已知函数fx=x2+1,且gx=f[f(x)],G(x)=g(x)-a f(x)————高一数学！！！！！！！！！_百度知道
已知函数fx=x2+1,且gx=f[f(x)],G(x)=g(x)-a f(x)————高一数学！！！！！！！！！
已知函数fx=x2+1,且gx=f[f(x)],G(x)=g(x)-a f(x),试问，是否存在实数a，使得G（x）在（负无穷，-1]上为减函数，并且在（-1,0）上为增函数。
假设存在实数a，使得G(x)在（－∞，－1 ]为减函数，在（-1，0）上为增函数.f(x)=x²+1g(x)=f[f(x)]=[f(x)]²+1=(x²+1)²+1=x^4+2x²+2G(x)=g(x)-af(x)= x^4+2x²+2-a(x²+1)=x^4+(2-a)x²+2-a函数G(x)可看作是由函数u=t²+(2-a)t+(2-a)与函数t=x²复合而成，易知，函数t=x²在（－∞，0）上为减函数，要使G(x)在（－∞，－1 ]为减函数，在（-1，0）上为增函数则函数u=t²+(2-a)t+(2-a) 在（0，1）为减函数，在（1，＋∞）上为增函数∴-(2-a)/2=1，2-a= -2，a=4，故存在a=4，使得G(x)在（－∞，－1 ]为减函数，在（-1，0）上为增函数.
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This question is of course inspired by the question
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a bounded interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)].
I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known:
Q1) Say $g:\mathbf{R}\to\mathbf{R}$ is an arbitrary function. Is there always a function $f:\mathbf{R}\to\mathbf{R}$ such that $f(f(x))=g(x)$ for all $x\in\mathbf{R}$?
Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above?
The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do w there is a version of Q1 for every cardinal and it's really a question in combinatorics.
EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago:
Q3) if $g$ is now a continuous function $\mathbf{C}\to\mathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $x\in\mathbf{C}$?
EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist.
Related MO questions: , and .
21.8k385169
Q1: No. Let g(0)=1, g(1)=0 and g(x)=x for all $x\in\mathbb R\setminus\{0,1\}$.
Assuming $f\circ f=g$, let a=f(0), then f(a)=1 and f(1)=g(a)=a since $a\notin\{0,1\}$.
Then g(1)=f(f(1))=f(a)=1, a contradiction.
Q2: No. Let $g(x)=-x$ or, in fact, any decreasing function $\mathbb R\to\mathbb R$. Then f must be injective and hence monotone. Whether f is increasing or decreasing, $f\circ f$ is increasing.
24.8k60125
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Q2) has a negative answer. Namely, if, e.g., $g(x)=-x$ for all $x\in\mathbb{R}$,
then there is no
continuous
$f:\mathbb{R\rightarrow\mathbb{R}}$ such
that $f\circ f=g$.
As to Q3, see, e.g., Theorem 3 in .
2,30211225
Ulm invariants.
Surely someone still knows this?
Given $f \colon A \to A$
and $g \colon B \to B$, is there a bijection $\phi \colon B \to A$
such that $f(\phi(x))=\phi(g(x))$?
There is a system
of cardinal numbers, the Ulm invariants, associated with
$f$ so that the answer is ``yes'' if and only if $f$ and $g$
have the same invariants.
If $f$ is bijective, then the Ulm invariants are just counts of
how many cycles of each size there are (including the
infinite cycle size modeled by the integers with $n \mapsto n+1$).
But when not bijective, the system of invariants is more
complicated.
You need to count how many points map to each
fixed point, and how many points map to each of them, and so on.
And similarly for cycles of other sizes.
But I cannot tell you
the details, and this box is probably not the right place to do
it anyway.
So for a solution to the problem, consider what the Ulm invariants of $f(f(x))$ are
in terms of those of $f$.
Then compare to the Ulm inveriants of $\cos$.
Or whatever you
want to get.
Ulm himself may have originally done this to study isomorphism of
abelian groups.
Taking products, reduce to the case of
a $p$-group for a given prime $p$, then your map for study is
$x \mapsto x^p$.
Or something like that.
Ulm invariants
may also be given to characterize up to isomorphism linear
transformations (on possibly infinite-dimensional vector space).
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I don't have a full answer, but I can offer here a small improvement on my other answer.
Namely, what was important was not that it worked on a bounded interval, but rather, that it works outside a bounded interval.
Theorem. For any function g on the reals, there are
numerous functions f such that f(f(x)) = g(x), for all x
except those in a given fixed tiny interval.
Proof. Suppose g is a function on the reals and that I is a given interval, no matter how small. Let h be a bijection of R - I with I. Let f(x) = h(x), if x is outside I,
and f(x) = g(h-1(x)), if x is in I. Thus, f
first translates x into I, if it is outside I, and
otherwise, untranslates and computes g, if it is in I. It
follows that f(f(x)) = g(x) for all x outside I. There are
2|R| many such h's, and hence also this many
If g is continuous, then this f can be chosen also to be continuous.
By using a Cantor set instead of an interval, one can find a function f that solves f(f(x)) = g(x) except on a set of measure zero.
117k15328569
This type of equation is an "iterative functional equation."
A good starting point for the literature on this subject is the book Iterative Functional Equations by Kuczma, Choczewski, and Ger, Cambridge University Press, 1990.
The most frequently asked question of this type has $g(x) = e^x$.
A real-analytic solution in this case was constructed by H. Kneser, "Reelle analytische L?sungen der Gleichung $\varphi(\varphi(x))=e^x$ und verwandter Funktional-gleichungen", J. Reine Angew. Math. 187 (1949), 56-67.
Other useful keywords include "fractional iteration" and "iterative square root" (or more generally "iterative root").
20.7k4117209
This is a repost, and partial rewrite of an earlier deleted answer by Anixx. If you want to discuss the wisdom of that deletion, let's keep this post focused on math only. This answer is community wiki, so that others can improve it.
If $a_k$ is any sequence of real numbers, indexed by the nonnegative integers, then define $\Delta^m(a) = \sum_{k=0}^{m} (-1)^k \binom{m}{k} a_k$. Then, for integer $n$, we have $a_n = \sum_{m=0}^{\infty} \binom{n}{m} \Delta^m(a)$. Note that the sum is finite, because all but finitely many binomial coefficients vanish. One can then try defining
$$A(x) = \sum_{m=0}^{\infty} \binom{x}{m} \Delta^m(a).$$
If this sum converges, it defines a function $A$ which interpolates $a_n$. This is sometimes called .
Anixx points out that, $a_n = \sin^{[n]}(x)$ this method appears to give a good answer, but for $\cos^{[n]}(x)$, it appears not to.
The following is inspired by the chapter Partition Polynomials in Riordan's Combinatorial Identities where $g(x)$ is analytic and $g(0)=0$. Find the Taylor series of $f(x)$ at zero by evaluating the derivatives of $f(f(x))=g(x)$ at $0$ in succession.
Set $f(0)=0$, giving $f(f(0))=g(0)=0$.
The first derivative gives $f'(x) f'(f(x))=g'(x)$, therefore $f'(0)=\sqrt{g'(0)}$ and $f'(0)=-\sqrt{g'(0)}$. Set $f'(0)=\sqrt{g'(0)}$ for this example.
The second derivative $f'(x)^2 f''(f(x))+f'(f(x)) f''(x)=g''(x)$ produces $f''(0)=\frac{g''(0)}{g'(0)+\sqrt{g'(0)}}$.
The first few term of the Taylor series are
$f(x)=\sqrt{g'(0)}x+\frac{ g''(0)}{2
\left(g'(0)+\sqrt{g'(0)}\right)}x^2$ $+\frac{
\left(-3 g''(0)^2+g^{(3)}(0) g'(0)^{3/2}+2
g^{(3)}(0) g'(0)+g^{(3)}(0) \sqrt{g'(0)}\right)}{6
\left(\sqrt{g'(0)}+1\right)^2 g'(0)
\left(g'(0)+1\right)}x^3+O(4)$.
This solution is based on $g(x)$ having a fixed point. Technically the question stated is whether there is always a solution and is focused on counter examples instead of a broad general answer as I have provided. But it does raise the question of whether there is a connection between $g(x)$ not having a fixed point and the counter examples that the other authors have provided.
I could have sworn that there was an old Monthly article that discussed precisely this question in some detail, but the closest that I could find in a few minutes on Mathscinet is the following article addressing the case $g(x) = 1/x$:
MR1641972:
Cheng et al, &When does $f^{-1} = 1/f$?&, Amer. Math. Monthly 105, number 8.
A variant of the question is: Suppose that $g$ is a diffeomorphism. Can you embed $g$ into a flow? If yes, then there is $f$ with $f\circ f\circ\dots\circ f=g$ ($n$ times for any $n$).
So let $Diff_c(M)$ be the regular Lie group of all diffeomorphisms of a smooth manifold $M$.
Its Lie algebra is the space $\mathfrak X_c(M)$ of all smooth vector fields with compact support, with the negative of the usual Lie bracket. The exponential mapping is the flow mapping which maps a vector field $X$ to its flow $t\mapsto Fl^X_t\in Diff_c(M)$.
It satisfies $T_0\exp = Id$ but:
It is not locally surjective near $Id_M$. This has been shown by Freifeld 1967 and by Koppell 1970. The strongest result is by Grabowski, 1988, who showed the following:
Suppose that $\dim M\ge 2$. Then there exists a smooth curve through $Id$ in $Diff_c(M)$ such that the points of this curve (sauf the identity) are free generators for a free subgroup of $Diff_c(M)$ (on $2^{\aleph_0}$ generators) which meets the image of the exponential mapping only at the identity. This free subgroup is arcwise connected.
See page 456 of , also for exact references.
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MathOverflow works best with JavaScript enabled设二次函数f(x)=ax^2+bx+c在区间[-2,2]上的最大值、最小值分别是M，m，集合A={X/f(x)=x}._百度知道
设二次函数f(x)=ax^2+bx+c在区间[-2,2]上的最大值、最小值分别是M，m，集合A={X/f(x)=x}.
若A={2}，且a&=1，记g(a)=M+m，求g(a)的最小值。求解啊！！谢谢啊！！
提问者采纳
解：f(x)=ax²＋bx＋c=x只有一个解x=2 则f(x)-x=a(x-2)²
即:二次函数顶点在(2,2) 所以f(x)=a(x-2)²+x=ax²-(4a-1)x+4a
对称轴为直线x=(4a-1)/(2a)=2 - 1/(2a)
由于a≥1,所以1/(2a)∈(0,1/2],对称轴x=2 - 1/(2a)∈[3/2,2)
二次函数f(x)开口向上，所以在区间[-2,2]上的最大值M=f(-2)=16a-2
最小值m=f[2 - 1/(2a)]=……=2-1/(4a) 所以g(a)=M-m=16a-2-2+1/(4a)=16a+1/(4a) -4 易证当a≥1时，g(a)为增函数，所以g(a)的最小值为g(1)=16+1/4 -4=47/4
只做第二问吧
因为，a>=1, 所以f(x)=ax^2+bx+c是一个开口朝上的抛物线， 因为A={2}只有一个元素，所以f(x)=x只有一个解，那么应该是y=x与y=ax^2+bx+c 相切于x=2.。所以f'(2)=4a+b=1. 又因为f(2)=2, 带入方程得到4a+2b+c=2
因为抛物线的对称轴 x0=-b/(2a)=(4a-1)/2a=2-(1/2a) 因为a>=1, 所以 3/2
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出门在外也不愁函数界限问题 f(x)&=g(x) x∈D 求证sup f(x)&=sup g(x) ;inf f(x)&=inf g(x), 谢谢帮助_百度知道
函数界限问题 f(x)&=g(x) x∈D 求证sup f(x)&=sup g(x) ;inf f(x)&=inf g(x), 谢谢帮助
你好 可以用反证法先假设sup f(x)& sup g(x) 当x∈D因为有理数的稠密性，两个上确界之间必然有一个有理数a所以sup f(x) & a & sup g(x)因为 sup f(x) & a，根据上确界的性质，必有x0∈D,使得f(x0)&a因为 sup g(x) & a，根据上确界的性质，g(x0) &= sup g(x) 所以g(x0)&a综合起来，有f(x0)&g(x0)，与f(x)&=g(x) x∈D矛盾，所以命题正确至于下确界的命题也是同理话说其实我是数分初学者~好开心
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求k的取值范围
f(x)是定义在R上的减函数。若不等式f(1+kx-x^2)&f(k+2)与不等式f(3k-1)&f(1+kx-x^2)在x&[0,1]上成立，求k的取值范围。
解：(1)∵f(1+kx-x²)＞f(k+2), f(x)为减函数
∴1+kx-x²＜k+2 ==& x²-kx+k+1＞0对x∈[0,1]成立
设g(x)=x²-kx+k+1, 则
Δ＜0; f(0)＞0, k/2≤0; f(1)＞0, k/2≥1; 解得k∈(-1,+∞)
(2)∵f(1+kx-x²)＜f(3kx-1), f(x)为减函数
∴1+kx-x²＞3kx-1 ==& x²+2kx-2＜0对x∈[0,1]成立
设q(x)=x²+2kx-2, 则
Δ＞0, q(0)＜0, q(1)＜0; 解得k＜1/2
∴k∈(-1,1/2).
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