(b+c+d)^2 <= 3(b^2+c^2+d^2)thus, delta = (b+c+d)^2 - 4(b^2+c^2+d^2) <= 3(b^2+c^2+d^2) - 4(b^2+c^2+d^2) which is in turn <= -(b^2+c^2+d^2) < 0 (as b, c, d not all 0)Thus (*) will always have no real solution for x for any value b, c, dHence (a, b, c, d) = (0, 0, 0, 0) is the only solution.BTW, do you have a degree related to math? Or just interested in math?Good. 不過我個個唔用Cauchy-Schwarz inequality 14/1/2020 09:45
Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abc

14/1/2020 13:26
 Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abcLet a,b and c are non-negative real numbers. Prove that (a+b+c)(ab+bc+ca)>=9abc 14/1/2020 19:36
 Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abcProve that if the product of n positive real numbers is 1, then their sum is at least n 14/1/2020 19:38
 Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abcProve that if the product of n positive real numbers is 1, then their sum is at least nI would like to see proofs of these problems without using AM>=GM 14/1/2020 19:51
Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abc

Let a,b and c are non-negative real numbers. Prove that (a+b+c)(ab+bc+ca)>=9abc

0
=-6abc+a(2bc)+b(2ac)+c(2ab)
<=-6abc+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)
=-9abc+(3abc+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2))
=-9abc+(a+b+c)(ab+bc+ca)
Hence (a+b+c)(ab+bc+ca)>=9abc

14/1/2020 20:07
 證明任何10個連續正整數中，必定有一個與其餘九個互質(互素，relatively prime)。 14/1/2020 20:15
I would just focus on the prime divisors 2, 3, 5, 7

14/1/2020 21:39
 1+1=2 14/1/2020 22:11
Find all positive integers x such that (x^2-x) is divisible by 10000.

16/1/2020 20:53
 唔好意思, 呢個post只限中學程度同奧數差唔多 17/1/2020 12:06
 唔好意思, 呢個post只限中學程度同奧數差唔多邊條唔係中學程度 17/1/2020 13:00
 唔好意思, 呢個post只限中學程度同奧數差唔多邊條唔係中學程度數論係中學程度? 17/1/2020 13:28

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