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13

$ m \mid n$ means that $m$ divides $n$. We could also say that $n$ is divisible by $m$ AND we could also say that $n$ is a multiple of $m$,so $n= a \cdot m, a \in \mathbb{Z}$.


11

Hint Let $g(x)=f(x)e^x$. Then $$g''=(f+2f'+f'')e^x \geq 0 \,.$$ That means that $g$ is.... How does this solve the problem?


7

Since $$f(x)+f(1-x)=\frac{4^x}{2+4^x}+\frac{4^{1-x}}{2+4^{1-x}}=\frac{4^x}{2+4^x}+\frac{2}{4^x+2}=1$$ you have: $$\begin{eqnarray*}S&=&\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)=f(1)+\sum_{i=1}^{1002}\left(f\left(\frac{n}{2005}\right)+f\left(\frac{2005-n}{2005}\right)\right)\\&=&f(1)+1002=\frac{2}{3}+1002.\end{eqnarray*}$$


7

Note that $\sec^2 x- \tan^2 x= 1$ for all $x$ except those in the form $x = \dfrac{\pi}{2} + \pi k$ (where $\sec x$ and $\tan x$ are undefined. Thus, $f(x) = 3^{\sec^2 x- \tan^2 x} = 3^1 = 3$ for all $x$ except those in the form $x = \dfrac{\pi}{2} + \pi k$, which are evenly spaced every $\pi$ units. Thus, $f(x)$ has period $\pi$.


7

You have to keep in mind what $\epsilon$ is requiring the function to do. Take a constant function $f(x) = 2$ as you asked. The $\epsilon-\delta$ definition demands that, for any $\epsilon > 0$ we can find a $\delta >0$ so that if $x$ is within $\delta$ of $a$, $f(x)$ is within $\epsilon$ of $f(a)$ But in the case of our constant function, ...


5

We have $\lim_{y\to-\infty}\phi(y)=0$ and $\lim_{y\to\infty}\phi(y)=1$. Moreover, the function $\phi$ is continuous. Thus, by the Intermediate Value Theorem, the function $\phi$ is surjective.


4

Consider $$f(x)=\left\{\matrix{\frac{\sin{(x^2)}}{x}&\hbox{if}&x\ne0\cr 0 &\hbox{if}&x=0}\right.$$


4

The easiest and most direct way to say that $f$ is a function is by stating it as such. Actually, even if you use the letter $f$, you should always explicitly state that $f$ denotes a function. For example, this statement: If $f$ is a differentiable function, then $f$ is continuous is, by my oppinion, much better than If $f$ is differentiable, ...


3

Building on your own answer, one sees that the crucial part is getting some lower bound on $f(t_{max}-\delta)$. The differential equation yields $$ \frac{df}{dt} = f-f^3-\alpha f(t-\delta) \leq (1+\alpha)f_{\max} $$ By Taylor expansion, $\exists \tau,\, f(t_{\max})-f(t_{max}-\delta) =\delta f'(\tau)$. Thus, using your global bound, ...


3

Hint The equation you have in question is $y=\sqrt{1-x^2}$. What would happen if you square both sides of the equation? You get a very familiar geometric object. Once you determine what it is, you can conclude that $y=\sqrt{1-x^2}$ is just the upper part of it. And, $y=-\sqrt{1-x^2}$ is the lower part of it.


2

The equation $y=\sqrt{1-x^2}$ can be rewritten as $\sqrt{(x-0)^2+(y-0)^2}=1$. The LHS describes the distance from the point $(0,0)$ to a point $(x,y)$, and the RHS says that this distance is a constant. Therefore the distance from the origin to any point on the function is $1$. The geometric shape that has a constant distance from a particular point is a ...


2

The equation $\log(a^b)=b\log a$ is valid only for positive values of $a$ (as your example clearly shows).


2

Your friend is right. Surjective means $f(\Bbb R) = \Bbb R$, or for any $y\in \Bbb R$ there exists $x\in \Bbb R$ such that $f(x) = y$. Now, if $f(x) = x$ and $g(x) = -x$, they are both surjective (why?). But $h(x) = f(x) + g(x) = 0$ is not surjective, since $h(\Bbb R) = \{0\}$, for example there does not exist $x\in \Bbb R$ such that $h(x) = 1$ since $h(x) = ...


2

Hint: The function is convex in each of the variables, hence the extreme values occur at the end points of the interval. I hope that helps.


2

$f(1+a)=a^2\sin(n\pi(1+a))=a^2\sin(n\pi+n\pi a)=a^2(\sin n\pi\cos n\pi a+\cos n\pi\sin n\pi a)$ $=a^2\cos n\pi\sin n\pi a$, but $f(1-a)=a^2\sin(n\pi(1-a))=a^2\sin(n\pi-n\pi a)=a^2(\sin n\pi\cos n\pi a-\cos n\pi\sin n\pi a)$ $=-a^2\cos n\pi\sin n\pi a$; so the function does not appear to be even about 1. For example, if $n=1$, $f(3/2)=-1/4$ while ...


2

Yes. In fact, one can show that if there is an injective function from $A$ to $B$ and an injective function from $B$ to $A$, then there is a bijection from $A$ to $B$. This is known as the Cantor–Bernstein–Schroeder theorem. (Of note is that its proof does not depend on the Axiom of Choice.)


2

You can simplify your argument. You should look at $g(i)$ compared to $g(i_0)$, which reduced the number of cases to 2: if $g(i_1)=g(i_2)=g(i)<i_0$, then $i<i_0$ as $g$ is strictly increasing. Then you know that $i_1=g(i_1)=g(i_2)=i_2$. same goes for $g(i_1)=g(i_2)=g(i)>i_0$ $i_1+1=g(i_1)=g(i_2)=i_2+1$. there is no case $g(i_0)=i_0$ by definition ...


2

Let $C$ be a Lipschitz constant for $g$ and let $\epsilon > 0$. Then, since $\lim_{x \to \infty}f(x)=k$, there is some $M>0$ such that $$x > M \quad \implies\quad |f(x)-k|<\frac{\epsilon}{C}.$$ It follows that $$x > M \quad \implies\quad |g(x,f(x))-g(x,k)| \leq C|f(x)-k| < C\frac{\epsilon}{C}= \epsilon,$$ i.e. $\lim_{x \to ...


2

As the approach that I'd made in this question is a bit lengthy, hence I'd write only it's procedure here. So what I firstly assumed that the second derivative (i.e $f''(x)$) of the function is $$f''(x)=a(x-1)$$ I made this assumption from the information that says local minima of $f'(x)$ is at $x=1$. And further integrating $f''(x)$ to $f'(x)$, ...


2

Probably not what you want, but consider this: Let $h : \mathbb R[X] \to \mathbb R[X]$ be defined by $$h(P)= X P'(X) \,.$$ Define now $h_1=h$ and recursively $$h_r=h_{r-1} \circ h \,.$$ Then, $h, h_r$ are independent of $q$ and satisfy $$h_r(X^q)=q^rX^q$$ Take $f_r(P(X)) = \frac{h_r(P(X))}{X}$. Note that $h$ is a linear function and $h(\mathbb P_n) ...


1

Building on Martin's post: Supposing earth is a perfect ellipsoid with equation: $\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{b^2} = 1$. It is usually considered that you don't need a different parameter for $y$ for the earth ellipsoid, which implies that the sections by planes $z=c$ (ie constant latitude) are circles. So, given a latitude of $\theta ...


1

This can be accomplished by using Thaddeus Vincenty's formulas for finding the ellipsoidal distance between two given points. Vincenty's formulas are defined as: $\alpha$ length of the semi-major axis of the ellipsoid $\beta$ length of the semi-minor axis of the ellipsoid $\gamma=\frac{1}{\alpha}(\alpha-\beta)$ flattening of the ellipsoid $x_1, x_2$ ...


1

It is a matter of doing things in an orderly manner. Yes, this covers all cases: either the image is contained in $[k]$, or it maps to $k+1$. Induction is made over $k$; not $m$. Careful. This is wrong. Induction is being made over $k$, and your induction hypothesis is that if $[m]\to [k]$ is an injection, $m\leqslant k$. You don't know anything about ...


1

$$ \dfrac{1}{x} - 1 \geq 0 \quad \Rightarrow \quad \dfrac{1 - x}{x} \geq 0 $$ Below we elaborate diagram of signals for the functions $f(x) = 1 - x$ and $g(x) = x$. The signs of the third row were obtained by dividing the signal by the first signal from the second row in each sector. Thus, $$ S = \{x \in \mathbb{R} : 0 < x \leq 1\} $$


1

We need to consider positive values for $x$ because $x<0\implies1/x-1<0$ and $x=0$ means that $1/x-1$ is undefined. So $$1/x-1\ge0$$ $$1/x\ge1$$ $$x\cdot1/x\ge x\cdot1$$ $$1\ge x$$ $$x\le1$$ thus (because $x\gt0$) $$0\lt x\le 1$$


1

Since $f\circ f=id$, we know that $f$ is a bijection. Now assume that $f\ne id$, i.e., there exists a point $x_0$ such that $$f(x_0)=y_0\ne x_0.$$ We know that also $$f(y_0)=x_0.$$ W.l.o.g. we can assume that $x_0<y_0$. From the intermediate value property we can get some information about values of $f$. We know that all values from the interval ...


1

(n + m x 200) x 838.8608 will give you 20000 numbers well spread out in the RGB space. Incrementing n will increase the green by two units and the blue by 71; incrementing m will increase the red by two or three units and green/blue by larger increments. In practice, you will see no correlation between (n, m) and the color.


1

The difference is at the "$\delta$" level in the $\epsilon$-$\delta$ definition of continuity. Let $f$ be a function on $(a,b)\subseteq \mathbb R$. Then $f$ is continuous on $(a,b)$ if $\forall x_0\in(a,b),~ \forall \epsilon>0~~\exists \delta=\delta(\epsilon, x_0)~:~\forall x~:~|x-x_0|\Rightarrow |f(x)-f(x_0)|<\epsilon,$ while $f$ is uniformly ...


1

$$(x-1) \mid f(x^n) \implies f(1)=0 \implies (x-1) \mid f(x) \implies (x^n-1) \mid f(x^n) $$


1

Hint: Use Lagrange Multipliers



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