# Tag Info

0

Let $\max_x \min_y f(x, y) = \max_x g(x) = g(a)$ and $\min_y \max_x f(x, y) = \min_y h(y) = h(b)$. We have $$\max_x \min_y f(x, y)= g(a) \le f(a, b) \le h(b) = \min_y \max_x f(x, y)$$ So for equality, we need an element $(a, b)$ to exist s.t. $\min_y f(a, y) = f(a, b) = \max_x f(x, b)$.

0

If there exists limit $$\lim_{n\to+\infty} \dfrac{f(n)}{g(n)}=C,\qquad |C|<\infty,$$ then we will assume that functions have the same order, if $0<|C|<\infty$, function $f(n)$ has lower order than $g(n)$, if $C=0$. Use these well known limits: $$\lim_{n\to+\infty}\dfrac{\ln n}{n^a}=0, \qquad a>0;$$ $$... 0 You are quite correct - your answer is wrong. I assume that you are talking about Big O notation. To answer the question: Read the article, your notes and the textbook Classify the functions by the notation the fall into - to help you have some linear, polynomial, exponential, logarithmic and quasilinier. Within each classification decide which will grow ... 0 Not true. Consider f(n) = 2^n,\ \ f(cn) = 2^{cn} \neq c_1 \cdot 2^n for arbitrary c. Therefore f(cn) \notin \Theta(f(n)) 0 First question: Y is the codomain of f: X \rightarrow Y. In this case you are taking any subset V \subset Y. Second question: Not necessarily, it will be a subset of the range if f is surjective. Third question: We say that A \subset B when \forall a\in A \Rightarrow a \in B. So taking an arbitrary y \in V and showing that y \in ... 3 As stated in the comments, we have x=y=0 implies 2f(0)=4f(0) so f(0)=0. Also we must have an even function since with x=0 we see that f(y)+f(-y)=2f(y) implying f(y)=f(-y). Now for k\in\mathbb N we can prove by induction that f(ky)=k^2f(y). This is clearly true for k=1. So if we assume this holds for k and proceed to k+1 we see that ... 1 Let y\in C then since h is onto there's x\in A such that h(x)=f(g(x))=y so we have z=g(x)\in B and f(z)=y. We proved that f is onto. 1 The series  1+t+t^2+\cdots has radius of convergence 1, so converges if |t|\lt 1 and diverges if |t|\gt 1. Thus our series 1+x^4+x^8+\cdots converges if x^4\lt 1, and diverges if x^4\gt 1. So we have convergence if |x|\lt 1, divergence if |x|\gt 1. Remark: Your sums should start at n=0, not at n=1. And g(x)=3+3x^4+3x^8+ ... 5 We say f:A\to B is a function if, for any a\in A there exists exactly one b\in B such that f(a)=b. Since \infty is not an element of \mathbb Z^+, if you want f:\mathbb Z^+\to \mathbb Z^+, you can't have f(n)=\infty for any positive integer n. 2 Continuity of f can be seen as "if you move just a bit, the image through f will move a bit too". In your case f^{-1} being not continuous would imply that you can find two points on the codomain "close enough" such that their f-counter image are actually far away. Think of a rectangle and glue two opposite sides together, and think of the obvious ... 2 Using L'Hospital rule for \infty/\infty format, if \lim_{x\to\infty}f(x)=\infty:$$\lim_{x\to\infty}\frac{f(x)}{x}=0\implies\lim_{x\to\infty}\frac{f'(x)}{1}=0$$See if you can use the same for second part similiary. 1 The function$${\rm sinc}(\pi x):=\cases{{\sin(\pi x)\over\pi x}\quad&$(x\ne 0)$\cr 1&$(x=0)$\cr}$$is an entire function which is =1 at 0 and =0 at all integers \ne0. 0 Start noting, using recursion, that$$f(n)=2f(n-1)+(n-1)=2^2 f(n-2)+2(n-2)+(n-1)=2^3 f(n-3)+2^2(n-3)+2^1(n-2)+(n-1)=...$$So you should be able to prove (using induction, maybe) that$$f(n)=2^{n-1} f(1)+\sum_{i=1}^{n-1} 2^{i-1}(n-i)$$Now$$\sum_{i=1}^{n-1} 2^{i-1}(n-i)=\sum_{i=1}^{n-1} 2^{i-1} n -i2^{i-1}=n\sum_{i=1}^{n-1} 2^{i-1}-\sum_{i=1}^{n-1} ...

2

Assume you can do it for $n=2^k$, specifically $n=4$, then $f(\frac{(x_1+x_2+x_3+(x_1+x_2+x_3)/3)}{4})\leq \frac{f(x_1)+f(x_2)+f(x_3)+f((x_1+x_2+x_3)/3)}{4}$ rearrange to get n=3 case. Same approach can be generalized to any non $2^k$ number.

2

You have $$A\cap B\subset A\implies f(A\cap B)\subset f(A),\\ A\cap B\subset B\implies f(A\cap B)\subset f(B)$$ so it follows that $f(A\cap B)\subset f(A)\cap f(B)$.

0

Because $$a^2+b^2=(a-b)^2+2ab,$$ if $a^2+b^2$ is even, then $a$ and $b$ must share parity. But if $a$ and $b$ are both odd, then $(a-b)^2$ is divisible by $4$ whereas $2ab$ is not, so it must be that $a$ and $b$ are both even.

2

$$(2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2$$ This means that $\left(a^2 + b^2\right) = 4 \lambda + 2$, meaning that $4$ could not possibly divide $\left(a^2 + b^2\right)$, since there's always a remainder of $2$. ...by the way, this means that for $4$ to divide $a^2 + b^2$, both $a$ and $b$ must be even, since certainly if one is odd ...

0

While $f=g$, it is in general not true that $f(x)=g(u)$. Notice that $f(x)=x+\sqrt{2-x}$. Notice also that $g(x)=x+\sqrt{2-x}$ $\dagger$. Thus, $f(x)=g(x)$, and $f=g$. $\dagger$ Why does $g(x)=x+\sqrt{2-x}$? Well, we know that $g(u)=u+\sqrt{2-u}$, for all $\boldsymbol u$. Thus, if we left $u=x$, we get $g(x)=x+\sqrt{2-x}$.

2

An equivalent question, if I understand this correctly, would be if we can find a smooth function h=f-g so that h(n)=0 on all naturals n except one. I think we can do this with an appropriately scaled bump function. http://en.wikipedia.org/wiki/Bump_function Take h=Ψ as defined on the wiki page, and then let g=f+h.

0

Yes, you can use a linear interpolation to do that. http://en.wikipedia.org/wiki/Interpolation#Linear_interpolation

0

To show onto, you must prove that every element in your range gets used. To do this, pick $y \in \Bbb{R}-\{5\}$. We must now prove that we can always find some $x \in \Bbb{R}-\{2\}$ such that $f(x)=y$. This would mean $$\frac{5x+1}{x-2}=y \\ \implies 5x+1 = yx-2y \\ \implies 5x-yx = -2y-1 \\ \implies x(5-y) = -(2y+1) \\ \implies x = -\frac{2y+1}{5-y} \\ ... 0 You should use LaTex to write maths. Anyway, f:X\to Y is said 'onto' if Y=f(X), that is$$\forall y \in Y, \exists x\in X : y=f(x)$$That is, for every number in the codomain you can find a number in the domain which is mapped by f to the first one. Let y be a real number \not =5. You have to find x such that$$\frac{5x+1}{x-2}=y$$Linearize ... 1 Let b\in\mathbb{R}\backslash \{5\}. We need to find a\in\mathbb{R}\backslash \{2\} so that f(a)=b We need \frac{5a+1}{a-2}=b ab-2b=5a+1 a=\frac{2b+1}{b-5} You still need to confirm that a\ne 2. 0 A function f : X \longrightarrow Y is onto or surjective if range f = Y, ie. for every y \in Y there exists x \in X such that f(x) = y. Does this help? -1 We can't answer your question since it depends on the domain of f and g. For example if we assume that the domain of f is [2,+\infty) and the domain of g is [3,+\infty) then f\ne g. 0 Even though the RHS are similar(just replace u by x)enter preformatted text here, it still depend on the domain and co-domain of the functions If two function have thesame domain and co-domain,also f(x)=g(x) for any x inside the domain, f=g. 0 You should generally look for asymptotes and known values. For example, \frac{1}{x-2} has asymptotes at x=2, and y=0. \lim_{x\rightarrow\infty}\arctan(x) = \pi/2 \approx 1.6 \lim_{x\rightarrow-\infty}\arctan(x) = -\pi/2 \approx -1.6 \arctan(0) = 0 \arctan(1) = \pi/4 \approx 3/4. e^0 = 1 e^1 \approx 2.7 \lim_{x\rightarrow\infty}e^{-x} = ... 2 The first part should be read "Suppose that f is a function from the set A into the set B." The second part doesn't make sense, as written. It should say "Show that if f(x)=f(y) for arbitrary x,y\in A, then x=y." That is, if f(x)=f(y) implies that x=y for any elements x and y of A (not necessarily different elements, just with ... 2 It reads as: "Show that if f of x equals f of y for arbitrary x and y in A with x different than y, then x equals y". The comma as nothing to do with pairs, it is just a comma from common language. Moreover f \colon A \to B should be read as: f is a function from the set A into the set B. 3 If you say x^2+y=\text{constant}, you're defining a curve in the xy-plane, which is a parabola y=-x^2+\text{constant}. If you move along that curve, both x and y are changing. But if you write z=y^2+x, defining z as a function of those two variables, both of which can vary freely, then the expression \partial z/\partial x means the rate of ... 1 You could also show an explicit inverse, f^{-1}: \mathbb{R} -\lbrace 5 \rbrace \rightarrow \mathbb{R} -\lbrace 2 \rbrace, where f^{-1}(x)= \frac{2x+1}{x-5} and use that$$f\ \ \text{is one-to-one if, and only if, has a left inverse}$$That is, f^{-1} \circ f(x) = f^{-1}(f(x))=x \in \mathbb{R} - \lbrace 2 \rbrace$$f\ \ \text{ is onto if, and only ...

0

There is no maximum volume, since the problem specifies that the sum must be less than $100$. So we assume that the regulations say $\le 100$. To maximize volume, it is best to use the full $100$. For height $h$ and radius $r$, we have the constraint $h+2\pi r=100$. This is because the perimeter of the base is $2\pi r$. The volume is $\pi r^2 h$. So we ...

3

In this context, $\mathbb{R}-\{x\}$ means the set of all real numbers that are different from $x$. Injective: if $f(s)=f(t)$ then $\frac{5s+1}{s-2}=\frac{5t+1}{t-2}$ so that $$(5s+1)(t-2)=(5t+1)(s-2)\implies 5st-10s+t-2=5st-10t+s-2$$ which simplifies to imply that $s=t$. Surjective: suppose that $y$ is any number in $\mathbb{R}-\{5\}$, let us demonstrate ...

1

If we have a cosine graph with amplitude A, then its largest value will be A and its smallest value will be -A. If we shift this vertically by s units, then the largest value will be A+s and the smallest will be −A+s; so adding these together and dividing by 2 gives the value of s.

1

A non-constructive example of an unbounded additive function can be achieved as follows. Consider $\mathbb{R}$ to be a vector space over $\mathbb{Q}$ (a continuum-dimensioned one). By the axiom of choice, it has a Hamel basis. Take any of the coordinate functions corresponding to this basis. It is additive as any coordinate function, but it cannot be ...

1

Dejan's answer is absolutely correct, let me elaborate a little bit, starting with the more basic question: 'What is a function?' A function (from the set of real numbers with values in the set of real numbers), is a rule, which assigns to each real number another real number. (For simplicity, I am assuming that the domain of the function is the entire ...

1

When you write $f(x)=x^2$ you usually want to say something like "$f(x)$ is equal to $x^2$". Here, $x$ might be a particular number, like $3$, or $x$ might stand for any possible number, i.e. the meaning of "for all $x$, $f(x)$ is equal to $x^2$" might be intended. Instead of just "$f(x)$ is equal to $x^2$" it might also mean "$f(x)$ is defined as $x^2$". ...

0

$$\frac{\sqrt{x^4+3x^2}}{x}=\frac{\sqrt{x^2(x^2+3)}}{\sqrt{x^2}} \\ =\sqrt{\frac{x^2(x^2+3)}{x^2}} \\ = \sqrt{x^2+3}$$

0

$$\frac{\sqrt{x^4+3x^2}}{x}=\frac{\sqrt{x^2(x^2+3)}}{x}=\frac{|x| \sqrt{x^2+3}}{x}$$

3

Is this what you're looking for? $$\min(a,\,b) \; = \; \frac{a\,+\,b\; - \; |a-b|}{2}$$ $$\max(a,\,b) \; = \; \frac{a\,+\,b\; + \; |a-b|}{2}$$

1

Yes. It's not hard to prove. Let $f(n)\in \mathcal{O}(F(n)), g(n)\in \mathcal{O}(G(n))$, and say the bounding is explicitly $$|f(n)|\le C_f F(n),\quad |g(n)|\le C_g G(n)$$ for constants, $C_f,C_g>0$. Then $$|f(n)+g(n)|\le |f(n)|+|g(n)$$ $$\le C_fF(n)+C_gG(n)$$ setting $M=\max\{C_f,C_g\}$ we see this is $$\le M(F(n)+G(n))$$ ...

1

To see your property, take $g_1 = f_1$ and $g_2 = f_2$ in your known theorem. Then your conjecture holds. Or you can see it as follows. Clearly $O(\max(f_1,f_2))$ is no greater than $O(f_1 + f_2)$, and $O(f_1 + f_2)$ is no greater than $O(2 \max(f_1,f_2))$.

0

The definition of continuous is always relative. If you have a function $\mathrm{f} : X \to Y$ between topological spaces $X$ and $Y$, then you need to know which topology you are using on the source $X$ and which topology you are using on the target $Y$ before you are able to decide if $\mathrm{f}$ is continuous or not. We say that $\mathrm{f}$ is ...

0

I think perhaps your prof used poor wording and meant piecewise continuous, and they are looking for the function given by gebruiker at the top. Had this happen to me in Real Variables and confused me. It is a function and piecewise continuous (over an infinitely large set too!) and if your studying Lebesgue measure/integrals then that is what you want. ...

1

The answer to your question is that the numerator $\sqrt{2}|\sin{x}|$ is always positive while the denominator $x$ is negative to the left of the origin and positive to the right. Thus the quotient changes sign around the origin. I think there is some confusion about the $\pm$ in how you expressed the numerator. Before you simplify, you have ...

-3

Everyone seems to be mistaken about what continuity of a function actually means. The Dirac Delta function answers the question (as others have already suggested). The Dirac delta is a continuous function; as defined by its integrability i.e. the n-dimensional integral of any n-dimensional Dirac delta function is unity (it would be naturally non-integrable ...

0

I think you could do in this way: $f(x)=x^2+[\Theta(x-1)](x^3-x^2)$ where $\Theta(x-1)= \begin{cases}0 \text{ if }\ x<1 \\ 1 \text{ if }\ x>1\end{cases}$ is the shifted unit step function.

1

This can indeed be done in $O(n^2 \log n)$ time. Here's a rough sketch. Firstly, sort the array using any of the comparison based sort algorithm (like Merge Sort, Quick Sort etc.), this will take $O(n \log n)$ time. Then, by using 2 nested loops going over the array (which will take $O(n^2)$ time) for each combination. Now, calculate sum of elements for ...

1

Try all $N(N+1)/2$ pairs $(A[i], A[j])$ and look for $Value-A[i]-A[j]$ by dichotomic search in the array (sorted).

5

The $\to$ arrow points from the domain of a function to its codomain or target set. The $\mapsto$ arrow (fittingly called \mapsto in TeX) shows what an individual element of the domain will be mapped to, i.e. it shows what the function does while $\to$ shows where it operates (so to say). So, the elaborate way to write a function definition is ...

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