# Tag Info

1

There are many functions for which this condition holds, in general for function $f(x)$, and $0\leq\theta\leq 1$, if: $$\theta f(z)+(1-\theta) f(y)\geq f(\theta z+(1-\theta)y)$$ then the function is called convex. In your case, $\theta=1/2$, $z=x+1$, $z=x$. You can find a list of convex functions here.

0

Your condition says that the function's value at $x+1$ is less than the average of its values at $x$ and $x+2$. In other words, at $x+1$, the function value must be below the "straight line" values between $x$ and $x+2$. Functions with this sort of property are called "convex". Basically, this just means that the graph of the function looks like a valley, ...

0

This condition is satisfied by any convex function, i.e., a function satisfying $$f\big(tx+(1-t)y\big)\le tf(x)+(1-t)f(y),$$ for all $t\in (0,1)$. If $f$ is twice differentiable, then it is convex if and only if $$f''(x)\ge 0,$$ for all $x$. HENCE, try a function with non-negative 2nd derivative.

1

$f(x) = x^2$ will do the job. \begin{align} \frac{f(x+2)+f(x)}{2} &= \frac{(x+2)^2 + x^2}{2} \\ &= \frac{x^2 + 4x + 4 + x^2}{2} \\ &= x^2 + 2x + 2 \\ &> x^2 + 2x + 1 \\ &= (x+1)^2 \\ &= f(x+1). \\ \end{align}

0

Yet another Counter example, this time non constant $$f(x) = atan(x) + c$$ Quite obviously it's mean should be $c$

0

Take $f(t)$ to be a constant, say $c$. Then the integral equals $c$ for all $T$, thus the limit and $M$ are equal to $c$.

0

Trying special cases is always useful. $f(t) = 1$ is one of the simplest options. And there's a rather intuitive idea of what its mean should be too! If $f$ has a limit at $+\infty$ and $-\infty$, then I'm pretty sure the mean turns out to be $$\frac{f(+\infty) + f(-\infty)}{2}$$ whenever defined. (i.e. this formula makes no prediction about the mean of ...

0

As far as I know, there is no closed form for the indefinite integral. It can be expressed as an infinite series of hypergeometric functions.

0

I like $f(x) = 1+k x^r$ for $k>0$ and $r>1$. It's derivative is positive on $(0,\infty)$. The only place it takes values near $1$ is near $x=0$, so it will not have a subsequence converging to $1$ anywhere else. And it trivially satisfies the limit... It's also a power law and not an exponential.

0

Note: not a full answer, but something which may lend a hand: Let $t=4m,\ s=4n,\ x=d,$ and define $$g(m,x)=(1-x)(1-x^{2m-1}),\quad h(n,x)=(1-x^n)^2.$$ Then your function is $$f(x)=\frac{g(m,x)}{g(m,x)+h(n,x)}=\frac{1}{1+h(n,x)/g(n,x)}.$$ Then $f(x)$ will increase or decrease iff $h/g$ respectively decreases or increases, so to keep things easier we can ...

0

Hint: per definition you know that $(f^{-1})^{-1}\circ f^{-1}=\mathrm{id}_Y$.

0

Assume $(f^{-1})^{-1}=g \$ where $g$ is some function. Therefore, you have $gf^{-1}=f^{-1}g=e \$ where $e$ is the identity function such that for any function $h$, $eh = he = h$. Now, you can do: $$f^{-1}g = e \\ ff^{-1}g = fe \\ g = f$$ There you have it.

0

The additional term comes from the relation $(x-y)k(x+y)-(x+y)k(x-y)=0$ for any $k$

4

$\Rightarrow \frac{f(x+y)}{x+y} -\frac{f(x-y)}{x-y} =4xy=(x+y)^2-(x-y)^2$ or, $\frac{f(x+y)}{x+y}-(x+y)^2=\frac{f(x-y)}{x-y}-(x-y)^2$ make the substitutions, $x\to \frac{x+y}{2}$ and $y \to \frac{x-y}{2}$ Then the above becomes $\frac{f(x)}{x}-(x)^2=\frac{f(y)}{y}-(y)^2=k(say)$ There you have $f(x)=x^3+kx$

0

Who says that won't be equal to x? $9x^2+6x+1=(3x+1)^2$. Plugging that in, we see that the last line is equal to x!

0

It is exactly equal to $x$, because $9x^2+6x+1=(3x+1)^2$

1

Do you mean a polynomial $P(x,y)$ of total degree $3$ such that the curve $P(x,y) = 0$ is only in one half-plane? For example, $x y^2 + x - 1 = 0$ is only in the first and fourth quadrants, while $x^2 y + y - 1$ is only in the first and second.

3

For a function to not have a $y$-intercept, the number $0$ would have to not be in its domain. $0$ is the the domain of every polynomial, including every cubic function, so they all have $y$-intercepts.

1

Let $g:A\to B, f:B\to C$. Then, $f\circ g:A\to C$. This is also $f:(g(A))\to C$. So, if $x\in A$, then $f\circ g$ is a function of $x$, since $g(x)$ is a function of $x$. However, if we were to take $x\in B$, then $f\circ g=(f\circ g)(y)$, $y\in A$, is a function of $g(y)$ (note the abuse of notation), as $g(y)\in B$.

1

The right way to see this notation is that, evaluated in $x$, $f\circ g$ takes the value $$f\circ g(x) = f(g(x))$$ that is, $f$ evaluated in $g(x)$. $f$ is not a function operating on functions, but on numbers. This is why you note: $$f:E\to F$$ meaning that $E$ is a way to associate an element of $F$ for each element of $F$. An function operating on ...

0

It means that all of the given statements are actually false in general, and the question asks you to find a case (for example via a specific choice for $f$ and $g$) such that the statement is false. For example, for (i), you can pick $f(n) = n$ and $g(n) = \frac{n}{2}$ to see that the implication does not hold.

0

Take $X=\{1,2\}$, $Y=\{1\}$, and $f(1)=f(2)=1$. Not true in general. Then $$\mathscr P(X)=\big\{\varnothing,\{1\},\{2\},\{1,2\}\big\}$$ and $$\mathscr P(Y)=\big\{\varnothing,\{1\}\big\}$$ Clearly $$f^{-1}(\varnothing)=\varnothing\quad\text{and}\quad f^{-1}(\{1\})=\{1,2\}.$$ Thus $\{1\}$ and $\{2\}$ DO NOT belong to the range of the inverse image ...

0

If $t<s$ then the integral is trivially zero. Let us assume from now on that $t\geq s$. Then $$\int_{-\infty}^{\infty}\!dx~ e^{2a(x-y)}~1_{|x-y|<t-s} ~\stackrel{z=x-y}{=}~\int_{-\infty}^{\infty}dz~ e^{2az}~1_{|z|<t-s}~=~\int_{s-t}^{t-s}dz~ e^{2az}$$ $$~=~\left\{\begin{array}{ccc}\left[ \frac{e^{2az}}{2a}\right]_{z=s-t}^{z=t-s} ... 0 First, you have an error in the expression for derivative. It should be$$f'(x)=1+A\frac{x^2+2}{x^2}-B\frac{x^2-6}{x^4}$$Second, consider$$g(x)=x^3f(x)=(1+A)x^4+(B-2A)x^2-2B$$Since we assume f'(\sqrt{2})=f''(\sqrt{2})=0:$$g'(\sqrt{2})=(3x^2f(x)+x^3f'(x))|_{x=\sqrt{2}}=6f(\sqrt{2})=6\sqrt{2}$$... 0 Solved it via the method suggested in the body of the question. the solution is: A=\frac{-5}{8}, B=\frac{1}{4}, and p=2. 3 Try f defined by f(x)=\frac{1}{2} for all x\in\mathbb{R}. Note that by definition f(x)=f(-x) for all x. 1 A simple example:$$f(x) = \begin{cases} 0, & x < 0 \\ \frac12, & x = 0 \\ 1, & x > 0 \end{cases}$$If you want the function to be continuous, maybe:$$f(x) = \frac12\left(1+\frac x{1+|x|} \right)$$0 Function f I'd add a 1 as a fourth coordinate (homogenous coordinates) so that you can include the addition in your matrix multiplication. Then you'd have$$M = \begin{pmatrix} 1/\sqrt2 & 1/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ -1/\sqrt2 & 1/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ 0 & -2/\sqrt6 & 1/\sqrt3 & -\pi/4 \\ 0 & 0 & 0 ...

1

You have $$g\left(\frac{2y}{1-y}\right)=\frac{\frac{2y}{1-y}}{\frac{2y}{1-y}+2}=\frac{\frac{2y}{1-y}}{\frac{2y+2(1-y)}{1-y}}=\frac{2y}{2y+2(1-y)}=\frac{2y}{2}=y.$$

1

Must be some calculation error. The inverse is actually: $\frac{2y}{1-y}$. If you substitute this and check, then you will find it satisfying the equation :$g({\frac{2y}{y-1}})=y$

0

Let $s=\sin(t)$, $c=\cos(t)$, $a=6\exp(-14t)$, $r=r(t)$. We have $r'=(a'c-as,a's+ac)$. Now observe that $a'=-14a$, hence $r'=a(-14c-s,-14s+c)$. From here $\|r'\|^2=197a^2$. Finally the unit normal is $$\frac{1}{\sqrt{197}}(-14c-s,-14s+c).$$

1

It is enough to expand the definitions, for example: \begin{align} (f\cdot g)(x) &= f(x) \cdot g(x) \tag{$\spadesuit$} \\ (f \circ g)(x) &= f\big(g(x)\big) \tag{$\clubsuit$} \\ \big((f\cdot g)\circ h\big)(x) &\stackrel{\clubsuit^\to}= (f \cdot g)\big(h(x)\big) \\ &\stackrel{\spadesuit^\to}= f\big(h(x)\big)\cdot g\big(h(x)\big) \\ ...

0

Two functions $f$ and $g$ are defined to be equal when $f(x) = g(x)$ for every $x$ in their domain. In your exercise, to show that $(f + g) \circ h = f \circ h + g \circ h$, you need to show that $$((f+g) \circ h)(x) = (f \circ h + g \circ h)(x)$$ for all $x \in \mathbb{R}$. If you systematically use the general definitions of $f+g$, $fg$ and $f \circ g$, ...

0

Let $B^A$ denote the set of all the maps from $A$ to $B$ Claim : If $A,B$ are finite then $\#( B^A)= \#B^{\#A}$ Proof: We induct on the cardinality of $A$. If the set is empty only we have the empty function (why?) and so $\#( B^A)= \#B^{\#A}$. Now suppose we have already proven the assertion for all sets of size $n$. Let $\#A=n+1$. The set is non-empty, ...

0

According to the definition of Arora, trigonometric functions are not good counter-examples since they are not functions from $\mathbb{N}$ to $\mathbb{N}$. However, the function $UC : \mathbb{N} \rightarrow \{0,1\} \subset \mathbb{N}$ defined by $UC(n) = 1 \Longleftrightarrow M_n(n) \neq 1$ is not time-constructible, since not constructible at all! (for a ...

2

The only functions which have this property are the constant functions. Why? Suppose $f(x) \geq f(m)$ for all $x$. Then $x=m$ is the global minimum of $f$. Likewise, if $f(x) \leq f(M)$ for all $x$. Then $x=M$ is the global maximum of $f$. So if $f(M)=f(m)$, we have $f(m) \leq f(x) \leq f(M)=f(m)$ for all $x$. This forces $f(x)=f(m)=f(M)$ for all $x$ so ...

2

If $a$ is such that $f(a) \leq f(x)$ and $f(x) \leq f(a)$ for all $x$ in the domain of $f$ we have $f(a) \leq f(x) \leq f(a)$ which implies $f(a) = f(x)$, that is $f$ is constant.

2

Hint. Suppose that the absolute maximum value is $M$. By definition this means that $f(x)\le M$ for every value of $x$. If the same $M$ is the absolute minimum value, this means that. . . Can you finish this and determine what $f(x)$ is?

3

$$\large{\large{\large{\large{\large{-}}}}}$$

1

The first part is simple. Suppose $f(x) = ax^n$for an even integer $n$. $f(-x) = a(-x)^n = ax^n = f(x)$ since $n$ is even. As for the second part, suppose $h(x) = f(x) + g(x)$ for each $x$ in the domain of $h$, where $f$ and $g$ are even functions. Then, $h(-x) = f(-x) + g(-x) = f(x) + g(x) = h(x)$ If you are unsure about the definition of an even ...

3

Part 1) $$f(x)=ax^n$$ So, let $n=2k$. Then, $$f(x)=ax^n=a(x^2)^k$$ Obviously, $x^2$ is an even function ($(-x)^2=x^2$). So, $$f(x)=ax^n=a(x^2)^k$$ is an even function. Part 2) Let $f(x),g(x)$ be two even functions. Then, $$f(x)=f(-x),\qquad g(x)=g(-x)$$ Adding the two gives $$f(x)+g(x)=h(x)$$ Using the above relation then gives ...

1

Let $K_0$ be the field of rationals. For any $n\ge 0$, let $K_{n+1}$ be the intersection of all the subfields of the reals that contain $f(K_n)$. Finally, let $K$ be the union of all the $K_n$. Note that $K$ is closed under $f$. For if $a\in K$, then $a\in K_n$ for some $n$, and therefore $f(a)\in K_{n+1}$. For countability, note that each $K_n$ is ...

0

Take the subfield generated by $\{f^n(1)\}_{n\in\mathbb N}\cup\{1\}$, where $f^n$ is the $n$-fold composition of $f$.

4

You have found the solution for the first derivative, the idea is then to use this solution to bootstrap yourself to the next derivative. First of all, you may assume that $f(0) =0$, since you may always translate the function vertically without affecting zeroes of the derivatives. Thus the question can be thought of equivalently as: If ...

0

Use $$\log(x)=\int_1^x\frac{dt}{t}.$$ For any $x>0$ the inequality $$\frac{1}{t}\leq 1-\frac{(t-1)}{x}$$ holds for all $t$ between $1$ and $x$ inclusive. Therefore if $x\geq 1$ then $$0\leq\log(x)\leq\frac{(x^{-1}+1)(x-1)}{2}=\frac{x-x^{-1}}{2}$$ and if $0<x\leq 1$ then $$0\geq \log(x)\geq\frac{x-x^{-1}}{2}.$$ The desired result for positive $x\neq ... 4 Under the assumption that$\epsilon$is constant across$a$, then no additional assumptions are needed. Suppose that$x_a$is the fixed point of$f_a$, and choose$e>0$. Then there is a$\delta$such that for all$b$within$\delta$of$b$, you have$e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert$. Now, repeatedly apply$f_b$to ... 2 You can use the contraction mapping estimates directly. You have the estimate$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} \|f_a(x_0) - x_0\|$, so we can see that if we let$B = \sup_{a \in B(\hat{a},1)} \|f_a(x_0) - x_0\|$, then$\|\bar{x}_a - f_a^{(k)}(x_0)\| \le {(1-\epsilon)^k \over \epsilon} B$for all$a \in B(\hat{a},1)$. So, ... 2 We can view$\bar{x}_a$as a minimizer of the continuous function$x\mapsto \|f_a(x)-f_a(f_a(x))\|$. If$f$is jointly continuous as a function of$\mathbb{R}^n\times\mathbb{R}$, then the argmin correspondence that maps$a$to the set of minimizers of this functions is upper hemi-continuous by Berge's maximum theorem (one has to show that locally all ... 1 HINT: For every$a_i$in$A$, there are$n$elements in$B$to which it can map. Try to proceed from here. SPOILERS BELOW Let$A=\{a_1,a_2, \cdots,a_m\}$and$B=\{b_1,b_2, \cdots,b_n\}$There are$n$possible values for$f(a_1)$, namely,$b_1,b_2, \cdots,b_n$. Similarly, there are$n$possible values for each$f(a_i)$. Thus net number of ways to map ... 1 The set of all functions$A\to B$is sometimes denoted$B^A$, exactly because when$A,B$are finite then there are$|B|^{|A|}$(in the question$n^m$) functions. This is so because for every element$a\in A$there are$|B|$possibilities for$f(a)$, and these$|A|$choices are entirely independent and together determine$f\$. Numbers of independent choices ...

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