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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 $d>0$.

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

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

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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$(y,z). For example \begin{align*} G\colon \mathbb R &\longrightarrow \mathbb R^2 \\ x &\longmapsto (2x, x+1) ... 0 Yes, parametrics can be an example of this. Such asr(t) = (t, 2t)$. 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} ThusN(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 $$Which implies 2a+1 = 2a+ 4 which is not true \forall a \in \mathbb{R} 1 Hint: \log(y) \geq 0 \iff y \geq 1. 1 Let T_a denote translation by a, so (T_af)(x)=f(x+a). Consider the complexification E\otimes\mathbb C. Since T_aT_b=T_{a+b}=T_bT_a, we find simultaneous eigenspaces of all T_a. Let W be such a simultaneous eigenspace and let \lambda(a) be the eigenvalue of T_a|_W. Then$$\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 ...

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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 $T=2$.

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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 ...

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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. ...

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

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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 $(-\infty,-3)$.

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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 $s(n) = s(n-2)+s(n-1)-s(n-4) =s(n-1)+s(n-2)-s(n-4)$.

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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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