# Tag Info

0

What you state shows that $L_1$ is the maximum, but it does not imply that $F$ is concave. In fact, $F$ is not concave on the entire interval $(0,1/a)$. Note that $$F''(1/a) = -2 \log(T) - 2a > 0,$$ since $T < e^{-a}$. So there is a point of inflection between $L_1$ and $1/a$.

0

$7$hrs and $5$minutes are $7\cdot 60+5=425$ minutes. Since you did $21$ batches that is $425/21=20.23$ minutes per batch, or equivalently $1$ batch every $20.23$ minutes. This means that you satisfy the goal of $1$ batch every $30$ minutes.

0

Well, first you need to see the range of $\frac{x^2+e}{x^2+1}$, which you can easily verify to be $(1,e]$. Then, the range of $\log$ in this domain is $(0,1]$. As pointed out in a comment: Since $\sin$ is increasing and $\cos$ decreasing in $(0,1]$, then $\sin(0,1] =(0,\sin(1)]$ and $\cos(0,1] =[\cos(1),1)$. Finally, since in $(0,\sin(1)]$ the cosine ...

0

Hmm, $x+1/x = 2 \cosh( \log (x))$ so what you have is also $$f(2 \cosh(\log(x))) = 2\cosh(3\log(x))$$ Perhaps you can then continue on your own...

0

This is to explain my motivation for asking the question. An analogy A programming language I am fond of (look me up if you're interested) has a small library of algorithms that defines a graph as having a set of nodes - comprising domain and codomain and a mapping from node to set of nodes, for some nodes. Without much trouble, I rewrote the library ...

0

The answer is, "it depends on your conventions". There are two very common conventions in mathematics, which I will summarize as: F-SET: A function "is" a set of ordered pairs in which no two pairs have the same first element. F-TRIPLE: A function "is" a triple $(A, B, \Gamma)$ where $\Gamma \subseteq A \times B$ and every element of $A$ occurs as the ...

2

Let $f:\mathbb R\rightarrow \mathbb R$ with $f(x)= x^{1/3}$. Then the domain of $f'(x)$ does not include $0$.

0

Yes, if you wish to explicitly write out the list of ordered pairs. However, this is usually not possible. So, to take for example the function given by 5xum, $f(x)=x^2$, how am I supposed to know the domain? Lets say I want the domain to be the unit interval as above, $[0,1]$. Then to list it as a set of ordered pairs, I need to write it as ...

0

Yes, you get a different function. It is not really clear to me, what you mean by "redundant" when you say that $A$ and $B$ are redundant in $f:A\to B$. One way to define a function is to define its graph, i.e. the set $(x,f(x))$. But to do so, you need to say where the set of pairs $(x,f(x))$ lie. In other words, the definition of a function consists of ...

2

Often you can find the definition that a function $f : A \to B$ is a subset of $A \times B$ such that for all $a \in A$ there is a unique $b \in B$ such that $(a,b) \in f$ (which is written as $b=f(a)$). Thus, if $B \subseteq C$, then $f$ is also a function $f : A \to C$. But often, it is very useful (especially in category theory, but also in order to ...

0

Technically speaking, yes. If you are very strict in your definitions, then $$f:[0,1]\to[0,1]\\ f: x\mapsto x^2$$ is not the same function as $$g:[0,1]\to[0,2]\\g:x\mapsto x^2$$

0

Suppose definition 1 holds and $y=f(a)=f(b) \in B$ for some $a,b\in A$, then we have $a=b$ by definition 1, which is the countrapositive of definition 2. Suppose definition 2 holds and $y=f(a)\in f(A)$ then we have $x\in A$ s.t. $f(x)=y$ since $f(a)$ is in the image, the uniqueness is guaranteed by definition 2.

1

If $f(x) = x^5 \sin x$. $$f(-x) = (-x)^5 \sin(-x) =(-1)^5x^5(-\sin\ x) = x^5 \sin x = f(x)$$ It follows by definition that $f$ is even. More generally, can you tell what happens with $f(x) = x^n \sin x$, when $n$ is even? And when $n$ is odd? Simply observe that $(-1)^n = \begin{cases} 1 &, n\ \ \text{is even}\\ -1 &, \ n\text{ is odd} ... 1 If$f(x) = x^5 \sin x$, then $$f(-x) = (-x)^5 \sin (-x) = -x^5 \left( - \sin x\right) = x^5 \sin x = f(x),$$ and so$f$is actually even. 0 $$f(x) = \frac{e^x+e^{-x}}{e^x-e^{-x}} = \frac{2\cosh(x)}{2\sinh(x)} = \frac{1}{\tanh(x)} = \coth(x)$$ therefore, $$f^{-1}(x)= \operatorname{arccoth}(x)$$ 2 The only$f$that is linear and satisfies$f(xy)=f(x)f(y)and has an inverse is the identity function. So you are asking to solve the equation \begin{align} x^2+\int_0^x(\sin(t)+a^2t^3+bt)\,dt&=\alpha\\ x^2+\left[-\cos(t)+\frac14a^2t^4+\frac12bt^2\right]_0^x&=\alpha\\ x^2-\cos(x)+1+\frac14a^2x^4+\frac12bx^2&=\alpha\\ ... 2 Hint: a bijective, continuous function is monotonic: it is either strictly increasing, or strictly decreasing. edit: As mentioned in the comments, monotonicity just gets you injective; \arctan(x) for instance is monotonic but not bijective. You also need surjectivity: you have to show your function actually does have all of \mathbb R as its range. 0 Hintf'(x)=a+\cos x$$so f is increasing iff a\geq 1. 1 Hints: f must one-to-one for the inverse function to exist. Consider what happens if it is onto S_2 or not. (See Bijection, injection, surjection) 1 Suggestions: On (0,\pi), \sin(x) takes on all values in (0,1]. So your problem is equivalent to finding the max and min of u^u for u\in (0,1]. By looking at derivatives, you should be able to do this using calculus techniques. 0 Hint: Our function is continuous. Evaluate at \frac{\pi}{2} and \pi, and use the Intermediate Value Theorem. You will also need to show that the function takes no values outside the interval [0,1]. Remark: The inverse function approach is not useful here. For one thing, our function is not one-to-one in our interval. Even if we restrict to an ... 0 Edit: (based upon edits to question) Ah, well, you clearly have more knowledge than you put into your original question, so perhaps you are holding out on more information? That said, there's still hope, you still need properties of y_2(w), but basically, you need to guarantee the following...$$ V''(w) \le 0$$Notice that y_1(w) is just a constant. ... 0 It's completely standard to say function, even when talking about proper classes. If you want to distinguish this fact, then you should say class function. I can't say that there's any difference in notation either, except when (for example) sets are denoted by a lowercase letter, and proper classes by capital letters, then a class function should probably ... 1 Let f\colon U\to V and g\colon V\to W. If you want to check the maps (g\circ f)^t and f^t\circ g^t are equal, you need to check they agree on an abritrary input \alpha\in W^*. So you need to check$$(g\circ f)^t(\alpha)=(f^t\circ g^t)(\alpha).$$As pointed out in the comments, this equality can already be proved directly. Alternatively, both ... 0 Ups, wikipedia (http://en.wikipedia.org/wiki/Map_%28mathematics%29) seems to help: Map (mathematics) In logic In formal logic, the term is sometimes used for a functional predicate, whereas a function is a model of such a predicate in set theory. Which is pretty clear. And thus I think a sufficient answer for my query, and also matches my ... 8 There is a pretty straightforward answer. Polynomials are very strict in what terms they allow. Why not try matching the first part x^3? What would you have to do with the term x+\frac{1}{x} to get x^3 as a first term? Then look at your result and try to remove whatever there is too much, again in terms of x+\frac{1}{x}. 2 Hint:$$\left(x + \frac1x\right)^3=x^3+3x^2\frac1x+3x\frac{1}{x^2}+\frac{1}{x^3}$$1 As \left|x+ \frac 1x\right| \ge 2, f is only constrained of \{y :|y|\ge 2 \}.$$ x + \frac 1x = y \\ x^2 - xy + 1 = 0 \\ x = \frac12\left( y \pm \sqrt{y^2 - 4} \right) $$Hence on this set:$$f(y) = \frac 18\left(y + \sqrt{y^2 - 4}\right)^3 + 8\left(y + \sqrt{y^2 - 4}\right)^{-3} $$When you expand the formula, it turns out that$$ f(y) = ... 5x^3+x^{-3}=(x + \frac{1}{x})^3-3(x + \frac{1}{x})$set$t=x + \frac{1}{x}, f(t)= t^3-3t$. 2 $$f\left(x + \frac1x\right)= x^3+x^{-3}$$ $$\left(x + \frac1x\right)^3 = x^3 + 3x + 3\frac1x+x^{-3}$$ $$\Rightarrow f(t) = t^3 - 3t$$ 0 Maybe $$f(x):=x^3-3x$$ Thus $$f(x+x^{-1})=x^3+3x+3x^{-1}+x^{-3}-3x-3x^{-1}=x^3+x^{-3}$$ 0 Let $$y=w-\frac{x}{w}$$ Then you get $$w^3-5x^3+1-\frac{x^3}{w^3}=0 \stackrel{\cdot w^3}{\iff} w^6+(-5x^3+1)w^3-x^3=0 \stackrel{\theta:=w^3}{\iff} \theta^2+(-5x^3+1)\theta-x^3=0 \iff \theta= \frac{5x^3-1\pm\sqrt{25x^6-6x^3+1}}{2} \iff y=\sqrt[3]{\theta}-\frac{x}{\sqrt[3]{\theta}}$$ Check your answers, the minus sign case on the quadratic doesn't satisfy ... 1 You have a depressed cubic equation$y^3+a y + b=0$in which$a$and$b$are functions of$x$. So consider the test (Cardano method) $$4a^3+27b^2=675 x^6-162 x^3+27$$ has no real root and so it is positive. I am sure that you can take from here. 0 Edit: Originally the question asked was for the equation$y^2 + 3x - 5x^3 + 1 = 0$, and so this is a response to that question. How you can go about trying to understand this: Note that the given equation is a quadratic equation in$y$, whose coefficients are polynomials in$x$. That is, we can rewrite it as $$y^2 + by + c = 0$$ where$b = 3x$and$c = ...

1

Answer to question before edit: It doesn't define $y$ as a function of $x$, because it doesn't pass the Vertical Line Test. Take for example $x=1$, which gives $y^2+3y-4=0 \Leftrightarrow y=1 \vee y=-4$.

1

First step: Assume that $f = g + h$, $h$ even and $g$ odd. Then $$f(x)= g(x) +h(x)\\ f(-x) = g(x) - h(x)$$ this gives you $$g(x) = \frac12(f(x) + f(-x))\\ h(x) = \frac12(f(x) - f(-x))\\$$hence $(g,h)$ is unique. Second step: just check the hypothesis, that is: $g$ is even, $h$ is odd. This proves existence of $g,h$. Conclusion: $$\exists ! (g,h)\ \ ... 0 If$$H(x)=\frac{f(x)+f(-x)}{2}F(x)=\frac{f(x)-f(-x)}{2}Note that H is even, and F is odd, also H(x)+F(x)=f(x) 1 The ingredient that you’re missing is the precise meaning of congruence modulo 3. Since 653=3k+2, where the particular value of k doesn’t concern us, and since f^3(x)=x, you get f^{653}(x)=f^{3k+2}(x)=(f^3)^k[f^2(x)]; but since f^3 is the identity, its k-fold iterate is identity too, and we can just erase that part of the expression: ... 1 The conclusion f^3(x)=x (=f^0(x)) applies whatever x is. Use induction on r to show that f^{3r+s}(x)=f^s(x) 2 for either case using formula on wiki, I'm stuck at these limits (putting \partial Z_0=1): \lim_{t \to \infty} \frac{\ln(-sint dt)}{t} and for the other one \lim_{t \to \infty} \frac{\ln(3 dt)}{t} Sorry? What is, in the second case, say, 3dt when t\to\infty? And why put \partial Z_0=1, as you say, when the limits are considered for ... 1 For each of the given functions, other than \tan \sqrt{x}, there is at least one fixed point between (k-1)\pi and k\pi for all positive integer k. So the partial sum for any range up to n\pi is greater than \sum_{k=1}^n \frac{1}{k\pi}, which we know diverges, so the sum cannot converge. For the case of f(x) = \tan \sqrt{x}, there is ... 0 Using Taylor's theorem to determine the first order polynomial \begin{align} L(x,y) &= f(0,0) + f_x(0,0)\, x + f_y(0,0) \, y \\ &= 6 + (2+8x)\rvert_{(x,y)=(0,0)} \, x + (3+10y)\rvert_{(x,y)=(0,0)} \, y\\ &= 6 + 2x + 3y \end{align} 1 For f(x)=\tan x, there is a fixed point of f in each interval ((n-\frac12)\pi,(n+\frac12)\pi), hence the series in question is essentially the harmonic series, diverging. For f(x)=\tan^2 x and f(x)=\tan 2x, the same argument applies. f(x)=\sqrt{\tan x} makes no difference compared to \tan x (except that the irrelevant negative parts become ... 1 Hint: for f(x)=\tan x, \tan x=x has at least one solution (which we call x_k) in each interval of the form (2k\pi-\pi/2,2k\pi+\pi/2). So\sum_{x\in S(f)} \frac1x \geq \sum_{x\in S(f)} \frac{1}{x_k} \geq \sum_{k} \frac{1}{2k\pi+\pi/2}...$$Similarly try for the other functions. 2 Wherever x appears in f(x), plug in the entire function g(x):$$f(g(x)) = \frac{1}{10(g(x)) + 17} + 13 = \frac{1}{10\left(\dfrac{1}{9x-6}\right) + 17} + 13 $$Multiply numerator and denominator of the main fraction by 9x-6:$$f(g(x))= \frac{9x-6}{(9x-6)\cdot 10\left(\dfrac{1}{9x-6}\right) + 17} + 13 =\frac{9x-6}{10\left(1\right) + 17(9x-6)} + ...

1

After examining all formulas, I got the following diagram: With the following formulas: $\color{purple}{y =\sqrt{x-1}}$ $\color{red}{y = \sqrt{x-1}+\sqrt[3]{2+x}}$ $\color{blue}{y = \sqrt{x-1}+\sqrt{2+x}}$ $\color{green}{y = \sqrt[3]{x-1}+\sqrt{2+x}}$ $\color{orange}{y = \sqrt[3]{x-1}+\sqrt[3]{2+x}}$ It shows that all formulas ...

1

First, find $f^{-1}(x)$, $g^{-1}(x)$. $f^{-1}(x):\quad$ Express $x$ as a function of $y$: $y = 23x+ 27\iff \dfrac{y-27}{23} = x$. Switch $x$ and $y$: $$y = \frac{x-27}{23} \implies f^{-1}(x) = \frac{x-27}{23}$$ Similarly found, $g^{-1}(x) = \dfrac{x+d}{12}$. Now compose $g^{-1}(f^{-1}(x))$ and $f^{-1}(g^{-1}(x))$ and set the two compositions equal to one ...

2

$$f(x)=\left\{\begin{array}{cl} x&\text{ if }x\leq 0\\ 0&\text{ if }x\geq 0 \end{array}\right.$$ $$g(x)=x^2$$ $$f\circ g(x)=0$$ $$g\circ f(x)=\left\{\begin{array}{cl} x^2&\text{ if }x\leq 0\\ 0&\text{ if }x\geq 0 \end{array}\right.$$

2

$f$ and $g$ don't need be constant. For example, consider $f= \begin{cases} 0 \quad &\text{ if } x \geq 0 \\ x &\text{ if } x \leq 0 \end{cases}$ and $g = \begin{cases} 0 \quad &\text{ if } x \leq 0 \\ x &\text{ if } x \geq 0 \end{cases}$

1

Note that if, $F:\mathbb{R}^2\mapsto\mathbb{R}$ and $G:\mathbb{R}\mapsto (0,1)$, with $F$, $G$ bijections then $G \circ F :\mathbb{R}^2\mapsto (0,1)$ is a bijection. http://mathoverflow.net/questions/126069/bijection-from-mathbbr-to-mathbbr2

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