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0

This should do the trick function mse Kc = 0.865; n = 2.08; dc = 3*1e-6; d = linspace(0,10,100); % divide the interval (0,10) into 100 steps Td = 1-exp(-Kc*(d/dc).^n); plot(d,Td); end Note, that the plot looks quite boring. Which is due to the fact, that $d/dc$ is very large, and then $\exp(-Kc(d/dc)^n)$ is almost zero, if $d>0$.


0

A fixed point of $f$ is a root of a particular defining function $g(x)=f(x)-x$ along with it inverse function.


2

A fixed point of $f$ is a root of $g(x)=f(x)-x$ and vice versa.


1

Look at the set $(a,b)-A$. WLOG we can assume the enumeration $\{q_n\}$ of $A$ is increasing, so that $q_n<q_{n+1}$ for all $n$. Then, $(a,b)-A = (a,q_1)\sqcup (q_1,q_2)\sqcup\ldots$. Now, the function given is constant on each of the intervals $(q_n,q_{n+1})$, so is continuous on each interval individually, and jumps by $\frac{1}{2^{n+1}}$ at $x_0 = ...


0

if $\pi_1:(s,t) \mapsto s$ and $\pi_2:(s,t) \mapsto t$ are the two co-ordinate projection maps from your $\mathbb{R^2}$-frame, then the statement is saying that: $$ y = \pi_1 \circ G(x) \\ z = \pi_2 \circ G(x) \\ $$ (emended, thanks to an error spotted by Christoph)


2

When $G$ is map from $\mathbb R$ to $\mathbb R^2$, we write $G\colon \mathbb R\to\mathbb R^2$. This means that $G$ takes any real number $x\in\mathbb R$ and depending on $x$ returns an element of $\mathbb R^2$, i.e. a pair of real numbers $(y,z)$. For example \begin{align*} G\colon \mathbb R &\longrightarrow \mathbb R^2 \\ x &\longmapsto (2x, x+1) ...


0

Yes, parametrics can be an example of this. Such as $r(t) = (t, 2t)$. So $r:\mathbb{R} \to \mathbb{R}^2$ since $(t, 2t) \in \mathbb{R}^2, \forall t \in \mathbb{R}$


0

This is due to some bad software and/or asking software to do something it wasn't meant to do. Suppose $y=\gcd(x/y,xy)$. Then $y$ must divide $x$, else $x/y$ is not an integer, and the gcd would not be defined. Say $x=yk$ for some integer $k$. Then $x/y$ = $k$, and $\gcd(x/y,xy)=\gcd(k,y^2 k) = k$. So $k=y$, i.e., $x=y^2$. In fact, $y=\gcd(x/y,xy)$ ...


0

Here's one that has $a_n$ strictily increasing to $\infty$ with $a_{2n}-a_n \to 0:$ Set $a_1 = 0, a_n =\sum_{k=2}^n \frac{1}{k\ln k}, n>1.$


2

Let $a_{2^n} = 1$ and $a_k = 0$ otherwise (i.e. when $k \neq 2^n$ for all $n \in \mathbb{N}$). The condition is fulfilled but no limit exists, as the sequence oscillates between $0$ and $1$.


0

By the generalized binomial theorem,(a_n + 1)^(1/3) - (a _n)^(1/3), = (a_n)^(1/3) + (1/3)(a-n)^(-2/3) +....smaller terms - (a_n)^(1/3),so lim n approaches infinity (a_n) + 1)^(1/3) - (a_n)^(1/3) = lim n approaches infinity (1/3)(a_n)^(-2/3) = 0 for a_n) approaching infinity. Edwin Gray


1

Well, I guess you can graph it. You might need to think carefully about the endpoints, though - or at any point where we just touch an integer value.


0

For $x>0$ we have $$\tag1 \sqrt[3]{x+1}-\sqrt[3]x=\frac{1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x^2+x}+\sqrt[3]{x^2}}<\frac1{3\sqrt[3]{x^2}}$$ Given $\epsilon>0$, let $M:=\sqrt{(3\epsilon)^3}$. Then $x>M$ implies $\sqrt[3]{x+1}-\sqrt[3]x<\epsilon$ according to $(1)$. As $a_n\to\infty$, there exists $N$ such that $n>N$ implies $a_n>M$. We conclude ...


1

We have $$x-y = \dfrac{x^3-y^3}{x^2+xy+y^2}$$ This gives us $$\sqrt[3]{a_n+1}-\sqrt[3]{a_n} = \dfrac{a_n+1-a_n}{(a_n+1)^2+(a_n+1)a_n + a_n^2} = \dfrac1{(a_n+1)^{2/3}+(a_n+1)^{1/3}a_n^{1/3} + a_n^{2/3}}$$ Can you finish it from here? Since $a_n \to \infty$, given any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n>N$, we have ...


2

Think of a particle moving along the curve. At any point where $x'(t) = 0$ and $y'(t) = 0$ the velocity of the particle is zero. After coming to a stop it can begin moving again in an entirely different direction. Thus even if $x'$ and $y'$ are continuous, the direction of the tangent vector can change abruptly at any time $t$ that both $x'$ and $y'$ both ...


1

It is not hard to evaluate this manually: \begin{align} f(425268)&=18.6\dots\\ f^{(2)}(425268)&=f(18.6\dots)=4.22\dots\\ f^{(3)}(425268)&=f(4.22\dots)=2.07\dots\\ f^{(4)}(425268)&=f(2.07\dots)=1.06\dots\\ f^{(5)}(425268)&=f(1.06\dots)=0.07\dots\\ f^{(6)}(425268)&=f(0.07\dots)=0\end{align} Thus $N(425268)=6$. In general, we have ...


0

$$\frac{\partial y_t}{\partial t} = \frac{\frac{\partial x_t}{\partial t}(1+x_t) - x_t\frac{\partial x_t}{\partial t}}{(1+x_t)^2} = \frac{\frac{\partial x_t}{\partial t}}{(1+x_t)^2}.$$


0

By definition, the generating function is $$H(z)=\sum_{n=0}^\infty h_nz^n=\sum_{n=0}^\infty (-2)^n n^2 z^n=\sum_{n=0}^\infty n^2(-2z)^n. $$ From $$\frac1{1+2z} = \sum_{n=0}^\infty (-2z)^n,$$ $$\frac{\mathsf d}{\mathsf dz}\left[\frac1{1+2z}\right] = \frac{-2}{(1+2z)^2} = \sum_{n=0}^\infty -2(n+1)(-2z)^n,$$ and $$\frac{\mathsf d^2}{\mathsf ...


1

We have $T(1)=0, T(x)=1, T(x^2)=2x+x\times2+x^2\times2=2x^2+4x$ So the matrix $A_T$ of T is $ A_T= \left[ \begin{array}{} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 4 & 2 \\ \end{array} \right] $


1

By contradiction: Suppose $\alpha(a)=2$. Then $$ 2=\frac{2x+1}{x+2}\Longleftrightarrow 2x+4=2x+1\Longleftrightarrow 4=1, $$ which is clearly absurd. Hence, $\forall a\in A, \alpha(a)\neq 2$. $\blacksquare$


1

As Ilham said, I think the formatting is a bit off. I believe you meant to say that $\alpha(x) = \frac{2x + 1}{x+2} $. If so assume $\alpha(a) = 2$ is true, then: $$ \frac{2a + 1}{a+2} = 2 $$ Which implies $ 2a+1 = 2a+ 4$ which is not true $\forall a \in \mathbb{R}$


1

Let $T_a$ denote translation by $a$, so $(T_af)(x)=f(x+a)$. Consider the complexification $E\otimes\mathbb C$. Since $T_aT_b=T_{a+b}=T_bT_a$, we find simultaneous eigenspaces of all $T_a$. Let $W$ be such a simultaneous eigenspace and let $\lambda(a)$ be the eigenvalue of $T_a|_W$. Then $$\tag1\lambda(a+b)=\lambda(a)\lambda(b)$$ and $\lambda(0)=1$. We verify ...


0

The notation is indicating that we are thinking of the expression $\phi_t(x)$ as a function of $t$ (with $x$ fixed; or rather that there is a family of functions of $t$ parameterized by $x$). The notation is read "$t$ maps to $\phi_t(x)$." One may also see something such as "Define $g_x:\mathbb{R}\to\mathbb{R}^n$ by $t\mapsto \phi_t(x)$ to be the solution ...


1

Use the pigeonhole principle, to see that you can't have the $|B|$ pigeonholes of $B$ having only one "pigeon" of $A$ in them without filling them all up since $|B| = |A|$. Thus injectivity implies surjectivity. The other direction is a dual statement. Now let for each $b \in B$, let $g(b)$ be the number of distinct elements of $A$ mapped to $b$ by $f$. ...


2

Hint: $$Y^{-n}\cdot X < Z \implies Y^{-n} <\frac{Z}{X}$$ What happens if you take the natural logarithm of both sides? Further, recall that $$\ln(A^B) = B\cdot\ln(A)$$


1

Let the function $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=-3x+4$. We say that a function $f$ is injective if $\forall x_{1},x_2 $ $f(x_{1})=f(x_2)$ implies $x_1 = x_2$. Hence, after some algebra we can see that $-3x_{1}+4= -3x_2+4$ implies $x_1=x_2$. So our function is injective. Now, we say a function is surjective if for all $y\in ...


2

You allready received nice answers focused on injectivity and surjectivity. Here a slightly different route. Can you find a function $g:\mathbb R\rightarrow \mathbb R$ such that $f(g(x))=x$ for each $x\in\mathbb R$? If you have found such $g$ then check whether it is also true that $g(f(x))=x$ for each $x\in\mathbb R$. If so then you are ready because you ...


0

One-to-one: Suppose there are two values $x_1,x_2$ such that $f(x_1)=f(x_2)$. Show that $x_1=x_2$ necessarily. Onto: Take $y=-3x+4$ and try to obtain $x$ in terms of $y$.


1

Surjective: For any $y\in \Bbb R$, there exists $x=\frac{4-y}{3}$ such that $f(x)=y$. Injective: For any $a\not=b$, does it $-3a+4=-3b+4$ hold?


1

Consider the function $f_t:[0,1]\to\Bbb R$, with $t>1$. $$f_t(x)=\frac1{t(1-x)+t-1}\\ \begin{align} \implies F_t &= \int_0^1f_t=\frac 1t\log\left(\frac{2t-1}{t-1}\right)\\ \implies G_t &= \max\{f_t(x)(1-x)\mid x\in[0,1]\}=\frac 1{2t-1}\\ \end{align}$$ As $t$ approaches $1$ from the right, $f_t(x)$ approaches $\frac1{1-x}$. The maximum area of a ...


0

Think of it like this, $2xy=2x\:\text{x}\:y$ and $-y= -1\:\text{x}\:y$, so if we multiply the term $2x-1$ by $y$, we get $2xy-y$. This works because $2xy$ and $y$ are both divisible by $y$. As others have said, the technical term is the distributivity of multiplication over addition, but I think you may be looking for the term 'Factorisation'. It works ...


1

The distributive property (of multiplication with respect to addition). To go from left to right you'd call it "factoring". To go from right to left it's called the distributive property.


1

Using Polar coordinates. Let $x = r\cos(\theta)\ and \ y=r\sin(\theta)$ hence the function becomes $\rightarrow$ $\frac{\ln( 1 +r^3\cos^3(\theta)sin^3(\theta))}{r}$ . As $r^3\cos^3(\theta)\sin^3(\theta)$ $\rightarrow$ $0$ as $r$ $\rightarrow$ $0$ hence $\ln(1 + r^3\cos^3(\theta)\sin^3(\theta))$ $\sim$ $r^3\cos^3(\theta)\sin^3(\theta)$ and so the ...


1

A smooth function on a $\textit{closed and bounded}$ interval is bounded. In fact the condition can be relaxed to continuous functions by the extreme value theorem.


1

Since $f$ is $1-1,\ f(1)$ can be choosen in $6$ ways, $f(2)$ can be choosen in $5$ ways, $f(3)$ can be choosen in $4$ ways and, $f(4)$ can be choosen in $3$ ways. Hence the number of $1-1$ functions from $A \rightarrow B$ is $6 \times 5 \times 4 \times 3 \ (= 360).$ {Thanks to multiplicative rule in combinatorics} As @Paul said, It is easy find the number ...


3

$ax+\frac bx$ is minimized when $x=\sqrt{b/a}$, and equals $2\sqrt{ ab}$ at that point.


0

Since it looks like you are going to measure some physical quantity, it seems fair to assume that $x=t$ is a local maximum or minimum for $f$. This would not be the case with a function like $$f(x)=x^3{\sin{\frac{1}{x}}}$$ that is even, differentiable but doesn't exhibit a maximum or a minimum at $x=0$. If you have a rough idea of where the simmetry axis ...


2

Consider $$ g(A, s) = \int_{-A}^A [h(s-x) - h(x)] dx $$ where $h = f + \epsilon$ is your observed function. Assuming for the moment that that $\epsilon = 0$, $A \mapsto g(A, s)$ is a function that's identically zero when $s = t$, and unlikely to be zero elsewhere. As you add epsilon back in, the function $ A \mapsto g(A, s)$, at $s = t$, looks more ...


5

You should just walk through without thinking too much if you have a problem: If $x < 0$, $f(x) = -x > 0$. This means we need to account for the cases $0 \le -x \le 1$ and $-x > 1$, the former giving $f(f(x)) = -x$ for $-1 \le x < 0$ and the latter $2-(-x) = 2+x$ for $x < -1$. This already gives us the components $$f(f(x)) = \begin{cases} 2+x ...


10

Not all smooth functions are bounded. For example, $f(x)=e^x$ is as smooth as they come, but is not bounded. Even if you are looking at functions on a bounded interval, $\frac 1x$ is smooth, but unbounded on $(0,1)$. You can produce certain restrictions for which $f$ will be bounded, however. For example, any continuous function on a compact set (for ...


1

From the question, you only need to evaluate the pointwise limit of $e^{-nx}$ as $n\to \infty$. Then if $x=0$, $e^{-n \cdot 0}=1, f(0)=1$. If $x>0$, $f(x)=\lim_{y\to -\infty}e^{-y}=0$. And if $x<0$, $f(x)=\lim_{y\to \infty}e^{y}=\infty$.


0

A periodic function $a(x)$ satisfies $a(x)=a(x+k)$ for some $k$ and all $x$. $f(x)$ clearly satisfies this in this instance ($k=2$) and is therefore periodic. Therefore, if there exist $m,d$ such that $h(x+m)=h(x)+d$ for all $x$, then $f\circ h$ will be periodic (with period LCM($d,k$), as a matter of fact). $h(x)$ satisfies this ($m=3$) and therefore $f ...


0

If $T$ is a period of $f$, then $T$ is a period of $g\circ f$. Since $f(x+T)=f(x) \forall x\in\mathbb{R}$, then $g\circ f(x+T)=g(f(x+T))=g(f(x))=g\circ f(x)$. In your case, set $T=2$.


1

Yes. Since $f$ is Lipschitz, it's uniformly continuous on $A$ hence it can be extended to $\bar{A}$ such that $f(x)=\lim_{x_n\to x}f(x_n)$, with $(x_n)\in A$. Then definition doesn't depend on choice of $x_n$ based on the uniform continuity. Then $x_n,y_n\in A, x_n\to x,y_n\to y$ and suppose $|f(x_n)-f(y_n)|\le C|x_n-y_n|$. Let $n\to \infty$, we see ...


2

It depends a little bit on the specifics of the vector. One place such expressions come up is in statistical mechanics, where what you have written is the Gibbs distribution on $n$ states with energy $-v_i$ at a cold temperature proportional to $\frac{1}{\lambda}$. In this physical situation, often one state has significantly less energy than the others. ...


0

Hint. Apply the MVT to $f(t)=\sqrt t$ on the interval $[x,y]$.


0

For first one, let $f(x)=ax+b$ be the mapping that maps $-3$ to $41$ and $7$ to $100$. Then $-3a+b=41$ and $7a+b=100$ Solve the 2 and we get $a=5.9, b=58.7$ and $f(x)$ is clearly $1-1$. For the second, $f(x)=8-\dfrac{1}{x+3}$ will map $-\infty$ to $8$ and $-3$ to $\infty$ and clearly $1-1$ on $(-\infty,-3)$.


0

If $ s(n)=s(n−2)+s(n−3) $, then $ s(n-1)=s(n−3)+s(n−4) $ so $ s(n-3)=s(n−1)-s(n−4) $ so that $s(n) = s(n-2)+s(n-1)-s(n-4) =s(n-1)+s(n-2)-s(n-4) $.


0

The condition $\int_{-\infty}^\infty f(x)\>dx=0$ is, apart from convergence issues, a very weak condition on $f$. Fix a function $u_0$ with $u_0(x)\geq0$ for all $x\in{\mathbb R}$ and $\int_{-\infty}^\infty u_0(x)\>dx=1$, e.g., $u_0:=$ the standard normal distribution. Let $g:\>{\mathbb R}\to{\mathbb R}$ be any function with $\int_{-\infty}^\infty ...



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