I saw a question that asked for the possible values of $f(-1)$ if $(f\circ f)(x)=4x-12$. The solution uses the fact that if the degree of $f$ is $n$, the degree of $f\circ f$ is $n^2$, and concludes that $f$ must be linear i.e. be of the form $ax+b$. The rest is easy. I understand that this is the case if $f$ is a polynomial, but the question doesn’t state that. What if there exists a weird function with logarithms, trigonometric functions etc. that is not linear but its composition with itself is linear? In other words, maybe the composition “cancels out” the weirdness. Obviously, involutions are a trivial example since the composition is $x$. My question is: Is there a function $f$ that is not linear, but $f\circ f$ is linear and is not the identity function?
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I'm thinking of $f$ as a function either from the reals to the reals or from the naturals to the naturals.   Edit: Okay, what if $f$ is from the naturals to the naturals?
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The question is, how many times does $f(C_r)$ go around $0$ for $C_r = \{x \in \mathbb{C}: |x|=r\}$ and a polynomial with complex coefficients $f$.    The answer is clear to me when $f(x) = ax^n + b$ but I don't see how exactly middle terms affect the result. How to see this without a graph? Can this be found for any $r$?    Many thanks
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Let $f(x)=x^n+5x^{n-1}+3$ where $n\geq1$ is an integer. Prove that $f(x)$ can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree $\geq1$.   If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that $f(x)$ is irreducible over real numbers. But since this condition is given,therefore, we can't exclude the case when $f(x)$ is reducible over real numbers but not with polynomials of integer coefficients. Anyone with idea how to proceed? Thanks in advance!!
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The question is:     When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?   The most obvious solution is a linear function of the form $f(x)=ax+b$. Is this the only solution?  Edit  I should add that $f:\mathbb R\to\mathbb R$ to the question.
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This is from Putnam:     If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.   Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.  Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.  Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.  Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.
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I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:   $f(x)-1$ is divisible by $(x-1)^3$ $f(x)$ is divisible by $x^3$   To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$. This is divisible by $x^3$, so $d, e, f $  must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$  This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.  From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
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I've read that function composition "generally does not commute."  Not counting compositions involving the identity function, and compositions of a function and its inverse, are there examples of functions on the reals (for example) $f, g$ where $fg = gf$ outside of these cases?
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Does there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x,y \in\mathbb{R}$; $f(1)=1$; and $f(\sqrt 2)=3$?  Intuitively, I believe the answer is no. However, I'm having a hard time proving this. My initial thought was a proof by contradiction. Somehow using  $f(1)=1$ and $f(\sqrt 2)=3$ to show that $f(x+y)\neq(x)+f(y)$ for some choice in $x$ and $y$. This didn't prove fruitful. My next thought was to use properties of functions satisfying the addition property of homomorphisms to prove that $f(\sqrt 2)\neq3$. For example, one such property is $f(nx)=nf(x)$ for all $n \in \mathbb{Z}, x \in \mathbb{R}$. With this, one could theoretically construct a proof by induction that $f(x)=x$ for all $x \in \mathbb{R}$. Once again, this didn't prove fruitful. After these failed attempts at showing such a function doesn't exist, I considered how to show that one does exist. For reference, this question was asked in the context of a Field Theory course. As such, the thought for how to construct a proof is to satisfy the addition property with some linear transformation and show that a correct selection of basis spans all of $\mathbb{R}$. This idea seems feasible, however I am not sure where to begin. Any advice on how to approach this problem is greatly appreciated.
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I'd like to characterise the functions that ‘have square roots’ in the function composition sense. That is, can a given function $f$ be written as $f = g \circ g$ (where $\circ$ is function composition)?  For instance, the function $f(x) = x+10$ has a square root $g(x) = x+5$.  Similarly, the function $f(x) = 9x$ has a square root $g(x) = 3x$.  I don't know if the function $f(x) = x^2 + 1$ has a square root, but I couldn't think of any.  Is there a way to determine which functions have square roots? To keep things simpler, I'd be happy just to consider functions $f: \mathbb R \to \mathbb R$.
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If we define $f(x)$ as some arbitrary function,  then we can define $f'(x)$ or $f^{(1)}(x)$ as the first order and $f''(x)$ or $f^{(2)}(x)$ as the second order.  My question is:  Is there sure thing as a $f^{(-1)}(x)$ notation? Could it be an antiderivative or an integration? And no, I am not talking about $f^{-1}(x)$ that represents an inverse of $f(x)$.
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Suppose that $[F(\alpha):F]$ is odd. Show that $F(\alpha)=F(\alpha^{2})$. Generalize this statement if possible.    Clearly, we always have that $F(\alpha^{2}) \subset F(\alpha)$ by definition of the field generated by $F$ and $\alpha$ (assuming that we have a field extension $K/F$ where $\alpha \in K$). However, how can we show the other inclusion? If the degree is odd, then it follows that $\alpha$ will solve a polynomial of odd degree in $K[x]$ (for the coefficients in $F$). What would be a way to proceed from here? (and as well to generalize the statement?)  I was thinking of using the formula: $[F(\alpha):F]=[F(\alpha):F(\alpha^{2})][F(\alpha^{2}):F]$ and maybe proceed from there.   Thanks for the help.
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Prove that if $[F(\alpha):F]$ is odd then $F(\alpha)=F(\alpha^2)$.  My justification for this question is as follows;  Suppose $F(\alpha^2)\subsetneq F(\alpha)$, we have $F \subsetneq F(\alpha^2) \subsetneq F(\alpha)$.  As $[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ and $[F(\alpha):F(\alpha^2)]=2$ we would be end up with the case that $[F(\alpha):F]$ is a multiple of 2 i.e., even which contradicts given condition of $[F(\alpha):F]$ being odd.   So, $F(\alpha^2) =F(\alpha)$.  After doing this, i have seen for a solution on some webpage where he/she used criterion of minimal polynomial.  Please let me know Is my justification clear or there are any gaps in between?
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I'm studying abstract algebra for the first time.  let $F$ be a field. I learned that polynomials over field $F$ means polynomials with coefficients in $F$ (denoted by $F[x]$). But when I keep studying, I get to confused the word, 'polynomial functions from $F$ to $F$'.  Firstly, I consider this as induced functions from $F[x]$, so I guess all coefficients of these functions are in $F$. But let $F_{4}$ be a finite field consisting of four elements $\{0, 1, a, b\}$. and let $f: \{0,1\} \rightarrow \{0,1\}$ be a function defined by $$ f(x)=ax^2-ax$$ Since $\{0,1\}$ is a subfield of $F_{4}$ and $f(0)=0, f(1)=0$, I think $f$ is a example polynomial functions from $F$ to $F$ of which coefficients are not in $F$. Am I wrong? if so, please help me.  I encounter this problem while solving 'the set of all polynomial functions from $\Bbb{Z_{3}}$ to $\Bbb{Z_{3}}$ cannot possibly be isomorphic to $\Bbb{Z_{3}}[x]$'.
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Let $f$ be a $k$ times continuously differentiable function defined on a neighborhood of $0 \in \mathbb{R}$. Show that if $f(-x) = -f(x) \forall x \in \mathbb{R}^n$, then the coefficients of the taylor series expanded about $0$ has all even coefficients equal $0$.  So this is an odd function, it makes sense. let's write a $k$-degree taylor polynomial:  $$\sum\limits_{i=0}^k \frac{f^{i}(0)}{i!}x^i = f(0) + f'(0)x+\frac{f''(0)x^2}{2} + . . .$$  $f(0) = 0$ we can easily conclude since $f$ is odd, but how can we say anything about any other terms? (since they all involve derivatives)
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$f:\mathbb{R}^3\to\mathbb{R}, f(rx)=r^{\alpha}f(x)$ for some $\alpha&gt;0$, $\forall x\in\mathbb{R}$, for any $r\in\mathbb{R}$, could any one tell me which of the following is true?  $1.$ If $f(x)=f(y)$ whenever $||x||=||y||=\beta&gt;0$, then $f(x)=\beta ||x||^{\alpha}$  $2.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)= ||x||^{\alpha}$  $3.$ If $f(x)=f(y)$ whenever $||x||=||y||=1$, then $f(x)=c||x||^{\alpha}$ for some constant $c$  $4.$ If $f(x)=f(y)$ whenever $||x||=||y||$, then $f(x)$ must be a constant function. Thank you.
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The problem is as follows:     Suppose $f$ entire satisfying   $$ |f(z)| \leq A + B |z|^{3/2} $$   for some fixed $A,B &gt; 0$. Prove that $f$ is a linear polynomial.   I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how.
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My lecturer was talking today (in the context of probability, more specifically Kolmogorov's axioms) about the additive property of functions, namely that:  $$f(x+y) = f(x) + f(y)$$  I've been trying to find what functions satisfy this. Intuition says that, for functions over $\mathbb{R}$, the only functions should be of the form $f(x) = ax$ for some real a. Unfortunately I've only shown this is true when the domain of the function is the rational multiples of a given real number.   My question is if it is possible to extend this result (that $f(x) = ax$ given additivity) to the real numbers, possibly without assuming the continuity of f. It seems to me that additivity introduces so many constrains on a function that nothing but the trivial case would be able to sneak through. The following is a summary of my thoughts to date, though they're obviously long and not 'compulsory reading'. :)  When x is rational - Preliminary Investigation  It is not hard to see that:  $$f(x + x + x) = 3f(x)$$  and more generally, for $a\in \mathbb{N}$,  $$f(ax) = af(x)$$ It is not too hard to prove (well, it took half a bus trip ... ) that this also applies first for $a\in \mathbb{Z}$ and then for $a\in \mathbb{Q}$, (for the latter you just need to consider $a=m/n$ and then note that:  $$f\Big(\frac{m}{n}x\Big)=mf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot nf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot f\Big(n\frac{x}{n}\Big)=\frac{m}{n}\cdot f(x)$$   The reason this little equation is cool is that we can set $x = 1$ and get:  $$f(a)=a\cdot f(1)$$  which is equivalent to what was expected intuitively, namely (after changing $a$ to $y$ and $f(1)$ to $a$)  $$f(y) = a\cdot y$$  as long as y is rational   $y$ is a rational multiple of a real number  But we can do a bit better than that. If we substitute in $x = \sqrt{2}$ or any other real number in $f(ax) = af(x)$ (which we know for rational $a$), you can conduct the exact same argument above and show that, for instance   $$f(y) =  \Big(\frac{f(\sqrt{2})}{\sqrt{2}}\Big)\cdot y=a\cdot y$$  Whenever $y = \frac{m}{n}\sqrt{2}$ i.e. whenever $y$ is a rational multiple of $\sqrt{2}$. Note however, that the value of the coefficient $a$ (i.e. the slope of the line) is apparently completely unrelated to the value taken in the case where $y$ is purely rational.   What I'm actually asking  We still haven't shown that $$f(x) = ax$$ for all $x \in \mathbb{R}$, as the slope of the line may change depending on what real number we are taking rational multiples of. As far as I've shown now, we might have $f(x) = x$ when $x$ is rational, $f(x) = 3x$ when $x$ is a rational multiple of $\sqrt{2}$, etc.  I still feel that $f(x) = ax$ for all $x \in \mathbb{R}$. One reason for thinking this comes from noting that $$f(2) = f(2-\sqrt{2})+f(\sqrt{2})$$  $2$, $2-\sqrt{2}$ and $\sqrt{2}$ are not rational multiples of each other, however the equation above gives a restraint on the slopes of the lines formed by their rational multiples (which we'll call $a_1, a_2$ and $a_3$ for the slopes on the rational multiples of $2, 2-\sqrt{2}$ and $\sqrt{2}$ respectively). We have $2a_1 = (2-\sqrt{2}) a_2 + \sqrt{2} a_3$  There's so many constraints here - all the rational multipes have the same coefficient, whenever 2 (or more) numbers which aren't rational multiples of each other are added together we get another constraint on their coefficients. The trivial solution is just that$$f(x) = ax$$  over $x \in \mathbb{R}$ and I really struggle to see how any other solution could possible squeeze through all these constraints.   Is there an additive function on $\mathbb{R}$ not of the form $f(x) = ax$?