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2

Once you get the first two steps, it is straightforward. Note that $$a_1a_2 \leq \left[\frac{a_1+a_2}{2}\right]^2$$ and $$(a_1a_2)(a_3a_4)\leq\left[\frac{a_1+a_2}{2}\right]^2\left[\frac{a_3+a_4}{2}\right]^2 = \left[\left(\frac{a_1+a_2}{2}\right)\left(\frac{a_3+a_4}{2}\right)\right]^2\\ \leq ... 0 The trick is to take advantage of the fact that the number of elements is a power of 2 by splitting the factors into two equal parts and using the induction hypothesis on both halves. We want to show that$$(a_1a_2\cdots a_{2^{n+1}})^\frac{1}{2^{n+1}}\leq \frac{a_1+a_2+\ldots+a_{2^{n+1}}}{2^{n+1}}.$$On the other hand, by the induction hypothesis we ... 1 Let us start by putting things in a nicer form.$$\begin{eqnarray*}A_{n,j}&=&3(-1)^j 2^{n-j+1}\frac{(2n-2j-4)!}{(n-j)!(n-j-2)!}\binom{j+2}{2}\frac{n^{5/2}}{8^n}\\&=&\frac{6\cdot 2^{n-j}}{8^n}(-1)^j\frac{(n-j)(n-j-1)n^{5/2}}{(2n-2j)(2n-2j-1)(2n-2j-2)(2n-2j-3)}\binom{2n-2j}{n-j}\binom{j+2}{2}\end{eqnarray*}but due to Stirling's formula, ... 0 My attempt: f(t)=U(t)e^t −U(t−1)e^t \begin{align} U(t)-U(t-1) & = \begin{cases} 1 & : t \ge 0 \\ 0 & : t\lt 0\end{cases}-\begin{cases} 1 & : t \ge 1 \\ 0 & : t\lt 1\end{cases} \\ & = \begin{cases} 0 & : t \ge 1 \\ 1 & : 0\le t \lt 1 \\ 0 & : t\lt 0 \end{cases} \\ \color{gray}{ \operatorname{\bf ... 0 First of all, "x/a + y/b" is just an expression and not an equation. The full answer is "\dfrac{x}{a}+\dfrac{y}{b} = 1". Second, not every point P which is co-linear with M and N satisfies MN = MP+NP. Try looking at the case where M(1,0), N(0,1), and P(2,-1). Finally, note that P being colinear with M and N simply means that M, ... 0 Hint. If you draw a diagram (please do this before reading further) you will see that the gradient of the line PM must be the same as the gradient of the line PN. This will give you the equation you need. 0 f(t) = U(t) e^t +U(t-1) U(1-t)e^t-U(t-1) e^t  I keep almost the same function as you have, just modifying it's behaviour at the point x=1. 0 e^t U(1-t) U(t) $$(with 1-t, not t-1). {{{{{{{{{{{{{{{}}}}}}}}}}}}}}} 0 Simple. f(t) = e^tU(-t+1)U(t) EDIT Your function f has the value of 0 on the entire x domain, in the exception of [0, 1] where it has the value of e^t. You can start with f(t) = e^t. The problem now is that for t < 0 or t > 1 the value of f(t) is still e^t. We start by clearing out f(t) if t > 1. We know that the Heaviside function U(t) is 1 ... 0 This is great! So here's some feedback on the presentation: Proof of (G1): This reads backwards in my opinion: "let A=\begin{bmatrix}a&0\\c&d\end{bmatrix} and B=\begin{bmatrix}e&0\\g&h \end{bmatrix} where A, B \in S..." I suggest writing: "let A, B \in S. Then, by definition, A=\begin{bmatrix}a&0\\c&d\end{bmatrix} ... 1 Using the power rule: \frac{d}{dx} ax^n = nax^{n - 1}. 1 = 1x^0, and so \frac{d}{dx} 1 = 0 \cdot 1 \cdot x^{-1} = 0. Alternately, with the definition of the derivative: \frac{d}{dx} f(x) = \lim\limits_{h \to 0} \frac{f(x) - f(x + h)}{h}. In this particular case, f(x) is the constant function 1. This gives \lim\limits_{h \to 0}\frac{1 - 1}{h} ... 4 Using the Euclidean division of x^5-3x^4+5x^3-7x^2+6x-2 and x-1 we get:$$x^5-3x^4+5x^3-7x^2+6x-2=(x^4-2x^3+3x^2-4x+2)(x-1)$$Then apply the Euclidean division of x^4-2x^3+3x^2-4x+2 and x-1. Then we get x^4-2x^3+3x^2-4x+2=q(x-1). Then apply the Euclidean division of q and x-1 and so on. 4 Hint \  If a polynomial \,f(x)\, has power series \,c_k x^k + \cdots +c_{k+j} x^{k+j},\,\ c_k\ne 0,\, then the highest power of \,x\, that divides \,f(x)\, is \,k,\, the order of the power series at \,x = 0.\, An analogous remark holds for divisibility by \,x-1\, using a series at \,x = 1.\, Computing its derivatives then evaluating them at ... 1 Two sides are parallel if the sum of the angles between them is a multiple of 180 degrees This is measured by starting from the ending point of the initial side to starting point of the terminal side Consider a square. We can select a side and move clockwise to the next side after encountering an angle of 90 degrees. After moving clockwise again we would ... 0 So here is my answer: $$q(x,t)=\frac{-V_t(1+\delta f[c,g(x)])}{P(x,t)}\left(\frac{dP_o}{dt}\right)$$ substitute into Darcy's relation: q_x=\frac{-k_x A}{\mu}\frac{dP}{dx} $$\frac{-V_t(1+\delta f[c,g(x)])}{P(x,t)}P'_o=\frac{-k_x A}{\mu}\frac{dP}{dx}$$ Using Klinkenberg's relation for gas ... 1 Let's just consider the one-dimensional case z=f(x). It follows from Taylor's theorem that$$ f(x+\Delta x)=f(x)+\frac{\partial f}{\partial x}(x)\Delta x + \epsilon(x)\Delta x $$for some function \epsilon with \lim_{\Delta x\to 0} \epsilon(x+\Delta x)=0. Now rearrange$$ \underbrace{f(x+\Delta x)-f(x)}_{\Delta z} = \frac{\partial f}{\partial ...

0

Hint: The origin is $O(0,0,0)$. Given two points: $P(x,y,z)$ and $Q(x',y',z')$ then the distance between those points, $d(P,Q)$, is: $$d(P,Q) = \sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}$$ So what you want is $d(A,O) = d(B,O) = d(C,O)$. Now, remember that a point on an axis has zeroes for two of it's coordinates. (your $A$, $B$ and $C$ are not on any axis)

1

You want to solve for $\sin(c)$ from $$\sin(a) \sin(b) \sin(c) + \cos(a) \cos(b) = 1$$ There are multiple approaches, but let's algebra first. Just rearanging we get $$\sin(c) = \frac{1-\cos(a) \cos(b)}{\sin(a) \sin(b)}$$ From here I would use product-to-sum identities, noting that $$\cos(a) \cos(b) = \frac{\cos(a-b) + \cos(a+b)}{2}$$ and $$\sin(a) ... 2 Using that 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta), we can write$$4\cos x\cos 2x\cos 3x=2\cos 2x\cdot \left(\cos4x+\cos 2x\right)=\cos6x+\cos2x+\cos4x+1,which immediately gives 2 as the final result. Added: the second equality follows from \begin{align} 2\cos 2 x\cos 4x&=\cos(4x+2x)+\cos(4x-2x),\\ 2\cos 2x \cos ... 1 Try the other way - 2\cos a\cos b=\cos(a+b)+\cos(a-b) 1 See the comment of André (at the end \frac{\lambda^{n}e^{-\lambda}}{n!} instead of \frac{\lambda^{x}e^{-\lambda}}{x!}). Let N,X_{1},X_{2},\dots be independent rv's with N\sim Poisson\left(\lambda\right) and X_{i}\sim Bernouilli\left(p\right). Now define X:=X_{1}+\cdots+X_{N}. Note that X\sim Binom\left(n,p\right) under condition ... 2 What you need is a simple transformation: Let j=n−x. Then P(X=x) = \sum_{j=0}^{\infty} p^x(1−p)^j \dfrac{\lambda^{x+j} e^{-\lambda}}{x!j!} And proceed. Hint: pull out the "x" terms. i remember similar from somewhere... 3 You have:a(y-b)=by+c$$So:$$ay-ab=by+c$$Putting all the terms with y in the same side:$$ay-by=ab+c(a-b)y=ab+c$$So, if a \neq b:$$y = \frac{ab+c}{a-b}$$To review some stuff you can take a look here: https://www.khanacademy.org/ The videos are short, everything is well explained and you can choose the specific topic you need. 0 find an equation to a plane that crosses the axes at points equidistant to the origin P[0,0,0]. Found it: x+y+z=1. It's a plane that crosses the axes at points equidistant to the origin (namely, at (1,0,0), (0,1,0), and (0,0,1)). No need to think about A,B,C at all, since no relation between the points and the plane is requested. Your steps ... 0$$\sec A = 2$$using an uncommon Pythagorean Identity\dots$$\sec^2A+\csc^2A=\sec^2A\csc^2A\csc^2A = \frac{\sec^2A}{\sec^2A-1}=4/3\therefore\csc A=2/\sqrt{3}$$This allows us to build this right triangle. from the triangle with opposite, adjacent, and hypotenuse sides 2, 2/\sqrt{3}, 4/\sqrt{3} we have \dots$$\sin A = \frac{\sec A}{\sec A\csc ...

6

$\forall t \in \mathbb{R}$ it holds that $t^2 \geq 0$ and $t^2 = 0 \Leftrightarrow t = 0$. So it follows that if either $x$ or $y$ is not zero, then $2x^2 + 3y^2 > 0$. Therefore $x = y = 0$ so $3x + 2y = 0$

6

Informally: You're taking the sum of the row sums of $\ \ \ \displaystyle{1\over 5^{\phantom 1}}$ $\ \ \ \displaystyle{1\over 5^{ 3}} \ \ \ \ \displaystyle{1\over 5^{ 3}}\ \ \ \ \displaystyle{1\over 5^{ 3}}$ $\ \ \ \displaystyle{1\over 5^{ 5}} \ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ ... 11 We have $$\sum_{k=1}^{n}k^3 = 1 + 8 + 27 + \ldots + n^3 = \\ \underbrace{1}_{1^3} + \underbrace{3+5}_{2^3} + \underbrace{7 + 9 + 11}_{3^3} + \underbrace{13 + 15 + 17 + 19}_{4^3} + \ldots = \\ \underbrace{\underbrace{\underbrace{1}_{1^2} + 3}_{2^2} + 5}_{3^2} + \ldots$$ which is $$\big( \sum_{k=1}^{n}k \big)^2$$ 8 Maybe this will help you visualize it: Source. or this one which is clearer:$\phantom{XXXXXXXX}$8 Let $$f(x)=\sum_{n=0}^\infty(2n+1)x^{2n+1}=x\sum_{n=0}^\infty(2n+1)x^{2n}=x\frac{d}{dx}\left(\sum_{n=0}^\infty x^{2n+1}\right)=x\frac{d}{dx}\left( \frac{x}{1-x^2}\right)=x\frac{x^2+1}{(1-x^2)^2}$$ and notice that the desired sum is$f\left(\frac15\right)$. 20 Hint : Let $$S=\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots\tag1$$ Dividing$(1)$by$5^2$, we obtain $$\frac{S}{5^2}=\frac{1}{5^3}+\frac{3}{5^5}+\frac{5}{5^7}+\frac{7}{5^9}+\frac{9}{5^{11}}+\cdots\tag2$$ Subtracting$(2)$from$(1), we obtain $$S-\frac{S}{5^2}=\frac{1}{5}+\color{blue}{\text{infinite geometric ... 3 We have identities for sums of powers like these. In particular:$$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$The rest is just a bit of arithmetic. 6 I am afraid that what you did is wrong : you are not integrating a polynomial expression. Just for your curiosity,$$\frac{d}{dt}\Big(\frac{e^{t+t^2}}{t^2/2+t^3/3}\Big)=\frac{6 e^{t+t^2} (t+2) \left(4 t^2-3\right)}{t^3 (2 t+3)^2} \neq e^{t+t^2}$$To compute$$I=\int e^{t}.e^{t^2} dt=\int e^{t^2+t} dt$$first complete the square for the exponent and perform a ... 1 It is useful to know at least some basic trigonometric identities. See List of trigonometric identities at Wikipedia for a very complete list. a) Since \cos A=\frac1{\sec A}, you get \cos A=\frac12. Can you get possible values of A from there? b) If$$\cos A=\frac{m^2-n^2}{m^2+n^2}=\left(\frac m{\sqrt{m^2+n^2}}\right)^2-\left(\frac ... 0 Honestly, I suspect that equation b is a red herring. \begin{align} \cos{A} &=\frac{1}{\sec{A}}\\ &=\frac{1}{2} \end{align} \begin{align} \sin{A} &=\sqrt{1-\cos^2{A}}\\ &=\frac{\sqrt{3}}{2} \end{align} \begin{align} \tan{A} &=\frac{\sin{A}}{\cos{A}}\\ &=\sqrt{3} \end{align} \begin{align} \csc{A} &=\frac{1}{\sin{A}}\\ ... 0 for case a) $$\sec A = 2$$ $$\cos A=\frac{1}{\sec A}=\frac{1}{2}$$ $$\sin A=\sqrt{1-\cos^2A}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$$ $$\tan A=\frac{\sin A}{\cos A}=\sqrt 3$$ $$\operatorname{cotg} A=\frac{1}{\tan A}=\frac{1}{\sqrt 3}$$ and for b) follow Martin Sleziak 0 LetS=2/3-2/9+2/27-2/81+\cdots$Let this be equation 1 Multiply, both sides by$\large\frac{-1}{3}\large \frac{-S}{3}=-2/9+2/27-2/81+\cdots$Let this be equation 2 Subtract 2 from 1$\large {S+\frac{S}{3}=\frac {2}{3}}\large \frac {4S}{3}=\large \frac{2}{3}$Or,$S=\large \frac{1}{2}$0 Let the point of intersection betweeen the tangent line and the curve be$P(a,\ 1-a^2)$. The slope of the tangent line is then $$D_x(1-x^2)|_{x=a}=-2a$$ Hence the equation of the tangent line in terms of$a$is $$y=-2ax+1+a^2$$ Using the given information that the line passes through$(0,2)$, we get the equation $$a^2+1=2$$ So $$a=\pm1$$ You want a positive ... 0 At$\left(x_1, \dfrac{2x_1+5}{x_1+2}\right)$, the slope of the curve$y=\dfrac{2x+5}{x+2}$is$\dfrac{-1}{(x_1+2)^2}$. So the equation of a line passing through that point with that slope (in point-slope form) is $$y-\frac{2x_1+5}{x_1+2}=\frac{-1}{(x_1+2)^2}\left(x-x_1\right)$$ If this line has$y$-intercept$2$, then it passes through$(0,2)$. Use that, ... 0 To diagonalize a matrix, only "distinct eigen values" is not enough... We must consider the field over which, we are going to diagonalize. You can easily get an example of a matrix, which has distinct eigen values but "not diagonalizable over Real numbers, But diagonalizable over complex numbers". Also there is a necessary and sufficient condition for ... 0 Hint : Apply determinant in both sides of the equation$STU = I$to get$S,T$and$U$are invertible. Also$I = (STU)^{-1} = U^{-1}T^{-1}S^{-1}$In the last equation substitute$T^{-1} = US$to verify that actually$T^{-1}$should be equal to$US$as inverse is unique. 0 Hints.$T$is invertible if and only if it has rank$n=\dim V$the rank of$ST$is at most equal to the maximum of the ranks of$S$and$T$. 0 To 'find' the number of distinct real roots: You need to think in terms of its graphs. Consider the general shape of the graph of$1+e^{-x}$and$13x^3$and guess the number of times the graph will cut each other. To 'prove' your answer: Show that the function (the L.H.S) is monotone increasing (differentiation?). Also find two points where the function ... 0 Let $$f(x)=13x^{13}-e^{-x}-1$$ then $$f'(x)=169x^{12}+e^{-x}$$ which is positive. Therefore f is strictly ascending. The root of the equation is the point of intersection of the graphs of the functions $$g(x)=13x^{13}$$ and $$h(x)=e^{-x}-1$$ Or observe that$f(0)<0$and$f(1)>0$, since$f$is continuous,$f(x)=0$has a root in$(0,1)$. 1 If you know about the derivative, this can be done directly. Let$f(x) = 13 x^{13} - e^{-x} - 1$; then $$f'(x) = 169 x^{12} + e^{-x}$$ which is always positive; hence$f$is strictly increasing, and there is at most$1$real solution. Finally, since$f(0) < 0$and$f(1) > 0$, the real solution is between$0$and$1$(Wolfram gives about$.844$). ... 0 First of all:$f(x) = 13x^{13}-e^{-x}-1$has at most one real root, because$f'(x) = 169x^{12}+e^{-x} > 0$. It has one real root$0<x_0< 1$by inspection. 1 Restrict$T$to the finite-dimensional subspace of polynomials with degree$\leq d$. This restriction is a bijection, since$T$is injective. Suppose some polynomial of degree$d$is mapped to a polynomial of degree$<d$, then$T$cannot be surjective, since there is a polynomial of degree$<d$also mapped to that polynomial. So$T$maps polynomials of ... 0 A singular matrix is one which is not full rank, meaning that there is at least one row (or column) that is a linear combination of the others, since a matrix represents a linear basis-space. If one of the eigenvalues has been given, then as shown, you can find a second eigenvalue of 4. Since the first two "discovered" eigenvalues are non-zero and the ... 2 "Singular" means that$0$is a third eigenvalue. With three distinct eigenvalues$0,-3,-4$,$P$must be diagonalizable.$P^2 + 3P$is also diagonalizable with respect to the same basis as$P$. 0 Every line parallel to the xz axis has the form$r(t) = (x_0,y_0,z_0) + (a,0,c)t$. Note that,$\vec{v}\cdot \vec{j} = (a,0,c)\cdot (0,1,0) = 0$. Ie, the vector$j$is orthogonal director vector$r$. Therefore,$r$is parallel to the$xz\$ plane.

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