Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Use stirlings approximation to prove inequality. I have come across this statement in a text on finite elements. I can give you the reference if that will be useful. The text mentions that the inequality follows from Stirling's formula. I can't prove it to myself but I think it is true from checking values with Math... | For
$\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}}
\le 1
$,
$\begin{array}\\
(p-1/2)^{2p-1}
&=p^{2p-1}(1-1/(2p))^{2p-1}\\
&\approx p^{2p-1}(1/e)(1-1/(2p))^{-1}
\quad\text{since }(1-1/(2p))^{2p} \approx 1/e\\
\end{array}
$
so
$\begin{align*}\\
\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}}
&\app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$. | The integral being considered is, and is evaluated as, the following.
\begin{align}
I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\
&= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\
&= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 4
} |
$z^3=w^3 \implies z=w$? I've reached this in another problem I have to solve: $z,w \in \Bbb {C}$. $z^3=w^3 \implies z=w$?
I've scratched my head quite a bit, but I completely forgot how to do this, I don't know if this is correct:
$$
z^3=|z^3|e^{3ix}=|w^3|e^{3iy}
$$
I know the absolute values are equal, so I get $3ix... | We assume $z \ne 0 \ne w$; it is obvious that in the other circumstance $z = w = 0$ is the only possible solution.
Now, since
$z^3 = w^3 \Leftrightarrow z^3 - w^3 = 0, \tag{1}$
then
$(z - w)(z^2 + zw + w^2) = z^3 - w^3 = 0; \tag{2}$
we see that if
$z^2 + zw + w^2 \ne 0, \tag{3}$
then
$z = w; \tag{4}$
furthermore, if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(... | using Bernoulli $$x \to 0 \\ {\color{Red}{(1+ax)^n \approx 1+anx} } \\\sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\ \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4} $$ so by putting them in limit :
$$\lim_{x \to 0} \frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.
I tried to find the mini... | Put $\sqrt{1+\cos x}$ +$\sqrt{1-\cos x} = A$
$A^2 = 2\pm 2 \sin x ,\quad A^2 - 2 =\pm 2 \sin x$
$ -2\leq A^2 - 2\leq 2,\quad -2\leq A\leq2$
So $f(x) = \frac{5 - A^2}{A}$ or $\frac{A^2 + 1}{A}$
Find the minimum and maximum of $f(x)$ in the two conditions with $-2\leq A\leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$? How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?
I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cd... | From your calculation :
$$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$
$$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$
From here, show that both the limits are zero by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$.
For the first lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$ Prove that
$$
\begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix}
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} c \\ 2c \\ \end{pmatrix}
$$
has solution
$$
\begin{pmatrix} x \\ y \\ ... | The fast approach, as I see it is (in case $a \neq 0$):
you have $ax+by = c$ , just plug in $y=0+ta$, and you get
$$ax + bta = c$$
$$x= \frac ca - bt$$.
alternatively, in case you don't have the solution in advance, use same approach and you will get: $$ x = \frac ca - \frac ba y$$, so the general solution will be $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$
My work this far:
we take the left side
$$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac... | Notice, $$LHS=\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}$$
$$=\frac{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}-\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}+\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Binomial expansion. Find the coefficient of $x$ in the expansion of $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$.
I've used the way that my teacher teach me.
I've stuck in somewhere else.
$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6=\left(2-\frac{4}{x^3}\right)\left(x^6\left(1+\frac{2}{... | Notice, we have
$$\left(x+\frac{2}{x^2}\right)^6=^6C_0x^{6}\left(\frac{2}{x^2}\right)^{0}+^6C_1x^{5}\left(\frac{2}{x^2}\right)^{1}+^6C_2x^{4}\left(\frac{2}{x^2}\right)^{2}+^6C_3x^{3}\left(\frac{2}{x^2}\right)^{3}+^6C_4x^{2}\left(\frac{2}{x^2}\right)^{4}+^6C_5x^{1}\left(\frac{2}{x^2}\right)^{5}+^6C_6x^{0}\left(\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\... | Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following:
$$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$
We take the derivative of the numerator and denominator again:
$$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Quadratic formula - check my simplificaiton I am trying to solve this equation using the quadratic formula:
$$x^2 + 4x -1 = 0$$
I start by substituting the values into the quadratic formula:
$$x = {-(4) \pm \sqrt {(4)^2 - 4(1)(-1)} \over 2}$$
which becomes
$$x = {-4 \pm \sqrt{20} \over 2}$$
This is the answer the textb... | Note that
$$\frac{B+C}{A}=\frac{B}{A}+\frac{C}{A}$$
$$x=\frac{-4\pm 2\sqrt{5}}{2}=\frac{-4}{2}\pm\frac{2\sqrt{5}}{2}=-2\pm\sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Algebraic Aproach For this word problem How can the followin question be solved algebraically?
A certain dealership has a total of 100 vehicles consisting of cars and trucks. 1/2 of the cars are used and 1/3 of the trucks are used. If there are 42 used vehicles used altogether, how many trucks are there?
| $
\newcommand{\xu}{x^{\text{used}}}
\newcommand{\yu}{y^{\text{used}}}
\newcommand{\xn}{x^{\text{new}}}
\newcommand{\yn}{y^{\text{new}}}
$
The key in converting text problems into algebraic expressions is to write an expression for every sentence or phrase which contains quantifiable information.
For example, consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}... | Notice, the mid=point of the line joining $(4, 0)$ & $(-6, -3)$ is given as $$\left(\frac{4+(-6)}{2}, \frac{0+(-3)}{2}\right)\equiv \left(-1, -\frac{3}{2}\right)$$ The slope of the perpendicular bisector
$$=\frac{-1}{\text{slope of line joining}\ (4, 0)\ \text{&}\ (-6, -3)}$$
$$=\frac{-1}{\frac{-3-0}{-6-4}}=-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
evaluate the integral Evaluate the integral from:
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$
The way I approach this problem is
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$
and
$$ \text{Res}_{i\sqrt3}(f(z)) ... | Here is another approach:
$$f(a)=\int_0^{\infty} \frac{x \cdot \sin(ax)}{x^2+3}dx$$ take a Laplace transform with respect to $a$ to obtain
\begin{align}
\mathcal{L}(f(a))&=\int_0^{\infty} \frac{x^2}{(x^2+3)(x^2+s^2)}dx\\
&=\frac{\pi}{2\sqrt3+2s}
\end{align}
now take an inverse Laplace to obtain
$$f(a)=\frac{\pi}{2} e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Question related to elliptical angles Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $ be an ellipse and $AB$ be a chord. Elliptical angle of A is $\alpha$ and elliptical angle of B is $\beta$. AB chord cuts the major axis at a point C. Distance of C from center of ellipse is $d$. Then the value of $\tan \frac{\alpha}{2}\tan \... | Notice, we have the coordinates of the points $A$ & $B$ as follows $$A\equiv(a\cos\alpha, b\sin \alpha)$$
$$B\equiv(a\cos\beta, b\sin \beta)$$ Hence, the equation of the chord AB
$$y-b\sin\beta=\frac{b\sin\alpha-b\sin \beta}{a\cos\alpha-a\cos \beta}(x-a\cos \beta)$$
$$y-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| The Cauchy-Schwarz-inequality yields
$$|a + b + c|^2 \le (a^2 + b^2 + c^2)\cdot 3 = 144$$
and therefore $a + b + c \le 12$. Plugging in $a = b = c = 4$ shows that this value is actually the maximum.
Alternatively, you could use the convexity of the function $x \mapsto \sqrt{x}$. By Jensen's inequality we have
$$a + b ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| Or without using L'Hospital, you can do this:
$$
\lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1+\sqrt [ 3 ]{ x } -1 }{ x-1 } } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding roots of the polynomial $x^4+x^3+x^2+x+1$ In general, how could one find the roots of a polynomial like $x^4+x^3+x^2+x^1+1$? I need to find the complex roots of this polynomial and show that $\mathbb{Q (\omega)}$ is its splitting field, but I have no idea of how to proceed in this question. Thanks in advance.
| Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac... | You may observe that: $$\frac{2 \cos (x)+1}{(\cos (x)+2)^2}=\frac{\cos (x)}{\cos (x)+2}+\frac{\sin ^2(x)}{(\cos (x)+2)^2}=\frac{\frac{d}{dx}\sin(x)}{\cos (x)+2}-\frac{\sin(x)\frac{d}{dx}(\cos x +2)}{(\cos (x)+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Area of shaded region circle help
Find the area of the shaded region
Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$
Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$.
Then subtract the former from the latter: $\frac{4\pi}{3} - 2$
Therefore the answer is $~2.189$?
Is this correct?
| Here, aperture angle $\theta=120^\circ=\frac{2\pi}{3}$
Area of shaded portion $$=\text{(area of the sector)}-\text{(area of isosceles triangle)}$$
$$=\frac{1}{2}(\theta)(r^2)-\frac{1}{2}(r^2)\sin\theta$$
$$=\frac{1}{2}\frac{2\pi}{3}(2)^2-\frac{1}{2}(2)^2\sin\frac{2\pi}{3}$$
$$=2.456739397$$
Edit:
Note: Area of an iso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Angle between segments resting against a circle Motivation:
A couple of days ago, when I was solving this question, I had to consider a configuration like this
Now, I didn't intentionally make those two yellow bars stand at what appears to be a $90^{\circ}$-angle but it struck me as an interesting situation, so much s... | Set Cartesian coordinates on the plane so that the center of the circle
is $(0,0)$ and the black line $\overline{AB}$ is parallel to the $x$-axis.
The coordinates of $A$ are $(-(a - b), c)$ and the distance $OA$ is
$\sqrt{(a - b)^2 + c^2}$. Therefore
\begin{align}
\angle OAB & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
two variable nonhomogeneous inequality Let $$x\ge 0,y\ge 0,x\neq 1,y \neq 1$$Prove the inequality
$$\dfrac {x}{(y-1)^2} +\dfrac {y}{(x-1)^2} \ge \dfrac {x+y-1}{(x-1)(y-1)} $$
| hint: The endpoints case you can handle with ease, for more general case that: $x, y > 1\to x(x-1)^2 +y(y-1)^2 \geq (x+y-1)(xy-(x+y-1))\iff x(x^2-2x+1)+y(y^2-2y+1)\geq xy(x+y-1)-(x+y-1)^2\iff (x^3+y^3)-2(x^2+y^2)+(x+y)\geq xy(x+y-1)-(x^2+y^2+1+2xy-2x-2y)\iff f(x,y)=x^3+y^3-(x^2+y^2)+1+3xy-(x+y) -xy(x+y)\geq 0$. Taking ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Division in Summations Suppose $a_n=\dfrac{2^n}{n(n+2)}$ and $b_n=\dfrac{3^n}{5n+18}$.
I need to find the value of:
$$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{b_n}$$
I think this problem is meant for me to compute each sum differently and then divide. Is this some property of summations that we need to utilize here... | $$\begin{equation}
\begin{split}
\sum_{n=1}^\infty \frac{a_n}{b_n}
&
= \sum_{n=1}^\infty \frac{2^n(5n+18)}{3^nn(n+2)} \\
&
= \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{5}{n+2} + 9\left(\frac{1}{n} - \frac{1}{n+2} \right) \right) \right] \\
&
= \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\p... | $\bf{Another\; Solution::}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx.$$
Put $\displaystyle x = \frac{2}{t}\;,$ Then $\displaystyle dx = -\frac{2}{t^2}dt$ and Changing Limits, We get
$$\displaystyle I = \int_{\infty}^{0}\frac{\ln\left(\frac{2}{t}\right)}{\frac{4}{t^2}+\frac{4}{t}+2}\cdot -\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 -... | $$\int \left(e^x\sin^2(x)\right)\text{d}x =$$
$$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$
$$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$
$$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$
$$\frac{1}{2} \left(\int e^x \text{d}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:
How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?
Answer is $2.73$
Simplify into Standard Form:
$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$
$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$
Hyperbola ... | $\frac{(x + 2)^2}{2} - \frac{(y - 1)^2}{1}; C = (-2, 1)$
$C = \sqrt{2 + 1} = \sqrt{3}$
Answer: $1 + \sqrt(3) = 2.73$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first par... | A hint:
Multiply $(1)$ by $x$, $(2)$ by $y$, and $(3)$ by $z$ and look at the three equations you got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)... | From $abc(a+b+c)=3$ we have $$a^2+a(b+c)=\frac {3}{bc}.$$ Therefore we have $$(a+b)(b+c)(c+a)\geq 8\iff$$ $$(a+b)(a+c)\geq \frac {8}{b+c}\iff$$ $$a^2+a(b+c)\geq -bc+ \frac {8}{b+c)}\iff$$ $$(\bullet ) \quad \frac {3}{bc}\geq -bc+\frac {8}{b+c}.$$ Now $b+c\geq 2\sqrt {bc},$ so $$\frac {8}{b+c}\leq \frac {4}{\sqrt {bc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Is the following solvable for x? I have the following equation and I was wondering if I can solve for x given that it appears both as an exponent and a base:
$[\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}-0.5\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}.\frac{4.(\frac{x-N}{\sqrt {2}.D})}{\sqrt {\pi}.e^{-\fra... | Let's simplify the equation a bit so we can better see its form:
$$C_{1}e^{X_{1}} + C_{2}e^{X_{1}}\cdot \frac{X_{2}}{C_{3}e^{X_{3}} \sqrt{C_{4}e^{X_{3}} + C_{5}X_{3}}} + C_{6}e^{X_{4}} + C_{7}e^{X_{4}}\cdot \frac{X_{5}}{C_{8}e^{X_{6}} \sqrt{C_{9}e^{X_{6}} + C_{10}X_{6}}} = C_{11}$$
With constants as $C_{n}$ above and e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$
MyAttempt
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$
I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sig... | $\displaystyle A \sin \ x + B \cos \ x = \sqrt{A^2 + B^2}\sin( x + \phi )$, where $\displaystyle \cos \phi = \frac{A}{\sqrt{A^2 + B^2}}$.
In our case $\displaystyle \cos\phi=\frac{1}{\sqrt{3}} \Rightarrow \phi\in\left(0;\frac{\pi}{2}\right)$.
Thus: $\displaystyle \begin{aligned}\int\limits_0^\pi\left|\sqrt{2}\sin x+2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different ... | Let, $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt x)}{x^2-x+1}dx\tag 1$$
Now, using property of definite integral $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get
$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{(1-x)^2-(1-x)+1}dx$$
$$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx\tag 2$$ Now, adding (1) & (2), we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Pleas... | The integral can also be found using a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$.
Here we see that $x = \dfrac{1 - u}{1 + u}$ such that $dx = -\dfrac{2}{(1 + u)^2} \, du$.
Writing the integral as
$$I = -\int \frac{1 - x}{(1 + x) x \sqrt{x + 1 + \frac{1}{x}}} \, dx,$$
if we observe that
$$x + 1 + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Please note that you have: $\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$, which on dividing numerator and denominator by $x^2$ becomes:
$\int\frac{1+1/x^2}{(x^2+1/x^2)+3+3(x-1/x)}dx$=$\int\frac{1+1/x^2}{(x-1/x)^2+5+3(x-1/x)}dx$
Now put x-1/x =t so that (1+1/$x^2$)dx=dt and thus you get:
$\int\frac{1}{t^2+5+3t}dt$;which can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$
Let the center of two circles be $O$ and $O'$ and ... | Let $A$ & $B$ be the centers of the circles with radii $a$ & $b$ respectively such that they have a common chord $MN=2x$ & intersecting each other at an angle $\theta$.
Let $O$ be the mid-point of common chord $MN$. Point $O$ lies on the line AB joining the centers of circle then we have $$MO=ON=x$$
In right $\triangl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I ... | $\begin{array}\\
f(x)
&=\frac{1}{1+e^x}\\
&=\frac{1}{2+(e^x-1)}\\
&=\frac12\frac{1}{1+(e^x-1)/2}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(e^x-1)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(x+x^2/2+x^3/6+...)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+x/2+x^2/6+...)^n\\
&=\frac12\sum_{n=0}^{\infty}(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$
Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?
| HINT
$$
\sqrt{ x + \sqrt{ x^2 + 1 } }
= \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } }
$$
That would be enough simple to solve the integral...
We get
$$
\begin{eqnarray}
\int \sqrt{ x + \sqrt{ x^2 + 1 } } dx
&=& \int \left\{ \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } }
\right\} d x\\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\left|(|x|-|y|)\right|\leq|x-y|$
Prove that $\left|(|x|-|y|)\right|\leq|x-y|$
Proof:
$$\begin{align} \left|(|x|-|y|)\right| &\leq|x-y| \\
{\left|\sqrt{x^2}-\sqrt{y^2}\right|}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$}\\
\sqrt{\left(\sqrt{x^2}-\sqrt{y^2}\right)^2}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{... | Because of the triangle inequality, there is: $\left| x+y \right| \le \left| x \right| +\left| y \right| $. Using this fact:
\begin{align}
\left| x \right| &= \left| (x-y)+y \right| \\
&\le \left| x-y \right| +\left| y \right| \\
\left| x \right| -\left| y \right| &\le \left| x-y \right| \tag{1}
\end{align}
Proceedin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
The number of ordered pairs of positive integers $(a,b)$ such that LCM of a and b is $2^{3}5^{7}11^{13}$ I started by taking two numbers such as $2^{2}5^{7}11^{13}$ and $2^{3}5^{7}11^{13}$. The LCM of those two numbers is $2^{3}5^{7}11^{13}$. Similarly, If I take two numbers like $2^{3-x}5^{7-y}11^{13-z}$ and $2^{3}5^{... | Let $a = 2^{x_{1}}\cdot 5^{y_{1}}\cdot 11^{z_{1}}$ and $b = 2^{x_{2}}\cdot 5^{y_{2}}\cdot 11^{z_{2}}\;,$ Then Given $\bf{LCM(a,b)} = 2^{3}\cdot 5^{7}\cdot 11^{13}$
So Here $0\leq x_{1},x_{2}\leq 3$ and $0\leq y_{1},y_{2}\leq 7$ and $0\leq z_{1},z_{2}\leq 13.$
Here ordered pairs of $(x_{1},x_{2}) = \left\{(0,3),(1,3),(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does $\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$ diverge? Does the limit of this summation diverge?
$$\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$$
Thanks!
| $$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln k}}{{1 + {{(\frac{k}{n})}^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln (\frac{k}{n})}}{{1 + {{(\frac{k}{n})}^2}}}} + \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln n}}{{1 + {{(\frac{k}{n})}^2}}}} $$
When $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$ Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about
$$\approx 111.024457130115028409990464833072173251135063166330638343951498... | $$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$
By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}... | You just need to expand the square:
$$
\int_{4}^{8}{\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}dx}=\int_{4}^{8}{\sqrt{1+\frac{x^4-8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\frac{x^4+8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\left(\frac{x^2+4}{4x}\right)^2}dx}=\int_{4}^{8}{\frac{x^2+4}{4x}dx}=\int_{4}^{8}{\frac{x}{4}+\frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
calculating the characteristic polynomial I have the following matrix: $$A=\begin{pmatrix}
-9 & 7 & 4 \\
-9 & 7 & 5\\
-8 & 6 & 2
\end{pmatrix}$$
And I need to find the characteristic polynomial so I use det(xI-A) which is $$\begin{vmatrix}
x+9 & -7 & -4 \\
9 & x-7 & -5\\
8 & -6 & x-2
\end{vmatri... | You could use $\displaystyle\begin{vmatrix}x+9&-7&-4\\9&x-7&-5\\8&-6&x-2\end{vmatrix}=\begin{vmatrix}x+2&-7&-4\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$ $\;\;\;$(adding C2 to C1)
$\displaystyle\hspace{2.6 in}=\begin{vmatrix}0&-x&1\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$$\;\;\;$(subtracting R2 from R1)
$\displaystyle\hspace{2.6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can the trigonometric equation be proven? This question :
https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle
has the answer $10°$. This follows from the equation
$$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$
which is indeed true , which I checked with Wolf... | $$\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Inequality problem: Application of Cauchy-Schwarz inequality Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that:
$$
\sqrt {a-1} + \sqrt {b-1} + \sqrt {c-1} \leq \sqrt {a+b+c}.
$$
This is supposed to be solved using the Cauchy inequality; that is, the scalar product inequality... | We apply the Cauchy Schwarz Inequality to the vectors $x = \left({\sqrt {\dfrac{a-1}{a}} , \sqrt {\dfrac{b-1}{b}} , \sqrt {\dfrac{c-1}{c}}}\right) $ and $y = \left({\dfrac{1}{\sqrt{bc}},\dfrac{1}{\sqrt{ac}}, \dfrac{1}{\sqrt{ab}} }\right)$ in $\Bbb R^3$.
Then, $x \cdot y \le \lVert x\rVert \lVert y\rVert$ yields,
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the closed form of the Euler Sums We know that the closed form of the series
$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}\left( {\begin{array}{*{20}{c}}
{2n} \\
n \\
\end{array}} \right)}}} = \frac{1}{3}\zeta \left( 2 \right).$$
but how to evaluate the following series
$$\sum\limits_{n = 1}^... | Using the main result proved in this question,
$$ \sum_{n\geq 1}\frac{x^n}{n^2 \binom{2n}{n}}=2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)\tag{1}$$
it follows that:
$$ \sum_{n\geq 1}\frac{H_n}{n^2 \binom{2n}{n}}=\int_{0}^{1}\frac{2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)-\frac{\pi^2}{18}}{x-1}\,dx \tag{2}$$
and:
$$ \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| If you wish a more direct application of induction, we have that if
$$
1^2+2^2+\cdots+(n-1)^2 < \frac{n^3}{3}
$$
then
$$
1^2+2^2+\cdots+n^2 < \frac{n^3+3n^2}{3}
< \frac{n^3+3n^2+3n+1}{3} = \frac{(n+1)^3}{3}
$$
Similarly, if
$$
1^2+2^2+\cdots+n^2 > \frac{n^3}{3}
$$
then
$$
1^2+2^2+\cdots+(n+1)^2 > \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Largest Number that cannot be expressed as 6nm +- n +- m I'm looking to find out if there is a largest integer that cannot be written as $6nm \pm n \pm m$ for $n,m$ elements of the natural numbers. For example, there are no values of $n,$m for which $6nm \pm n \pm m = 17.$ Other numbers which have no solution are $25... | This question is equivalent to the Twin Prime Conjecture.
Let's look at the three cases:
$$6nm+n+m =\frac{(6n+1)(6m+1)-1}{6}\\
6nm+n-m = \frac{(6n-1)(6m+1)+1}{6}\\
6nm-n-m = \frac{(6n-1)(6m-1)-1}{6}
$$
So if $N$ is of one of these forms, then:
$$6N+1=(6n+1)(6m+1)\\
6N-1=(6n-1)(6m+1)\\
6N+1=(6n-1)(6m-1)$$
If $6N-1$ and ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}
$$
The result should be a number.
I try this:
$$
\frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sq... | $(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} )^3 \\
=(\sqrt[3]{2 + \sqrt{5}})^3+(\sqrt[3]{2 - \sqrt{5}} )^3+3(\sqrt[3]{2 + \sqrt{5}} ) (\sqrt[3]{2 - \sqrt{5}} )(\sqrt[3]{2 + \sqrt{5}} +\sqrt[3]{2 - \sqrt{5}} ) $
S0 $s^3=4-3s$ From this we get S = 1
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\sum\limits_{n=1}^{\infty}\frac{n^4}{4^n}$ So, yes, I could not do anything except observing that in the denominators, there is a geometric progression and in the numerator, $1^4+2^4+3^4+\cdots$.
Edit: I don't want the proof of it for divergence or convergence only the sum.
| \begin{align}
n^4 = {} & \hphantom{{}+{}} An(n-1)(n-2)(n-3) \\
& {} + B n(n-1)(n-2) \\
& {} + C n(n-1) \\
& {} + D n
\end{align}
Find $A,B,C,D$. Then
\begin{align}
n^4 x^n & = An(n-1)(n-2)(n-3) x^n + \cdots \\[10pt]
& = A\frac{d^4}{dx^4} x^n + B \frac{d^3}{dx^3} x^n + \cdots
\end{align}
Then add over all values of $n$... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$ Find $x$ from logarithmic equation:
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$
This is how I tried:
$$x^2-8x+16>0$$
$$ (x-4)^2>0 \implies x \not = 4$$
then
$$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$
because of base $\frac{x}{... | Given $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\;,$$ Here function is defined when $\displaystyle \frac{x}{5}>0$ and $\displaystyle \frac{x}{5}\neq 1$
and $(x-4)^2>0$. So we get $x>0$ and $x\neq 5$ and $x\neq 4$
If $$\displaystyle \; \bullet\; \frac{x}{5}>1\Rightarrow x>5\;,$$ Then $$\displaystyle \log_{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422858",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\... | Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int \sqrt{2-x-x^2}\cdot \frac{1}{x^2}dx\;, $$ Now Using Integration by parts
$$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\int\frac{1+2x}{2\sqrt{2-x-x^2}}\cdot \frac{1}{x}dx $$
So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\underbrace{\int\frac{1}{\sqrt{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Determining values of a coefficient for which a system is and isn't consistent. Given the system : \begin{array}{ccccrcc}
x & + & 2y & + & z & = & 3 \\
x & + & 3y & - & z & = & 1 \\
x & + & 2y & + & (a^2-8)z & = & a
\end{array}
Find values of $a$ such that the system has a unique solution, infinitely many solutions, or... | What strikes me immediately
is that if
$a^2-8 = 1$,
the first and last equations
have the same LHS.
Since choosing $a=3$
makes these equations
identical,
this becomes only two equations
in three unknowns,
so there are an
infinite number of solutions.
On the other hand,
if you choose
$a=-3$,
then the first and third equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the $n$th derivative of trigonometric function.. My maths teacher has asked me to find the $n$th derivative of $\cos^9(x)$. He gave us a hint which are as follows:
if $t=\cos x + i\sin x$,
$1/t=\cos x - i\sin x$,
then $2\cos x=(t+1/t)$.
How am I supposed to solve this? Please help me with explanations bec... | De Moivre taught us that if $t=\cos x + i\sin x$ then $t^n = \cos(nx) + i\sin(nx)$ and $t^{-n} = \cos(nx) - i\sin(nx)$ so
$$
t^n + \frac 1 {t^n} = 2\cos(nx).
$$
Then, letting $s=1/t$, we have
\begin{align}
& (2\cos x)^9 =(t+s)^9 \\[10pt]
= {} & t^9 + 9t^8 s + 36t^7 s^2 + 84 t^6 s^3 + 126 t^5 s^4 + 126 t^4 s^5 + 84 t^3 ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find the sum of the following series to n terms $\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$ Find the sum of the following series to n terms $$\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$$
My attempt:
$$T_{n}=\frac{n^2}{(2n-1)(2n+1)}$$
I am unable to represent to proceed furthe... | The $n$th term is $$n^2/(4n^2-1) =$$ $$ \frac{1}{4}.\frac {(4n^2-1)+1} {4n^2-1}=$$ $$\frac{1}{4} + \frac{1}{4}.\frac {1}{4n^2-1}=$$ $$\frac{1}{4}+ \frac {1}{4}. \left(\frac {1/2}{2n-1}- \frac {1/2}{2n+1}\right).$$ Is this enough?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Inequation: quadratic difference equations Given:
$$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$
Step 1:
$$(x + 3)(x - 3) > (x + 4)(x - 4)$$
Step2 : Solving step 1:
$$x^2 - 3^2 > x^2 - 4^2$$
*Step 3:
$ 0 > -16 + 9$ ???
As you see, I can delete the $x^2$, but there is no point in doing that.
What should be the nex... | What you did in step 1 amounts to multiply both sides by $(x-4)(x+3)$. Unfortunately, you have to reverse the inequation if this expression is negative, and leave it as is if it is positive. And as you don't know the sign of this product…
You can simplify solving this inequation writing both sides in canonical form:
$$... | {
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"source": "stackexchange",
"question_score": "1",
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Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28 This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents.
Prove that if an integer $a$ is both a square and a cube then $a \equiv 0,1,9, \textrm{ or } 28 (\textrm{ mod}\ 36)$
An outline of ... | What you did is correct, but yes, a lot of the work (especially the computer check) could have been avoided.
Firstly, if $a$ is both a square and a cube, then it is a sixth power.
This is because, for any prime $p$, $p$ divides $a$ an even number of times (since it is a square), and a multiple of 3 number of times (sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $
My attempt to solve this is as follows:
$ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{... | Your calculus is correct.
A first family of characteristic curves comes from $\frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}$ which solution is $z=\frac{1}{c_1}y=c'_1y$
$$\frac{z}{y}=c'_1$$
A second family of characteristic curves comes from
$$\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$$
The solution of this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - ver... | Can you help with O-symbols? It's all right here?
$$f(x) = \sqrt[3]{3x^2}\left(1 + \frac{4}{9x} + O\left(\frac{1}{x^2}\right) - 1 - \frac{1}{x} -O \left(\frac{1}{x^2}\right)\right)= \sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right). $$
Hence
$$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $g$ is continuous at $x=0$ Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?.
My work: Let $\epsilon >0$ be given.... | Hint. See my comment to your post.
After doing your algebra, as you have shown,
$$\left|\dfrac{1}{1-x}+1-2\right| = \left|x\right|\left|\dfrac{1}{1-x}\right|\text{.}$$
We have $|x| < \delta$. Let $\delta = 1/2$, then
$$|x| < \delta \implies -\delta < x < \delta \implies 1-\delta < 1-x < 1 + \delta \implies \dfrac{1}{1... | {
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"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$
I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks
| Here is what I got from Wolfram Alpha, condensed a little bit
$$\int tan^6(x)sec^3(x)dx$$
$$= \int (sec^2-1)^3sec^3(x)dx$$
$$= \int \Big(sec^9(x) -3sec^7(x)+3sec^5(x)-sec^3(x)\Big)$$
$$= \int sec^9(x)dx -3\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$
Since $\int sec^m(x) = \frac{sin(x)sec^{m-1}(x)}{m-1} + \frac{m-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435755",
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"source": "stackexchange",
"question_score": "1",
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Assume this equation has distinct roots. Prove $k = -1/2$ without using Vieta's formulas. Given $(1-2k)x^2 - (3k+4)x + 2 = 0$ for some $k \in \mathbb{R}\setminus\{1/2\}$, suppose $x_1$ and $x_2$ are distinct roots of the equation such that $x_1 x_2 = 1$.
Without using Vieta's formulas, how can we show $k = -1/2$ ?
Here... | The problem is find $k$ not $x_1$ isn't it? If so take the equation you arrived at before you started doing computer algebra:
$$(1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$$
and multiply by $x_1$ to get
$$(1-2k)x_1^2 - (3k+4) x_1 + (1 -2k) = 0$$
but you know that
$$(1-2k)x_1^2 - (3k+4) x_1 + 2 = 0$$
so you must have $1 - 2k = ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Application of Differentiation Prove that if the curve $y = x^3 + px + q$ is tangent to the x-axis, then
$$4p^3 + 27q^2 = 0$$
I differentiated $y$ and obtained the value $3x^2 + p$. If the curve is tangent to the x-axis, it implies that $x=0$ (or is it $y = 0$?). How do I continue to prove the above statement? Thanks.... | Notice, we have $$y=x^3+px+q$$ $$\frac{dy}{dx}=3x^2+p$$ Since, the x-axis is tangent to the curve at some point where $y=0$ & slope $\frac{dy}{dx}=0$ hence, we have $$\left(\frac{dy}{dx}\right)_{y=0}=0$$ $$3x^2+p=0\iff x^2=\frac{-p}{3}\tag 1$$
Now, at the point of tangency with the x-axis we have $$y=0\iff x^3+px+q=0$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$ I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square bo... | Note that when you square something like $a\sqrt{b}$ you get $a^2b$.
Thus, you should get:
$\begin{align}
x^2-6x+9 &= 64(x-10)\\
x^2-6x+9 &= 64x-640\\
x^2-70x+649 &= 0\\
(x-11)(x-59) &= 0\\
\therefore \boxed{x=11,59}.
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluation of $\int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$
Evaluation of $\displaystyle \int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$
$\bf{My\; Try::}$ Let $$I= \int_{0}^{\infty}\frac{1}{\left[x^4+(1+... | I think your way of evaluating the integral is very short and elegant. Nevertheless, you can do partial fraction decomposition. You'll end up with (as long as I did not do any mistakes)
$$
\frac{1}{\sqrt{2+a}}\biggl[\arctan\Bigl(\frac{\sqrt{2-a}+2x}{\sqrt{2+a}}\Bigr)+\arctan\Bigl(\frac{\sqrt{2-a}-2x}{\sqrt{2+a}}\Bigr)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate limit without L'hopital $$\lim_{x\to2}\dfrac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}$$
Please help me evaluate this limit. I have tried rationalising it but it just can't work, I keep ending up with $0$ at the denominator... Thanks!
| $$\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}\times \frac{\sqrt{x^2+1}+\sqrt{2x+1}}{\sqrt{x^2+1}+\sqrt{2x+1}}\times \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^3-x^2}+\sqrt{x+2}}$$
So we get $$\displaystyle \lim_{x\rightarrow 2}\left[\frac{x^2-2x}{x^3-x^2-x-2}\times \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int \frac{\sin x}{\sin 4x}\mathrm dx$ $$\int \frac{\sin x}{\sin 4x}\mathrm dx$$
I have tried to do this by expanding $\sin 4x$ but it was worthless.
| Using $\displaystyle \bullet\; \sin 2x = 2\sin x\cdot \cos x$ and $\; \bullet\; \cos 2x = 1-2\sin^2 x$
Let $$\displaystyle I = \int\frac{\sin x}{2\sin 2x\cdot \cos 2x}dx = \int\frac{\sin x}{4\sin x\cos x\cdot \cos 2x}dx = \frac{1}{4}\int\frac{1}{\cos x\cdot \cos 2x}dx$$
Now $$\displaystyle I = \frac{1}{4}\int\frac{\cos... | {
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"url": "https://math.stackexchange.com/questions/1445491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}... | I think you overlooked this multiplication $(1+(4-3x)^{1/3})(1-(4-3x)^{1/3})$ which equals $1-(4-3x)^{2/3}$ not $1-(4-3x)$
| {
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"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Integral $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$ Here is an integral I derived while evaluating another. It appears to be rather tough, but some here may not be so challenged :)
Show that:
$$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\frac{11}{9}\zeta(3)-\frac{\pi}{72\sqrt{3}}\left(5\psi_{1}\left(\frac13\righ... | Asset at our disposal: $$\sum\limits_{n=0}^{\infty} \frac{x^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}} = 4(\arcsin (x/2))^2$$
Differentiation followed by the substitution $x \to \sqrt{x}$ gives:
$\displaystyle \sum\limits_{n=0}^{\infty} \frac{x^{n}}{(2n+1)\binom{2n}{n}} = \frac{2\arcsin (\sqrt{x}/2)}{\sqrt{x}\sqrt{1-(\sqrt{x}/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Tile a 1 x n walkway with 4 different types of tiles... Suppose you are trying to tile a 1 x n walkway with 4 different types of tiles: a red 1 x 1 tile, a blue 1 x 1 tile, a white 1 x 1 tile, and a black 2 x 1 tile
a. Set up and explain a recurrence relation for the number of different tilings for a sidewalk of lengt... | Call the number of tilings of length $n$ $t_n$, then to get a tiling of length $n$, you take one of length $n - 1$ and add a red, a white or a blue tile (3 ways); add a black tile to one of length $n - 2$. I.e.:
$\begin{equation*}
t_{n + 2}
= 3 t_{n + 1} + t_n
\end{equation*}$
Directly we find $t_0 = 1$, $t_1 = 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x^2+y^2+xy=1\;,$ Then minimum and maximum value of $x^3y+xy^3+4\;,$ where $x,y\in \mathbb{R}$
If $x,y\in \mathbb{R}$ and $x^2+y^2+xy=1\;,$ Then Minimum and Maximum value of $x^3y+xy^3+4$
$\bf{My\; Try::} $Given $$x^2+y^2+xy=1\Rightarrow x^2+y^2=1-xy\geq 0$$
So we get $$xy\leq 1\;\;\forall x\in \mathbb{R}$$
and ... | Using your second last line,
$$f(x,y) = \frac{17}{4} - (xy-\frac 12)^2 $$
now let $\displaystyle xy=u$,
$x^2 + y^2 + xy = 1$ becomes $(x+y)^2 = 1+u$
Therefore $x,y$ are roots of the quadratic $k^2 \pm \sqrt(1+u) k + u = 0.$
If $x, y$ are real, discriminant is non negative, solving this gets $\displaystyle u\leq \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A non-linear homogeneous diophantine equation of order 3 I'm a math teacher and one of my student have come to me with several questions.
One of them is the following;
Prove that there is no positive integral solution of the equation
$$x^2y^4+4x^2y^2z^2+x^2z^4=x^4y^2+y^2z^4.$$
I have tried several hours but failed.
Hel... | Let $f(x,y,z) = xy(x+y) +z^2(x-y)$. As you already have in the post, we need to prove that $f(x,y,z)=0$ implies $xyz=0$ over integers.
First, rewrite $f(x,y,z) = y x^2 + (y^2+z^2) x - z^2 y$ and consider $f(x,y,z)=0$ as a quadratic equation in $x$. Then the discriminant of the quadratic equation is
$$
D=(y^2 + z^2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Stuck solving a logarithmic equation $$\log _{ 2 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } -\log _{ 2 }{ 2x } $$
Steps I took:
$$\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } =\log _{ 4 }{ 4x^{ 6 } } -\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } $$
$$2\log _{ 4 }{ 2x } +2\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
$$... | $$
\log_{2} 2x = \log_{4} 4x^{6} - \log_{2}2x \quad \text{iff} \quad \frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2};\\
\frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2} \quad \text{iff} \quad 2\log 2x = \log 4x^{6} - 2\log 2x;\\
2\log 2x = \log 4x^{6} - 2\log 2x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ ... | Notice, $\ \ \large \log_{a^n}(b)=\frac{1}{n}\log_a(b)$
Now, we have $$2\log_8x=\log_2x-1$$
$$2\log_{2^3}x=\log_2x-1$$
$$\frac{2}{3}\log_{2}x=\log_2x-1$$ $$\frac{1}{3}\log_{2}x=1$$
$$\log_2x=3\implies x=2^3=\color{red}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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The possible number of blue marbles is A boy has a collection of blue and green marbles. The number of blue marbles belong to the sets $\{2,3,4,\ldots,13\}$. If two marbles are chosen simultaneously and at random from this collection, then the probability that they have different colour is $\frac{1}{2}$. The possible n... | Suppose, there are $m$ blue and $n$ green marbles.
There are $\binom{m+n}{2}=\frac{(m+n)(m+n-1)}{2}$ ways to choose $2$ marbles.
There are $mn$ ways to choose $2$ marbles with different colors.
The probability of getting two marbles with different colours is therefore
$$\frac{2mn}{(m+n)(m+n-1)}$$
So, the probability is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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An example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge Give an example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge conditionally.
I've come up with an example.
$\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\... | Consider , $a_n=\frac{(-1)^n}{n}$. Then $a_n$ is conditionally convergent. But , $a_n^3=\frac{(-1)^n}{n^3}$ is NOT conditionally convergent ; as it is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is the name for the mathematical property involving addition or subtraction of fractions over a common denominator? For any number $x$ where $x\in\Bbb R$ and where $x\ne0$, what is the mathematical property which states that:
$${1-x^2\over x} = {1\over x} - {x^2\over x}$$
| Just the computation rules for fractions:
$$
\frac{a - b}{c} = \frac{a + (-b)}{c} \\
\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c} \\
\frac{(-b)}{c} = -\frac{b}{c}
$$
for $a = 1$, $b = -x^2$ and $c = x \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is $\frac{4x^2 - 1}{3x^2}$ an abundancy outlaw? Let $\sigma(x)$ denote the sum of the divisors of $x$. For example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$.
We call the ratio $I(x) = \sigma(x)/x$ the abundancy index of $x$. A number $y$ which fails to be in the image of the map $I$ is said to be an abundancy outlaw.
My ques... | A partial answer.
Let $I(y)=\frac{4x^2 - 1}{3x^2}$.
If $\gcd(4x^2 - 1, 3x^2) = 1$ then $3$ is a factor of $y$ contradicting the fact that $I(y)<\frac{4}{3}$.
Therefore $\gcd(4x^2 - 1, 3x^2) = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Incorrect General Statement for Modulus Inequalities $$|x + 1| = x$$
This, quite evidently, has no solution.
Through solving many inequalities, I came to the conclusion that,
If,
$$|f(x)| = g(x)$$
Then,
$$f(x) = ±g(x)$$
And this was quite successful in solving many inequalities. However, applying the above to this pa... | The definition of the absolute value is
$$
\lvert x \rvert =
\begin{cases}
x & \text{if $x\geqslant 0$} \\
-x & \text{if $x < 0$},
\end{cases}
$$
so
$$
\lvert f(x) \rvert = g(x)
\iff
\begin{cases}
f(x) = g(x) \\
f(x) \geqslant 0
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$ How to find the solution of this trigonometric equation
$$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$$
I have used the formulae $$\cos(x+60^\circ)\cos(x-60^\circ)=\cos^2 60^\circ - \sin^2x$$
$$\sin(x+60^\circ)\sin(x-60^\circ)=\sin^2x-\sin^260^\cir... | $$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$$
$$\frac{\cot(x+110^\circ)}{\cot x} = \cot(x+60^\circ)\cdot \cot(x-60^\circ)$$
$$\frac{\cos(x+110^\circ)\cdot \sin x}{\sin (x+110^\circ)\cdot \cos x} = \frac{\cos(x+60^\circ)\cdot \cos(x-60^\circ)}{\sin(x+60^\circ)\cdot \sin(x-60^\circ)}$$
Now Using Componendo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many roots are rational?
If $P(x) = x^3 + x^2 + x + \frac{1}{3}$, how many roots are rational?
EDIT:
$3x^3 + 3x^2 + 3x + 1 = 0$, if any rat roots then,
$x = \pm \frac{1}{1, 3} = \frac{-1}{3}, \frac{1}{3}$, and none of these work. Complete?
| The polynomial $3P(x) = 3x^3 + 3x^2 + 3x + 1$ has integer coefficients. So, if $p/q$ with $\operatorname{gcd}(p,q) = 1$ is a rational root, necessary $p$ divides the constant term $1$, and $q$ divides the dominant coefficient $3$. So we have four candidates $-1$, $-1/3$, $1/3$ and $1/3$. Clearly, the roots of $P(x)$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first ine... | I just want to add another way to solve the problem.
$$\begin{align} & a^4 + b^4 + c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \\& \ge a^2 b^2 + b^2c^2 + c^2a^2 \text{ (From Cauchy-Schwarz)}\\&= (ab)^2 + (bc)^2 + (ca)^2 \\ & \ge abbc + bcca + caab\ (\because a^2 + b^2 + c^2 \ge ab + bc + ac) \\ &= abc(a+b+c), \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular. Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular.
Let the equation to the tangent be $$y=mx+\sqrt{a^2m^2+b... | Let $(x_1,y_1)$ be the generic point of the locus.
It is well known that $$\frac {xx_1}{a^2}+\frac {yy_1}{b^2}=1$$ represents the line passing through the points of contact of the tangents from $(x_1,y_1)$.
Now consider the equation $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\left(\frac {xx_1}{a^2}+\frac {yy_1}{b^2}\right)^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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representation of a complex number in polar from
Write the following in polar form: $\frac{1+\sqrt{3}i}{1-\sqrt{3}i}$
$$\frac{1+\sqrt{3}i}{1-\sqrt{3}i}=\frac{1+\sqrt{3}i}{1-\sqrt{3}i}\cdot\frac{1+\sqrt{3}i}{1+\sqrt{3}i}=\frac{(1+\sqrt{3}i)^2}{1^2+(\sqrt{3})^2}=\frac{1+2\sqrt{3}i-3}{4}=\frac{-2+2\sqrt{3}i}{4}=-\frac{1... | HINT:
$$\frac{1+i\sqrt{3}}{1-i\sqrt{3}}=$$
$$\left|\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right|e^{\arg\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)i}=$$
$$\frac{|1+i\sqrt{3}|}{|1-i\sqrt{3}|}e^{\left(\arg\left(1+i\sqrt{3}\right)-\arg\left(1-i\sqrt{3}\right)\right)i}=$$
$$\frac{\sqrt{1^2+\left(\sqrt{3}\right)^2}}{\sqrt{1^2+\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489339",
"timestamp": "2023-03-29T00:00:00",
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Taylor Series Expansion of $ f(x) = \sqrt{x} $ around $ a = 4 $ guys.
Here's the exercise: find a series representation for the function $ f(x) = \sqrt{x} $ around $ a = 4 $ and find it's radius of convergence.
My doubt is on the first part: I can't seem to find a pattern.
$ \circ f(x) = \sqrt x \rightarrow f(4) = 2 \\... | You were on the right track. We have for $n\ge 2$
$$\begin{align}
f^{(n)}(x)&=(-1)^{n-1}\frac12 \frac12 \frac32 \frac52 \cdots \frac{2n-3}{2}x^{-(2n-1)/2}\\\\
&=(-1)^{n-1}\frac{(2n-3)!!}{2^n}x^{-(2n-1)/2}\tag 1
\end{align}$$
where $(2n-3)!! = 1\cdot 3\cdot 5\cdot (2n-3)$ is the double factorial of $2n-3$. Evaluating ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $2^{3^{100}}$ (mod 5) and its last digit The question is to find $2^{3^{100}} \pmod 5$ and its last digit.
I think we have to find $2 \pmod 5$ and $3^{100}\pmod 5$ separately, right?
$$2 = 2 \pmod 5$$
$$3^4 = 1 \pmod 5$$
$$3^{100} = 1 \pmod 5$$
$$2^{3^{100}} = 2^1 = 2 \pmod 5$$
Is this solution correct? And to fin... | Yes, you solve modulo $10$ for the last digit. Note that the number is even, so its last digit is taken from $2, 5, 6, 8$--it's not divisible by $5$ so the last digit cannot be $0$. Your original solution is a bit off, though $3^4\equiv 1\mod 5$, but in the exponent it's every $4$ which gives $1$, so we want $3^2\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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(Combinatorial) proof of an identity of McKay Lemma 2.1 of this paper claims that for integer $s>0$ and $v \in \mathbb{N}$, we have
$$
\sum_{k=1}^s \binom{2s-k}{s} \frac{k}{2s-k} v^k (v-1)^{s-k}
= v \sum_{k=0}^{s-1} \binom{2s}{k} \frac{s-k}{s} (v-1)^k
$$
The author gives a combinatorial interpretation of the left hand ... | Suppose we seek to verify that
$$\sum_{k=1}^n {2n-k\choose n} \frac{k}{2n-k} v^k (v-1)^{n-k}
= v \sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n} (v-1)^k.$$
Now the LHS is
$$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-1} k v^k (v-1)^{n-k}.$$
Re-write this as
$$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-k} k v^k (v-1)^{n-k}.$$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Square root of complex number. The complex number $z$ is defined by $z=\frac{9\sqrt3+9i}{\sqrt{3}-i}$. Find the two square roots of $z$, giving your answers in the form $re^{i\theta}$, where $r>0$ and $-\pi <\theta\leq\pi$
I got the $z=9e^{\frac{\pi}{3}i}$. So I square root it, it becomes $3e^{\frac{\pi}{6}i}$. But the... | Notice, $$z=\frac{9\sqrt 3+9i}{\sqrt 3-i}$$ $$=\frac{9(\sqrt 3+i)(\sqrt 3+i)}{(\sqrt 3-i)(\sqrt 3+i)}$$
$$=\frac{9(\sqrt 3+i)^2}{3-i^2}=\frac{9(2+2i\sqrt 3)}{3+1}$$$$=9\left(\frac{1}{2}+i\frac{\sqrt 3}{2}\right)=9\left(\cos\frac{\pi}{3}+i\sin \frac{\pi}{3}\right)=9e^{i\pi/3}$$
hence, the square roots of $z$ are found a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the reflection that reflects in an arbitrary line y=mx+b How can I find the reflection that reflects in an arbitrary line, $y=mx+b$
I've examples where it's $y=mx$ without taking in the factor of $b$
But I want to know how you can take in the factor of $b$
And after searching through for some results, I came to... | One way to do this is as a composition of three transformations:
*
*Translate by $(0,-b)$ so that the line $y=mx+b$ maps to $y=mx$.
*Reflect through the line $y=mx$ using the known formula.
*Translate by $(0,b)$ to undo the earlier translation.
The translation matrices are, respectively,
$$
\begin{pmatrix}
1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates:
$A = (-1, 3)$
$B = (3, 1)$
$C = (x, y)$
$BC$ is perpendicular to $AB$. Find the coordinates of $C$
My attempt:
Grad of $AB$ =
$$\frac{3-1}{-1-3} = -0.5$$
Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).
Equation of $BC$... | The equation of $AC$ is $x-3y=-10$ as slope of any line is $-\frac{a}{b}$ where $a$ is $x$-coordinate and $b$ is $y$-coordinate so slope is $-\left(\frac{-1}{3}\right)$. $C$ is the point where $AC$ and $BC$ so meet we have two simultaneous equations $2x-y=5$ and $x-3y=-10$ solving them you get $x=5$ and $y=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Laplace's equation after change of variables
Show that if $u(r, \theta)$ is dependent on $r$ alone, Laplace's
equation becomes $$u_{rr} + \frac{1}{r}u_r=0.$$
My first reaction is to replace $r=x$ and $\theta=y$, but obviously it does not work. Then I recall $x=r\cos \theta$ and $y=r\sin \theta$. Then I obtain the f... | You have
$$ \frac{\partial f}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial f}{\partial y} = f_x \cos{\theta} + f_y \sin{\theta}. $$
Then differentiating again (and using that $\partial \theta/\partial r=0$),
$$ f_{rr} = \cos{\theta} (\cos{\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Can this function be integrated? Can't seem to figure out this integral! I'm trying to integrate this but I think the function can't be integrated? Just wanted to check, and see if anyone is able to find the answer (I used integration by parts but it doesn't work). Thanks in advance; the function I need to integrate is... | Probably you'd want to use the calculus of residues to do this.
But below I do it using first-year calculus methods.
The cumbersome part may be the algebra, and that's what I concentrate on here.
\begin{align}
x^5 + 2 & = \left( x+\sqrt[5]{2} \right) \underbrace{\left( x - \sqrt[5]{2} e^{i\pi/5}\right)\left( x - \sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Can one do anything useful with a functional equation like $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$? I got
$$g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$$
as a functional equation for a generating function. Is there a way to get a closed form or some asymptotic information about the Taylor coefficients from such an equation?
Here... | I get that
the only solution is
$g(x) = 0$
if we can write
$g(x)
=\sum_{n=0}^{\infty} a_n x^n
$.
Here is my proof:
We have
$g(x^2)
= \frac{4x^2-1}{2x^2+1}g(x)
$
or
$(2x^2+1)g(x^2)
= (4x^2-1)g(x)
$.
From this,
as copper.hat pointed out,
$g(0) = 0$.
If
$g(x)
=\sum_{n=1}^{\infty} a_n x^n
$,
(since
$g(0) = 0$)
the left s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof that given equation(quartic) doesn't have real roots $$
(x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0
$$
I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real r... | Your polynomial is
$$P(x) = ({x^2} - 9)(x - 2)(x + 4) + ({x^2} - 36)(x - 4)(x + 8) + 153\tag{1}$$
Now consider theses
$$\eqalign{
& f(x) = ({x^2} - 9)(x - 2)(x + 4) \cr
& f({x \over 2}) = \left( {{{\left( {{x \over 2}} \right)}^2} - 9)} \right)\left( {\left( {{x \over 2}} \right) - 2} \right)\left( {\left( {{x \o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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show that for any prime p: if $p|x^4 - x^2 + 1$, with $x \in \mathbb{Z}$ satisfies $p \equiv 1 \pmod{12}$? show that for any prime p: if $p|x^4 - x^2 + 1$ satisfies $p \equiv 1 \pmod{12}$
I suppose that if $p$ divides this polynomial we can see that:
$x^4 - x^2 + 1 = kp$ for some $k \in \mathbb{N}$. But then $x^4 - x^2... | If $p\mid x$, then $p\mid x^4-x^2+1\implies p\mid 1$, contradiction. Therefore $p\nmid x$.
$(2x^2-1)^2\equiv -3\pmod{p}$ and $\left(\left(x^2-1\right)x^{-1}\right)^2\equiv -1\pmod{p}$. We can't have $p=3$, because $2x^2-1\equiv 0\pmod{3}$ has no solutions. We also can't have $p=2$, because $x^4-x^2+1$ is always odd, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix?
I know in matrix $3 \times 3$
$$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$
but how to work with a $5\times 5$ matrix?
| Multiplying the 1st row by $3$ and then adding it to the 4th row, and then multiplying the 3rd row of the resulting matrix by $-\frac 1 4$ and adding it to the 5th row, we obtain
$$\det \begin{bmatrix} 1 & 2 & 3 & 4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ -3 & -6 & -9 & -12 & 4\\ 0 & 0 & 1 & 1 & 1\end{bmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculating two specific limits with Euler's number I got stuck, when I were proving that
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \frac {5}{2}$$
$$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \frac {1}{3}$$
First one I tried to solve like
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n... | $$(a)\;\;\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n} =\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n}\times \frac{\sqrt{n^2+5}+n}{\sqrt{n^2+5}+n}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+2}+n} $$
So we get $$=\lim_{n\rightarrow \infty}\frac{5}{2}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square ... | \begin{align}
&\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} = 1\\ \implies &\sqrt{(x-1)-4\sqrt{x-1} + 4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1\\ \implies &\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1\\ \implies &|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = 1\tag{1}
\end{align}
This calls for casework:
1. $\quad\sqrt{x-1}\geq 3$
$(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
In a triangle prove that $\sin^2({\frac{A}{2}})+\sin^2(\frac{B}{2})+\sin^2(\frac{C}{2})+2\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})= 1$ Let ABC be a triangle. Thus prove that
$$\sin^2\left({\frac{A}{2}}\right)+\sin^2\left(\frac{B}{2}\right)+\sin^2\left(\frac{C}{2}\right)+2\sin\left(\frac{A}{2}\right)\sin\left... | Let $x=\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$, $x+y+z=\pi/2$
$$\sin^2(x)+\sin^2(y)+\sin^2(z)+2\sin(x)\sin(y)\sin(z)\\=\sin^2(x)+\sin^2(y)+\sin^2(\pi/2-x-y)+2\sin(x)\sin(y)\sin(\pi/2-x-y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+2\sin(x)\sin(y)\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+(\cos(x-y)-\cos(x+y))\cos(x+y)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$
I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?
| When you're doing the trigonometric substitution, you write $x=a\sin\theta$, which is good; you should also remember how to get back from $\theta$ to $x$, that is,
$$
\theta=\arcsin\frac{x}{a}=\arcsin\frac{x}{1/2}=\arcsin(2x)
$$
which actually should be the starting point, because it guarantees the angle $\theta$ is be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrating $\sqrt{x^2+a^2}$ I'm trying to integrate this function wrt $x$, substituting $x = a \tan \theta$
$$ \int \sqrt{x^2+a^2} dx = a^2 \int \frac {d\theta}{\cos^3\theta} = $$
$$= a^2 \cdot \frac 12 \left( \tan\theta \sec\theta + \ln\lvert \tan\theta + \sec\theta \rvert \right) = \frac 12 \left( x \sqrt{a^2 + x^2... | $$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)\\=\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)-\frac{a^2\ln|a|}{2} $$
and
$$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$
are di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.