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$$\textrm{Volume} = \int_{\theta = 0}^{2\pi} \int_{r=0}^1 r \left(\int_{z = \sqrt{3(x^2+y^2)}}^{\sqrt{4 -(x^2+y^2)}} 1 \ dz \right) dr \ d\theta$$ $$\\$$ They key, yes is to use cylindrical coordinates. You know the region is bounded above by the sphere and below by the cone so you can get the bounds for $z$. Next, find the intersection which will give you ...

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It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis). In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described by $\rho^2 = z$. The solid bounded ...

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Note $z=x^2+y^2>0$, therefor $$V=\int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\int_{x^2+y^2}^{\sqrt{2-x^2-y^2}}dzdydx$$ Now apply cylinder coordinate.

2

They're not asking for all the coefficients, just which ones are zero. Your friend seems to be making arguments to linearity. In the first two (top) graphs, they split into the "1" part and the "sin" part, each 0 on the half of the period. Then he writes the first part as a sum again: a constant 1/2, and an alternating-sign 1/2. (The first two graphs on the ...

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The solution is not hard: I think: $$a_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1cos(nx)dx + \int\limits_{0}^{\pi}sin(x)cos(nx) dx) = \frac{1}{\pi} \int\limits_{0}^{\pi}sin(x)cos(nx) dx = \frac{1}{\pi} \frac{cos(n\pi)+1}{1-n^2}$$ $$b_{n} = \frac{1}{\pi} (\int\limits_{-\pi}^{0}-1sin(nx)dx + \int\limits_{0}^{\pi}sin(x)sin(nx) dx)=\frac{1}{\pi} (\frac{... 0 Note: The left hand side of OPs equations are lower and upper Darboux integrals whereas the right hand side are lower and upper Darbous sums. These are quite different things as we will see below. Let us consider the bounded function f(x)=x^2 on [0,1] and the partition P=\{0,\frac{1}{4},\frac{1}{3},\frac{2}{3},\frac{8}{9},1\}. At first we want ... 0 If a nxn matrix A(t)=A_{ij}(t) is differentiable, then d(det(A(t))/dt can be performed as follows (for brevity, we shall take the particular case n=3). The determinant of a matrix A(t)=A_{ij}(t) is given by$$det(A_{ij}(t))=e_{ijk}A_{i1}(t)A_{j2}(t)A_{k3}(t),$$where e_{ijk} is the Cartesian alternator. Let us take the derivative of this ... 0 No: Recalling the quotient rule$$\frac{d}{dx}\dfrac{f(x)}{g(x)}=\dfrac{\frac{df}{dx}g(x)-f(x)\frac{dg}{dx}}{g(x)^2}=\dfrac{1}{g(x)}\dfrac{df}{dx}-\frac{f(x)g'(x)}{g(x)^2},$$we conclude that \dfrac{1}{g(x)}\dfrac{d}{dx}f(x) \neq \dfrac{d}{dx}\dfrac{f(x)}{g(x)} unless g(x)= const. 0 You could always make a substitution in your surface equations to get the curve of intersection. Then parameterize this curve and take the derivative of it. This will give you a direction vector which you can then use to construct a tangent line with the given point. If you do this with gradients as you have tried, the gradient will give you a vector normal ... 1 When x<1 we need$$(m-1)x+m\geq 0.\qquad (*)$$If m-1>0 then the LHS of (*) is linearly increasing and so negative for small enough x. Thus we need m\leq 1, in which case the LHS is weakly decreasing and so (*) is satisfied for all x<1 as long as it is satisfied at x=1. At x=1, inequality (*) becomes$$2m-1\geq0.$$So we also ... -1 HINTS: The second part of your function definition is a quadratic expression, so you need to find its minimum value, which will depend on m. The general quadratic can be rewritten by completing the square,$$q(x)=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$So if a is positive, the expression has the minimum value of c-\... 3 You can rephrase your question as finding c to minimize E[|e^X-c|] where X is a uniform random variable. In general, The c that minimizes E[|Y-c|] is the median of the distribution of Y. Some intuition can be found in this answer. Here, we see the median of X is 1/2, and since x \mapsto e^x is increasing, the median of e^X is e^{1/... 1 Hint: use as substitution hyperbolic cosine$$ x=\sqrt { \frac { 3 }{ 7 } } \cosh { t } ,dx=\sqrt { \frac { 3 }{ 7 } } \sinh { t } dt\\ \\ \frac { 27 }{ \sqrt { 7 } } \int { \sqrt { { \left( \cosh ^{ 2 }{ t } -1 \right) }^{ 5 } } } \sinh { t } dt=\frac { 27 }{ \sqrt { 7 } } \int { \sinh ^{ 6 }{ t } } dt$$3 The question specifically asks to find an equation for the constant c that minimizes this expression. The typical method here is to let this expression be a function of c and find the minimum of this function with respect to c. Thus, we have$$ f(c) = \int_0^1 |e^x - c| \ dx$$We have to be careful here, as we are not sure if the expression inside of ... 0 We prove that \dfrac{1}{2} \le m \le 1 is the answer. Pick x = 0 \implies m = f(0) \ge 0, and if m > 1 then f(x) < 0 for x large and negative. Thus 0 \le m \le 1. For 0 \le m \le 1, and for x \ge 1, f'(x) = 2(x-1) + m \ge 0 \implies f(x) \ge f(1) =3-m \ge 0. Also if x < 1 \implies f'(x) = m - 1 < 0 \implies f(x) > f(1) = 2m -... 1 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 0 Hint...... Just use the formula, \lim_{x \rightarrow a}f(x)^{g(x)} =\lim_{x \rightarrow a} e^{g(x)(f(x)-1)} 0 The Matrix Cookbook is the most comprehensive one. Can't post the link due to my missing rep., but it's first hit on Google. See these, too: Matrix Calculus - Colorado and Matrix Algebra - LMU - German. 3 Since \lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e (or \\ \lim _{ x\rightarrow \infty }{ { \left( 1+\frac { 1 }{ x } \right) }^{ x } } =e\\ )$$\lim _{ x\to 3 } \left( \frac { x-1 }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\lim _{ x\to 3 } \left( 1+\frac { 3-x }{ 2x-4 } \right) ^{ \frac { 1 }{ x-3 } }=\\ =\lim _{ ...

0

$$(a,b,c)\times (d,e,f)=(bf-ce,cd-af,ae-bd)$$

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Note that it asks "in what direction" hinting at a directional derivative. For that we need the gradient of $f$ which is $$\nabla f=<y,x>$$ Then we are looking for a unit vector such that \begin{align} \nabla f(2,0)\cdot\hat u &= <0,2>\cdot\hat u \\ &= -1 \end{align} Since $\hat u$ is a unit vector it is of the form \hat u=\frac{<... 0 Hint: Instead of differentiation you could also apply the formula for the binomial series expansion with \alpha=-2 \begin{align*} (1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1\tag{1} \end{align*} We obtain \begin{align*} \frac{1}{(7+x)^2}&=\frac{1}{7^2}\cdot\frac{1}{(1+\frac{x}{7})^2}\\ &=\frac{1}{7^2}\sum_{n=0}^\... 0 If a unique value be assigned to f(a), then f(x) is said to determinate at x=a, otherwise f(x) is said to be indeterminate at x=a For example ,f(x)=\frac{x^2-9}{x-3}$$f(4)=7 , which is unique Hence f(x) is said to be determinate at x=4 But$$f(3)=\frac{9-9}{3-3}=\frac{0}{0}$$Hence we can't assign a unique value to f(3) \frac{0}{0}=... 1 Use the sub u = a-x so that$$\lim_{x\to a^{-}} (a-x)\ln(a-x) = \lim_{u\to 0^{+}} u \ln u$$then u = e^{-y} gives$$\lim_{u \to 0^+} u\ln u = -\lim_{y \to \infty} ye^{-y}$$since e^y > y^2/2! for y>0 we obtain the required limit via squeezing. 2 Apply L'HÃ´pital's rule :$$\lim _{ x\to a- } \left( a-x \right) \ln { \left( a-x \right) } =\lim _{ x\rightarrow { a }- }{ \frac { \ln { \left( a-x \right) } }{ \frac { 1 }{ \left( a-x \right) } } } =\lim _{ x\rightarrow { a }- }{ \frac { -\frac { 1 }{ a-x } }{ \frac { 1 }{ { \left( a-x \right) }^{ 2 } } } } =\lim _{ x\rightarrow a- }{ -\left(...

1

You can interpret it as a quotient and use L'HÃ´pital: $$(a-x)\log(a-x) = \frac{\log(a-x)}{(a-x)^{-1}}$$ Differentiate to get: $$\frac{-1}{(a-x)^{-1}} = -(a-x).$$ As this goes to zero you are done.

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Sketch: Typically one does $$\lim_{x\to a^-}(a-x)\ln(a-x)=\lim_{x\to a^-}\frac{\ln(a-x)}{\frac1{a-x}}\stackrel{\color{blue}{\text{l'HÃ´pital}}}= \lim_{x\to a^-}\ \frac{\color{blue}?}{\color{blue}?}$$

0

Let $c >0$. Then Weierstrass theorem implies that there is a smooth $g_1$ so that $|f(t) - g(t)| <c/4$ for all $t\in [-1,1]$. Since $f(0) = 0$, there is $0< \delta <1/2$ so that $|f(t)| < c/4$ for all $t\in (-\delta, \delta)$. Now let $g_2$ be a smooth function so that $0\le g_2\le 1$ and $$h_2(x) = \begin{cases} 1 & \text{if } t\in [-\... 0 Here is a graph: (Large version) 2 They're not correct and also not equivalent. (1) is close. In fact f is piecewise C^k if there is a partition such that f is C^k on each (x_j,x_{j+1}), and the k-th derivative f^{(k)} has one-sided limits at each endpoint (which implies that f^{(j)} also has one-sided limits for 0\le j < k). If we were talking about a bounded interval (... 1 QUESTION.- Do you want your curves necessarily be all concave? If not, you have a nice example to add to the concave ones with the Witch of Agnesi, whose equation is y=\frac{8a^3}{x^2+4a^2} where a is the radius of the circle that generates the curve (so you have infinitely many examples). In the figure you have (with  a = 300) a "witch" ... 0 Hints:$$\frac {\partial {F(x,y)}}{\partial x}=\frac {\partial {F_1(X,Y)}}{\partial X}\frac {\partial X}{\partial x}+\frac {\partial {F_1(X,Y)}}{\partial Y}\frac {\partial Y}{\partial x}=\frac {\partial {F_1(X,Y)}}{\partial X}\phi_x+\frac {\partial {F_1(X,Y)}}{\partial Y}\psi_x$$1$$A=s^2\implies {dA\over dt}={d\over dt}(s^2)=2s{ds\over dt}=2\times 10\times 5=100$$0 Notice that$$-\log\left(1-\frac1kx\right)=-\log\left(\frac1k\left(k-x\right)\right)=-\log(k-x)-\log\left(\frac1k\right),$$and that k is constant with respect to x, which shows that your antiderivatives differ by a constant, namely -\log\left(1/k\right). For more information look up the fundamental theorem of calculus. 0 Since$$\frac1k=-\left(\frac{N_t}{k}\right)'\implies -\int\frac{-\frac1k}{1-\frac{N_t}k}dN_t=-\log\left(1-\frac{N_t}k\right)+C=\log\frac k{k-N_t}+C$$2 Any function of the form f(x)=g(|x|) where g is an increasing concave function with g(0)=0 and \lim_{x\to\infty}g(x)=a will work. Letting, g(x)=a[1-h(x)] we need h convex with h(0)=1 and \lim_{x\to\infty}h(x)=0. For example, you could take h(x)=\frac{1}{1+x}, so that$$f(x)=a-\frac{a}{1+|x|}.$$0 A solution of a somewhat different flavor: Set a=Y/A \leq b=Y/B and c=X/Y\geq 1. Define$$v_t=\left( \begin{matrix} \log(1+ta) \\ \log(1+tb) \end{matrix} \right) \ \ \mbox{with} \ \ \ v_t' = \left( \begin{matrix} \frac{a}{1+at} \\ \frac{b}{1+bt} \end{matrix} \right) .$$The original problem amounts to showing that S=\det\left( v_c,v_1\right) \geq 0. ... 1 An object's velocity can only be measured relative to another object. For example, suppose you are on a bus. You are not moving on the bus, but the bus itself is moving. Therefore, you can say you are at rest "relative" to the bus, but you are moving relative to the ground/pavement. 4 We can write the interal in (4) as$$I=-\int_{0}^{1}\frac{1-x+\log\left(x\right)}{\log^{2}\left(x\right)}dx $$now define$$I\left(\alpha\right)=-\int_{0}^{1}\frac{x^{\alpha}\left(1-x+\log\left(x\right)\right)}{\log^{2}\left(x\right)}dx,\,\alpha\geq0 . $$We have$$I''\left(\alpha\right)=-\int_{0}^{1}x^{\alpha}\left(1-x+\log\left(x\right)\right)dx=-\...

2

The computation of sum of squares is wrong. $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$

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Through the substitution $x=e^{-t}$ the original integral equals: $$I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$ where the blue integral is yet manageable through Frullani's theorem (...

2

For any $y\in\mathbb{R}$ such that $|y|<1$ we have: $$\log(1+y) = \int_{0}^{y}\frac{dx}{1+x} \tag{1}$$ but if $|x|<1$ we also have: $$\frac{1}{1+x} = 1-x+\frac{x^2}{1+x}\tag{2}$$ and: $$\log(1+y) = \color{red}{y-\frac{y^2}{2}}+\color{blue}{\int_{0}^{y}\frac{x^2}{1+x}\,dx} \tag{3}$$ where, by the Cauchy-Schwarz inequality: $$\left|\color{blue}{\... 3 One approach directly based on a definition of \log x is as follows. We use the following definition of \log x$$\log x = \lim_{n \to \infty}n(x^{1/n} - 1)Replacing x by (1 + x) and for |x| < 1 using binomial theorem we get \begin{align} \log (1 + x) &= \lim_{n \to \infty}n((1 + x)^{1/n} - 1)\notag\\ &= \lim_{n \to \infty}n\left(\frac{... 1 Hint: For any x with \|x\|_1=1, we have \|x\|_\infty \leq 1. 0lim_{x \to c} x^2 = c^2$$Let \epsilon>0 , we know that$$x^2-c^2=(x-c)(x+c)|x^2-c^2|=|x-c||x+c|$$If$$|x-c|<2 \implies -2<|x|-|c|<2 (by\ \ triangular \ \ property)|c|-2<|x|<2+|c|$$By triangular property, we have$$|x+c|\le |x|+|c|<2+|c|+|c||x+c|<2+2|c|$$So,$$|x^2-c^2|<2.(2+2|c|)|x^2-c^2|<(4+4|c|)...

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An alternative is to assume that $\ln (1+y) \approx a + by + cy^2$ and then differentiate twice and use $y=0$ to get $\ln (1+y) \approx y - \frac{y^2}{2}$.

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\begin{align} & \sum_{n=0}^{100}2^{100-n}(-1)^n\binom{100}{n} \\ & =\binom{100}{0}2^{100}-\binom{100}{1}2^{99}+\binom{100}{2}2^{98}- \ldots +\binom{100}{98}2^{2}-\binom{100}{99}2^{1}+\binom{100}{100}2^{0} \\ & =(1-2)^{100} \\ & =(-1)^{100} \\ & =1 \end{align} Please refer to binomial expansion.

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$$\forall n \in \mathbb{N}, \qquad \sum_{k=1}^n \dbinom{n}{k} 2^{n-k} (-1)^k = \left(2 - 1\right)^n = (-1)^n.$$ Put $n = 100$ and you're done.

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I'm sure you know the geometric series $$\sum_{n=0}^{\infty}(-y)^n=1-y+y^2-y^3+\ldots=\frac{1}{1+y},\qquad |y|<1\tag{1}$$ Now integrate both sides of $(1)$, and the result follows. For small $y$, the higher power terms become negligible.

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The ratio test will prove that the series is absolutely convergent if the absolute value of the result is less than $1$. This is true because the series will eventually grow smaller than an arbitrary convergent geometric series. Therefore, since we know that series a and b converges with $|x|<r_a,r_b$ respectively, we also know that as $lim_{k>\infty}$,...

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