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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: \begin{equation} q(x,t)=\frac{-V_t(1+\delta f[c,g(x)])}{P(x,t)}\left(\frac{dP_o}{dt}\right) \end{equation} substitute into Darcy's relation: $q_x=\frac{-k_x A}{\mu}\frac{dP}{dx}$ \begin{equation} \frac{-V_t(1+\delta f[c,g(x)])}{P(x,t)}P'_o=\frac{-k_x A}{\mu}\frac{dP}{dx} \end{equation} 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

Let $S=2/3-2/9+2/27-2/81+\cdots$ 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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