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HINT: Observe that $$f(a)=\frac19=3^{-2}$$

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If you are asking for a simple mathematical identity in addition, subtraction, multiplication and division, then I am afraid that I do not know of any other than those already mentioned. Since you mentioned, however, that you were asking "merely for curiosity's sake", I will attempt to satisfy at least some of that curiosity. Since bitwise operations are, ...

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Such a function must be constant. Consider $g(z) = \exp(-f(z))$. We have $$\left|g(z)\right| = \exp(-\Re(f(z)) \le \exp(-\Re(f(z_p)).$$ Since $g$ is entire and bounded, it must be constant. Hence $f$ is constant too.

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Well, the real part of a holomorphic function is a harmonic function, which satisfies the mean value property, hence the maximum principle. So, if such a $z_p$ exists, then your function is a constant (which clearly does satisfy the requirement).

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Part (a) is just an application of the chain rule: this function is of the form $f(x) = g(h(x))$ where $g(u) = \ln(u)$ and $h(x) = x^2 - 2x + 2$. Then the chain rule informs us: $$f'(x) = g'(h(x)) h'(x) = \frac{1}{h(x)} (2x - 2) = \frac{2x-2}{x^2 - 2x + 2}.$$ For part (b), you need to find the values of $x$ so that $f'(x) = 0$, i.e. for what $x$ is $$... 3 F_Y(y) = P(Y<y)=P(1-X^2<y)=P(X^2>1-y)=P(X>\sqrt{1-y})+P(X<-\sqrt{1-y})= 1 - P(X<\sqrt{1-y})+P(X<-\sqrt{1-y})=1-\int^{\sqrt{1-y}}_{-1}\frac{x+1}{2}dx+\int^{-\sqrt{1-y}}_{-1}\frac{x+1}{2}dx 0 In some cases, the function s placed to the right of the variable even in modern notation, as in n!. 0 Or if you are unwilling to assume differentiability outside x=0, consider the sequence a_k=f(x_0+k\,h) for some fixed x_0 and h. Then you get the linear recursion a_{k+1}-qa_k-a_{k-1}=0 where q=f(h)-f(-h)\approx 2h\,\log a. 2 Take y = h, and divide both sides by h. Add and subtract f(x) on LHS, Add and subtract f(0) on the RHS. You will end up with$$ \frac{f(x+h) -f(x)}{h} + \left(\frac{f(x-h) - f(x)}{-h}\right) = f(x)\left [\frac{f(h) -f(0)}{h} + \left(\frac{f(-h) - f(0)}{-h}\right)\right] $$Assuming f is differentiable everywhere, take a limit to get$$ f'(x) = ...

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Putting the operator to the right is indeed how it is done in reverse polish notation. I once saw a claim in Abraham Robinson's book that the origin of the $f(x)$ notation derives from Newton. Certainly this would have to originate from late 16th or early 17th centuries. Note: thanks to @Michael Hoppe for pointing out that the notation $f(x)$ is due to ...

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Seems like a simple function of the form :$$\dfrac{-k}{x^{2}+y^{2}}$$ Graph example

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$$\ln(|x|)$$ where $\ln$ is the logarithmic function to the base $e$, and $|x|$ is the absolute value of $x$.

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I'm sorry your teacher was in a bad mood when writing this problem. I would not want to look for solutions of $f'=0$ myself. Begin by sketching the argument of sine: it's an even function that for positive $x$ begins at $y=1$, is initially increasing (because $5x^2$ grows more rapidly than $x^4$ when $x$ is small), reaching $y=3$ at $x=1$. It will grow a ...

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Answers: When $\;V_{\Bbb F}\;$ is a vector space over a field $\;\Bbb F\;$ , a linear transformation $\;T:V\to \Bbb F\;$ is called A linear functional . $\;P_n\;$ is the vector space of all polynomials of degree less than or equal $\;n\;$ with coefficients from some field. Its dimension is $\;n+1\;$ $\;p\;$ represents a polynomial in $\;P_3\;$ . The next ...

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Another example is the map $\mathbb{R} \rightarrow \mathbb{R}-\{0\}$ with $x \mapsto e^x$. It misses $(-\infty,0)$.

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Nope, let $X=\mathbb{Z}$ and $Y=2\mathbb{Z}=\{2n\mid n\in\mathbb{Z}\}$. Let $f\colon X\to Y$ be given by $f(k)=4k$. This is injective but misses, for instance $2\in Y$ and so isn't surjective. I should add, an infinite set needed to be chosen for $X$ because there exist no injective functions from a finite set to a proper subset. Similarly, $Y$ needed to be ...

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In the search for a bijection, consider an "infinite hotel" map where $2p\over q$ is mapped to $p\over q$ and $2p+1\over q$ is mapped to $-p\over q$. Proving whether this mapping has surjective and injective properties should be fairly straightforward. Note that $p\over q$ is not required to be in reduced form. Edit: Efforts to rectify the shortcomings ...

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Hint $\ f$ is $1$-$1$ iff $f(a,b) = f(c,d) \Rightarrow (a,b)=(c,d),\$ i.e. $\ 2^a 3^b = 2^c 3^d\Rightarrow a=c, b=d$ This is an immediate consequence of the uniqueness of prime factorizations. If you can't use that then prove this: if $m,n$ are odd and $\,2^a m = 2^c n\,$ then $a = c\$ (hint: cancel $2$'s then use parity). Applying this above, we can ...

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EDIT: I believe now I am able to fully solve your problem, that is explicitly show a bijection between $\mathbb {Q}_{\geq0}$ and $\mathbb{Q}$. First we build a bijection $g:\mathbb {Q}_{\geq0} \to \mathbb{Q_{>0}}$, consider the set $A=\left\{{0, 1/2,1/3,1/4,...}\right\}$ of the sequence $(a_{n})=(0, 1/2,1/3,1/4,...)$ and let $g(x)=\begin{cases} a_{n+1}, ... 1 How about this other idea? You can get from$(a,b)$to$(0,1)$, using a composition of two operations, each of which is a bijection: 1)scaling the length of any one of the intervals, so that the lengths are equal , and 2) Translating,i.e., or moving the scled interval so that it overlaps the other interval.Then you have a composition of two bijections, which ... 1 Hint:$f(a)=0$,$f(b)=1$,$f'(x)=1/(b-a)>0$. 0 $$F(i_1,j_1)= F(i_2,j_2) \implies 2^{i_1}3^{j_1} = 2^{i_2}3^{j_2} \implies \dots \implies i_1=i_2,\,j_1=j_2 .$$ 3 Hints: $$2^i3^j=2^n3^m\iff 2^{i-n}=3^{m-j}$$ Suppose$\;i\ge n\;$, then the left side in the last equality above is an integer. Use now the Fundamental Theorem of Arithmetic.... 3 Hint 1: There are obvious inclusions $$\mathbb{N} \hookrightarrow \mathbb{Q}_{\ge 0} \hookrightarrow \mathbb{Q}$$ so if you can find an injection$\mathbb{Q} \to \mathbb{N}$then you'll have proved that all three sets have the cardinality. Hint 2: (hover mouse over to see) Hint 3: (hover mouse over to see) 1 You have at least two ways of solving this kind of exercise. You can use purely algebraic methods, that is, you evaluate your linear combination at various points, giving relations between the$k_i$, and, with sufficiently many relations, you will find that the only solution is that all$k_i$are zero. This can be long and tedious. But you can also use ... 0 Hint check whether Wronskian is non-zero. 0 Consider$x^\alpha\sin(x^{-1})$for a suitable$\alpha$. 3 Check the function $$f(x) = x^{1/2 + n} ,$$ Which is differentiable$n$times, but not$n+1$. 1 Basically the problem is the number of permutations where the first three numbers are not$1,2$or$3$. First select$f(1),f(2),f(3)$. Which is selecting the first three digits. Now note that any permutation of the remaining elements gives a different permutation. There are$(5\cdot4\cdot3)$ways to do this. Now select an appropriate permutation of the ... 0 You can write the function for the ramp between$1$and$5$: $$y(t)=\frac{x}{5}$$ $$Y(s)=\frac{1}{5s^2}$$ You can do this with the formula:$y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)$, between the point:$A(x_1=0 ; y_1=0)$and$B(x_2=5 ; y_2=1)$. Now you have to consider the ramp just in the interval$[0,5]$, and to do so you use the frequency shift ... 7 Hint Use the squeeze theorem knowing $$\frac{b}{x}-1\le\left[ {\frac{b}{x}} \right]\le \frac{b}{x}$$ 2 HINT Use the fundamental limits$\lim_{x\to 0}\frac{\sin x}{x}=1$and$\lim_{x\to 0}\frac{e^x-1}{x}=1$and for$x\to\infty$observe that$\sin x$is bounded. 1 To ensure the validity of a function, you need to check for two things: Division by Zero Passing the Vertical Line Test Division by zero is obvious: if in the domain of the function$f(x)$you have some value of$x$that implies$f(x) = \frac{\ldots}{0}$, then the function is undefined there, and thus is invalid. Examples of this include:$f(x) = ...

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Let $f:[0,1)\to \mathbb{R}$, $f(x)=tan(\pi x/2)$. This function is continuous and strictly increasing but not absolutely continuos. Just to show that this function is in fact not absolutely continuous. Take $\epsilon=1$, and suppose there is a $\delta>0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_{k},y_{k})$ of $[0,1)$ ...

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Hint Assume that $b>0$, prove that $$\mathop {\lim }\limits_{x \to 0^+ } \frac{a}{x}\left[ {\frac{x}{b}} \right]=0$$ and $$\mathop {\lim }\limits_{x \to 0^{-} } \frac{a}{x}\left[ {\frac{x}{b}} \right]=\infty$$

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Just add the identity function, $\text{id}(x) = x$, to the Cantor function, $\text{c}$. The sum of continuous functions are continuous, and the sum of an increasing function with a strictly increasing one is strictly increasing. As in the proof that $\text{c}$ is not absolutely continuous choose $\epsilon < 1$. For every $\delta > 0$ there is a ...

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By induction all numbers of the form $-(2^x\cdot 3^y)$ for non negative integers $x$ and $y$ are in $X$). Proof: $-(2^0\cdot 3^0)$ is in $X$. now suppose the number $-(2^a\cdot 3^b)$ is in X. Then the numbers $-(2^{a+1}\cdot 3^b)$and $-(2^a\cdot 3^{b+1})$are also in $X$ Yes, you can easily see that only negative numbers are there and $10$ is not of the ...

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Counterexample number $8.30$ of "Counterexamples in Analysis" by Gelbaum and Olmsted (which can be found here) provides a continuous, strictly increasing function on $[0,1]$ which is singular. Since it is not constant, it can't also be absolutely continuous.

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Observe that $J_1(t)$ and $Y_1(t)$ are the Bessel function of the first kind and the second kind respectively, and $H_1^{(1)}(t)=J_1(t)+iY_1(t)$ is the Hankel function of first kind. For small argument we have \begin{align} J_1(t)&\sim \frac{t}{2}\tag 1\\ Y_1(t)&\sim \frac{2}{\pi}\left[\log\left(\frac{t}{2}+C\right)\right]\frac{t}{2}-\frac{2}{\pi ... 2 Since the function x\mapsto \ln x is increasing we have\ln(n!)=\sum_{k=1}^n\ln k\sim_\infty\int_1^n\ln t dt\sim_\infty n\ln n$$Added Using the monotonicity of the \ln  function we have$$\int_{k-1}^k\ln tdt\le\ln k\le\int_k^{k+1}\ln t dt\;\;\forall k\ge2$$then by summing for k=2 to n and using some algebra we find easily$$\sum_{k=1}^n\ln ...

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Your proof of injectivity is perfect. I would approach surjectivity slightly differently. Take some $m\in\Bbb R,$ and then let $k=(b-a)m+a.$ Then $k\in\Bbb R,$ and you can show that $f(k)=m.$ "Let $m=\frac{k-a}{b-a}$" basically assumes what you're trying to prove, though you'll certainly want to start there with your calculations. As an alternate approach, ...

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You certainly can't get $\sin(\phi)\sin(\theta)/9 + \cos(\phi)\cos(\theta)/16 = 1$ with real $\phi$ and $\theta$. According to Maple: with(Groebner): e1:= x/3*st+y/4*ct; e2:= x/3*sp+y/4*cp; e3:= st*sp/9 + ct*cp/16; B:= Basis([e1-e2,e2-e3,e3-1,ct^2+st^2-1,cp^2+sp^2-1], plex(st,ct,sp,cp,x,y)): factor(B[1]); $$\left( 10\,{y}^{2}-25+17\,{x}^{2} \right) ... 1 It is the only function that meets the criteria. I don't understand what you mean by "formal justification". 1 Functions are defined to be equal if they agree on all inputs (and have the same domain and range). If you prescribe the function's value on all elements of the domain, there can therefore be only one function that takes these values. 1 HINT: Using Weierstrass substitution observe that the roots of the Quadratic Equation$$\frac x3\frac{2t}{1+t^2}+\frac y4\frac{1-t^2}{1+t^2}=1$$(in t) are \tan\frac\theta2,\tan\frac\phi2 Then use Vieta's formula to find \displaystyle\tan\frac\theta2+\tan\frac\phi2 and \displaystyle\tan\frac\theta2\cdot\tan\frac\phi2 Finally use Weierstrass ... 2 I think that there is no contradiction between the post and the answer given by Ramana Venkata. I propose to go back to what the OP describes and to enter in more details just for clarfication purposes. Let us consider two points along the curve : (x0,y0) and (x1,y1). Then, the square of the hypothenuse is given by (x1 - x0)^2 + (y1 - y0)^2. Since we ... 2 Your intuition is on point. I do not know your math background, but it sounds like you are at least familiar with basic calculus. The name for the line which you describe is arc length. We can pretty much approximate the arc length of any function, and obtain the exact value for quite a few types of functions. There are some pathological cases for which we ... 3$$\int\limits_{a}^{b}\sqrt{(f'(x))^2+1}dx$$Explanation: For every dx distance you move along X-axis you move dy distance along Y-axis since the length along curve is very small it can be treated as straitght line so it is approximately \sqrt{(dy)^2 + (dx)^2}. 0 Hints: for \;a\in\Bbb R\;$$\frac{1-x^2}{1+x^2}=a\iff(1+a)x^2=1-a\iff\ldots$$Can you see what the condition(s) must be to solve the above for \;x\; ? Likewise with the other function:$$\sqrt{\frac{1-x}{1+x}}=a\iff 1-x=a^2+a^2x \iff (a^2+1)x=1-a^2\iff\ldots$$But in this case you also must take into account that the function is defined only when ... 0 Suppose you have some bound say a>0$$\frac{1-x^2}{1+x^2}\leq a \Rightarrow 1-x^2\leq a+ax^2 \Rightarrow (a+1)x^2 +(a-1) \geq 0 Do you think this holds true for all $x\in \mathbb{R}$. For second question as you are guessing range includes $[0,\infty)$ what you could do is : Take any $a >0$ then try finding some $x\in (-1,1]$ such that ...

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