Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to show that this difference of products is $O \left( \frac{1}{n^2} \right) $ Let $k \leq n$. Consider the following difference of products:
$$ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right)$$
For $n=1,2,3$, this is clearly $O \left( \frac{1}{n^2} \right) $ for... | $$\begin{align}\prod_{i=1}^{k-1}{\left (1-\frac{i}{n+1} \right )} &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n} + \frac{i}{n^2} \right )} + O \left (\frac1{n^3} \right )\\ &= \prod_{i=1}^{k-1}{\left (1-\frac{i}{n} \right )} + \sum_{j=1}^{k-1}\frac{j}{n^2} \prod_{i=1,i \ne j}^{k-1}{\left (1-\frac{i}{n} \right )}+ O \left ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
| We have
$$\sqrt{x^2+5} - \sqrt{x^2-3} = 2
\sqrt{x^2+5} = 2+ \sqrt{x^2-3}\\
x^2+5 = 4+ x^2-3+4\sqrt{x^2-3}\\
4 = 4\sqrt{x^2-3}\\
1 = \sqrt{x^2-3}\\
1 = x^2 -3\\
x^2 = 4\\
x = \pm 2
$$
So that
$$\sqrt{x^2+5} + \sqrt{x^2-3} = 3 + 1 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$
| We have
$$
\int \frac{x}{(7x - 10 - x^2)^{3/2}}\,dx.
$$
Let $u=-x^2 +7x-10$, so that $du = (-2x+7)\,dx$ and $\dfrac{-du} 2 = \left( x - \dfrac 7 2 \right)\, dx$. Then the integral becomes
$$
\int\frac{x - \frac 7 2}{(7x-10+x^2)^{3/2}} \,dx + \frac 7 2 \int \frac{dx}{(7x-10+x^2)^{3/2}}
$$
The substitution above handles ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance... | If $z=5e^{i\theta}$, and we use $1+2i=\sqrt5e^{i\phi}=z_0$, then we have
$$|z^5-z^3z_0|=125|z^2-z_0|$$
as the thing to maximise, but this implies we need $z^2$ on the diameter through $z_0$, so we need $\theta=\frac{\phi+\pi}{2}$.
Then, $$z^4=625e^{4i\theta}=625e^{2i\phi+2i\pi}=625e^{2i\phi}=625\cos(2\phi)+625i\sin(2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $x^2+y^2$ is constant for all values of $\theta$. Given that $x=3\sin \theta-2 \cos \theta$ and $y=3\cos \theta+2 \sin \theta$
i)Find the value of the acute angle $\theta$ for which $x=y$
ii)Show that $x^2+y^2$ is constant for all values of $\theta$.
My attempt,
$3\sin \theta-2 \cos \theta=3 \cos \theta+2 \s... | hint: $x^2+y^2 = 9(\cos^2 \theta+\sin^2\theta)+4(\sin^2\theta + \cos^2\theta) = ....$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.
| Notice, we have
$$(1+x)^n=^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+\ldots +^nC_n.1^0.x^n$$ Given condition $$\color{red}{\text{coefficient of}\ x^3=4\times \text{coefficient of}\ x^2}$$ $$\implies ^nC_{3}=4\times ^nC_{2}$$ $$\frac{n!}{(n-3)!3!}=\frac{4\times n!}{(n-2)!2!}$$ $$\frac{1}{3}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}... | There is such a triangle $\triangle\subset{\mathbb R}^3$ with vertices
$${\bf 0}=(0,0,0), \quad{\bf a}:=(x,y,0), \quad {\bf b}:=(0,y,z)\ .$$
Its area is given by
$${\rm area\,}(\triangle)={1\over2}\bigl|{\bf a}\times{\bf b}\bigr|={1\over2}\sqrt{y^2z^2+z^2x^2+x^2y^2}\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Minimum value of reciprocal squares I am bit stuck at a question. The question is : given: $x + y = 1$, $x$ and $y$ both are positive numbers. What will be the minimum value of: $$\left(x + \frac{1}{x}\right)^2 + \left(y+\frac{1}{y}\right) ^2$$
I know placing $x = y$ will give the right solution. Is there any other ... | We have $(x+1/x)^2+(y+1/y)^2=x^2+y^2+1/x^2+1/y^2+4=x^2+y^2+\frac{x^2+y^2}{x^2y^2 }+4\geq 1/2+16/2+4=25/2$. Here note that $x+y=1$ implies $xy\leq 1/4$ then $1/(xy)^2\geq 16$ and $x+y=1$ implies $x^2+y^2=1-2xy\geq 1-(x^2+y^2)$ so $x^2+y^2\geq 1/2$. For $x=y=1/2$ we have equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Proving an Inequality using a Different Method Is there another way to prove that: If $a,b\geq 0$ and $x,y>0$
$$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$
using a different method than clearing denominators and reducing to $(ay-bx)^2 \ge 0$?
| $\frac{a^2}{x} + \frac{b^2}{y}
\ge \frac{(a+b)^2}{x+y}
$
Let
$u = a^2/x$
and
$v = b^2/y$,
so
$a = \sqrt{ux}$
and
$b = \sqrt{vy}$.
The inequality then becomes
$u+v
\ge \frac{(\sqrt{ux}+\sqrt{vy})^2}{x+y}
= \frac{ux+vy+2\sqrt{uxvy}}{x+y}
$
or
$ux+uy+vx+vy
\ge ux+vy+2\sqrt{uxvy}
$
or
$uy+vx
\ge 2\sqrt{uxvy}
$
or
$(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Sums of Fourth Powers While fooling around on my calculator I found:
$$7^4 + 8^4 + (7 + 8)^4 = 2 * 13^4$$
$$11^4 + 24^4 + (11 + 24)^4 = 2 * 31^4$$
I'm intrigued but I can't explain why these two equations are true. Are these coincidences or is there a formula/theorem explaining them?
| I'll start at
Will Jagy's hint.
If
$c^2
=a^2+ab+b^2
$,
$\begin{array}\\
c^4
&=a^4+a^2b^2+b^4+2a^3b+2a^2b^2+2ab^3\\
&=a^4+3a^2b^2+b^4+2a^3b+2ab^3\\
\text{so}\\
2c^4
&=2a^4+6a^2b^2+2b^4+4a^3b+4ab^3\\
&=a^4+b^2+a^4+6a^2b^2+b^4+4a^3b+4ab^3\\
&=a^4+b^4+a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
&=a^4+b^4+(a+b)^4\\
\end{array}
$
Yep.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n}$ with induction Problem: Prove with induction that \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n} \end{align*} for every natural number $n \geq 2$.
Attempt at proof: Basic step: For $n = 2$ we have $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ which is correct.
Indu... | Notice that :
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n}\sqrt{n} + 1 $
$\sqrt{n}\sqrt{n+1} + 1 > n + 1 $
$ \sqrt{n}\sqrt{n+1} + 1 > \sqrt{n+1}\sqrt{n+1} $
$ \frac{\sqrt{n}\sqrt{n+1} + 1}{\sqrt{n+1}} > \sqrt{n+1}. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\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{(... | Multiplying it by $$\frac{{1+\sqrt{1-x^2}}}{1+\sqrt{1-x^2}}\cdot\frac{2+\sqrt{4-x^2}}{{2+\sqrt{4-x^2}}}\ (=1)$$gives$$\begin{align}&\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ (1-\sqrt{1-x^2})}{\sqrt{1-x}\ (2-\sqrt{4-x^2})}\\&=\lim_{x\to 0}\frac{\sqrt{2(2-x)}\ \color{red}{(1-\sqrt{1-x^2})}}{\sqrt{1-x}\ \color{blue}{(2-\sqrt{4-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?
| Finding the last two digits necessarily implies $\pmod{100}$
As $(14^n,100)=4$ for $n\ge2$
Let use start with $14^{5532-2}\pmod{100/4}$ i.e., $14^{5530}\pmod{25}$
As $14^2\equiv-2^2\pmod{25}$
Now $2^5\equiv7,2^{10}\equiv7^2\equiv-1\pmod{25}$
$\implies14^{10}=(14^2)^5\equiv(-2^2)^5=-2^{10}\equiv-1(-1)\equiv1$
As $5530\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\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... | $\bf{My\; Solution::}$ Given $\displaystyle \frac{1+\sin \alpha-\cos \alpha}{1+\sin \alpha+\cos \alpha} = \frac{\left(1-\cos \alpha\right)+\sin \alpha}{\left(1+\cos \alpha\right)+\sin \alpha}$
$\displaystyle = \frac{2\sin^2 \frac{\alpha}{2}+2\sin \frac{\alpha}{2}\cdot \cos \frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}+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": 1
} |
Question about Euler's formula I have a question about Euler's formula
$$e^{ix} = \cos(x)+i\sin(x)$$
I want to show
$$\sin(ax)\sin(bx) = \frac{1}{2}(\cos((a-b)x)-\cos((a+b)x))$$
and
$$ \cos(ax)\cos(bx) = \frac{1}{2}(\cos((a-b)x)+\cos((a+b)x))$$
I'm not really sure how to get started here.
Can someone help me?
| $$\sin { \left( ax \right) } \sin { \left( bx \right) =\left( \frac { { e }^{ aix }-{ e }^{ -aix } }{ 2i } \right) \left( \frac { { e }^{ bix }-{ e }^{ -bix } }{ 2i } \right) } =\frac { { e }^{ \left( a+b \right) ix }-e^{ \left( a-b \right) ix }-{ e }^{ \left( b-a \right) ix }+{ e }^{ -\left( a+b \right) ix } }{ -4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not rel... | Hint:
Let, each side of the regular polygon $A_1A_2A_3\ldots A_n$ be $a$ then we have $$A_1A_2=a$$
$$A_1A_3=2a\sin\left(\frac{(n-2)\pi}{2n}\right)$$
$$A_1A_4=a-2a\cos\left(\frac{(n-2)\pi}{n}\right)$$
Now, we have
$$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$$
$$\frac{1}{a}=\frac{1}{2a\sin\left(\frac{(n-2)\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"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}{... | In case you're interested, here's another way of doing this kind of question: remove fractional terms to the outside first...
$$(2-\frac{4}{x^2})(x+\frac{2}{x^2})^6=\frac{1}{x^{14}}(2x^2-4)(x^3+2)^6$$
So now you just need the coefficient of $x^{15}$ in the expansion of $$(2x^2-4)(x^3+2)^6$$ towards which the $2x^2$ ter... | {
"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": 0
} |
Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,... | Notice, in general, the equation of the line passing through the point $(x_1, y_1)$ & having slope $m$ is given by the point-slope form: $$\color{blue}{y-y_1=m(x-x_!)}$$
We have, equation of line A: $3x+6y-1=0\iff \color{blue}{y=-\frac{1}{2}+\frac{1}{6}}$ having slope $-\frac{1}{2}$
1.) slope of the line passing throug... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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:
$${ m=\frac{y^2 - y^1}... | First you computed midpoint of $X$ wrongly.
Later you evaluated $b$ wrongly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Calculate the volume bounded by the surfaces Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$
I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.
Then I did a double integration
$$\displaystyle
\iint_{x^2... | Three volumes.
Say $r^2 = x^2 + y^2$, cut through the volume gives the following image.
Cylinder #1
$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^4 dz r = 16 \pi
$$
Paraboloid #1
$$
V_{P_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^{r^2} dz r = 8 \pi
$$
Cylinder #2
$$
V_{C_1} = \int_0^{2\pi} d\phi \int_0^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Branch cuts for the contour integral $\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx$ How can the branch cut be handled in the contour integral, for $|b| \leq 1, \, a > 1$,
$$\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx \quad ?$$
If $a=1$ can the value of the integral be shown to be zero?
... | The problem of the integrand is that, if we use the principal branches then the singular points somehow overlap, which makes it hard to deal with directly. So if you want to apply complex analysis technique, we need some modification.
Notice that the integral
$$ I = \mathrm{PV}\!\int_{-1}^{1} \frac{\log(x+a)}{(x+b)\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumptio... | Here is what I did:
$$\frac{1}{2}bh \equiv \frac{1}{2}c^2\frac{\sin A \sin B}{\sin (A+B)}$$
Assume that $b$ = $c$.
Then, $c\frac{\sin A \sin B}{\sin (A+B)} \equiv \frac{c}{\cot A + \cot B}$. But $\cot A \equiv \frac{b \cos A}{h}$ and $\cot B \equiv \frac{a\cos B}{h}$. Therefore $\frac{c}{\cot A + \cot B} \equiv \frac{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding the area between three curves in which one is a piecewise function Let $f(x)$ be a continuous function given by $$f(x)=\begin{cases} 2x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\leq1\\ x^2+ax+b, \ |x|>1\end{cases}$$
Find the area of the region in the third quadrant bounded by the curves $x=-2y^2$ and $y=f(x)$ lying... | It should be
$$\int\limits^{\frac{-1}8}_{-1} \left(-\sqrt{\frac{-x}2}-2x\right)\,dx+\int\limits^{-1}_{-2} \left( -\sqrt{\frac{-x}2}-(x^2+2x-1)\right) \,dx$$
The curve $x=-2y^2$ is on top in both of the two parts, so should be the first term. Also you are picking the branch of it where $y$ is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't orthogonally diagonalise this symmetric matrix. So I have a symmetric matrix A = $\begin{bmatrix} 2 & 2 & -4 \\ 2 & -1 & -2 \\ -4 & -2 & 2 \end{bmatrix}$ and I want to orthogonally diagonalise it. I know that there are eigenvalues $-2$ and $7$ (-2 with an algebraic multiplicity of two, and a basis $$\left\{{\pmat... | If you know there are only eigenvalues $-2$ and $7$, then you can eliminate those with eigenvalue $-2$ and be left with the others. This is done by considering
$$
(A+2I) = \left[\begin{array}{ccc} 4 & 2 & -4 \\ 2 & 1 & -2 \\ -4 & -2 & 4\end{array}\right]
$$
The columns have to be eigevectors with eigenvalue $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Line for set of three-dimensional vectors If there is a set for 3D vectors $v$ where
$ v \times \begin{pmatrix} -1 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ -27 \\ 8 \end{pmatrix}$
is a line, what is this line's equation?
I'm not sure how to solve this, except for setting $v = \begin{pmatrix} a \\ b \\ c \end{pmat... | $
\newcommand{\i}{\hat{\mathbf{i}}}
\newcommand{\j}{\hat{\mathbf{j}}}
\newcommand{\k}{\hat{\mathbf{k}}}
\newcommand{\v}{\vec{\mathbf{v}}}
$
Let $v = \v$ and
$$
\v = \begin{bmatrix} a \\ b \\ c \end{bmatrix},
\quad
\i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix},
\quad
\j = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"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 -... | Your solution is correct. To reach to required form See here,You allready Got this
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
Now multiply and divide your result by $2$, you will get
$$\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x}{10}+C\\
=\frac{2e^x \sin^2 x -4e^x \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:
How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?
Answer is $2.73$
Simplify into Standard Form:
$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$
$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$
Hyperbola ... | Please check the formulas as you have two foci. $x^2 - 4x - 4 - 2y^2 - 4y = 0\to (x-2)^2 -2(y+1)^2 =6 \to \dfrac{(x-2)^2}{(\sqrt{6})^2}-\dfrac{(y+1)^2}{(\sqrt{3})^2}=1\to a = \sqrt{6}, b = \sqrt{3}\to c = \sqrt{6+3} = 3\to F= (2\pm 3,-1)=(-1,-1), (5,-1)\to \text{ distance to x-axix = 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| As Zhanxiong answered, Taylor expansion always works.
Consider, for large value of $x$, $$A(x)=\big(1+\frac 1x\big)^x$$ $$\log\big(A(x)\big)=x\log\big(1+\frac 1x\big)=x\big( \frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+\cdots\big)=1-\frac{1}{2 x}+\frac{1}{3 x^2}+\cdots$$ $$A(x)=e \, e^{-\frac{1}{2 x}+\frac{1}{3 x^2}}$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first par... | Another possibility. First,
$$\tag{$*$}
xy^2z^4 = 32 \Longrightarrow y^2 = \frac{32}{xz^4}.
$$
Notice that we can arbitrarily divide by $x,y$, or $z$ since any point with a coordinate equal to $0$ do not belong to your surface.
Now, consider the squared norm of a generic point $(x,y,z)$ and use $(*)$:
$$
x^2+y^2+z^2 =... | {
"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": 1
} |
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)... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9uv^2-w^3\geq8$, which is a linear inequality of $v^2$
and the condition does not depend on $v^2$.
Thus it remains to prove our inequality for an extremal value of $v^2$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ ... | {
"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": 2
} |
Tangent planes to $2+x^2+y^2$ and that contains the $x$ axis I need to find the tangent planes to $f(x,y) = 2+x^2+y^2$ and that contains the $x$ axis, so that's what I did:
$$z = z_0 + \frac{\partial f(x_0,y_0)}{\partial x}(x-x_0)+\frac{\partial f(x_0,y_0)}{\partial y}(y-y_0) \implies \\ z = 2 + x_0^2 + y_0^2 + 2x_0(x-... | So far you have worked out that the tangent plane to the surface at $(x_0,y_0,x_0^2 + y_0^2+2)$ has equation
$$
2xx_0 + 2yy_0-z-x_0^2-y_0^2+2=0
\tag{1}
$$
If this plane contains the $x$-axis, it contains all points $(x,y,z)$ with $y=z=0$. So the equation reduces to
$$
2x x_0 = x_0^2 + y_0^2 -2
\tag{2}
$$
Again, we are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value. I am trying to solve this question,
If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value.
I've tried this,
$$x^2-2x + z^2-2z + 2 + y^2 = 0$$
$$ (x-1)^2 + (z-1)^2 + y^2 = 0$$
T... | The OP wrote:
The left hand side can only become $0$ if $x=1$, $y=0$ and $z=1$, so the only solution is $x=z=1$ and $y=0$, which gives $x^3 + y^3 + z^3 = 2$.
I would add a justification that these values for $x$, $y$, and $z$ are unique.
We know that each of the three terms, $(x-1)^2$, $(z-1)^2$ and $y^2$ are po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows tha... | What about Feynman's way? If we take:
$$ f(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{x(1+x^2)^5}\,dx $$
we have $f(0)=0$ and:
$$ f'(a)=\int_{0}^{+\infty}\frac{dx}{(1+x^2)^5 (1+a^2 x^2)} $$
that is a manageable integral through partial fraction decomposition / the residue theorem.
We have:
$$\begin{eqnarray*} f'(a) &=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
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 ... | $$
\begin{align}
\int_0^1\frac{\sin^{-1}\left(\sqrt{x}\right)}{x^2-x+1}\,\mathrm{d}x
&=\int_0^1\frac{\frac12\cos^{-1}(1-2x)}{x^2-x+1}\,\mathrm{d}x\tag{1}\\
&=\int_{-1}^1\frac{\cos^{-1}(x)}{3+x^2}\,\mathrm{d}x\tag{2}\\
&=\frac12\int_{-1}^1\frac{\cos^{-1}(x)+\cos^{-1}(-x)}{3+x^2}\,\mathrm{d}x\tag{3}\\
&=\frac\pi2\int_{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\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)}}$$
Then I got stuck. Pleas... | Let $$\displaystyle I = -\int\frac{x-1}{\left(x+1\right)}\cdot \frac{1}{\sqrt{x^3+x^2+x}}dx = -\int\frac{(x^2-1)}{\left(x+1\right)^2\cdot \sqrt{x^3+x^2+x}}dx$$
So $$\displaystyle = -\int\frac{(x^2-1)}{(x^2+2x+1)\sqrt{x^3+x^2+x}}dx = -\int\frac{\left(1-x^{-2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}d... | {
"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": 1
} |
Prove that $\int_{0}^{2\pi}\frac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$ Prove that $$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$$
My Attempt:
$$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\int_{0}^{2\pi}x^2\frac{\sin x}{8+\sin^2x}$$
I applied integration by parts,consideri... | Just a guideline:
Use geometric series to get
$$
I=\frac{1}{8}\sum_{n=0}^{\infty}\frac{(-1)^n}{8^n}\underbrace{\int_{0}^{2 \pi}x^2 \sin^{2n+1}(x)}_{I_n}
$$
Show that the inner integral is equal to (use integration by parts and find a recurion relation):
$$
I_n=2\pi ^2\frac{\sqrt{\pi }\Gamma (n+1)}{\Gamma(n+\frac{3}{2})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\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$$
But this does not seem to be solving.Please help.
| Don't take this answer too seriously, but I could not resist. I leave details to you to fill in.
First, you factor the denominator into second degree factors, as
$$
\begin{aligned}
x^4+3x^3+3x^2-3x+1&=\bigl(x^2+\frac{1}{2} \sqrt{7+2 \sqrt{37}} x+\frac{3 x}{2}+\frac{1}{4} \sqrt{46+10 \sqrt{37}}+\frac{\sqrt{37}}{4}+\frac... | {
"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": 1
} |
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 ... | Here's another way to derive it term by term:
$$y^{-1}=1+e^x\Rightarrow -y^{-2}y'=e^x=y^{-1}-1\Rightarrow y'=y^2-y$$
Now you can continue to differentiate with the minimum of fuss:
$$y''=2yy'-y'$$
$$y'''=2yy''+2y'y'-y''$$
$$y^{(4)}=2yy'''+6y''y'-y'''$$
And so on, evaluating at each step.
| {
"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": 3
} |
Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$ Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$
I tried to solve it,i got range $\frac{\pi}{2}$ but the answer is ${\frac{-\pi}{2},\frac{\pi}{2}}$
$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt=\int... | $$f(x)=\int_{-1}^{1}\frac{\sin x}{t^2-2t\cos x+1}\ dt$$
$$=\int_{0}^{1}\frac{\sin x}{t^2-2t\cos x+1}+\frac{\sin x}{t^2+2t\cos x+1}\ dt$$
$$=2\sin x\int_{0}^{1}\frac{t^2+1}{(t^2+1)^2-4t^2\cos^2 x}\ dt$$
$$=2\sin x\int_{0}^{1}\frac{1+\frac {1}{t^2}}{\biggl(t+\frac {1}{t}\biggr)^2-4\cos^2 x}\ dt$$
$$=2\sin x\int_{0}^{1}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove an improper double integral is convergent I need to prove the following integral is convergent and find an upper bound
$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{1+x^2+y^4} dx dy$$
I've tried integrating $\frac{1}{1+x^2+y^2} \lt \frac{1}{1+x^2+y^4}$ but it doesn't converge
| Continuing from zhw.'s answer,
$$ \int_{0}^{+\infty}\frac{\sqrt{r}}{2(1+r^2)}\,dr = \int_{0}^{+\infty}\frac{u^2\,du}{1+u^4}=\frac{\pi}{2\sqrt{2}} $$
by the residue theorem, while:
$$ \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin\theta}}=\int_{0}^{1}\frac{du}{\sqrt{u(1-u^2)}}=\frac{1}{2\sqrt{2\pi}}\,\Gamma\left(\frac{1}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| We have $$\cos\left(x\right)+\sin\left(x\right)\tan\left(\frac{x}{2}\right)=1\Leftrightarrow\cos\left(x\right)\cos\left(\frac{x}{2}\right)+\sin\left(x\right)\sin\left(\frac{x}{2}\right)=\cos\left(\frac{x}{2}\right)
$$ and, using product to sum identity and the fact that cos is a even function$$\cos\left(x\right)\cos\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 6
} |
Rearrangement and Cauchy Let $a_1, \ldots, a_n$ be distinct positive integers. I want to prove that
$$\frac{a_1}{1^2} + \frac{a_2}{2^2} + \cdots + \frac{a_n}{n^2} \geq \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}.$$
I've been considering using Rearrangement and Cauchy Schwarz, but cannot make any progress
| Use the following: If $a_1, \ldots, a_n$ are distinct positive integers, then we can sort them according to their size $a_{n_1} \leq a_{n_2} \leq \ldots \leq a_{n_n}$ and get
$$k \leq a_{n_k} ~ \forall k = 1, \ldots, n.$$
Hence, using Cauchy Schwarz, we compute
\begin{align*}
\left( \sum_{k=1}^n \frac{1}{k} \right)^2 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$... | Just a different way to highlight what has already been correctly noted, and to explain why the answer could be $66$, as stated in one of the initial comments by the author of the OP. To avoid confusion, I interpreted the problem as the question of how many values of $x$ lead to a positive $z$ that is divisible by $6$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find cartesian coordinates of the incenter $A(a_1,a_2)$, $B(b_1,b_2)$ and $C(c_1,c_2)$ form the triangle $ABC$. What are the cartesian coordinates of the incenter and why?
| I'll argue up to the formula, so the "why" will be clear. Let's use this diagram of $\triangle ABC$ with the Cartesian coordinates moved so that points $A$ and $B$ are on the $x$-axis and $C$ has a positive $y$-coordinate.
Point $I$ is the incenter, segments $ID$, $IE$, and $IF$ are perpendiculars from $I$ to sides $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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}$$
| \begin{align}
1^2+2^2+...+&(n-1)^2 \quad\quad= \frac{n(n-1)(2n)}{6} = \frac{n^2(n-1)}{3} &< \frac{n^3}{3}
\\
1^2+2^2+...+&(n-1)^2+n^2 =\frac{n(n+1)(2n+1)}{6} =\frac{2n^3+3n^2+n}{6} =\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} &>\frac{n^3}{3}
\end{align}
Cleary our inequality is proven.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\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... | Let $s=a+b$ be our sum, where $a=\sqrt[3]{2+\sqrt{5}}$ and $b=\sqrt[3]{2-\sqrt{5}}$. Note that
$$s^3=a^3+b^3+3ab(a+b)=a^3+b^3+3abs.$$
Thus since $a^3+b^3=4$ and $ab=\sqrt[3]{-1}=-1$, we have
$s^3=4-3s$. This has the obvious root $s=1$ and no other real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\frac{x^2}{\sqrt{16-x^2}}$ using trig substitution During our integration of the following integral, using $x = 4 \sin \theta$
$$\int \frac{x^2}{\sqrt{16-x^2}} dx$$
We eventually come to the following point:
$$\int \frac{16 {\sin ^2 \theta} }{4 \sqrt{\cos ^2 \theta}} dx$$
At this point, we say that $\sqrt{\... | Let $$\displaystyle I = \int\frac{x^2}{\sqrt{16-x^2}}dx\;,$$ Let $x=4\sin \phi\;,$ Then $dx = 4\cos \phi d\phi$
So Integral $$\displaystyle I = \int\frac{16\sin^2 \phi}{4|\cos \phi|}\cdot 4\cos \phi d\phi = \pm 8\int 2\sin^2 \phi d \phi$$
Now Using $\bullet\; 2\sin^2 \phi = 1-\cos 2\phi$ and $\bullet\; \sin 2\phi = 2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Riemann integral on trigonometric functions I have to calculate Riemann integral of function $g:[0;\pi/4]\rightarrow\mathbb{R}$ (on interval $[0;\pi/4]$) given as $g(x)=\frac{\tan(x)}{(\cos(x)^2-4)}$. Function $g$ is continous on interval $[0;\pi/4]$ so it is enough to calculate $\int_0^{\frac{\pi}{4}}{\frac{\tan(x)}{(... | Hint: let $t=\tan(\frac{x}{2})$, then
$$\sin(x) = \frac{2t}{1+t^2}, \cos(x) = \frac{1-t^2}{1+t^2}$$
and
$$x=2\arctan(t) \implies dx = \frac{2 dt}{1+t^2}$$
so
$$\begin{aligned}
&\int \frac{\tan(x)}{(\cos(x)^2-4)} dx = \int \frac{\sin(x)}{\cos(x)(\cos(x)-2)(\cos(x)+2)} dx\\
&= \int \frac{2t}{1+t^2} \cdot \frac{1+t^2}{1-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\... | We have $z^4=e^{3\pi i/2+2k\pi}$.
So $z=(e^{3\pi i/2+2k\pi})^{1/4}=e^{3\pi i/8+k\pi/2}$
As $e^{i\theta}=e^{i(\theta+2\pi)}$, the $4$ values of $z$ are $e^{3\pi i/8}, e^{7\pi i/8}, e^{11\pi i/8}, e^{15\pi i/8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $5\times 5$ determinant The following $5\times 5$ det. comes from a Russian book. I don't want to expand the det. rather than do some operations on it and extract the result.
Prove:
$$\begin{vmatrix}
-1 &1 &1 &1 &x \\
1& -1 &1 &1 &y \\
1& 1 & -1 & 1 &z \\
1& 1 & 1 & -1 & u\\
x& y & z & u &0
... | I am not sure if this is the most efficient way to do it but doing the following row opperations:
$R_2 \to R_2 + R_1, R_3 \to R_3 + R_1, R_4 \to R_4 + R_1, R_5 \to R_5 + xR_1$
Assuming i didnt make any silly mistakes (as is always the case with there types of problems)
$\begin{vmatrix}
-1&1&1&1&x\\
1&-1&1&1&y\\
1&1&-1&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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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)\... | Hint: $\sqrt{2-x-x^2}=\sqrt{\frac94-(x+\frac12)^2}$
Substitute $x+\frac12=t; dx=dt$
$$\int \frac{\sqrt{\frac94-t^2}}{(t-\frac12)^2}\ dt$$
Put $\frac32\sin\theta=t;\frac32\cos\theta d\theta=dt$
$$\int \frac{\frac32\cos\theta.\frac32\cos\theta d\theta}{(\frac32\sin\theta-\frac12)^2}$$
$$9\int \frac{\cos^2\theta d\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Determining values of a coefficient for which a system is and isn't consistent. Given the system : \begin{array}{ccccrcc}
x & + & 2y & + & z & = & 3 \\
x & + & 3y & - & z & = & 1 \\
x & + & 2y & + & (a^2-8)z & = & a
\end{array}
Find values of $a$ such that the system has a unique solution, infinitely many solutions, or... | Just from the first reduction you can find:
$$
(a^2-9)z=a-3
$$
Now you simply have to discuss such equation ( the others does not contain the parameter). And you see that this equation has a solution only if $a^2-9 \ne0$ i.e $a \ne \pm 3$.
For $a=3$ the equation becomes: $ 0=0$ that is an identity and this means that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any $x$ for which $x ^2= (x-1)(x+1)$?
$x^2 = (x-1)(x+1)$.
Does anybody know if this is true for any numbers (specifically the larger the better)?
Because: $256^2 = 255 \cdot 257 + 1$
It's very very close, just an interesting thing I noticed when dealing with 2-byte integers in computer programming.
| This identity, $(x-1)(x+1) = x^2 - 1$, or more generally, $(x-a)(x+a)=x^2 - a^2$, occurs a lot in algebra. For example
\begin{align}
199 \times 201 = (200-1)(200+1) = 200^2 - 1^2 = 40000 - 1 &= 39999 \\
(\sqrt 7 - \sqrt 2)(\sqrt 7 + \sqrt 2) = 7 - 4 &= 3
\end{align}
You can also use it to solve the "sum-product p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Where I'm wrong? Why my answer is different from the book? The question is to integrate
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometim... | With the help of Bernard's answer you can further transform your term
\begin{align}
\int x \cos^{-1}(x)dx&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x)\\
&=\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac 14 x\sqrt{1-x^2}+c \ \text{ but this does not equal}\\
&\neq (2x^2-1)\frac{\cos^{-1}x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does the determinant of a matrix made up of column vectors being non-zero imply that the vectors are independent? Let's say we have 3 vectors and we make up a matrix where we depict the vectors as the columns of the matrix. If we calculate the determinant of the matrix and we get a non-zero number, does that mean that ... |
Let's say we have 3 vectors
$$
\left(
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right),
\left(
\begin{matrix}
d\\
e\\
f
\end{matrix}
\right),
\left(
\begin{matrix}
g\\
h\\
i
\end{matrix}
\right)
$$
and we make up a matrix where we depict the vectors as the columns of the matrix
$$
\left(
\begin{matrix}
a & d & g\\
b &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Rearange an expression I am learning induction. At one step I have to show that:
$$ 1 - \frac{1}{(1+n)} + \frac{1}{(n+1)(n+2)} $$
can be transformed to
$$ 1 - \frac{1}{n+2} $$
Theese are the steps for the transformation, but I cant understand them:
$$ 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)... | The first step is as follows:
$$1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} = 1 - \frac{(n+2) -1}{(n+1)(n+2)}$$
Consider the fractional part of the first expression; this is:
$$- \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$
Notice that there is a common factor of $1/(n+1)$, so we will factor this out and obtain:
$$\frac{1}{(n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mathematical Inequality Proof There are $4$ positive real numbers $\,a,b,c,d\,$ and one positive integer $\, p>0 $.
Prove:
$$
\left\lvert\sqrt[p]{a^p+b^p}-\sqrt[p]{c^p+d^p}\right\rvert^p
\leq \left\lvert a-c\right\rvert^p + \left\lvert b-d\right\rvert^p
$$
I have proved the cases $\,p=1$ and $\,p=2$. But the general ca... | The inequality is equivalent to
$$\left|(a^q+b^q)^{1/q}-(c^q+b^q)^{1/q}\right|\le \left(|a-c|^q+|b-d|^q\right)^{1/q}.$$
Up to a change of notation (|a-c|=x, |b-d|=y) and by symmetry, we only need to show
$$((x+a)^q+(y+b)^q)^{1/q}\le (x^q+y^q)^{1/q}+(a^q+b^q)^{1/q},$$
which is a simple case of the Minkowski Inequality. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $8$ does not divide $x^2-1$, then $x$ is even; prove by contrapositive If $8$ does not divide $x^2-1$, then $x$ is even
proof by contrapositive
the contrapositive of this is :
if $x$ is odd, then $8$ divides $x^2-1$
proof by contrapositive:
Assume $x$ is odd
by definition of odd $∃k∈ℤ$ such that $x=2k+1$
Well, $x^2-... | 8 does not divide $x^2-1$ is the same as
"if 8 divides $x^2-1$, then x is odd"
which is the same as:
$$8|(x^2-1)\rightarrow 2|(x+1)$$
but
$$x^2-1=(x-1)(x+1)$$
Then since $8=4*2$
Two divides both $x-1$ and $x+1$, and we proven our case
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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What is $\sum_n n^6$ series? I went to OEIS but I didn't quite understand the formula for sum of this series. Could someone type it in normal shape and answer me?
and if it's no bother please do this for 7th, 8th, 9th and 10th powers?
| $$
\sum_{k=1}^n k^6 = \frac{1}{42} n (n+1) (2 n+1) (3 n^4+6 n^3-3 n+1)\\
\sum_{k=1}^n k^7 = \frac{1}{24} n^2 (n+1)^2 (3 n^4+6 n^3-n^2-4 n+2)\\
\sum_{k=1}^n k^8 = \frac{1}{90} n (10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3)\\
\sum_{k=1}^n k^9 = \frac{1}{20} n^2 (n+1)^2 (n^2+n-1) (2 n^4+4 n^3-n^2-3 n+3)\\
\sum_{k=1}^n k^{10} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2... | ${x^2+y^2\over x+y}=4 \implies x^2+y^2=4x+4y \implies x^2+y^2-4x-4y=0 \implies (x-2)^2+(y-2)^2=(2\sqrt{2})^2$ which is a circle with center $(2,2)$ and radius $2\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\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)}... | You mixed up square roots and cube roots:
$$(1-\sqrt[2]{a})*(1+\sqrt[2]{a})=1-a$$
$$(1-\sqrt[3]{a})*(1+\sqrt[3]{a})≠1-a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Minimum value of the expression given below. Is $a,b,c$ are three different positive integers such that :
$$
ab+bc+ca\geq 107
$$
Then what is the minimum value of $a^3+b^3+c^3-3abc$.
I expanded this expression to $(a+b+c)((a+b+c)^2-3(ab+bc+ca))$ and tried to find the minimum value of $(a+b+c)$ by AM-GM inequalities but... | You can see $a^3+b^3+c^3-3abc=\frac{1}{2}[(a-b)^2+(a-c)^2+(c-b)^2](a+b+c)$
For $[(a-b)^2+(a-c)^2+(c-b)^2]$, since they are three different integers, this is greater or equal to $[(0-1)^2+(0-2)^2+(1-2)^2]=6$;
For $(a+b+c)$, $(a+b+c)=\sqrt{(a+b+c)^2}\geq \sqrt{3(ab+bc+ca)}\geq\sqrt{3*107}>17$. Since the're integers, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule because I don't know how to use derivates yet. So I tried rationalizing the denominator and numerator but it didn't work.
$\lim_{x ... | Notice, use rationalization as follows $$\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x^2}}{\sqrt{1+x}-1}$$
$$=\lim_{x\to 0}\frac{1+x-(1-x^2)}{1+x-1}\times \left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}\frac{x^2+x}{x}\left(\frac{\sqrt{1+x}+1}{\sqrt{1+x}+\sqrt{1-x^2}}\right)$$
$$=\lim_{x\to 0}(x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Flea on a triangle "A flea hops randomly on the vertices of a triangle with vertices labeled 1,2 and 3, hopping to each of the other vertices with equal probability. If the flea starts at vertex 1, find the probability that after n hops the flea is back to vertex 1."
Could someone provide a hint to help me start this?
| After hop $i$, if the flea is at vertex $1$, then it has $0$ probability to be at vertex $1$ next; if the flea is not at vertex $1$, then it has $0.5$ probability to be at vertex $1$ next.
Let $p_i$ be the probability of the flea at vertex $1$ after $i$ hops, and $q_i = 1-p_i$.
$$\begin{align*}
\begin{pmatrix}p_{i}\\q_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Based on a coordinate system centered in a sphere where is $M(x, y, z) = 6x - y^2 + xz + 60$ smallest? I am trying to work through a few examples in my workbook, and this one has me completely dumbfounded.
Suppose I have a sphere of radius 6 metres, based on a coordinate system centred in that sphere, at what point on... | Using a Lagrange Multiplier
You want to minimize
$$
M(x,y,z) = 6x -y^2 + xz + 60
$$
under the constraint
$$
R(x,y,z) = 6
$$
where
$$
R(x,y,z) = \sqrt{x^2+y^2+z^2}
$$
We define the Lagrange function
$$
\Lambda(x,y,z, \lambda) = M(x,y,z) + \lambda(R(x,y,z)-6)
$$
for some constant $\lambda$ and derive its gradient
$$
\mbo... | {
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"url": "https://math.stackexchange.com/questions/1452177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Product $abc$ if $a$,$b$,$c$ $\in$ $\mathbb{R}$ and if $$a+\frac{1}{b}=\frac{7}{3}$$ $$b+\frac{1}{c}=4$$ $$c+\frac{1}{a}=1$$ Then find the value of $abc$
I multiplied the three equations with $bc$, $ca$ and $ab$ respectively we get
$$abc+c=\frac{7bc}{3}$$ $$abc+a=4ac$$ $$abc+b=ab$$ Again multiplying above thre... | Hint: Multiplying the first three expressions we get $$abc+\left(a+\dfrac{1}{b}\right)+\left(b+\dfrac{1}{c}\right)+\left(c+\dfrac{1}{a}\right)+\dfrac{1}{abc}=\dfrac{28}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right.
| Notice, the binomial expansion $$(2x-y+5)^8=^{8}C_0(2x-y)^85^0+\color{red}{^{8}C_1(2x-y)^75^1}+^{8}C_2(2x-y)^65^2+^{8}C_3(2x-y)^55^3+^{8}C_4(2x-y)^45^4+\dots +^{8}C_8(2x-y)^05^8$$
In the given term $x^3y^4$, the sum of powers $3+4=7$ hence in the above expansion there is only one term $^{8}C_1(2x-y)^75^1$ which has su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Long Inequality problem for $a, b, c $ positive real numbers $$ \left( a+\frac{1}{b} -1\right) \left( b+\frac{1}{c} - 1\right) +\left( b+\frac{1}{c} -1\right) \left( c+\frac{1}{a} -1\right) +\left( c+\frac{1}{a} -1\right) \left( a+\frac{1}{b} -1\right) \geq 3$$
How we can prove the inequality above. Actually it take... | Maybe this could help:
If you denote $x=a+\frac{1}{b}>0, \, y=b+\frac{1}{c}>0, \, z=c+\frac{1}{a}>0$
you will have $x+y+z=a+1/a +b+1/b+c+1/c\ge 2+2+2=6$ and the equality is achieved iff $a=b=c=1$
You get $(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)\ge 3$ which is equivalent to
$$xy+yz+zx-2(x+y+z)\ge 0$$
So you have to prove the ... | {
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"url": "https://math.stackexchange.com/questions/1456300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Calculate $\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx $ Calculate
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx
$$
My try:
$$\int {\frac{{\sqrt {x + 1} - \sqrt {x - 1} }}{{\sqrt {x + 1} + \sqrt {x - 1} }}} dx = \left| {x + 1 = {u^2}} \... | $$\begin{eqnarray*}\int\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\,dx &=& \int\frac{2x-2\sqrt{x^2-1}}{2}\,dx\\ &=&\; C+\frac{x^2}{2}-\frac{x}{2}\sqrt{x^2-1}+\frac{1}{2}\log\left(x+\sqrt{x^2-1}\right).\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(... | $(3^{k+1}+7^{k+1}-2)=3^k+7^k-2+2\cdot3^k+6\cdot7^k$
So, it is enough to show that $8$ divides $2\cdot3^k+6\cdot7^k=6\cdot3^{k-1}+6\cdot7^k$
Also, $3^{odd}=3$mod$(8)$ , $3^{even}=-1$mod$(8)$
$7^{odd}=-1$mod$(8)$, $7^{even}=1$mod$(8)$.
implies $6\cdot3^{k-1}+6\cdot7^k=0$mod$(8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction area of koch snowflake For all $n>=0$ prove that the area of a Koch snowflake is $a_n = a_0(\frac{8}{5}-\frac{3}{5}(\frac{4}{9})^n)$ where $a_0=\frac{\sqrt{3}}{4}$
I'm trying to get $P(n+1)$ from $P(n)$ but I'm not sure how to proceed.
For $P(n)$ I now have $\frac{2\sqrt{3}}{5}-\frac{3\sqrt{3}}{20}(... | Do you have to use induction?
Let's say that the flake after $j$ steps has $n_j$ sides of length $s_j$, now since you replace each side with four of length $s_j/3$ and thereby adding a equilateral triangle of side $s_j$ you will end up with $s_{j+1}$ = $s_j/3$ and $n_{j+1} = 4n_j$ and the area added in this step to be ... | {
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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 ... | You already showed that $$f(x,y) = xy(x^2+y^2)+4=xy(1-xy)+4,$$
which is a quadratic function in $xy$. It remains to find the exact range for $xy$. For which you showed that $xy\ge -1$, equality happens when $(x,y)=(1,-1)$ or $(x,y)=(-1,1)$.
The other restriction $xy\le1$, while correct, never reaches equality. Instead... | {
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"timestamp": "2023-03-29T00:00:00",
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Upper bound on $\ln(\frac{1}{1-x})$ for $0\leq x\leq 1/2$ Prove that $$\ln\left(\frac{1}{1-x}\right)\leq x+2x^2$$ for $0\leq x\leq 1/2$.
I thought about the Taylor series $\ln(1+x)=x-x^2/2+x^3/3-\ldots$. For small $x$, the values $1+x$ and $1/(1-x)$ are very close to each other, so the inequality should hold since in t... | There are many ways. You suggested series, so let's do it that way.
We have
$$\ln\left(\frac{1}{1-x}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots.$$
The part from $\frac{x^2}{2}$ on is $\le \frac{x^2}{2}\left(1+x+x^2+x^3+\cdots\right)$. Note that $1+x+x^2+\cdots=\frac{1}{1-x}$ and $\frac{1}{1-x}\le 2$ on ... | {
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"source": "stackexchange",
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Need what the final answer is How do I integrate:
$\int \frac{1}{(1+x\tan x)^2}dx$
Tried the following: substituting $\tan x = \frac{\sin x}{\cos x}$ and got stuck
Next tried $\tan x = \frac{1}{\cot x}$ ,took LCM and did substitution.
Don't know the final answer is correct or not
| your answer is correct
$$\int \frac{1}{(1+x\tan x)^2}dx =\dfrac{\tan x}{x\tan x+1}+Cst $$ and Substitute $$ \tan x = \dfrac{1}{\cot x}$$
\begin{align}
\int \frac{1}{(1+x\tan x)^2}dx&=\dfrac{\tan x}{x\tan x+1}+Cst\\
&=\dfrac{\dfrac{1}{\cot x}}{x\dfrac{1}{\cot x}+1}+Cst \\
&=\dfrac{1}{x+\cot x}+Cst \\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fundamental Identities I was given a task to prove:
$$\frac{\sec\theta}{\sec\theta\tan\theta} = \sec\theta(\sec\theta + \tan\theta)$$
Then I replaced them with their Ratio and Reciprocal Identities
\begin{align*}
\sec\theta & = \frac{1}{\cos\theta}\\
\tan\theta & = \frac{\sin\theta}{\cos\theta}
\end{align*}
so I came u... | Well the identity you mentioned is not correct:
Let $\theta = \frac{\pi}{4}$. Them your identity becomes $$\frac{\sec{\frac{\pi}{4}}}{(\sec{\frac{\pi}{4}})(\tan{\frac{\pi}{4}})}=1$$But the result you mentioned gives a different answer.
| {
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How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=... | Hint:
For brevity, let us write $R_1:=|z-1|, R_2:=|z+1|$, which are square roots of a quadratic polynomial.
Squaring the sum,
$$(R_1+R_2)^2=R_1^2+2R_1R_2+R_2^2=a^2.$$
Then
$$4R_1^2R_2^2=(a^2-R_1^2-R_2^2)^2
=a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2+2R_1^2R_2^2,$$
$$a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2-2R_1^2R_2^2=0,$$
$$a^4+(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find powers of 3 mod 17 Show that powers of $3\ (\text{mod}\ 17)$ are $3,9,10,13,15,11,16,14,8,7,4,12,2,6,1$.
What is meaning of power of something, how do I proceed this question? Just provide me hint to get start in this problem.
| Let us try a very naive approach. I should add that this is precisely the method suggested in André Nicolas's comment.
You start by $3^0=1$ and multiply both sides by $3$ repeatedly
$$3^0 \equiv 1 \pmod{17}\\
3^1 \equiv 3 \pmod{17}\\
3^2 \equiv 9 \pmod{17}$$
The next result would be $3^3 \equiv 27 \pmod{17}$. But $27\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465079",
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Question on Converting between Base Number Systems Question on Converting Between Base Numbers
The people of Jupiter use Base 13. Therefore, their numerals are 0,1,2,3,4,5,6,7,8,9,A,B,C.
The people on Saturn use Base 7. Therefore, their numerals are 0,1,2,3,4,5,6.
A person on Jupiter has A906BC Jupits on Jupiter. This ... | Jovian bank account: $10 \cdot 13^5 + 9 \cdot 13^4 + 6 \cdot 13^2 + 11 \cdot 13^1 + 12 = 3971148 \text{ Jupits}$
Saturnian bank account: $6 \cdot 7^7 + 5 \cdot 7^6 + 3 \cdot 7^5 + 2 \cdot 7^4 + 5 \cdot 7^3 + 4 \cdot 7^2 + 6 = 5586643 \text{ Jupits}$
The remaining parts should be easy for you.
| {
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The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not s... | Using $\displaystyle \bf{A.M\geq G.M}$
$$\displaystyle \frac{4\sin^2 x+\csc^2 x}{2}\geq \sqrt{4\sin^2 x\cdot \csc^2 x}\Rightarrow 4\sin^2 x+\csc^2 x\geq 4$$
and equality hold when $$4\sin^2 x=\csc^2 x$$
and $$\displaystyle \frac{\tan^2 x+\cot^2 x}{2}\geq \sqrt{\tan^2 x\cdot \cot^2 x}\Rightarrow \tan^2 x+\cot^2 x\geq 2$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rank of a matrix of binomial coefficients This question arose as a side computation on error correcting codes.
Let $k$, $r$ be positive integers such that $2k-1 \leqslant r$ and let $p$ a prime number such that $r < p$. I would like to find the rank of the following $k \times k$ matrix with coefficients in the field $F... | Not true, unfortunately.
Indeed, by adding the $2$nd row to the $1$st, $3$rd to $2$nd, ... , $k$th to $(k-1)$th etc, you end up with the matrix
$$
\begin{pmatrix}
\binom {r+1}{k} & \binom {r+1}{k+1} & \dotsm & \binom {r+1}{2k-1}\\
\vdots & \vdots & \ddots & \vdots \\
\binom {r+1}{2} & \binom {r+1}{3} & \dotsm & \bino... | {
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Solution to this recurrence? Is there exists a solution to this recurrence.
$$F(N,1) = N, N≥1$$
$$F(N,K) = \frac {1}{\lfloor\frac 1{F(N-1,K-1)} -\frac 1{F(N,K-1)}\rfloor} \;\;\;\;2≤K≤N$$
I tried to simplify the equation but i am not able to find F(1,2) and thus unable to proceed.
I am new to this site,so please let me ... | The original recurrence relation is equivalent to
$$
\frac{1}{F(N,K)}
=
\frac{1}{F(N-1,K-1)}
-
\frac{1}{F(N,K-1)}.
$$
For $K = 2$, we have
\begin{align}
\frac{1}{F(N,2)}
&=
\frac{1}{F(N-1,1)}
-
\frac{1}{F(N,1)} \\
&=
\frac{1}{N-1}
-
\frac{1}{N}
\\
&=
\frac{1}{N\,(N-1)}.
\end{align}
Next, for $K = 3$, we have
\begin{ali... | {
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The names of 8 students are listed randomly. What is the probability that Al, Bob, and Charlie are all in the top 4? My attempt at a solution: There are $C_{4,3}$ ways to arrange Al, Bob, and Charlie in the top 4, and $C_{5,4}$ ways to arrange the other 5 people. All of this is over $C_{8,4}$, giving $$\frac{C_{4,3}C_{... | Here is a slightly different approach:
There are $\dbinom{8}{4}$ ways to select the top 4, and
there are $\dbinom{5}{1}$ ways to choose the top 4 including Al, Bob, and Charlie
$\;\;\;$(since 1 person out of the remaining 5 must be selected),
so the probability is $\displaystyle\frac{5}{\binom{8}{4}}=\frac{5}{70}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximizing $\sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta$ I need to maximize
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \tag{1}$$
where $\alpha, \beta \in [0, \frac{\pi}{2}]$.
With numerical methods I have found that
$$ \sin \beta \cos \beta + \sin \alpha \co... | Let $$f = \sin \alpha \cos \alpha + \sin \beta \cos \beta -\sin \alpha \sin \beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta) -\sin \alpha \sin \beta$$
$$\frac{\partial f}{\partial \alpha} = 0 \implies \cos 2\alpha - \sin \beta \cos \alpha = 0$$
$$\frac{\partial f}{\partial \beta} = 0 \implies \cos 2\beta - \sin \alpha \... | {
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$\lim_{x\to 0} \frac{1-\cos (1-\cos (1-\cos x))}{x^{a}}$ is finite then the max value of $a$ is? I know the formula $\frac{1-\cos x}{x^{2}}=\dfrac{1}2$
But how do i use it here and tried l hospital and no use and doing l hospital twice will make it very lengthy how do i approach ?
| Using Taylor series: when $x\to 0$, we have
$\cos x = 1-\frac{x^2}{2} + o(x^2)$.
Step by step, we can use this as follows:
$$
1-\cos x = \frac{x^2}{2} + o(x^2)
$$
so that, plugging it back:
$$
\cos(1-\cos x) = 1-\frac{1}{2}\left(\frac{x^2}{2} + o(x^2)\right)^2 = 1-\frac{x^4}{8} + o(x^4)
$$
and therefore
$$
1-\cos (1-\c... | {
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Number Theory: How to solve $x^2\equiv 4\pmod{143}$? I'm not sure how to solve this quadratic congruence:
$x^2 \equiv 4\pmod{143}$
Thanks!
So here's what I did, but the solution seems a bit long the way that I did it:
The four solutions are
1: $x\equiv 2\pmod{11}, x\equiv 2\pmod{13}$:
$x = 2 + 11k\implies 2+11k\equiv 2... | The four possible solutions are
$x\equiv 2\ (mod\ 11)$ and $x\equiv 2\ (mod\ 13)$
$x\equiv 2\ (mod\ 11)$ and $x\equiv -2\ (mod\ 13)$
$x\equiv -2\ (mod\ 11)$ and $x\equiv 2\ (mod\ 13)$
$x\equiv -2\ (mod\ 11)$ and $x\equiv -2\ (mod\ 13)$
The solutions $2$ and $143-2=141$ are easy to get, a bit more
difficult are the ... | {
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$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}$$ or new --
first ine... | Late answer since it was tagged as duplicate there and this way of solving is neither here nor there.
\begin{eqnarray*} abc(a+b+c)
& \stackrel{GM-AM}{\leq} & \left(\frac{a+b+c}{3}\right)^3(a+b+c)\\
& = & 3\left(\frac{a+b+c}{3}\right)^4 \\
& \stackrel{x^4 \; is \; convex}{\leq} & 3\cdot \frac{1}{3}(a^4+b^4+c^4) = a^4+b... | {
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"source": "stackexchange",
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How to find the inverse Laplace transform of this? I am looking for the inverse Laplace transform of $$\frac{1}{s-1}e^{-\sqrt{s}x}$$
This is for an introductory partial differential equations class so I am thinking that it should not be too hard. If it wasn't for the $s-1$ in the denominator and we had just $s$ instead... | You could simply convolve the ILT with that of $1/(s-1)$, which is $e^t$. But I prefer using the definition of the ILT in the complex plane.
In this case, consider the contour integral
$$\oint_C dz \frac{e^{-x \sqrt{z}}}{z-1} e^{t z}$$
where $C$ is a Bromwich contour in the left half-plane with a detour along the nega... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x= \sin(k- x)$? Is there a way to solve $x = \sin (k-x)$ without a computer, that is with a pocket calculator or pencil and paper?
| Let $f(x) = x - \sin(k-x)$. It is trivial that there is no zero when $|x| > 1$. We also know that $f(0) = -\sin(k) \ne 0$.
The problem with the Newton's method is that it will fail when $f'(x) = 0$.
Let $x-k=y$, $g(y) = f(y+k) = y + k + sin(y)$. Let's also constrain $k \in [-pi/2, pi/2]$.
Using the Taylor's series for ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coordinates of the points on the curve $y=2x^4 - 3x^2 + x - 7$ where the gradient is parallel to the line $y=3x$ I got that $\frac{dy}{dx}= 3$
(because of the gradient of the parallel line)
and then, I found the derivative of the equation of the curve to be $8x^3 -6x +1= 3$.
This simplifies as $8x^3 - 6x -2 =... | Gradient of $y=3x$ is $3$.
So by differentiating, we have $\frac{dy}{dx}=8x^3-6x+1$
Now solve $8x^3-6x+1=3$.
Clearly $8x^3-6x+1=3\Rightarrow 8x^3-6x-2=0$.
$\Rightarrow 8x^3-6x-2=(x-1)(8x^2+8x+2)=0\Rightarrow x=1$.
Thus coordinates of the point $=(1,-7)$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the maximum value of $ \sin x +\sin {2x} (0I would appreciate if somebody could help me with the following problem
Q: What is the maximum value of $$ \sin x +\sin {2x} (0<x<\frac{\pi}{2})$$
I have done my work here
$$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 =0$$
$$\cos x=\fr... | $$f (x)= \sin x +\sin {2x} $$
$$f'(x)= \cos x +2\cos {2x}=\cos x + 4\cos^2 x-2 = 0 \implies
\cos x = \frac{-1 \pm \sqrt {33}}{8}$$
Since we are interested in the maxima only, discard the negative root. In general, $$x = 2k\pi \pm \cos^{-1}\frac{\sqrt{33}-1}{8} = 2k\pi \pm 0.93592946$$
However, in this case $$x = 0.9359... | {
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representation of a complex number in polar from
Write the following in polar form: $\frac{1+\sqrt{3}i}{1-\sqrt{3}i}$
$$\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... | You forget the brackets in your wolfram formula, the correct one is this and as you can see your result is correct.
| {
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Use of property of ratio and factorization method Recently I have faced a problem related to algebraic ratio.I have tried much but I can not find any clue.Can anyone give me some hints.
Here is the problem.
I am given $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\neq0$$
I have to prove: $(a+b+c)(x+y+z)=ax+by+cz$... | Define $k$ by $$\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}=k\neq0,$$
so $1/k\ne0$.
Now, finding the product, we have
$$\begin{align}(a+b+c)(x+y+z)=&\;ax+by+cz\\&+a(y+z)\\ &+b(x+z)\\&+c(x+y);\end{align}$$
thus we need to show that $a(y+z)+b(x+z)+c(x+y)=0$.
Since
$$\frac{x^2-yz}{k}\cdot(y+z)=a(y+z)\,$$
$$\frac{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\int _0^{2\pi }(1-\cos t)(\frac{5}{4}-\cos t)^{-3/2}\,dt$
How to compute
$$ \int_{0}^{2\pi}\dfrac{1-\cos(t)}{\biggl(\dfrac{5}{4}-\cos(t)\biggr)^{\dfrac{3}{2}}} dt $$
I'm interested in more ways of computing this integral.
My thoughts:
I'm tired to use Bioche rules :
Let $x=\operatorname{tg}(\frac{t... | First we can see that the integral is twice the integral over $[0,\pi]$ and further using $\cos t=1-2\sin^2(t/2)$ and substitution $t=2x$ we get $$I=64\int_{0}^{\pi/2}\frac{\sin^2x}{(1+8\sin^2x)^{3/2}}\,dx=64\int_{0}^{\pi/2}\frac{\sin^2x}{(9-8\cos^2x)^{3/2}}\,dx$$ and this further leads to $$I=\frac{64}{27}\int_{0}^{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine co-ordinates of point?
How do you work out the co-ordinates of $P_{t+1}$ if you're given the co-ordinates of $P_0, P_1, P_2$ ? example (1,1);(2,5);(6,6)
The angle alpha remains constant between $(P_1, P_2)$ and $(P_2, P_{t+1})$
| Slope of $P_0P_1 = \frac{1-5}{1-2} = 4$
Slope of $P_2P_1 = \frac{6-5}{6-2} = \frac{1}{4}$
Slope of $P_{t+1}P_2 = \frac{y-6}{x-6} = \tan \alpha$
$\alpha=\tan^{-1}\left(\frac{4-\frac{1}{4}}{1+4\cdot \frac{1}{4}}\right) = \tan^{-1}\left(\frac{\frac{15}{4}}{2}\right) = \tan^{-1}\left(\frac{15}{8}\right)$
By the problem, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 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... | Geometrical Approach:
In general (see derivation), when a given point $P(x_0, y_0)$ is reflected about the line: $y=mx+c$ then the co-ordinates of the point of reflection $P'(x', \ y')$ are calculated by the following formula
$$\color{blue}{(x', y')\equiv \left(\frac{(1-m^2)x_0+2m(y_0-c)}{1+m^2}, \frac{2mx_0-(1-m^2)y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Show that a matrix $A=\pmatrix{a&b\\c&d}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$
Let $A=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
,a,b,c,d\in\mathbb{R}$ . Prove that every matrix $A$ satisfies the condition $$A^2-(a+d)A+(ad-bc)I=O .$$ Find $$ \begin{bmatrix}
a & b \\
... | The short answer is: The equality can be proofen by simply inserting A, performing a matrix multiplication and adding up the matrices. For the second part of your question, one may find a recursion formula.
The long answer including the calculation follows here:
Assuming $I=\begin{bmatrix}1& 0 \\ 0 &1\end{bmatrix}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Why is $n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} $ never zero? Here $n_i$ are integral numbers, and not all of them are zero.
It is natural to conjecture that similar statement holds for even more prime numbers. Namely,
$$ n_1 \sqrt{2} +n_2 \sqrt{3} + n_3 \sqrt{5} + n_4 \sqrt{7} + n_5 \sqrt{11} +n_6 \s... | Assume that:
$$ n_1\sqrt{2}+n_2\sqrt{3}+n_3\sqrt{5}+n_4\sqrt{7}=0.\tag{1} $$
That implies:
$$ (2n_1^2+3n_2^2)+2n_1 n_2 \sqrt{6} = (5n_3^2+7n_4^2)+2n_3 n_4 \sqrt{35}\tag{2} $$
as well as (just keep rearranging and squaring):
$$ (2n_1^2+3n_2^2-5n_3^2-7n_4^2)^2 = 24 n_1^2 n_2^2 + 140 n_3^2 n_4^2 - 8n_1 n_2 n_3 n_4 \sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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solve: $z^3=\sqrt(3)-i$
Solve: $z^3=\sqrt(3)-i$
$r=\sqrt{\sqrt{3}^2+(-1)^2}=\sqrt{4}$
$\theta=tan^{-1}(\frac{-1}{\sqrt{3}})=\frac{-\pi}{6}$
0: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{-\pi}{18})+isin(\frac{-\pi}{18})]=1.24-0.96i$
1: $\sqrt[3]{z}=\sqrt[6]{4}*[cos(\frac{11\pi}{18})+isin(\frac{11\pi}{18})]=-0.43-1.18i$
2: $\... | $$z^3=\sqrt{3}-i \Longleftrightarrow$$
$$z^3=\left|\sqrt{3}-i\right|e^{\arg\left(\sqrt{3}-i\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{\left(\sqrt{3}\right)^2+1^2}e^{\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z^3=\sqrt{3+1}e^{-\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)i} \Longleftrightarrow$$
$$z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $p=4k+1$ then $p$ divides $n^2+1$ I am stuck in one step in the proof that if $p$ is congruent to $1 \bmod 4$, then $p\mid (n^2+1)$ for some $n$.
The proof uses Wilson's theorem, $(4k)!\equiv -1 \pmod p$.
The part I am stuck is where it is claimed that $(4k)!\equiv (2k)!^2 \bmod p$. Why is this so?
| Take a typical small prime of the form $4k+1$, such as $17$. So $k=4$.
Then
$$16!=\left[(1)(2)(3)(4)(5)(6)(7)(8)\right]\left[(9)(10)(11)(12)(13)(14)(15)(16)\right].\tag{1}$$
We have $16\equiv -1\pmod{17}$, $15\equiv -2\pmod{17}$, $14\equiv -3\pmod{17}$, and so on down to $9\equiv -8\pmod{17}$.
It follows that $(16)(15... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the area of the region bounded by the parabola $y = 2x^2$, the tangent line to this parabola at $(3, 18)$, and the $x$-axis. Is my answer correct?
$f(x) = 2x^2 \gets$ this is the parabola
$f(3) = 2 \times 9 = 18 \to$ the parabola passes through $A (3 ; 18$), so its tangent line does too.
$f'(x) = 4x \gets$ this ... | $y=2x^2$ has derivative $y'=4x$. So the derivative will be positive for all $x>0$. This means the tangent line will intersect the x-axis at some point $x_a<x$ for a given x.
This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $x<x_a$ which is made up of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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.