# Tag Info

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Yes, this is correct, this is indeed what is meant by memorylessness. In general, for all $s\geq 0$, $t \geq 0$, we have $$P(T>s+t\mid T>s) = P(T>t) = e^{-\lambda t}.$$ We can prove the memorylessness by transforming the expression on the LHS using the definition of conditional probability, as follows \begin{align} P(X>s+t\mid X>s) ... 0 If x is even and y is odd, then x=2k for some integer k and y=2l+1 for some integer l, so x+3y=2k+3(2l+1)=2(k+3l+1)+1 is odd. If x is odd and y is even, then x=2m+1 for some integer m and y=2n for some integer n, so x+3y=2m+1+3(2n)=2(m+3n)+1 is odd. 1 I think you need to use \epsilon-N definition. suppose given a \epsilon, then pick N=\lceil\log_r{\epsilon}\rceil, as we can see for every n>N, |r^n-0|<\epsilon, so \lim_{n\to \infty}r^n=0 1 If you suppose that (r^n)_{n\in\Bbb N} has un upper bound, then exists L\in\Bbb R such that r^n\rightarrow L\Rightarrow L>0 and L*r=L (taking limit to r^{n+1}=r*r^n) \Rightarrow r=1. 0 So you are given y=r^n with r>1 You could take the natural log on both sides: lny=lnr^n=nlnr Now if r>1 then ln1>0 and if n goes to infinity, multiplied by a quantity greater than zero, then lny is infinite as well, so y is infinite. 0 Verify the definition of limit: Fixed \varepsilon >0, you need to find an N such that |r^n| = |r|^n< \varepsilon for any n \geq N. If you solve the inequality in n, you get n > \log_{|r|}(\varepsilon), thus it is sufficient to take N = \left\lceil\log_{|r|}(\varepsilon)\right\rceil. (The direction of the inequality has been ... 0 I would take another approach. Hope it helps. TakeG(x,y,z)=4x^2+y^2-z^2-4$$The gradient of that function at the point is normal to the level curves (in this case we need G(x,y,z)=0)$$\nabla G(x,y,z)=(8x,2y,-2z)$$Evaluated at that point:$$\nabla G(1,-2,2)=(8,-4,-4)$$Then with the normal and a point we have the equation of the plane: ... 2 You seem to have determined (and be able to prove, if asked!) that the sequence r^n is descending. It is also clearly bounded from below (by zero), so by the theorem on monotonic sequences it converges to some limit A. But then the subsequence r^{n+1} alsot tends to A as n\to\infty. Therefore$$ ...

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Hint: Let $r=\frac1m , m>1 \to$ you need to show $$lim_{n\to +\infty} r^n = 0$$ Since $0<r<1$ $$lim_{n\to +\infty} r^n = lim_{n\to +\infty} \left(\frac1m\right)^n=lim_{n\to +\infty}\frac{1}{m^n} = \frac{1}{+\infty} = 0$$

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Write $r=\frac{1}{m}$ with $m>1$ Then $r^n=\frac{1}{m^n}$ But $m^n$ goes to infinity as $n$ goes to $\infty$ and so $$r^n\rightarrow\frac{1}{\infty}=0$$

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Hint: Show $p(x) = 8x^3 - 7x^2 + 1 > 0$ for all $x ∈ (0..1)$ by showing that $p$ has a positive minimum on $[0..1]$. Look at $p(0)$, at $p(1)$ and at $p'$ on $(0..1)$. I have carried this out below. First observe that the only zero of $p'$ in $(0..1)$ is at $x = 7/12$, since $p'(x) = x(24x-14)$ for $x ∈ (0..1)$. Now because $p(7/12) > 0$, the only ...

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We have, $\displaystyle \frac{x^2+1}{x^2(1-x)} = \displaystyle \frac{x+\frac{1}{x}}{x(1-x)}= \displaystyle \frac{2 + (\sqrt x - \frac{1}{\sqrt x})^2}{\frac{1}{4} - (x-\frac{1}{2})^2} > 8$. (inequality is strict since $x\in(0,1)$).

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If $\mu$ is an eigenvalue of $A$ and $x$ is an eigenvector associated to $\mu$ then $$(A+\lambda I)x=(\mu+\lambda)x$$ hence we see that $$\operatorname{sp}(A+\lambda I)=\operatorname{sp}(A)+\lambda=\{\mu+\lambda,\quad\mu\in\operatorname{sp}(A)\}$$

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Note: There is a difference between $f(x,y)$ and $f(x+y)$. The first is a function of two variables and the second is a function of one variable $z=x+y$. For example $$f(x+0)=f(x)=x^2\implies f(x+y)=(x+y)^2\\f(x,0)=x^2\implies \, \text{nothing}$$Also there is a meaning of $\frac{\partial f(x,y)}{\partial x}$ but it doesn't mean any thing for $f(x+y)$ ...

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Take an eigenvalue $\mu$ of the matrix $A$. Therefore, you have a vector $v$ such that $Av = \mu v$. Take a look at what multiplying $A+\lambda I$ with $v$ does.

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Hint: We can define the eigenvalues of a matrix $M$ to be the values $\mu$ for which $M - \mu I$ is singular (i.e. has a determinant of zero).

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Hint: To see what's happening, calculate explicitly let $f(x)=x^2$. Start with say $x_1=1$. You will find that the convergence is somewhat slow, or at least a lot slower than Newton "usually" is. Then generalize.

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I will summarize the proof from the book "Theory of Groups of Finite Order" by W. Burnside. It would not be very easy to think of this if you had not seen it before! With your notation, conjugation by $a$ is inducing a fixed-point-free automorphism of order $3$ of $G$. You know that $x$ commutes with $a^{-1}xa$ and similarly with $axa^{-1}$. If you ...

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$$\left|\frac{xy(x+iy)}{x^2+y^2}\right| = |z| \frac{|xy|}{x^2+y^2} \leq \frac{1}{2}|z|$$ because $|xy| \leq \frac{1}{2}(x^2+y^2)$.

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substitute $x=r\cos \theta$ and $y=r\sin \theta$, the expression $f(z)$ becomes $r\cos \theta \sin \theta e^{i\theta}$, continuity follows from boundedness of $\cos \theta \sin \theta e^{i\theta}$ as $r \to 0$

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The approach is this: 1) Calculate the expected utility $EU$ from not installing the new drill. (Compute W if catastrophy hits, compute utility for this case, and weight with probability of catastrophy; compute W if catastrophy does not hit, etc., and add the two.) 2) Now calculate the utility of replacing the drill, leaving the days it takes to install it ...

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Assume that $J \colon H \to \mathbb{R}$ is G-differentiable. By definition, given $u \in H$, there is a linear functional $J'(u) \in H^*$, the G-derivative at $u$. By the Riesz representation theorem, this element is isometrically identified to a vector $\nabla J(u) \in H$, called the gradient of $J$ at $u$. Now, consider thae map $$\phi \colon \tau ... 0 Let's be careful: The second condition leads to:$$\left.\frac{\partial u}{\partial t}\right|_{t=0}=-cf'(x)+cg'(x)=-2cf'(x)=\frac{x}{(1+x^2)^2}$$Thus$$f(x) = \frac{1}{4c} \cdot\frac{1}{1+x^2}+c_1,$$where c_1 is a constant, as$$ f'(x)=-\frac{1}{2c}\cdot\frac{x}{(1+x^2)^2}. $$9 Notice that the rows are perpendicular to each other and the length of the rows are all the same.If you call the matrix M then \det(M^tM)=\det(M^t)\det(M)=\det(M)^2 But$$ M^tM= (a^2+b^2+c^2+d^2)\left( \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array} \right)$$Therefor ... 0 Imagine that if X is odd then Alphonse wins the game, while if X is even then Beti wins the game. It is clear that with probability 1 one of Alphonse or Beti wins. Let p be the probability A wins. If A fails to win on the first trial, then the roles of A and B are reversed, and B has probability p of winning. It follows that$$p+\frac{9}{10}p=1.$$... 3 Our complementary solution has an e^{−t}. We would have normally chosen y_p=(a+bt)e^{-t}, but since we already have e^{-t} in the complementary, we need to multiply by a factor of t, else we would just get another complementary solution. Hence, y_p=t(a+bt)e^{-t}. I wrote my y_p slightly different than you, but they are actually the same. 1 It's Quite simple : Use the Unitary Method 300g = 4.29 1g = (4.29/300) {dividing both sides by 300} Now you can easily calculate the cost for 100g or 1Kg i.e (A and B) For C, do the other way round: 4.29 = 300g 1 = (300/4.29)g {dividing both sides by 4.29} Now for 20 how much can u buy? 1 The problem is with E \left[ \left(\sum X_i \right)^2 \right] Let Y=\sum X_i. We know E[Y]= n \mu. We also know \mathrm{Var} (Y)= n \sigma^2. That means E[Y^2]=n \sigma^2 + n^2 \mu^2. Then plugging into your equation we find \mathrm{Var} (\bar X ) = {1 \over n^2} \left[ n \sigma^2 + n^2 \mu^2 \right] - \mu^2 = {\sigma^2 \over n} 1 The intuitive answer is that you take a solution that goes around a times with 0 \lt a \lt 1. Now run the clock backwards. Unless the solution blows up to \infty, you can just continue backwards in time. Each time you pass the initial angle, read off the angle and velocity and you have an initial condition that will circle a given number of times. ... 2 Well since a,b\in\mathbb{R}^+ (positive reals), you know that a= \exp(x) and b=\exp(y) for some x,y\in\mathbb{R}. Then use the fact that \log is the inverse function to \exp in conjunction with properties of exponentials. To see the second one, note that n\log(a) = \underbrace{\log(a)+\cdots+\log(a)}_{\text{n times}}. I think the last one ... 1 The 1 comes from the closed-loop transfer function. See these WikiBook Routh-Hurwitz-Criterion notes for details. From G(s) = \dfrac{K}{s(s+1)(s+5)}, we form:$$1 + \dfrac{K}{s(s+1)(s+5)} = 0 \implies 1 + \dfrac{K}{s^3+6 s^2+5 s} = 0$$If we find a common denominator and multiply through, we arrive at:$$\tag 1 s^3+6 s^2+5 s + K = 0$$The Routh ... 1 By the definition of the derivative, f is differentiable at a iff there is a linear transformation B such that the following limit is zero:$$\lim_{h\to\vec0} \frac{f(a+h)-f(a)-B(h)}{\|h\|} = 0$$Since f is linear, you can rewrite the numerator as f(h)-B(h), which is clearly 0 if we let B just be the linear transformation f. Therefore, Df(a) = ... 0 Hint: Let n=2 How many majority elements can there be? Then try n=3 and n=4 0 Recall that if you take the nth root you should be looking for n solutions.$$\left(re^{ix}\right)^{\frac{1}{2}} = r^{1/2}e^{ix/2}$$This gives us one equation. Additionally, recall that e^{ix} = e^{ix+2i\pi n}, n \in \mathbb{Z}. We can find another solution by setting n = 1. (Why can't we find a third by setting n = 2?) ... 2$$P(\text{X is odd}) = P(X=1) + P(X=3) + P(X=5) + \ldots$$The pmf of a geometric distribution is q^{k-1}p. So$$P(\text{X is odd}) = q^{0} p + q^{2}p + q^{4} p + \ldots = p \sum_{k=0}^{\infty} q^{2k} = \frac{p}{1-q^{2}}$$EDIT: which equals  \displaystyle\frac{1}{q+1} as pointed out by Dilip Sarwate in the comments 0 On the first pick, the probability of it NOT being a senior is \frac{6}{10}. On the second pick (assuming we picked a junior on the first) the probability of it NOT being a senior is \frac{5}{9}. For the third pick the probability is similarly \frac{4}{8}. The probability of having chosen a committee with at least one senior is then$$ 1 - ...

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Two sets have the same cardinality if there exists a bijection between them; that is, a function that is a one-to-one correspondence between the two sets. All you have to do is prove that (1)$f(a)=f(b) \implies a=b \forall a \in S$ and (2) $\forall u \in U, \exists s \in S$ such that $f(s)=u$. This should be fairly straightforward, and for (2), I suggest ...

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Hint: The function is : f(x) = 2x

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Recall that two sets $X$ and $Y$ are of the same cardinality if there exists a bijection between them. That is, we may construct a 1-1 and onto function from $X$ to $Y$. There is certainly such a map between your two sets (in fact there a uncountably many) but one obvious one. I'll give a hint: its a very simple map and it takes the ends of the interval in ...

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X = event that committee has no senior, and Y = event that committee has at least 1 senior. P(X) = C(6,3)/C(10,3) = 20/120 = 1/6. So P(Y) = 1 - P(X) = 1 - 1/6 = 5/6

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\begin{align*} \frac{1}{h} & \left[ \langle \gamma(t+h),\, \eta(t+h)\rangle - \langle \gamma(t),\, \eta(t) \rangle \right] \\ & = \frac{1}{h} \left[ \langle \gamma(t+h),\, \eta(t+h)\rangle - \langle \gamma(t),\, \eta(t+h)\rangle \right] + \frac{1}{h} \left[ \langle \gamma(t),\, \eta(t+h)\rangle - ... 0 Integration by parts formula:\int u \ dv = uv - \int v \ du$$Our integral that we want to solve:$$\int 2x\cos(2x) \ dx$$Let:$$u=2x, \ du=2 \ dxdv=\cos(2x) \ dxv=\int \cos(2x) \ dx$$Use integration by substitution. Let u=2x, \ du = 2 \ dx$$v=\int \dfrac{1}{2}\cos(u) \ duv=\dfrac{1}{2}\int \cos(u) \ duv=\dfrac{1}{2}\sin(u)$$... 1 Yes, your instinct was right to include the first part of the problem! The left-hand side of the equality becomes \zeta(s) when you set f(k)=k^{-s} and let n go to infinity; so let's do the same to the right-hand side. But first: let B_2(y) = \int_1^y (x-\lfloor x\rfloor-\frac12)\,dx. Note that B_2 is a continuous periodic function with period 1 ... 0 \frac{K}{s(s+1)(s+5)}+1=0 FOIL bottom: \frac{K}{s(s^2+6s+5)}+1=0\rightarrow\frac{K}{s^3+6s^2+5s}+1=0 Move 1 to the right: \frac{K}{s^3+6s^2+5s}=-1 Move denominator to the right: K=-1*(s^3+6s^2+5s)=-(s^3+6s^2+5s) Divide the negative to the K: s^3+6s^2+5s=-K Move the K: s^3+6s^2+5s+K=0 Cheers! -Shahar 1 This limit can be tranformed to a defintion of the derivative of 2\sqrt{x} at x=5.$$ \lim_{n\to +\infty}n\cdot\left(\sqrt5-\sqrt{5-\frac2n}\right)=\lim_{n\to +\infty}2\cdot\frac{\sqrt{5-\frac2n}-\sqrt5}{-\frac{2}{n}}=2\lim_{h\to 0}\frac{\sqrt{5+h}-\sqrt5}{h}=2 (\sqrt{x})'_{x=5}\\=2\cdot\frac{1}{2}\cdot \frac{1}{\sqrt{5}}=\frac{1}{\sqrt{5}}. 2 Yes, it's true. It should be noted that from time to time it is worth to limit the context, to help the students focus on the things which are more important for the course. If I were the one writing these problems, then I agree that I would have given the general theorem. 1 Multiply by the conjugate : \begin{align} n \left( \sqrt{5} - \sqrt{5 - \frac{2}{n}}\right) &= n \left( \sqrt{5} - \sqrt{5 - \frac{2}{n}}\right) \cdot \frac{\left( \sqrt{5} + \sqrt{5 - \frac{2}{n}}\right)}{\left( \sqrt{5} + \sqrt{5 - \frac{2}{n}}\right)}\\ &= \frac{2}{\sqrt{5}+\sqrt{5-\frac{2}{n}}}. \end{align} What happens when n \to \infty ? 1 Try multiplying by the fraction\frac{\sqrt{5} + \sqrt{5-\frac2n}}{\sqrt{5} + \sqrt{5-\frac2n}} Then, after canceling, you should obtain an expression where you can let $n$ go to infinity and see what happens.

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Hint: To find a counterexample, try to construct two sets $A$ and $B$ such that the behavior of algorithms will differ. Then, their answers will be different, hence one will be smaller and so the other one must be wrong. The most useful method for proving greed algorithms is to take any optimal solution and transform it, while not loosing anything, into ...

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What the investment is worth in the end is computed this way: $2000*(1+16/100)*(1+2/100)*(1-5/100) = 2000 * 1.16 * 1.02 * 0.95 = 2248.08$ Thus, there is a gain of 248.08 in the end. To find average rate of return, you would do: $³√(1.16 * 1.02 * 0.95) - 1 = ³√(1.12404) - 1 = 1.0397459931939351874765829661942 - 1 = .0397$ 3.97% is the average rate of ...

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