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I find it useful to plot the function after solving for the constant from the ICs. We have: $$y(x) = -\dfrac{8}{e^{4 x}-2}$$ A plot of $y(x)$ shows: Of course, analytically, we care about where that denominator is $0$, so we have: $$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$ From the plot, you can also see what happens ...

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Take the $\ln$ of the bottom. We get $x\ln x$. Do it again. We get $\ln x+\ln\ln x$. Take the $\ln$ of the top. We get $(\ln x)^{\ln x}\ln\ln x$. Do it again. We get $\ln x\ln\ln x+\ln\ln\ln x$. The ratio of the $\ln\ln$'s is $$\frac{\ln x\ln\ln x+\ln\ln\ln x}{\ln x+\ln\ln x}.\tag{1}$$ This $\to\infty$ as $x\to\infty$. To see that, divide top and bottom ...

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First note that, we can evaluate the integral using the beta function $$\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^m}= 2\int_{0}^{\infty}\frac{dx}{(1+x^2)^m}={\frac {\sqrt {\pi }\,\Gamma \left( m-\frac{1}{2} \right) }{\Gamma \left( m \right) }},$$ where $\Gamma(n)$ is the gamma function. Now, you can see that $$C = {\frac {\,\Gamma \left( m ... 2 Suppose there is s \in (0,1) with f'(s) > A/2 and f(s) \ge 0. Then for t \in [s,1], f'(t) = f'(s) + \int_s^t f''(x)\ dx > A/2 - A(t-s), and$$f(1) = f(s) + \int_s^1 f'(x)\ dx > \int_s^1 (A/2 - A(x-s))\ dx = A s (1-s)/2 > 0$$which is impossible. Similarly in three other cases (in two of them you consider f(0) rather than f(1)). 2 You have reduced it to row echelon form. You can go further to reduced row echelon form as follows:$$ \begin{pmatrix} 1 & 0 & 1 & -2 &|&0 \\ 0 & 1 & -2 &2 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ 0 & 0 & 0 & 0 &|&0 \\ \end{pmatrix} $$You can read this matrix as: ... 2 The answer to the second question is "yes". Consider the affine curves given by the equations y^n-x over some field k. Their function field F is rational: F=k(x,y)=k(y). Hence the projective closure of all of these curves is birational to \mathbb{P}^1_k. However these projective closures are pairwise non-isomorphic: for n>1 the curves possess ... 2 HINT: At this point you’re expected to know the binomial expansion:$$(k+x)^n=\sum_{i=0}^n\binom{n}ik^{n-i}x^i\;.$$The first four terms are the terms for i=0,1,2,3:$$\binom{n}0k^n,\quad\binom{n}1k^{n-1}x,\quad\binom{n}2k^{n-2}x^2,\quad\text{and}\quad\binom{n}3k^{n-3}x^3\;.$$Since the coefficients of x^2 and x^3 are equal, you know that ... 1 The series$$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$$is indeed divergent, because it has infinitely many partial sums equal to 1, and infinitely many partial sums equal to 0. Inserting parentheses as ... 1 I agree with you nearly up to this point:$$ds \ = \ 3a \sin t \cos t \ , $$but because we are taking the square-root  \ \sqrt{\cos^2 t \ \sin^2 t} \  to obtain the infinitesimal arclength element , it would be better that we write the result as  \ ds \ = \ 3a \cdot | \sin t \cos t \ | \ = \frac{3}{2} a \ | \sin 2t \ | \ .  The curve itself is ... 1 For part a) you should check that the symmetry of P(x) with respect to x gives$$\int_{-\infty}^{\infty}Ne^{-|x|/a} dx=2\int_{0}^{\infty}Ne^{-x/a}dx$$and then do a change of variable \displaystyle u=\frac{x}{a} For part b) note that by definition \displaystyle\langle f(x) \rangle=\int_{-\infty}^{\infty}f(x)Ne^{-|x|/a}dx and that both x, x^2 ... 1 The notation in Dummit and Foote is actually pretty iffy. In the hint, they are referencing an earlier problem in which the setup is the same except that the K_i are called G_i. Thus, you should interpret g_i as an element of K_i. If you would like a further hint as to the actual solution, feel free to ask. 1 For part i)... i) For the maximum rate of change, try taking the gradient. The gradient vector is <2y^{1/2}, xy^{-1/2}>. The maximum rate of change will occur in the direction of <2*(4)^{1/2}, 3*(4)^{-1/2}> = <4, 3/2>. The maximum rate of change is then \sqrt {4^2 + (3/2)^{2}} = \sqrt {73}/2 As for part ii)... To find the ... 1 If Y = -X, f_{-X}(-x) = f_X(x) is the same as f_Y(t) = f_X(-t) where t = -x. Saying X and Y have the same distribution says that f_Y(t) = f_X(t). So this is exactly the same. And how could you possibly argue with  {\mathbb P}(X(\omega)\le -x) = {\mathbb P}(X(\omega)\le -x)? 1 Note that -2^2 = - (2 \times 2) = -4 and is not (-2) \times (-2) = 4. Hence,$$(-2^2)^3 = (-4)^3 = (-4) \times (-4) \times (-4) = - 64$$In general, when m is a positive integer, we have$$-a^m = - (\underbrace{a \times a \times \cdots \times a}_{m \text{times}})$$and is not$$(-a)^m = (\underbrace{(-a) \times (-a) \times \cdots \times (-a)}_{m ...

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