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0

How about adding each tick (max value-current value)/100, or whatever divisor makes you happy? Then subtract however much is used up in the current action.


0

Curve fitting of the sort you described makes sense for data that does not come from an explicitly given function. Here you already have an elementary function like $\sqrt{x}$ or $\log x$. Fitting $a/(1+bx^c)$ to this will likely to be a bad fit for no gain in simplicity. To answer your questions directly: write out the Taylor series expansion on both ...


0

It is a constant function. It is technically undefined at the points JimmyK4542 mentions. In order for the identity $f(x+p)=f(x)$ to hold, is is required that the domain $D$ of $f$ satisfies $D+p=D$ (where $D+p$ is defined as $\{x+p:x\in D\}$). For the domain in question, this occurs whenever $p$ is an integer multiple of $\pi$. Because the function is ...


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 $\pi$.


0

Consider the derivatives $$ f'(x)=3x^2-8x^3 $$ $$ f''(x)=6x-24x^2 $$ So we have critical points at $x=0$ and $x=3/8$. For $x\in (-\infty,0)$ we see that $f'(-1)=3-8(-1)=11>0$ so $f(x)$ is increasing. $f'(1)=3-8=-5<0$ so for $x\in (3/8, \infty)$ $f(x)$ is decreasing. We also see $f''(3/8)=6(3/8)-24*(3/8)^2=9/4-9/8=18/8-27/8=-9/8<0$ which ...


0

Your function is continuous and only non-negative on the interval $[0,0.5]$. Thus the function must take a maximum value on this interval (since it's continuous and the interval is closed), and so any positive value greater than the maximum is not achieved by your function so your function is not surjective.


0

If the function is only positive in $[0, 0.5]$, and it's continuous, it must have a maximum there. Hence, any value greater than this maximum is not $f(x)$, for any $x \in \Bbb R$. Finding this maximum is another problem, better dealt with calculus.


0

Yes, you're right: for $0<x<\frac{1}{2}$ $f(x)$ is positive. But for $x \in (0,\frac{1}{2})$ you have: $$x^3-2x^4=x^3(1-2x)<\frac{1}{2^3}(1-2x)$$ It's because $f(x)$ is positive and $0<x<\frac{1}{2}$. Next: $$\frac{1}{2^3}(1-2x)<\frac{1}{2^3}$$ Because for $0<x<\frac{1}{2}$ you have $0<1-2x<1$. So $f$ reach values less than ...


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) = ...


1

A function is continuous if and only if it is upper and lower semicontinuous. The function you defined is upper semicontinuous but not lower semicontinuous at $x=1$.


1

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.


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

Note : log{ base 1/2} w = - log {base 2 } w (2/3) ^(log { base 1/2} (x+2)² ) = (3/2)^(log {base 2} (x+2)² ) = (9/4)^(log {base 2 } (x+2) ) < (9/4)^log{base 2 } (x² - 3x -10) means (x + 2 ) < (x² - 3x - 10 ) OR 0 < x² - 4x -12) = ( x - 6 )(x+2)---> x in (-∞ , - 2 ) U (6 , ∞)


0

$dP/dt = (b-d)P$ To solve the equation put P on the LHS and dt on the RHS. $\frac{1}{P}dP=(b-d)dt$ Now integrate: $\int \frac{1}{P}dP=\int (b-d)dt$ $log(P)=(b-d)t+C$ Now solve the equation for P. But without knowing the table I cannnot proof, which values the parameters b,d and C might have.


0

If that is in a textbook, it's a horrible example. The cartesian product of the sets $A$ and $B$ is really the set of all possible combinations $(x \in A, y \in B)$. So I don't "see" the cartesian product in what is written. It's like they gave you two possible $(x,y)$ elements from the cartesian product, $f=\{(x_1,y_1),(x_2,y_2)\}$ and somehow called this a ...


0

In rough terms, if we have a function that maps elements of A to elements of B, the pre-image of B $f^{-1}(B)$ is defined to be the set of all elements of A such that the function maps them to elements of B (the vertical bar in set notation is read to mean "such that"). To use the x and Y notation you use near the end of your question, making x into an ...


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 ...


0

We have 20000 possible inputs, if we want to distribute these into the 3 channels we get $\sqrt[3]{20000} = 27.14...$ different values per channel. Now we just need a function $$ f: \mathbb R^2 \to \mathbb R^3 \text{ with } f([1,100],[1,200]) = [1,27]^3 $$ But you cannot find a function that maps these two sets really smoothly into eachother. I'd try for ...


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.


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.


1

Take $c=1$, since $g$ is obviously growing way faster. Let $n_{0}=1000000$. Now assume $n>n_{0}$. Notice the following property: $n^{\frac{10}{n}}=(n^{\frac{1}{n}})^{10}$ is decreasing, and since plug in $n=1000$ showed that $n^{\frac{10}{n}}<2$ we have $n^{\frac{10}{n}}<2$ for all $n>n_{0}$. Also $\log(n)>4e$ and $f(1000000)<g(1000000)$ ...


0

You might prove the following lemma and find it useful for #1: If $f,g$ are functions taking values greater than $2$, and $\log f(x) = o(\log g(x))$, then $f(x) = o(g(x))$. (Note that this is false in general if $o$ is replaced by $O$.) Since $f(x) = o(g(x))$ implies $f(x) = O(g(x))$, this can help for #1 (which is a situation where taking logs makes the ...


0

Reading this problem again, here is a simple construction to map the monomial $x^q$ to $q^{r-1}x^{q-1}$. Consider the formal derivative operator $D(x^q) =q x^{q-1}$; then $(Dx)(x^q)=(q+1)x^q$. Consequently $$(Dx)^r D(x^q) = (Dx)^r(qx^{q-1})=(Dx)^{r-1}q^2 x^{q-1}=\cdots = q^r x^{r-1}.$$ (Equivalently we may write this as $D(xD)^{r-1}$.)


1

It seems to me that the operator you seek would be: $$f(x^p)=\frac{1}{x}\int_0^x t^p dt$$ this has the desired properties. Note that this operator is actually independant of $p$, and can be written more generally, for some $g\in\Bbb R[x]$ $$f(g)=\frac{1}{x}\int_0^x g(t)dt$$


1

EDIT: I'm quite sure that I misunderstood the OP's intention. However, the comments to this answer may still be of interest and so have made this a wiki answer. No such function exists. We want $f(x^q)=x^q/(q+1)$ for all $q$; consider in particular $q=6$. Then we require \begin{align} f(x^{6}) &=f((x^2)^3))=f((x^3)^2))\\ ...


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, ...


0

Assuming your calculations to be correct, we need to have $x^2+4x+4=(x+2)^2, x^2-3x-10=(x-5)(x+2)>0$ The first one is already perfect square, so only needs $x+2\ne0\iff x\ne-2$ The second needs $x>$max$(5,-2)=5$ or $x<$min$(5,-2)$


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?


0

HINT: If not, $f$ must have a maximum in the interior of the interval.


0

Disclaimer: I don't agree on the conclusion that $f$ is not continuous at the origin unless you define $f(0,0)$ in some way. This said, I can't really understand your question. Is it Prove that $f$ is continuous at every point of $\mathbb{R}^2 \setminus \{(0,0\}$ or Provethat the only discontinuity point of $f$ is $(0,0)$? In the first case, ...


1

Use the identity $$xy=\frac{1}{4}\left((x+y)^2-(x-y)^2\right)=\frac{1}{4}\left(K^2-(x-y)^2\right).\tag{1}$$ Since $(x-y)^2\ge 0$, for given $K\ge 0$ the left side of (1) reaches a maximum when $x=y=\frac{K}{2}$, and that maximum value is ${K^2}{4}$. Suppose now that $K$ is an integer, and $x$ and $y$ are constrained to be integers. If $K$ is even, we ...


0

Vishwa's hint is the most general way of solving constrained optimization problem. From your question I'm assuming (though this was never stated explicitly) that you are looking for only integer solutions $x, y$ which complicates things. For your simple problem, the way you wrote is absolutely the right way to go. You can solve for $y$ so you might as well ...


1

Hint: Use Lagrange Multipliers


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 $f'(x)$, ...


0

Your proof is a bit jumbled. In its principle it is the same proof as sketched by William, but your proof is harder to read for two reasons. You are using $f$ for the a choice function and $h$ is in fact the function you prove to exist. That's fine, but it is confusing to the reader. The way you find the choice function is a bit cumbersome. You are correct ...


1

Here is a sketch of a proof using the axiom of choice: Let $R$ be a relation on $A$. For each $a \in A$, let $X_a = \{b \in A : b R a\}$. By the assumption on the relation $R$, $X_a \neq \emptyset$. Let $\mathcal{X} = \{X_a : a \in A\}$ ($\mathcal{A}$ is a set by the axiom of replacement). Since $\mathcal{X}$ is a family of nonempty sets, the axiom choice ...


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 ...


0

It depends on exactly what you mean by "basic properties." There are collections of conditions that do imply equality of two functions: For example, If we know that $f$ and $g$ are continuous on the same domain, and they can be shown to agree on a dense subset of that domain, then $f = g$ everywhere on the domain. If we know that $f' = g'$ everywhere, and ...


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.)


0

No, the limit function, $F$, need not even be differentiable if the original sequence $\phi_n \rightarrow F$ pointwise. For an example on $[0,1]$, take $F = 1_{[1/2,1]}$ (indicator function) and mollify it with an approximation to the identity $\eta_n$. This yields a sequence $\phi_n = F \ast \eta_n$ such that $\phi_n \rightarrow F$. $\phi_n$ are analytic ...


0

My attempt: $f(t)=U(t)e^t −U(t−1)e^t$ $\begin{align} U(t)-U(t-1) & = \begin{cases} 1 & : t \ge 0 \\ 0 & : t\lt 0\end{cases}-\begin{cases} 1 & : t \ge 1 \\ 0 & : t\lt 1\end{cases} \\ & = \begin{cases} 0 & : t \ge 1 \\ 1 & : 0\le t \lt 1 \\ 0 & : t\lt 0 \end{cases} \\ \color{gray}{ \operatorname{\bf ...


0

$f(t) = U(t) e^t +U(t-1) U(1-t)e^t-U(t-1) e^t $ I keep almost the same function as you have, just modifying it's behaviour at the point $x=1$.


1

$$ e^t U(1-t) U(t)\qquad \text{(with $1-t$, not $t-1$)}. $$ ${{{{{{{{{{{{{{{}}}}}}}}}}}}}}}$


1

Simple. f(t) = $e^tU(-t+1)U(t)$ EDIT Your function f has the value of 0 on the entire x domain, in the exception of [0, 1] where it has the value of $e^t$. You can start with f(t) = $e^t$. The problem now is that for t < 0 or t > 1 the value of f(t) is still $e^t$. We start by clearing out f(t) if t > 1. We know that the Heaviside function U(t) is 1 ...


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 $n= a \cdot m, a \in \mathbb{Z}$.


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

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 ...


0

Yes. It surely does. The $\epsilon$ band doesn't have to be a subset of the codomain. You can choose the value of $\epsilon > 0$ yourself. The task is to find a suitable $\delta$.


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, ...


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 ...



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