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-2

Hint: We do not need a double or triple angle formula to solve the problem. What angles $\phi$ between $0$ and $(3)(360)$ satisfy $\cos\phi=\frac{1}{2}$?

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The multivariable chain rule is not normally stated in terms of strange things like $\frac{dv}{du}$. It is stated in terms of total derivatives, i.e., linear transformations that approximate the map at a given point. In particular, it says that if $T(u_0,v_0)=(x_0,y_0)$, then the derivative of the composition $T^{-1}\circ T$ at $(u_0,v_0)$ is the ...

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$\max_{x,y} f(x,y) = \max_x \max_y f(x,y)$, where $x \in [0,1], y \in [-1,1]$, and similarly for $\min$. For example, $\max_y f(x,y) = f(x,1) = 3x+4$, and $\max_x (3x+4) = 3+4 = 7$.

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The statement $(a)$ is true. $X,Y$ are independent r.v if, and only if, the events $\{ X \le a \}$ and $\{ Y \le b \}$ are independent sets. Once the images are countable you can write $$\{ X \le a \} = \bigcup_{d \le a}\{X=d\}$$ $$\{ Y \le b \} = \bigcup_{c \le b}\{Y=c\}$$ Now, note that union is a disjoint union. So, $$P(\bigcup_{c \le ... 0 You're basically done! Setting x(t) = 2\cos t, \tag{1} and y(t) = -2\sin t, \tag{2} note that x^2(t) + y^2(t) = 4 \cos^2 t + 4 \sin^2 t = 4(\cos^2 t + \sin^2 t) = 4, \tag{3} which shows that (1)-(2) describe a circle of radius 2 centeted at (0, 0). Note also that, starting at t = 0 and increasing t, the point (x(t), y(t)) rotates ... 0 The hint is not all that useful. The hidden subtext of the problem is that for the first few odd n, the expressions ((x+y)^n - (x^n + y^n)) factorize into easily understood pieces of degree 1 and 2. Spivak is essentially asking for the factorization when n=5. In addition to the obvious x and y there is a and that uniquely determines the ... 0 You need to use the Implicit function theorem$$ F(x,y)=ln(x+2y)+32x^3y^2-\frac{1}{4}=0 \\ \frac{\partial F}{\partial y}=\frac{2}{x+2y}+64x^3y\\ \frac{\partial F}{\partial y}(\frac{1}{2},\frac{1}{4})\ne 0 $$Therefore y defines a function of x.$$ \frac{\partial F}{\partial x}=\frac{1+2y'}{x+2y}+96x^2y^2\\ \frac{\partial F}{\partial ...

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You're done. Or maybe you want to write $$r(t) = (2 + 3t, 1 + 3t), 0 \le t \le 1,$$ to make it a little more complete.

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As you pointed out, he is equating what he got from simplifying $(x+y)^5=x^5+y^5$ and the binomial expansion of $(x+y)^3$. The first one is an equation, of which you're trying to find the solution. The second is an identity, which is always valid. The combination of both gives the said condition. Explicitly, $(x+y)^3 = x^3+3x^2y+3xy^3+y^3 = x^3 + ... 2 $$(x+y)^3 = x^3+3x^2y+3xy^2+y^3.$$ Suppose$x^3 + 2x^2y+2xy^2+y^3 = 0$. Add to both sides$x^2y+xy^2$: $$x^3+2x^2y+2xy^2+y^3 + \color{red}{x^2y+xy^2} = \color{red}{x^2y+xy^2} \\ x^3+3x^2y+3xy^2+y^3 = x^2y+xy^2 \\ (x+y)^3 = xy(x+y).$$ 0 The differential is a linear application$Df(A)[\cdot]$such that$f(A+B)-f(A)=Df(A)[B]+o(\|B\|)$for "small" matrices$B$. In your case$L\in \mathcal L(M_n(\Bbb R)). i.e. it is not a matrix, but a linear application from the space of matrices to itself (it is a tensor of fourth order, if you want). We can write $$f(A+B)-f(A)= BA^{m-1} ... 0 The series \displaystyle\sum_{n\ge0} a_n is convergent if and only if the partial sum sequence \left(\displaystyle\sum_{k=0}^n a_k\right)_n is convergent which's equivalent to the convergence of the sequence \left(\displaystyle\sum_{k=n_0}^n a_k\right)_n for all n_0\in\Bbb N hence the convergence of a series doesn't depend of its first few terms. ... 3 This is a possible way (I assume a\geq 0, b\geq 0): 1) substitute x= a \sin t . 2) integrate the resulting integral by parts to get rid of \ln 3) you should end up with$$-\int_0^{\pi/2}\frac{2 a^2 b \sin^2 t\,dt }{b^2 -a^2 \cos^2 t} .$$4) it is possible to integrate the last integral by elementary means or using the residue theorem. You should ... 2 It is enough to compute \int \ln(a+\sqrt{1-x^2}) dx. Do a change of variables x=\sin t and integrate by parts:$$\int \ln (a+\cos t) \cos t \, dt = \ln(a+\cos t) \sin t + \int \frac{\sin^2 t}{a+\cos t} \, dt$$Use the substitution u = \tan t/2 so that \sin t = 2u/(1+u^2), \cos t = (1-u^2)/(1+u^2) and dt = 2/(1+u^2) du. This is known as the ... 3$$\begin{align*} I&=\int^1_0\frac{1-x^2+(1+x^2)\log x}{x+1}\frac{dx}{x\log^3x}\\ &=\left.\frac{-1}{2\log^2x}\frac{1-x^2+(1+x^2)\log x}{x+1}\right|^1_0-\int^1_0\frac{-1}{2\log^2x}\frac{\partial}{\partial x}\left(\frac{1-x^2+(1+x^2)\log x}{x+1}\right)dx\\ &=\int^1_0\frac{1}{2\log^2x}\frac{\partial}{\partial x}\left(1-x+\frac{(1+x^2)\log ... 0 Hint: It is a system of linear equation withn_1$= No of units of Fertilizer Type I$n_2$= No of units of Fertilizer Type II$n_3$= No of units of Fertilizer Type III Linear system of equations: $$10n_1+20n_2+50n_3 = 2000$$ $$30n_1+30n_2 = 900$$ $$60n_1+50n_2+50n_3 = 3100$$ Solve it using Matrix Algebra: The solution is$n_1$= 10,$n_2$=20, ... 1 Hint Just name$x_1$,$x_2$and$x_3$the quantities of product$I$,$II$and$III$which will be made. Now, what do they contain ? Concerning$A$, you have $$10 x_1+20 x_2 +50 x_3=2000$$ Do just the same for$B$and$C$. So you have three equations for three unknowns and you have to solve it. I am sure that you can take from here. 0 In order to prove that$f$is differentiable at$(1,-1)$we have to investigate the difference$f(1+X,-1+Y)-f(1,-1)$in terms of the increment variables$X:=x-1$and$Y:=y-(-1). Computation gives \eqalign{f(1+X,-1+Y)-f(1,-1)&=(1+X)^2(-1+Y)-(-1) \cr &=-2X+Y +2XY-X^2+X^2 Y \cr &=-2X+Y + r(X,Y)\ ,\cr} wherer(X,Y):=2XY-X^2+X^2Y$satisfies ... 2 You plot each function on the same pair of axes. Then for each$x$value of interest, you plot$(f+g)(x)$by measuring$g(x)$above the$f(x)$line. It results in the same thing as making a table of values of$(f+g)(x)$and plotting them. I believe "graphical addition" refers to the process of adding the functions on the graph paper. 1 Graphical addition means$(f+g)(x)=f(x)+g(x)$for all domain elements. So you need to graph$(f+g)(x)=x+\frac{1}{x}$. 1 Here is how I did it. Here is how I did it: 4 $$x + \frac{x^2}{\sqrt{x-1}} = \dfrac{x\sqrt{x-1} + x^2}{\sqrt{x - 1}}$$ But note that this does not make the function a rational function, because a rational function requires that the numerator and denominator are polynomials, which is not the case here. 0 I'm assuming we are using this definition of differential of a function. So the problem is, what can be the candidate linear function required in the definition above? First of all it is necessary, but not sufficient, that the partial derivatives of the function exist. However, if these partial derivatives exist and are continuous, then$f$is ... 0 Another proof that the limit in$(0,0)$doesn't exists. You can write: $$\frac{\sin(2xy)}{x^2+y^2} = \frac{\sin(2xy)}{2xy} \cdot \frac{2xy}{x^2+y^2}.$$ The first member always goes to$1$, while the second causes some troubles. In fact if you use polar coordinates$(x,y) = (\rho \cos \vartheta, \rho \sin \vartheta)$, you get $$\frac{2xy}{x^2+y^2} = ... 0 \frac{\partial}{\partial x}f(x,y) = \lim_{(x,y_{0})\rightarrow (x_{0},y_{0})}\frac{f(x,y_{0}) - f(x_{0},y_{0})}{x-x_{0}} which is made easier in this case if we first substitute throughout that y_{0}=-1 and x_{0}=1. \lim_{(x,-1)\rightarrow (1,-1)} \frac{f(x,-1)-f(1,-1)}{x-1}=\lim_{x\rightarrow 1}\frac{x^{2}-1}{x-1}. Now that this is reduced to a ... 1 I would take another approach. Hope it helps. Take$$G(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)$$Then with the normal and a point we have the equation of the plane: ... 2 As Antonio Vargas said, it will help to first find the range of M = |\frac x {1 + x^2}|. Since an absolute value is never negative, if M can be zero then zero is the minimum. I believe you can find the value of x for which M is zero. Now we need to find out the maximum. How large can M get? Notice that the denominator is "bigger" than the ... 1 No, I don't think you can. Consider the limit of the function for (x, y) approaching the origin in some direction (x, mx): \lim_{x\rightarrow 0}\frac{sin(2mx^2)}{(x^2 + m^2 x^2)}. This tends to \frac {2m}{1+m^2}, taking any value in range [-1..1]. 1 Take x=0, y=\frac{1}{n}, then$$f(x,y)=f\left(0,\frac{1}{n}\right)=0\to 0.$$Take x=y=\frac{1}{n}, then$$f(x,y)=f\left(\frac{1}{n},\frac{1}{n}\right)=\frac{\sin \frac{2}{n^2}}{\frac{2}{n^2}}\to1.$$Hence the function f is discontinious at (0,0), independently of the value f(0,0). 1 By using the polar transformation, we have:$$ \lim_{(x,y)\rightarrow(0,0)}\frac{\ln(1+xy)}{\sqrt{x^2+y^2}}=\lim_{r\rightarrow 0}\frac{\ln(1+r^2\sin(\theta)\cos(\theta))}{r}=\lim_{r\rightarrow 0}\frac{r^2\sin(\theta)\cos(\theta)}{r}=0 $$Also:$$ \lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^4+y^2}=\lim_{r\rightarrow ... 0 In your diagram: delete the Imaginary axis, change Integer to "Polynomial with a finite number of terms", change Rational to Rational function, change Algebraic to Algebraic function. I'm not sure what Real corresponds to. On another axis you could have the C^1>C^2 ... as you mentioned. On another axis you could have the field of the polynomial, i.e. ... 0 Hint By AM-GM,$\dfrac{1+x^2}2 \ge \lvert x \rvert$. 0 I am giving this a go just to see if I have progressed enough to handle it. Please comment on/correct the answer but do not rate yet as it will unfinished. here goes.... $$I=\int_0^1\frac{1-x^2+\left(1+x^2\right)\ln x}{\left(x+x^2\right)\ln^3x}dx.\tag1$$ $$\frac {1-x^2+lnx+x^2lnx}{xlnx^3+x^2ln^3x}=$$ $$\frac{lnx+1-x^2+lnx^{x^2}}{lnx^{3x}+{lnx^{3x^2}}}=$$ ... 0 In spherical coordinates the intersection points$r=\sqrt 3/2$,$z=1/2$have colatitude$\varphi_0=\arctan\sqrt 3=\pi/3$and the second sphere is$\rho=2\cos\varphi: $$V= \int_0^{2\pi}\int_0^{\pi/3}\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta+ \int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^{2\cos\varphi}\rho^2\sin\varphi d\rho d\varphi ... 0 The theorem you cite is correct for when the exponent of the variable is just n. Here obviously we have 2n+1. 1 This power series contains only terms of odd powers, and hence its derivative contains terms of even powers only, and since the power series starts from the first power of x, then its derivative start with the zero power of x. 1 You can look at the problem as the solution of a quadratic equation in a. The solutions are$$a_{\pm}=1+\frac{1}{2} \left(b\pm \sqrt{3} \sqrt{-(b-2)^2}\right)$$Since a,b\in R, b must be equal to 2; then a=2. 0 Your condition can be translated into words as: f is bounded in every open ball around a This is quite different from being continuous! PS Not every function satisfies this condition, but the ones that don't are rather fierce. (In fact, in a strict sense, almost no function satisfies the condition, but the functions arising in analysis normally do.) 3 other solution: since$$a^2-(2+b)a+b^2-2b+4=0\Delta_{a}=(2+b)^2-4(b^2-2b+4)=-3b^2+12b-12=-3(b^2-4b+4)=-3(b-2)^2\le 0$$so$$b-2=0\Longrightarrow b=2\Longrightarrow a=2$$6 The expression can be written as, a^2-ab+b^2-2a-2b+4=\frac{3}{4}(a-b)^2 + \frac{1}{4}(a+b-4)^2=0 Since, (a-b)^2\ge0 and (a+b-4)^2\ge 0, So, the only possible real solutions are a-b=0 and a+b-4=0 That is a=b=2. a^2b=8. 0 Answer using Cylindrical Coordinates: Volume of the Shared region = Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = \sqrt{\frac{3}{4}}. Now the sphere is shifted by 1 in the z-direction, Hence Volume of the Shared region =$$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} ... 7 Here is a proof of Cleo's answer. Rewrite the integral as \begin{align*} I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \end{align*} The first integral can ... 1\frac{2^i}{n-i} = \frac{1}{2^{-i}(n-i)} = \frac{2^n}{2^{n-i}(n-i)}$So,$\sum_{i=0}^{n-1}\frac{2^i}{n-i} = \sum_{i=0}^{n-1}\frac{2^n}{2^{n-i}(n-i)} = 2^n \sum_{i=0}^{n-1}\frac{1}{2^{n-i}(n-i)} = 2^n \sum_{i=1}^{n}\frac{1}{i2^{i}}\tag1$Now,$\frac{d}{dx}\sum_{i=1}^n\frac{x^i}{i} = \sum_{i=1}^{n-1}x^i = \frac{1-x^n}{1-x}\tag2$To achieve a closed form of ... 0 Consider the sequence $$0,1,0,1/2,1,0,1/3,2/3,1,0,1/4,2/4,3/4,1,0,1/5,2/5,3/5,4/5,1,0,1/6,2/6,3/6,\dots.$$ Every real number in the interval$[0,1]$is the limit of a subsequence of the above sequence. Remark: We cannot get the half-open interval$(0,1]$as the set of subsequential limits, For take$n$very large. If$\frac{1}{n}$is a limit point of an ... 0 I think this works: Let$(x_n)$be an enumeration for$\mathbb{Q}\cap (0,1]$. Then each$x\in (0,1]$is a cluster point of$(x_n)$2 There is a kind of algebraic structure called a real closed field. It's a structure with constants$0$,$1$, and has operations$+$and$\cdot$and a relation$<$. The theory of real closed fields is interesting, because any (first-order) logical statement you can make using just$0,1,+,\cdot,<$is true in a single real closed field if and only if it ... 3 Let$t = \frac xy > 1. Then: \begin{align} \frac{\log x}x = \frac{\log t + \log y}{ty} < \frac{\log y}y &\iff \log t + \log y < t\log y\\ &\iff \log t < (t - 1)\log y\\ &\iff \frac{\log t}{t-1} < \log y\\ &\iff\frac{\log [(t-1) + 1]}{t-1} < \log y. \end{align} Putz = t-1 > 0$, then we prove:$\frac{\log(z+1)}z ...

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I think the notation $x^+=\max(x,0)$ is nice. The function $F(x)=\sum_{n=1}^\infty (x-a_n)^+$ is an antiderivative of your $f$, and therefore $$\int_a^b f(x)\,dx = \sum_{n=1}^\infty \bigg( (b-a_n)^+ - (a-a_n)^+ \bigg)$$

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I'm sure there is a more elementary method than the heavy sledgehammer I used here. Since I only know this one, here is the sledgehammer. Let $\displaystyle c = \frac{1+i\sqrt{7}}{2}$, we are going to prove $\Re(c^k) \to \infty$ as $k \to \infty$. Consider the sequence $(c_k)_{k\in\mathbb{N}}$ where $c_k = 2\Re(c^k) = \left(\frac{1+i\sqrt{7}}{2}\right)^k + ... 1 Here is another way using the Cauchy's criterion: Let$M>0$such that$|b_n|< M$. Given$\varepsilon>0$, choose$n_0$such that $$\sum_{n=p+1}^q|a_n|< \varepsilon/M$$ for all$q<p<n_0\$ Then $$\bigg|\sum_{n=p+1}^qa_nb_n\bigg|\le\sum_{n=p+1}^q|a_nb_n|\le M\sum_{n=p+1}^q|a_n|< \varepsilon$$

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