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"if $f$ injective and $g$ surjective can $g\circ f$ be both injective and surjective?" Sure, e.g. let $f:\{0\}\rightarrow\mathbb R$ and $g:\mathbb R\rightarrow\{0\}$. But not always, e.g. let $g:\{0,1,2\}\rightarrow\{0,1\}$ be prescribed by $0\mapsto0$, $1\mapsto1$, $2\mapsto1$ and let $f:\{0,1\}\rightarrow\{0,1,2\}$ be prescribed by $0\mapsto1$, ...

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Work from the inside outwards. So $f(g(tx))=f(t^kg(x))=(t^k)^kf(g(x))$.

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Take $\epsilon = 1/2$, and suppose there is a continuous function $f$. Call $f(1) = A, f(2) = B$. So $B > 2A$. Call $B/A = D$, and $\lceil\log_2(D)\rceil + 1 = d$. So in particular, $2^d > B/A$. Choose $d$ points $x_i$ between $1$ and $2$. Then we have that $f(x_2) > 2f(x_1)$ and $f(x_3) > 2f(x_2)$ and so on, so that $f(2) > 2^d f(1)$, or ...

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You are right. Let $\epsilon=\frac{1}{2}$, and let $f(1)=c\gt 0$. Then $f(2)\gt 2c$. By the same reasoning, $f(1+1/2)\gt 2c$, so $f(2)\gt 4c$. By the same reasoning, $f(1+1/3)\gt 2c$ so $f(1+1/2)\gt 4c$, so $f(2)\gt 8c$. And so on. Note that continuity, monotonicity are irrelevant.

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"Draw" the unit half-circle with $x > 0$ and the tangent line at $(1, 0)$. Any point on that line determines a unique line segment to the origin, which interects the half-circle in a unique point . Conversely, each point on the half-circle uniquely determines a ray from the origin, which intersects the line in a unique point. Thus there is a bijection ...

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Notice under the direct mapping, any point $p$ on $T$ get mapped to a point $f(p)$ on $S$ which is a scalar multiple of $p$. So for any point $q = (x,y,z) \in S$, the inverse mapping $f^{-1}$ will send $q$ to a point which is a scalar multiple of $q$. This means there exists a function $\lambda : S \to \mathbb{R}$ such that $$q = (x,y,z)\quad\mapsto\quad ... 5 You are definitely correct. You can regard the equation y^4 = 8x^2 as the relation$$xRy \iff y^4 = 8x^2,$$and R is a function of x if and only if, for each x, there is at most one y such that xRy. But that is clearly false in this case. For example, let x=2; then y^4 = 32 so that y = \pm\sqrt[4]{32} = \pm 2\sqrt[4]{2} (as well as some ... 3 y^4=8x^2 is an equation, not a function. Moreover, this equation does not implicitely define y as a function of x due to the fact that more y's correspond to one x. Also, x is not a function of y. On the other hand, you can consider a function f(x,y)=y^4-8x^2 of two variables and the domain of satisfiability of your equation equals the zero ... 2 y(x-1)=2x+3 thus x(y-2)=y+3, hence x=\dfrac{y+3}{y-2}. Now y^{-1}=\dfrac{x+3}{x-2} 1 An injection from A to B is: f(i,j) = i + j(j-1)/2 = n So solve x^2-x-2(n-1)=0 to find j=\lfloor x\rfloor and use that to find i . 2 I will try to create a bijection for you. What we do is we count by rows. So the row (1,1),\ldots,(1,N) ends on spots 1,\ldots,N. We then start the next row at (2,2) which needs to be put on N+1 and finish the row. So: (1,1)\mapsto 1 (2,2)\mapsto N+1 (3,3)\mapsto 2N (4,4)\mapsto 3N-2 (5,5)\mapsto 4N-5 (6,6)\mapsto 5N-9 etc. What we ... 2 Yes, that's exactly right - you just break up the absolute value into parts:$$f(x)=\frac{1}{|x-2|}-x=\begin{cases}\frac{1}{x-2}-x&\text{if }x>2\\\frac{1}{2-x}-x&\text{if }x<2\end{cases}$$From here, you can proceed as normal with the derivative arguments - that is breaking up the intervals into pieces where the derivative is positive and ... 4 Hint: Notice that$$f(x)=\begin{cases} -\frac{1}{x-2}-x&x<2\\ \frac{1}{x-2}-x&x>2 \end{cases}$$1 The function f(x) is increasing whenever the derivative f'(x) is positive, and the function is decreasing whenever the derivative is negative. You have the correct derivative (though there's a typo - it should be f'(x)=3x^2-3, but I assume it's a typo because you have the correct solutions for f'(x)=0). Now you just need to figure out where that ... 1 There is NO need to calculate f''. f'(x)=3x^2-3. When |x|>1 then f'(x)>0. So f is strictly increasing in (1,\infty)\cup (-\infty ,-1). Again , f'(x)<0 for |x|<1. That is , f is strictly monotone decreasing in (-1,1). 2 A vertical asymptote means that k-x^2 = 0 for x=\sqrt{5} Hence, k=5. More generally speaking: vertical asymptote : division my zero horizontal asymptote : function tends to this value for very large absolute values of x. 1 The domain for the second one is OK, because \frac1x is defined for all nonzero real numbers, and \sin \frac1x will then also be defined for all nonzero numbers because \sin x is defined for all real x. You are also correct in that \frac{1}{\sin x} is not defined for x=0, but you should not have stopped there.$$\frac{1}{f(x)}$$is undefined ... 6 Almost there. The first is not defined at integer multiples of \pi. If we agree on calling D_{\phi} the domain of the function \phi(x), then we can say D_{f\circ g} = \mathbb{R}−\{k\pi | k\in \mathbb{Z}\}. 0 The first definition of derivative people see is f'(x) = \lim\limits_{h\to 0} \frac{f(x+h) - f(x)}{h} for real-valued functions f of a real variable x. That last clause is important. Adding "real-valued" should suggest something: You can also define functions that are not real-valued, but take their values elsewhere, say in \Bbb R^3, or more ... 0 A function is surjective if you can get any value you want by giving it the adequate argument. For example, y=f(x)=2x+1 (x\in R) is surjective because for any value of y you can find a value of x such that f(x)=y. Another way to see it is that if you express x as an expression of y that expression exists for all the possible values of x (this ... 0 f\colon A\to B is surjective if every element b\in B has a pre-image in A. In other words: the equation \;f(x)=b \enspace(x\in A) always has a solution, whatever the value of b. 0 Let f:A\to B (i.e. f has domain A and codomain B). Then: f is injective iff every value of B is obtainable at most once. f is surjective iff every value of B is obtainable at least once. f is bijective iff every value of B is obtainable exactly once. 0 Basically, if a function f: A \to B is surjective, it means that every element of the set B can be "obtained" through the function f. Formally, like you said, it means that for every b \in B, there exists some a \in A such that f(a) = b. Phrased differently, f being surjective means that the range of f includes all of the set B. For an ... 1 (Prepare to facepalm) In simple cases, especially when A,B are finite sets, you may represent a function f\colon A\to B by a Venn diagram of two disjoint sets A and B and a number of arrows, each going from a point in A to a point in B. The fact that f is a function corresponds to the fact that each point in A is the origin of exactly one ... 0 You can restate it in terms of the range of f: a function f : A \to B is surjective if its range is B. We also say that "f maps onto B". Surjectivity is a little bit of a weird notion, actually, because every function maps onto its range. That is, given any function f : A \to B, there is a surjective function g : A \to \operatorname{range}(f) ... 0 Remember that a function f is defined by a domain A and a co-domain B, i.e.$$f: A\to B$$and some relation that defines the values f takes. Then a surjective function is essentially a function that "hits" all elements in the co-domain B. Or said in another way, there is no element y\in B that is not some function value of f. This is explained ... 4 A surjective function is a function that "hits everything": so, for example, the function f(x)=2x is surjective as a function from \mathbb{R} to \mathbb{R}, since - for any real a - {a\over 2} is also a real number, and we have f({a\over 2})=a. By contrast, the function g(x)=x^2 is not surjective as a function from \mathbb{R} to \mathbb{R}: ... 2 Solution 1. We introduce two sequences of functions T_n(x) and I_n(x) by$$ T_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!} \quad \text{and} \quad I_n(x) = \int_{x}^{\infty} \frac{t^n}{n!} e^{-t} \, dt. Applying integration by parts, we obtain the following recurrence relation: \begin{align*} I_n (x) &= \left[ -\frac{t^n}{n!}e^{-t} ... 0 Hint: Since sinz is entire hence by Picard's little theorem, it must assume all values on complex plane C with at most one possible exception. 1 Some hints. a) Put g(x)=1-f(x). Show that g(x+y)=g(x)g(y) for all x,y. b) Put y=x and show that g(y)\geq 0 for all y. c) Suppose that for a u\in {\mathbb R} g(u)=0, and show then that g(x)=0 for all x, and that this imply a contradiction. d) Now you can put h(x)=\log g(x), and find the form of the (continuous ) fonction h. e) I ... 3 In full generality, you can talk about a section of any vector bundle p : E \to M: it's a map s : M \to E such that p(s(x)) = x for all x \in M. By definition, a vector field is a section of the tangent bundle TM \to M, and so there's no need to specify that it's a "tangent vector field" because it's already implied. In your case, M = ... 5 This can be proved purely combinatorially using nothing more than the binomial theorem. In fact, I claim that the polynomial P_n(x)=\left(\sum\limits_{k=0}^{2n-1} \frac{x^k}{k!}\right)\left(\sum\limits_{\ell=0}^{2n-1}(-1)^\ell \frac{x^\ell}{\ell!}\right)-1 is even, has no constant term, and has no positive coefficients, which is sufficient to show that it ... 5 By Taylor series with Lagrange remainder e^x=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{(e^x)^{(2n)}|_t}{2n!}x^{2n}=\sum\limits_{k=0}^{2n-1}\frac{x^k}{k!}+\frac{e^t}{2n!}x^{2n} $$where |t|<|x|. And$$ ...

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You can find the matrix of this transformation: $$\begin{pmatrix} 2 & 0 & 0\\0 & 1 & 0 \\0&1 &3\end{pmatrix}.$$ Do you see this matrix is inverse?

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The Wiki article would get a 0 on a probability course.You cannot compute odds by dividing the number of favorable cases by the total number of cases UNLESS (1) No two of the cases can occur at once AND (2) Each case has equal probability. In the Birthday Problem, (1) is true but (2) is false. The solution in the article may be approximate ( depending on how ...

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1) They are using 366 instead of 365 to take February 29th into account. 2) $\displaystyle\frac{366!}{(366-N)!}$ does correspond to what you have for the number of injective functions, since $\;\;\displaystyle\frac{366!}{(366-N)!}=\frac{366\cdot365\cdot364\cdots(366-N+1)(366-N)(366-N-1)\cdots3\cdot2\cdot1}{(366-N)(366-N-1)\cdots3\cdot2\cdot1}$ $\hspace{.8 ... 2 Let$M$be an integer greater than$1/\epsilon$and let$f(x)=M^{x-1}$. 1 The steps are right. Just don't put the 3 in the denominator at last, multiply 3. 2 Almost it is: $$x=3(\ln{(\frac{y}{3})}-1)$$ You should have multiplied by$3$in the last step. 2 You are correct that$f'(x)$has no (real) roots, but you've miscalculated the point of inflection. Also, there's no particular purpose (that I can think of) to calculating the point of inflection, in the first place. Instead, one could easily use the Intermediate Value Theorem on$(-0.75,-0.5),$since$f$has only one real root (why?). Also,$(0.75,0.5)$... 1 Your reasoning is clearly wrong. You could say the same thing for practically all nonreal-valued function. When someone is asking for the maximum of a nonreal-valued object, it is implied that you have to consider the absolute value (or, more generally, some kind of norm). The proof that$\sin z$is unbounded on$\mathbb C$can be traced as follows (there ... 1 "Does a tangent exist at x=0 to y=sgn(x)?." It all depends on how you are defining the tangent. If you are calling the line with the slope equal to the derivative at a point as a tangent, then the answer would be no, because now sgn(x) is discontinous. If you are calling the limiting chord a tangent then the Y-axis is tangent at x=0. 1 Yes, the easiest way to prove this statement is to show that that$\alpha$and$\beta$both have an inverse function. As stated, they are actually each others inverse, so you must prove that$\beta\alpha = 1_A$and$\alpha \beta = 1_B$both hold. The first identitiy is one of the assumptions, so you only need to prove the second one. Hint: Consider the ... 0$\tan^{-1}$would normally denote the FUNCTION$\mathbb{R}\longrightarrow (-\frac{\pi}{2}, \frac{\pi}{2})$given by sending$x$to the unique$y$satisfying$\tan(y)=x$, while$\tan(x)^{-1}$presumably stands for the real NUMBER, which is the multiplicative inverse$\frac{1}{\tan x}$and is well-defined whenever$\tan(x)\neq 0$, i.e. whenever$x$is not an ... 2$\tan^{-1}$denotes the inverse tangent function, AKA the arc tangent (the angle the tangent of which is the given number). When applied to an argument, you spell $$\tan^{-1}(x)=\arctan(x).$$ As far as I know, $$\tan(x)^{-1}$$ can be interpreted as the reciprocal of the tangent, i.e. the cotangent $$\frac1{\tan(x)},$$ and it is safer to write ... 0 If we know that the composition$\beta \alpha$of two functions$\alpha$and$\beta$is onto, then this implies that$\beta$has to be onto. Similarly, if we know that$\beta \alpha$is one to one, then$\alpha$has to be one to one. With these statements you should be able to prove what you want. As additional exercise, try to prove the statements I gave. 0 Note that the set of orthogonal matrices in$\mathbb{R}^3$with determinant$1$is called the rotation group (denoted$SO(3)$). By definition,$SO(3)$preserves Euclidean distance, meaning it is an isometry. Rotations are also linear transformations preserving the inner product on$\mathbb{R}^3$, hence homogeneity is satisfied by definition. 0 First look only at the shape, not the scale or position of the graph. These can later be adjusted by choosing the range of the coordinates. the front right side seems to be a parabola opening to the bottom so something like$z = -y^2$the shape on the right back wall is also a parabola (see the grid) but opening to the right$z = -\sqrt{x}$. the surface ... 1 Consider$g(k) = k + \lfloor \frac{k}{2^1} \rfloor + \lfloor \frac{k}{2^2} \rfloor + \lfloor \frac{k}{2^3} \rfloor + \ldots $, which is a strictly increasing sequence. Claim: If$ g(k) \leq n < g(k+1)$, then$f(n) = k$. 4 In some contexts, when direct products are defined (e.g. vector spaces, groups, modules...) given two maps $$f: A \to B \\ g : C \to D$$ one defines $$f \times g : A \times C \to B \times D$$ as$(a,c) \mapsto (f(a), g(c))$. Sometimes (when you work with modules or vector spaces) products are denoted with the symbol$\oplus\$ and are called direct sums. This ...

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