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2

Just found the answer: Replace $f\left(x,y\right)$ by $\left(u,v\right)$ resulting in: $x+y+1=u$ and $x-y-1=v$ Switch $x$ and $u$ and switch $y$ and $v$ resulting in: $u+v+1=x$ and $u-v-1=y$ Solve for $u$ and $v$ resulting in: $u=\frac{1}{2}x+\frac{1}{2}y$ and $v=\frac{1}{2}x-\frac{1}{2}y-1$ Replace $\left(u,v\right)$ with $f^{-1}\left(x,y\right)$ ...


-1

(i) Any uniformly continuous function, e.g. $f(x) = x$. (ii) $f(x) = \text{constant}$.


1

(A) $f(x)=x$ should work (B) Any constant function should work as then $f(x)=f(y)$ and $|x-y|\ge 0$


1

Your intuition is accurate. Following the hint, let $m=2x+y$ and $n=x-2y$. We know that $m$ and $n$ are natural numbers, and we’d like to use this information to say something about the possible values of $x$ and $y$. That’s reminiscent of a more familiar situation: if we actually had specific values for $m$ and $n$, we could solve those two equations for ...


1

I don't know about topological spaces. But given $f:X\rightarrow Y$ and $A,B \subseteq X$ here's an attempt for proving that $$ f \: \text{injective} \rightarrow f(A \cap B) = f(A) \cap f(B),$$ where $$ (f(A \cap B) = f(A) \cap f(B)) \leftrightarrow ( \underbrace{f(A \cap B) \subseteq f(A) \cap f(B)}_{(i)} ) \wedge ( \underbrace{f(A \cap B) \supseteq ...


0

A formula is a string of symbols, arranged according to mathematical grammar. A function is a mathematical object that plays a role in arithmetic operations like "evaluation" or "composition". A key point is that if $f$ and $g$ are expressions that denote two functions with the same domain and codomain, and we have $f(x) = g(x)$ for every $x$ in the domain, ...


1

A function is a map from one set to another. They are the same if both the domain and the 'formula' are the same. A formula on the other hand is a word physicists and chemists like to use for a function that expresses a relation between variables that arise in nature.


1

Depending on how the discrete function is defined, there might be simple ways to extend it to a continuous function on $\mathbb R$, e.g. if $f: \mathbb N\to\mathbb R$ is defined as the constant $65$, it's quite obvious to consider the function that is constant $65$ on $\mathbb R$ as an extension, but $65\sin(\pi x)$ is also a choice. In some cases the ...


1

In general, there any many ways for extending a function $f:\mathbb{Z}\to\mathbb{R}$ in such a way that $g$ is a continuous function and $g_{|\mathbb{Z}}\equiv f$. For the $\Gamma$ function, the uniqueness of the extension follows from requiring that for any $z\in\mathbb{R}^+$, $g(z+1)=z\, g(z)$ (the same functional identity satisfied by $f$ on $\mathbb{N}$) ...


3

To note that the only two possible values are $0$ and $1$ is a good start. Whatever you did to get the limits is false. Assume the function takes the value $0$ for some $x$ and $1$ for some $y$ and apply the intermediate value theorem to get a contradiction to the function taking only the values $0$ and $1$. Then you will know the function is either ...


6

A continuous function $\mathbb{R} \to \mathbb{R}$ whose image consists of finitely many points is constant. Indeed, if $f: \mathbb{R} \to \mathbb{R}$ is continuous and not constant, there are $a,b \in \mathbb{R}$ with $f(a) < f(b)$. By the intermediate value theorem, $f$ will take all values in the interval $[f(a),f(b)]$, which are infinitely many ...


1

It follows from an application of Jensen's inequality to the random variable $g(X)$ (provided that $g$ is Borel measurable) instead of $X$.


0

From $g(x)=f(x)-x$, we deduce that $g$ is continuous on $[a,b]$. Moreover, we see that $g(a)=f(a)-a\ge 0$, $g(b)=f(b)-b\le 0$, and so $g(a).g(b)\le 0$. This implies that there exists a $c\in [a,b]$ such that $g(c)=0$, or $f(c)=c$.


1

Hint: Suppose you had found such an $a$. Then $f$ is continuous, as is $x\mapsto \cos (ax)$. Thus, $x\mapsto \frac{f(x)}{\cos(ax)}=\lfloor x \rfloor$ is continuous everywhere where $\cos(ax)\neq 0$. Use this to determine $a$ (or show that the condition can't be fulfilled by any $a$!)


0

Your proof that there are two solutions is correct (if you remember to argue that you can use the mean value theorem. From your calculations you know that one of the solutions are in the interval $[0,1]$, use the bisection methos on that.


0

We have $$|f(x)|=|x(x-1)|\xrightarrow{x\to a}0=f(a)$$ where $a$ is $0$ or $1$. Hence we proved that $f$ is continuous at $0$ and $1$. Now for $a\not\in\{0,1\}$ and using the sequential characterization of the limit we prove easily that $f$ hasn't a limit on $a$.


3

Your choice of $f$ is fantastic. We just need to show that $\lim_{x \to c} f(x)$ doesn't exist whenever $c \neq 0$ and $c \neq 1$. To this end, suppose that $c \in \mathbb{R} \setminus \{0, 1 \}$. Let $s_n$ be a sequence of rationals converging to $c$, and $t_n$ be a sequence of irrationals converging to $c$. What can we conclude? Hint: Consider the ...


-2

may be answer can be( n!*m!/n!(m-n)!*n^m-n)/2


0

http://en.wikipedia.org/wiki/Logit Looking for the same thing too. This looks promising.


4

Notice that: $\begin{align} f(1-x)=&\dfrac{f(1-(1-x))-f(1-x)}{1-2(1-x)}\\ =&\dfrac{f(x)-f(1-x)}{1-2+2x}\\ =&\dfrac{f(1-x)-f(x)}{1-2x}=f(x) \end{align}$ Which would show that the only solution to that equation would be $f\equiv0$.


3

By convention, $\sqrt x$ doesn't just mean any number whose square is $x$ -- it means the positive (or zero) number whose square is $x$. In particular $\sqrt 4$ is $2$ and only that. It is true that $(-2)^2$ is also $4$, but because $-2$ is not positive, it doesn't satisfy the condition for being $\sqrt 4$.


0

In this case it is so called principal square root, i.e. the nonnegative root of equation $x^2=y$. See also http://en.wikipedia.org/wiki/Square_root.


0

Functions send each $x$ to only one $y$ value. So while it is true that the root of a real number can be positive or negative, in order to graph as a function, the convention is to choose the positive root. However, you'll notice the inverse of the function $y= {x}^{1/2}$ should also be a parabola, as it is formed by reflecting over the line $y=x,$ but one ...


0

Triangles ABB 'and AMM' are similar $\Rightarrow \frac{y-y1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$ is the equation of a line through the points A, B. M (x, y) is the generic point of the search line.


3

This set describes a line. The mapping $x \mapsto (x,5-x)$ is a bijection from $\mathbb{R} \to L$, so L has the same cardinality as $\mathbb{R}$. In other words, $|L|=\mathcal{c}$.


-2

If $f$ is Riemann integrable on $[a,b]$, then $f$ is Lebesgue integrable on $[a,b]$, and the Riemann and Lebesgue integrals are equal. It is fairly straightforward to show that a nonnegative measurable function has Lebesgue integral zero if and only if it is zero almost everywhere. Since we have the additional assumption that $f$ is continuous, if $f$ is ...


3

Suppose that $f(x) = \varepsilon 0$ for some $x \in [a,b]$. Because $f$ is continuous there exists $\delta$, that $f(y)=\frac{\varepsilon}{2}>0$ for $y \in [x-\delta,x+\delta]$, so: $$\int_{a}^{b}f(x)dx=\int_{a}^{x-\delta}f(x)dx+\int_{x-\delta}^{x+\delta}f(x)dx+\int_{x+\delta}^{b}f(x)dx$$ Now $\int_{a}^{x-\delta}f(x)dx\geq0$ and ...


6

Let $$\Phi(t)=\int_a^tf(x)dx$$ then $\Phi'(t)=f(t)\ge0$ hence $\Phi$ is non-decreasing but since $\Phi(a)=\Phi(b)=0$, this means $\Phi$ is constant and $\Phi'(t)=f(t)=0$.


-1

Other answers build a summation and it isn't necessary. Here is a solution exclusively using the repertoire method and in the same spirit as 1.18 in the book Let $$g(n)=A(n)α+B(n)γ+C(n)β_0+D(n)β_1 $$ Recall that $(\alpha, \gamma, \beta_0, \beta_1) \to (\alpha, 0, \beta_0, \beta_1)$ for $n = (b_mb_{m−1}...b_1b_0)_2$ is the radix changing solution ...


4

Yes, it’s continuous. Let $U$ be any open set in $\Bbb R$. If $k\notin U$, then $f^{-1}[U]=U\times\{0\}$, which is open in $\Bbb R\times\{0,1\}$, and if $k\in U$, then $f^{-1}[U]=(U\times\{0\})\cup(\Bbb R\times\{1\})$, which is also open in $\Bbb R\times\{0,1\}$.


2

The number of turning points is the number of solutions to the equation $${dy\over dx}=0$$ and the derivative of a polynomial has one less power of $x$ than the polynomial itself. The number of intersections with the x axis is the number of roots of the polynomial. The statement that a polynomial of order $n$ has $n$ roots is called the fundamental theorem ...


4

A degree $n$ polynomial has at most $n$ roots (intersections with the $x$ axis). A "turning point" is a place where the derivative of the polynomial is zero (though not every place the derivative vanishes is a turning point), and since the derivative of a degree $n$ polynomial is a degree $n-1$ polynomial, there are at most $n-1$ turning points.


1

Note that $f(x)=2x-3^{-x}$ is continuous and strictly increasing, so there'll be a (unique) solution if and only if $$f(0)\le 0\le f(1)$$ I believe you can check that.


1

$f(x) = 2x- 3^{-x} \text{ is continuous and } f(0) = - 1 < 0, f(1) = 2 - 3^{-1} = 5/3 > 0$. $\text{ By IVT}$ , $\exists c \in (0,1): f(c) = 0 \to 2c - 3^{-c} = 0 \to c \text{ is a solution in } [0,1]$. $\text{ For another example the OP wants to show that the equation } e^x - x^2 + 5x - 3 = 0 \text{ has a solution on } [0,1]. \text{ Its the same ...


1

Split the integral into intervals where $\lfloor x\rfloor$ is constant: $$\int_0^\infty 10^{-2\lfloor x\rfloor}dx = \sum_{n=0}^\infty\int_{n}^{n+1} 10^{-2\lfloor x\rfloor}dx = \sum_{n=0}^\infty\int_{n}^{n+1} 10^{-2n}dx = \sum_{n=0}^\infty 10^{-2n} = \frac{100}{99}$$


1

Using $f^{-1}\circ f=\operatorname{id}$ and applying the chainrule we find: $$g'\left(x\right)=\left(g\circ f^{-1}\circ f\right)'\left(x\right)=\left(g\circ f^{-1}\right)'\left(f\left(x\right)\right)f'\left(x\right)$$ Divide by $f'(x)$ to achieve: $$\left(g\circ f^{-1}\right)'\left(f\left(x\right)\right)=\frac{g'\left(x\right)}{f'\left(x\right)}$$ ...


1

Set $S \cap T$ is bounded. Let $x \in S \cap T$. Then $x \in S$, so $x \geq -1$ (it's because $(t-1)^=t^2-2t+1 \geq 0$ for all $t \in \mathbb{R}$) and $x \leq 101$ (because for $t>101$ we have $t^6-t^5=t^5(t-1)>100$). So $x \in [-1,101]$, thus $S \cap T \in [-1,101]$. Set $S \cap T$ is closed. Note that $S$ is preimage of set $(-\infty,100]$ under ...


0

Hint on being closed: $T$ can also be written as: $\{x^2-2x:x\in[0,\infty)\}$ since $0=2^2-2.2\in T$ So both sets $S$ and $T$ are preimages of closed sets under continuous functions. Hint on being bounded. If a set is bounded then so are its subsets.


2

$f'(2)=4$, so $\bigl(f^{-1}\bigr)'\bigl(f^{-1}(2)\bigr)=\dfrac{1}{4}$. Btw, there's a typo at the beginning of the formula: it's $\bigl(g\circ f^{-1}\bigr)'(2)$ that is being computed. More generally, you can prove that $$\bigl(g\circ f^{-1}\bigr)'(x)=\dfrac{g'\bigl(f^{-1}(x)\bigr)}{f'\bigl(f^{-1}(x)\bigr) }. $$ Indeed, the chain rule results ...


6

In general, if $f\in C^k,$ then the derivatives $f',\ f'',\ \dots,\ f^{(k)}$ exist and are continuous. Here is the reference.


1

$f$ is twice differentiable on $\left [a, b \right]$ and $f''$ is continuous on $\left [a,b \right]$. The derivatives at the boundaries are to be understood in the sense of one-sided derivatives.


4

It means that $f^{\prime\prime}$ is continuous on $[a,b]$.


8

$f(x)=0$ is a polynomial function, with degree $-\infty$ (by convention). In this way, $\deg(fg) = \deg(f) + \deg(g)$ and $\deg(f+g) \leq \max(\deg (f), \deg (g))$ are true for any polynomials $f$ and $g$.


1

See here. Take enough derivatives of each function to form a $nxn$ matrix (you have three functions, so take 2nd derivative). Form a matrix with the functions and derivatives, take determinant of that matrix. \begin{bmatrix} f_{1}(x) & f_{2}(x) & f_{3}(x)\\ {f_{1}'(x)} & {f_{2}'(x)} & {f_{3}'(x)} \\ {f_{1}''(x)} & {f_{2}''(x)} ...


1

Right, you have to show that if $x \mapsto ax^2 + b \cos(x) + ce^x$ is the zero function (i.e., $ax^2 + b \cos(x) + ce^x = 0$ for every real number $x$), then $a = b = c = 0$. Hint: One possible approach is to think about the asymptotic growth rates of the functions. What happens as $x \to \infty$?


2

Note that the equality $\;ax^2+b\cos x+ c\,e^x=0\;$ is an equality of functions, so that it remains for any value of $\;x\;$ you choose, so for example: $$x=0\implies b+c=0$$ $$\begin{align}&x=\frac\pi2\implies \frac{a\pi^2}4+ce^{\pi/2}=0\\{}\\&x=-\frac\pi2\implies \frac{a\pi^2}4+ce^{-\pi2}\end{align}$$ Substracting both equations above: ...


4

It seems the following. Let $1>\varepsilon>0$ be an arbitrary number. There exists a number $N>0$ such that $$|f(x+1)-f(x)-l|\le \varepsilon$$ for each $x\ge N$. Since the function $f$ is continuous, $$\sup \{f(y): y\in [N;N+1]\}=M<\infty.$$ Let $y\ge \max \{N+1, M, |l|N\}/\varepsilon $ be an arbitrary number. There exists a nonegative integer ...


4

Here is a similar but more interesting statement. Let $f$ defined on $(a,+\infty)$ and bounded on all bounded interval $(a,b)$. If $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ then $\lim_{x\to\infty}\frac{f(x)}{x}=l$ as well. Let $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ and introduce $M_n=\sup_{[n,n+1)} f(x)$ and $m_n=\inf_{[n,n+1)} f(x)$. The ...


1

I'm taking $A=B=\Bbb R$ here. Following your intermediate value theorem proof that no continuous function exists for $n=2$, you can prove similarly that none exists whenever $n$ is even. However, you can easily construct continuous functions for any $n$ odd. Although I've always done so graphically, I'm sure that you could find explicit formulas.


0

You would have to find a continuous function that violates the horizontal line test. However, it won't have an $n$-tuple at the critical point. Consider the inverse of the function $y = 4x^2 - 4x + 1$. This function's pre-image is a 2-tuple except at $0.5$. The same can be said for all pre-images of polynomials with degree $k: 2 \leq k \leq n$.



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