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超簡單自創數學題
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+2x

Step: f(0)=0^3f(1)-2(0)^4+2(0)^2=0
f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2
f(1)=0
f(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) >=3
Suppose 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 (*)= 2n
degree of R.H.S. of (*)= n+3
L.H.S.=R.H.S. implies 2n=n+3, n=3
Since 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 constant
By(*),
C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2
Hence -3C=-2-C and 2C=2 gives C=1,
f(x)=x^3-3x^2+2x Q.E.D.

#good#


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+2x

Step: f(0)=0^3f(1)-2(0)^4+2(0)^2=0
f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2
f(1)=0
f(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) >=3
Suppose 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 (*)= 2n
degree of R.H.S. of (*)= n+3
L.H.S.=R.H.S. implies 2n=n+3, n=3
Since 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 constant
By(*),
C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2
Hence -3C=-2-C and 2C=2 gives C=1,
f(x)=x^3-3x^2+2x Q.E.D.

#good#

你邊度搵架


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


我覺得拎走f(0)=/=0會有趣d


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#


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


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


我覺得拎走f(0)=/=0會有趣d

話到明超簡單


Find all real numbers a, b, c, d such that:
a^2+b^2+c^2+d^2=a(b+c+d)


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#yup#


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#yup#

Step 呢你係咪google出黎


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#yup#

Step 呢你係咪google出黎

Uniqueness都冇你玩野?


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#yup#

Step 呢你係咪google出黎

心算都得啦駛乜googleZ_Z


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 equation
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?


x^5+x^4+x^3+x^2+x^1+1=0
Let x^5+x^4+x^3+x^2+x^1=-1
-1+1=0
z^6-1-z^6+1=0
z^6-1-(z^6-1)=0
(z^6-1)(1-1)=0
z^6-1=0
z^6=1
z=1 or -1
Since 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, .., 5
Coincidentally, when k is 3, z is just -1.


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 equation
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?

Good. 不過我個個唔用Cauchy-Schwarz inequality


Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abc


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


Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abc

Prove that if the product of n positive real numbers is 1, then their sum is at least n


Proof if a,b,c are non-negative real numbers then a^3+b^3+c^3 >=3abc

Prove that if the product of n positive real numbers is 1, then their sum is at least n

I would like to see proofs of these problems without using AM>=GM


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


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)。


I would just focus on the prime divisors 2, 3, 5, 7


1+1=2 ?_?


Find all positive integers x such that (x^2-x) is divisible by 10000.


唔好意思, 呢個post只限中學程度
同奧數差唔多


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?_?


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?


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