再来一道题
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#6 Re: 再来一道题
Grok3 gives:
To find ( f(0) ) given that f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, we need to determine the function ( f(x) ) such that composing it with itself produces the given quadratic expression, and then evaluate ( f(0) ).Let’s denote the given condition as:f(f(x))=x2−x+1.f(f(x)) = x^2 - x + 1.f(f(x)) = x^2 - x + 1.
Our goal is to find ( f(x) ) and compute ( f(0) ). Since ( f(f(x)) ) is a quadratic polynomial, let’s explore the possibility that ( f(x) ) is a linear function, as composing two linear functions can yield a quadratic. Suppose:f(x)=ax+b.f(x) = ax + b.f(x) = ax + b.
Then, compute the composition:f(f(x))=f(ax+b)=a(ax+b)+b=a2x+ab+b.f(f(x)) = f(ax + b) = a(ax + b) + b = a^2 x + ab + b.f(f(x)) = f(ax + b) = a(ax + b) + b = a^2 x + ab + b.
Set this equal to the given expression:a2x+ab+b=x2−x+1.a^2 x + ab + b = x^2 - x + 1.a^2 x + ab + b = x^2 - x + 1.
For the polynomials to be equal, their coefficients must match:Coefficient of x2x^2x^2
: a2=1 ⟹ a=±1a^2 = 1 \implies a = \pm 1a^2 = 1 \implies a = \pm 1
.
Coefficient of ( x ): 0=−1 ⟹ 0 = -1 \implies0 = -1 \implies
This is a contradiction unless the linear assumption is incorrect.
Constant term: ab+b=1ab + b = 1ab + b = 1
.
The coefficient of ( x ) gives 0=−10 = -10 = -1
, which is impossible, suggesting that ( f(x) ) may not be linear. Let’s try assuming ( f(x) ) is a quadratic polynomial, say:f(x)=ax2+bx+c.f(x) = ax^2 + bx + c.f(x) = ax^2 + bx + c.
Then:f(f(x))=a(f(x))2+bf(x)+c=a(ax2+bx+c)2+b(ax2+bx+c)+c.f(f(x)) = a(f(x))^2 + b f(x) + c = a (ax^2 + bx + c)^2 + b (ax^2 + bx + c) + c.f(f(x)) = a(f(x))^2 + b f(x) + c = a (ax^2 + bx + c)^2 + b (ax^2 + bx + c) + c.
Expanding this is complex, so let’s simplify our approach by testing the functional equation directly or exploring a substitution method. Since f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, let’s try to find ( f(0) ) by setting ( x ) to specific values and looking for patterns.Suppose we set x=0x = 0x = 0
:f(f(0))=02−0+1=1.f(f(0)) = 0^2 - 0 + 1 = 1.f(f(0)) = 0^2 - 0 + 1 = 1.
Let f(0)=kf(0) = kf(0) = k
. Then:f(k)=1.f(k) = 1.f(k) = 1.
This tells us that applying ( f ) to ( f(0) ) yields 1. Let’s try another point. Set x=1x = 1x = 1
:f(f(1))=12−1+1=1.f(f(1)) = 1^2 - 1 + 1 = 1.f(f(1)) = 1^2 - 1 + 1 = 1.
Let f(1)=mf(1) = mf(1) = m
. Then:f(m)=1.f(m) = 1.f(m) = 1.
Notice that both f(k)=1f(k) = 1f(k) = 1
and f(m)=1f(m) = 1f(m) = 1
, suggesting that 1 might be a fixed point or a value that multiple inputs map to. Let’s try to hypothesize a form for ( f(x) ). Since f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, let’s test if f(x)=x2−x+1f(x) = x^2 - x + 1f(x) = x^2 - x + 1
, assuming ( f ) is its own inverse under composition (i.e., f(f(x))=xf(f(x)) = xf(f(x)) = x
), but that would imply:x2−x+1=x,x^2 - x + 1 = x,x^2 - x + 1 = x,
which is not true for all ( x ), so f(x)≠x2−x+1f(x) \neq x^2 - x + 1f(x) \neq x^2 - x + 1
.Let’s try a substitution approach. Assume y=f(x)y = f(x)y = f(x)
, so f(y)=x2−x+1f(y) = x^2 - x + 1f(y) = x^2 - x + 1
. We need ( f(y) ), and then we can try to express ( x ) in terms of ( y ). This is complex, so let’s focus on finding ( f(0) ) directly.Another approach is to assume ( f ) is a polynomial and use the degree of the composition. If deg(f)=n\deg(f) = n\deg(f) = n
, then deg(f(f(x)))=n2\deg(f(f(x))) = n^2\deg(f(f(x))) = n^2
. Since x2−x+1x^2 - x + 1x^2 - x + 1
has degree 2, we have:n2=2 ⟹ n=2,n^2 = 2 \implies n = \sqrt{2},n^2 = 2 \implies n = \sqrt{2},
which is not an integer, suggesting ( f ) may not be a polynomial unless it’s of degree 1, which we’ve ruled out. Let’s try a functional approach by setting values. Suppose we try to find a fixed point where f(x)=xf(x) = xf(x) = x
:f(f(x))=f(x)=x2−x+1.f(f(x)) = f(x) = x^2 - x + 1.f(f(x)) = f(x) = x^2 - x + 1.
Then:f(x)=x2−x+1=x.f(x) = x^2 - x + 1 = x.f(x) = x^2 - x + 1 = x.
Solve:x2−x+1=x ⟹ x2−2x+1=0 ⟹ (x−1)2=0 ⟹ x=1.x^2 - x + 1 = x \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1.x^2 - x + 1 = x \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1.
So, if f(1)=1f(1) = 1f(1) = 1
, then:f(f(1))=f(1)=1,f(f(1)) = f(1) = 1,f(f(1)) = f(1) = 1,
and check:12−1+1=1,1^2 - 1 + 1 = 1,1^2 - 1 + 1 = 1,
which holds. This confirms f(1)=1f(1) = 1f(1) = 1
, but we need ( f(0) ). Let’s try assuming ( f ) is an involution, i.e., f(f(x))=xf(f(x)) = xf(f(x)) = x
, but that contradicts our given equation unless misinterpreted.Instead, let’s try numerical or iterative substitution. Set f(0)=kf(0) = kf(0) = k
, so f(k)=1f(k) = 1f(k) = 1
. We need another relation. Try x=kx = kx = k
:f(f(k))=k2−k+1.f(f(k)) = k^2 - k + 1.f(f(k)) = k^2 - k + 1.
Since f(k)=1f(k) = 1f(k) = 1
, we have:f(1)=k2−k+1.f(1) = k^2 - k + 1.f(1) = k^2 - k + 1.
We already have f(1)=mf(1) = mf(1) = m
and f(m)=1f(m) = 1f(m) = 1
. Let’s hypothesize f(1)=0f(1) = 0f(1) = 0
:f(0)=k,f(k)=1,f(1)=0,f(0)=k ⟹ f(f(1))=f(0)=k.f(0) = k, \quad f(k) = 1, \quad f(1) = 0, \quad f(0) = k \implies f(f(1)) = f(0) = k.f(0) = k, \quad f(k) = 1, \quad f(1) = 0, \quad f(0) = k \implies f(f(1)) = f(0) = k.
Check:f(f(1))=12−1+1=1,f(f(1)) = 1^2 - 1 + 1 = 1,f(f(1)) = 1^2 - 1 + 1 = 1,
so:f(0)=k=1.f(0) = k = 1.f(0) = k = 1.
This suggests f(0)=1f(0) = 1f(0) = 1
. Let’s verify by assuming f(0)=1f(0) = 1f(0) = 1
:f(f(0))=f(1)=1,f(f(0)) = f(1) = 1,f(f(0)) = f(1) = 1,
and we need consistency. Try another point, say x=2x = 2x = 2
:f(f(2))=22−2+1=4−2+1=3.f(f(2)) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.f(f(2)) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.
So:f(f(2))=3.f(f(2)) = 3.f(f(2)) = 3.
Assume f(2)=pf(2) = pf(2) = p
, then f(p)=3f(p) = 3f(p) = 3
. We need a consistent function. Let’s try to construct ( f ). Suppose f(0)=1f(0) = 1f(0) = 1
, and check if we can build pairs. This is getting complex, so let’s finalize by testing f(0)=1f(0) = 1f(0) = 1
.Assume a linear form again or test values. Since linear didn’t work, let’s try a simpler hypothesis or numerical check. Given f(0)=1f(0) = 1f(0) = 1
, and knowing f(f(0))=1f(f(0)) = 1f(f(0)) = 1
, we have:f(1)=1.f(1) = 1.f(1) = 1.
Check other points to ensure consistency. Try x=3x = 3x = 3
:f(f(3))=32−3+1=9−3+1=7.f(f(3)) = 3^2 - 3 + 1 = 9 - 3 + 1 = 7.f(f(3)) = 3^2 - 3 + 1 = 9 - 3 + 1 = 7.
So:f(f(3))=7.f(f(3)) = 7.f(f(3)) = 7.
Without a closed form for ( f ), let’s confirm f(0)=1f(0) = 1f(0) = 1
by assuming it and checking consistency across multiple points, but our derivation suggests:f(0)=1f(0) = 1f(0) = 1
is plausible. To be rigorous, we’d solve for ( f ), but the pattern f(0)=1f(0) = 1f(0) = 1
, f(1)=1f(1) = 1f(1) = 1
, fits the equation at tested points.Thus, the value of ( f(0) ) is likely:1\boxed{1}\boxed{1}
This answer is based on the consistency of f(f(0))=1f(f(0)) = 1f(f(0)) = 1
and testing the functional equation, though a full function ( f ) would require solving the functional equation more thoroughly, possibly assuming a specific form or numerical iteration. If you need a deeper derivation, I can explore further!
To find ( f(0) ) given that f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, we need to determine the function ( f(x) ) such that composing it with itself produces the given quadratic expression, and then evaluate ( f(0) ).Let’s denote the given condition as:f(f(x))=x2−x+1.f(f(x)) = x^2 - x + 1.f(f(x)) = x^2 - x + 1.
Our goal is to find ( f(x) ) and compute ( f(0) ). Since ( f(f(x)) ) is a quadratic polynomial, let’s explore the possibility that ( f(x) ) is a linear function, as composing two linear functions can yield a quadratic. Suppose:f(x)=ax+b.f(x) = ax + b.f(x) = ax + b.
Then, compute the composition:f(f(x))=f(ax+b)=a(ax+b)+b=a2x+ab+b.f(f(x)) = f(ax + b) = a(ax + b) + b = a^2 x + ab + b.f(f(x)) = f(ax + b) = a(ax + b) + b = a^2 x + ab + b.
Set this equal to the given expression:a2x+ab+b=x2−x+1.a^2 x + ab + b = x^2 - x + 1.a^2 x + ab + b = x^2 - x + 1.
For the polynomials to be equal, their coefficients must match:Coefficient of x2x^2x^2
: a2=1 ⟹ a=±1a^2 = 1 \implies a = \pm 1a^2 = 1 \implies a = \pm 1
.
Coefficient of ( x ): 0=−1 ⟹ 0 = -1 \implies0 = -1 \implies
This is a contradiction unless the linear assumption is incorrect.
Constant term: ab+b=1ab + b = 1ab + b = 1
.
The coefficient of ( x ) gives 0=−10 = -10 = -1
, which is impossible, suggesting that ( f(x) ) may not be linear. Let’s try assuming ( f(x) ) is a quadratic polynomial, say:f(x)=ax2+bx+c.f(x) = ax^2 + bx + c.f(x) = ax^2 + bx + c.
Then:f(f(x))=a(f(x))2+bf(x)+c=a(ax2+bx+c)2+b(ax2+bx+c)+c.f(f(x)) = a(f(x))^2 + b f(x) + c = a (ax^2 + bx + c)^2 + b (ax^2 + bx + c) + c.f(f(x)) = a(f(x))^2 + b f(x) + c = a (ax^2 + bx + c)^2 + b (ax^2 + bx + c) + c.
Expanding this is complex, so let’s simplify our approach by testing the functional equation directly or exploring a substitution method. Since f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, let’s try to find ( f(0) ) by setting ( x ) to specific values and looking for patterns.Suppose we set x=0x = 0x = 0
:f(f(0))=02−0+1=1.f(f(0)) = 0^2 - 0 + 1 = 1.f(f(0)) = 0^2 - 0 + 1 = 1.
Let f(0)=kf(0) = kf(0) = k
. Then:f(k)=1.f(k) = 1.f(k) = 1.
This tells us that applying ( f ) to ( f(0) ) yields 1. Let’s try another point. Set x=1x = 1x = 1
:f(f(1))=12−1+1=1.f(f(1)) = 1^2 - 1 + 1 = 1.f(f(1)) = 1^2 - 1 + 1 = 1.
Let f(1)=mf(1) = mf(1) = m
. Then:f(m)=1.f(m) = 1.f(m) = 1.
Notice that both f(k)=1f(k) = 1f(k) = 1
and f(m)=1f(m) = 1f(m) = 1
, suggesting that 1 might be a fixed point or a value that multiple inputs map to. Let’s try to hypothesize a form for ( f(x) ). Since f(f(x))=x2−x+1f(f(x)) = x^2 - x + 1f(f(x)) = x^2 - x + 1
, let’s test if f(x)=x2−x+1f(x) = x^2 - x + 1f(x) = x^2 - x + 1
, assuming ( f ) is its own inverse under composition (i.e., f(f(x))=xf(f(x)) = xf(f(x)) = x
), but that would imply:x2−x+1=x,x^2 - x + 1 = x,x^2 - x + 1 = x,
which is not true for all ( x ), so f(x)≠x2−x+1f(x) \neq x^2 - x + 1f(x) \neq x^2 - x + 1
.Let’s try a substitution approach. Assume y=f(x)y = f(x)y = f(x)
, so f(y)=x2−x+1f(y) = x^2 - x + 1f(y) = x^2 - x + 1
. We need ( f(y) ), and then we can try to express ( x ) in terms of ( y ). This is complex, so let’s focus on finding ( f(0) ) directly.Another approach is to assume ( f ) is a polynomial and use the degree of the composition. If deg(f)=n\deg(f) = n\deg(f) = n
, then deg(f(f(x)))=n2\deg(f(f(x))) = n^2\deg(f(f(x))) = n^2
. Since x2−x+1x^2 - x + 1x^2 - x + 1
has degree 2, we have:n2=2 ⟹ n=2,n^2 = 2 \implies n = \sqrt{2},n^2 = 2 \implies n = \sqrt{2},
which is not an integer, suggesting ( f ) may not be a polynomial unless it’s of degree 1, which we’ve ruled out. Let’s try a functional approach by setting values. Suppose we try to find a fixed point where f(x)=xf(x) = xf(x) = x
:f(f(x))=f(x)=x2−x+1.f(f(x)) = f(x) = x^2 - x + 1.f(f(x)) = f(x) = x^2 - x + 1.
Then:f(x)=x2−x+1=x.f(x) = x^2 - x + 1 = x.f(x) = x^2 - x + 1 = x.
Solve:x2−x+1=x ⟹ x2−2x+1=0 ⟹ (x−1)2=0 ⟹ x=1.x^2 - x + 1 = x \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1.x^2 - x + 1 = x \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1.
So, if f(1)=1f(1) = 1f(1) = 1
, then:f(f(1))=f(1)=1,f(f(1)) = f(1) = 1,f(f(1)) = f(1) = 1,
and check:12−1+1=1,1^2 - 1 + 1 = 1,1^2 - 1 + 1 = 1,
which holds. This confirms f(1)=1f(1) = 1f(1) = 1
, but we need ( f(0) ). Let’s try assuming ( f ) is an involution, i.e., f(f(x))=xf(f(x)) = xf(f(x)) = x
, but that contradicts our given equation unless misinterpreted.Instead, let’s try numerical or iterative substitution. Set f(0)=kf(0) = kf(0) = k
, so f(k)=1f(k) = 1f(k) = 1
. We need another relation. Try x=kx = kx = k
:f(f(k))=k2−k+1.f(f(k)) = k^2 - k + 1.f(f(k)) = k^2 - k + 1.
Since f(k)=1f(k) = 1f(k) = 1
, we have:f(1)=k2−k+1.f(1) = k^2 - k + 1.f(1) = k^2 - k + 1.
We already have f(1)=mf(1) = mf(1) = m
and f(m)=1f(m) = 1f(m) = 1
. Let’s hypothesize f(1)=0f(1) = 0f(1) = 0
:f(0)=k,f(k)=1,f(1)=0,f(0)=k ⟹ f(f(1))=f(0)=k.f(0) = k, \quad f(k) = 1, \quad f(1) = 0, \quad f(0) = k \implies f(f(1)) = f(0) = k.f(0) = k, \quad f(k) = 1, \quad f(1) = 0, \quad f(0) = k \implies f(f(1)) = f(0) = k.
Check:f(f(1))=12−1+1=1,f(f(1)) = 1^2 - 1 + 1 = 1,f(f(1)) = 1^2 - 1 + 1 = 1,
so:f(0)=k=1.f(0) = k = 1.f(0) = k = 1.
This suggests f(0)=1f(0) = 1f(0) = 1
. Let’s verify by assuming f(0)=1f(0) = 1f(0) = 1
:f(f(0))=f(1)=1,f(f(0)) = f(1) = 1,f(f(0)) = f(1) = 1,
and we need consistency. Try another point, say x=2x = 2x = 2
:f(f(2))=22−2+1=4−2+1=3.f(f(2)) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.f(f(2)) = 2^2 - 2 + 1 = 4 - 2 + 1 = 3.
So:f(f(2))=3.f(f(2)) = 3.f(f(2)) = 3.
Assume f(2)=pf(2) = pf(2) = p
, then f(p)=3f(p) = 3f(p) = 3
. We need a consistent function. Let’s try to construct ( f ). Suppose f(0)=1f(0) = 1f(0) = 1
, and check if we can build pairs. This is getting complex, so let’s finalize by testing f(0)=1f(0) = 1f(0) = 1
.Assume a linear form again or test values. Since linear didn’t work, let’s try a simpler hypothesis or numerical check. Given f(0)=1f(0) = 1f(0) = 1
, and knowing f(f(0))=1f(f(0)) = 1f(f(0)) = 1
, we have:f(1)=1.f(1) = 1.f(1) = 1.
Check other points to ensure consistency. Try x=3x = 3x = 3
:f(f(3))=32−3+1=9−3+1=7.f(f(3)) = 3^2 - 3 + 1 = 9 - 3 + 1 = 7.f(f(3)) = 3^2 - 3 + 1 = 9 - 3 + 1 = 7.
So:f(f(3))=7.f(f(3)) = 7.f(f(3)) = 7.
Without a closed form for ( f ), let’s confirm f(0)=1f(0) = 1f(0) = 1
by assuming it and checking consistency across multiple points, but our derivation suggests:f(0)=1f(0) = 1f(0) = 1
is plausible. To be rigorous, we’d solve for ( f ), but the pattern f(0)=1f(0) = 1f(0) = 1
, f(1)=1f(1) = 1f(1) = 1
, fits the equation at tested points.Thus, the value of ( f(0) ) is likely:1\boxed{1}\boxed{1}
This answer is based on the consistency of f(f(0))=1f(f(0)) = 1f(f(0)) = 1
and testing the functional equation, though a full function ( f ) would require solving the functional equation more thoroughly, possibly assuming a specific form or numerical iteration. If you need a deeper derivation, I can explore further!
#8 Re: 再来一道题
(f(x))^2 - f(x) + 1 = f(f(f(x)) = f(x^2 -x + 1)
x=1: f(1)^2 - f(1) + 1 = f(1)
f(1) = 1
x=0: f(0)^2 - f(0) + 1 = f(1) = 1
f(0) = 0 or 1
if f(0)=0: 1=0^2-0+1=f(f(0))=f(0)
so, f(0)= 1
x=1: f(1)^2 - f(1) + 1 = f(1)
f(1) = 1
x=0: f(0)^2 - f(0) + 1 = f(1) = 1
f(0) = 0 or 1
if f(0)=0: 1=0^2-0+1=f(f(0))=f(0)
so, f(0)= 1
x3
#12 Re: 再来一道题
按给的公式： f[f(0)]=0^2-0+1=1 ， f[f(1)]=1^2-1+1=1
所以1==f[f(0)]==f[f(1)]
那么 f(1)=f{f[f(0)]}==f{f[f(1)]}
看右边这半, 因为函数嵌套是可以遵循结合律的, 把 f(1)看作一个整体，设f(1)=z
z= f(1)= f [f( f（1))]=f[f(z)] =z^2-z+1
解二元一次方程得z=1
所以f(1)=1
代回 f(1)=f{f[f(0)]},设f（0）=w
1= f[f(w）]=w^2-w+1
得f(0）=w=0 或者1
到底是0还是1？没想好。
所以1==f[f(0)]==f[f(1)]
那么 f(1)=f{f[f(0)]}==f{f[f(1)]}
看右边这半, 因为函数嵌套是可以遵循结合律的, 把 f(1)看作一个整体，设f(1)=z
z= f(1)= f [f( f（1))]=f[f(z)] =z^2-z+1
解二元一次方程得z=1
所以f(1)=1
代回 f(1)=f{f[f(0)]},设f（0）=w
1= f[f(w）]=w^2-w+1
得f(0）=w=0 或者1
到底是0还是1？没想好。
上次由 huangchong 在 2025年 7月 12日 14:21 修改。
#13 Re: 再来一道题
偷看了一眼kde同学的答案：huangchong 写了： 今天 14:15 按给的公式： f[f(0)]=0^2-0+1=1 ， f[f(1)]=1^2-1+1=1
所以1==f[f(0)]==f[f(1)]
那么 f(1)=f{f[f(0)]}==f{f[f(1)]}
看右边这半, 因为函数嵌套是可以遵循结合律的, 把 f(1)看作一个整体，设f(1)=z
z= f(1)= f [f( f（1))]=f[f(z)] =z^2-z+1
解二元一次方程得z=1
所以f(1)=1
代回 f(1)=f{f[f(0)]},设f（0）=w
1= f[f(w0]=w^2-w+1
得f(0）=w=0 或者1
到底是0还是1？没想好。
f(0)=0的话， 0=f（0）=f[f(0)]=0^2-0+1=1 ,矛盾了，
而f(0)=1的话， 1=f（1）=f[f(1)]=1^2-1+1=1 ,不矛盾，所以 答案只剩下1
那会不会有其他的矛盾呢？不知道。
不管怎么说，这是我这辈子第一次会做这种函数套函数题。本科时候看这种题就晕。
#14 Re: 再来一道题
a nice problem!
2023年12月，张雪峰在一个直播片段中表示“文科都是服务业，什么是服务业？总结称一个字就是‘舔’，就是‘爷, 我给你笑一个’”。这一言论引发网民争议。
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#15 Re: 再来一道题
这个证明确实很巧。kde23 写了： 今天 13:08 (f(x))^2 - f(x) + 1 = f(f(f(x)) = f(x^2 -x + 1)
x=1: f(1)^2 - f(1) + 1 = f(1)
f(1) = 1
x=0: f(0)^2 - f(0) + 1 = f(1) = 1
f(0) = 0 or 1
if f(0)=0: 1=0^2-0+1=f(f(0))=f(0)
so, f(0)= 1
但，这个函数是不是真的存在？
如果不是，那么这个题的意义只在于这个技巧本身。