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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 ...

5

Note that every nonnegative number $x$ is between the squares of two consecutive nonnegative integers. Say $n^2 \leq x < (n+1)^2$. Then $n \leq \sqrt{x} < n + 1$, so $\lfloor \sqrt{x} \rfloor = n$ in this situation. So you just have to make sure the left hand side is also $n$. Since $n^2$ is an integer, one has $n^2 \leq \lfloor x \rfloor \leq x$, so ...

3

1) Yes, your instincts are right. For the squeeze theorem, you need to find an upper bound and a lower bound for the function $\frac{3-\sin (e^{x})}{\sqrt{x^2+2}}$ so that both of these bounds converge to the same limit. Since $\sin (e^{x}) \geq -1$ for every $x$, one upper bound is $\frac{4}{\sqrt{x^2+2}}$. Now, does this upper bound converge to something, ...

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.

2

\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 - ... 2 Hint: for any positive integer n, \sqrt{n+1}-\sqrt n=\frac 1{\sqrt{n+1}+\sqrt n}\lt 1, hence\sqrt{x}\leqslant \sqrt{[x]+1}\lt 1+\sqrt{[x]},$$hence [\sqrt x]\leqslant \sqrt{[x]}. Since [\sqrt x] is an integer, we actually have$$[\sqrt x]\leqslant [\sqrt{[x]}].$$The opposite inequality is a consequence of increasingness of t\mapsto \sqrt t ... 2$$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}}$$EDIT: which equals  \displaystyle\frac{1}{q+1} as pointed out by Dilip Sarwate in the comments 2 Well since a,b\in\mathbb{R}^+ (positive reals), you know that a= \exp(x) and b=\exp(y) for some x,y\in\mathbb{R}. Then use the fact that \log is the inverse function to \exp in conjunction with properties of exponentials. To see the second one, note that n\log(a) = \underbrace{\log(a)+\cdots+\log(a)}_{\text{n times}}. I think the last one ... 2 The only continuous field isomorphisms \mathbb C\to\mathbb C are the identity and complex conjugation. Using the Axiom of Choice one can prove that additionally there exists infinitely many "wild" everywhere-discontinuous automorphisms of \mathbb C. Each of them acts either like the identity or like conjugation on all points with rational coordinates, ... 1 The 1 comes from the closed-loop transfer function. See these WikiBook Routh-Hurwitz-Criterion notes for details. From G(s) = \dfrac{K}{s(s+1)(s+5)}, we form:$$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 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 ... 1 Injectivity If you know that \exp' = \exp, use the fact that \exp'(x) = \exp > 0 \;\; \forall \, x. Alternatively, if you know that \exp (a + b) = \exp a \exp b, then assume \exp x = \exp y. Conclude \exp(x - y) = 1. Assume WLOG x \ge y, and then use the series expansion of \exp(x - y) to show x - y = 0, hence x = y. Surjectivity ... 1 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 ... 1 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 ... 1 Correct answer with simple proof: \frac{1}{2}. Notice that the entire situation is symmetric. Everything happens with the same probability to white as it does to red. Thus P(Red)=P(white)=\frac{1}{2} What's wrong with your answer: "We prove that if all that jugs have same amount of red balls and the same amount of white balls (each one), the probability ... 1 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 ... 1 I assume that you mean that x,y are two (fixed) elements of [0,1) and that M is the collection of all sets that either contain both x and y or contain neither x nor y. We will show that M is a \sigma-algebra. It is immediate that \Omega \in M as \Omega contains both x and y. Furthermore, if A \in M then A^c \in M since if A ... 1 Try adding the equations 2^x-2^y=1 and 2^x+2^y=\frac 53 To go further per your comment: This gives 2\cdot 2^x=\frac 83 Subtracting the equations gives 2\cdot 2^y=\frac 23 Now recall that \frac {2^a}{2^b}=2^{a-b} and divide in an appropriate way to make further progress. 1 Well, now it's really simple because you have \begin{cases} 2^x - 2^y = 1 \\ 2^x + 2^y = \frac53 \\ \end{cases} so$$ 2^x - 2^y + 2^x - 2^y = 2\cdot 2^x = 1+\frac53 = \frac83 \Rightarrow 2^x = \frac43 \Rightarrow x = log_2{\frac43}$$Finally, since 2^x = \frac43,$$ 2^y = \frac53 - \frac43 = \frac13 \Rightarrow y = log_2{\frac13}$$The solution is ... 1 It's Quite simple : Use the Unitary Method 300g = 4.29 1g = (4.29/300) {dividing both sides by 300} Now you can easily calculate the cost for 100g or 1Kg i.e (A and B) For C, do the other way round: 4.29 = 300g 1 = (300/4.29)g {dividing both sides by 4.29} Now for 20 how much can u buy? 1 Yes, your instinct was right to include the first part of the problem! The left-hand side of the equality becomes \zeta(s) when you set f(k)=k^{-s} and let n go to infinity; so let's do the same to the right-hand side. But first: let B_2(y) = \int_1^y (x-\lfloor x\rfloor-\frac12)\,dx. Note that B_2 is a continuous periodic function with period 1 ... 1 The intuitive answer is that you take a solution that goes around a times with 0 \lt a \lt 1. Now run the clock backwards. Unless the solution blows up to \infty, you can just continue backwards in time. Each time you pass the initial angle, read off the angle and velocity and you have an initial condition that will circle a given number of times. ... 1 Edit: The answer below is the answer to the original question. The question has been modified since then, and the calculation has been corrected. We keep the unmodified answer since it makes some methodological suggestions. You can check by differentiating whether you have done an integration correctly. In this case, differentiation will show that there ... 1 The matrix$$A=\pmatrix{0&1&0&0\cr1&0&1&0\cr0&1&0&1\cr0&0&1&0\cr}represents g, in the sense that if x is the column vector (x_1,x_2,x_3,x_4), then g(x)=x^tAx. The maximum value of g(x) on x of length 1 is the maximum eigenvalue of A. This is explained at length at ... 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 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 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}}. $$1 By the definition of the derivative, f is differentiable at a iff there is a linear transformation B such that the following limit is zero:$$\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) = ...

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