Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving a quadratic equation identity Continued question here
I encountered this question:
For quadratic polynomials such as $x^2\pm5x\pm6$ or $x^3\pm5x\mp6$, can be factorised over the integers. The main problem is to find a generator which can generate every polynomial that have this identity.
What have I tried:
For ... | Let $m$ and $n$ be two positive integers such that the two polynomials
$$x^2+mx+n\qquad\text{ and }\qquad x^2+mx-n,$$
both factor over the integers. Note that then also the two polynomials
$$x^2-mx+n\qquad\text{ and }\qquad x^2-mx-n,$$
factor over the integers. It means there exist integers $p$, $q$, $s$ and $t$ such t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the flux of $F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through surface $S$ ... $4y^2+z^2=4, \space x\in [-2,2]$ I am trying to find the flux of $\vec F=\langle\sin(xyz), x^2y, z^2e^{x/5}\rangle$ through the surface $S$ where $S$ consists of the elliptical cylinder defined by
$S$ ... $4y^2+z^2=4, \space x\i... | While you can apply divergence theorem, this one is quite straightforward for direct surface integral as well. Given it is a cylinder, it is easier to set up the integral in cylindrical coordinates. An elliptic cylinder with form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be parametrized as $x = a \cos\theta, y = b \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$ find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.
| Observe that
\begin{align}
I=&\ \int^\infty_0 \frac{\sin x}{2x(1-\frac{1}{2}\sin^2x)}\ dx = \frac{1}{4}\int^\infty_{-\infty} \frac{\sin x}{x}\sum^\infty_{k=0}\frac{\sin^{2k}x}{2^k}\ dx\\
=&\ \frac{1}{4}\sum^\infty_{k=0}\frac{1}{2^k}\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx
\end{align}
If you believe that
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$ I'm working on a diffraction simulator. In order to take into account for slit-width, I need to take this integral:
$$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$$
Where $x$ and $y$ are the coordinates of the p... | As you, I am more than skeptical about a closed form.
For an approximation, you could try this $1,400$ years old approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad-\frac \pi 2 \leq t\leq\frac \pi 2$$
This would give
$$\int \cos \left(\sqrt{(h+y)^2+x^2}\right)\sim \frac{5 \pi ^2 }{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\rig... | The inequality
$$
\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right)
$$
holds for all real numbers $x, y$ satisfying $0 < y < x$.
With the substitution $t=y/x$ this is equivalent to
$$ \tag{*}
\log\left( \frac{1+t}{1-t}\right)>2\left(\sqrt{1+t}-\sqrt{1-t}\right)
$$
for $0 < t < 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Compute the probability mass function of X There are two bags A and B. Bag A has 2 green marbles and 3 red marbles. Bag B has 2 green marbles and 3 blue marbles. Two marbles are drawn without replacement from bag A and put in bag B. Then two marbles are drawn without replacement from bag B and put in bag A. Let X be th... | You wrote:
take all green from A and put non-greens back
First sight first error found...
$$\mathbb{P}[X=0]=\frac{2}{5}\cdot\frac{1}{4}\cdot\frac{3}{7}\cdot\frac{2}{6}=\frac{1}{70}$$
At present I did not check the rest
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sum_{n \leq x} \tau (n) = 2(\sum_{n \leq \sqrt{x}} [\frac{x}{n}]) - [\sqrt{x}]^2$ Here $\tau(n)$ is the number of positive integers dividing $n$ and $[x]$ is the floor of $x$.
So I know if $n$ is not a perfect square, then half the positive integers dividing $n$ are less than $\sqrt{n}$, and the other half are ... | $$\begin{align}\sum_{n=1}^x\tau(n)&=\sum_{n=1}^x\#\{\,(a,b):ab=n\,\}\\
&=\#\{\,(a,b,n):ab=n\le x\,\}\\&
=\#\{\,(a,b):ab\le x\,\}\\&
=\#\{\,(a,b):ab\le x, a\le b\,\}+\#\{\,(a,b):ab\le x, a\ge b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\#\{\,(a,b):ab\le x, a\le b\,\}-\#\{\,(a,b):ab\le x, a= b\,\}\\&
=2\sum_{a\le\sqrt x}\#\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determine if $x^3+y^3+z^3+t^3 = 10^{2021}$ has a solution I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions
PowersRepresentations[10^2021, 4, 3]
return
PowersRepresentations::ovfl: Overflow occurred in computation.
FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x ... | Two solutions may be found starting from:
$$10^{2021}=10^5\times10^{2016}=12500\times2^3\times10^{3\times672}$$
Since $12500=19^3+17^3+8^3+6^3=18^3+17^3+12^3+3^3$ we have:
$$10^{2021}=(38\times10^{672})^3+(34\times10^{672})^3+(16\times10^{672})^3+(12\times10^{672})^3$$
$$10^{2021}=(36\times10^{672})^3+(34\times10^{672}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Point within the interior of a given angle The point $M$ is within the interior of given angle $\alpha$. Find the distance between $M$ and the vertex of the angle ($OM=?$) if $a$ and $b$ are the distances from $M$ to the sides of the angle.
We can see that $$OM^2=OP^2+PM^2=OK^2+KM^2$$
Let $OP=x;PK=y$. Then $$2OM^2=OP^... | Let $d=OM$
$$ \arcsin(a/d)+ \arcsin(b/d) = \alpha$$
$$ a \sqrt{d^2-b^2} + b \sqrt{d^2-a^2} = d^2\sin \alpha$$
Let Mathematica solve fourth order equation for $d$
$$sal= \sin \alpha$$
{d -> 0.5` Sqrt[(4.` (1.` + 1.` b^2))/sal^2 +
2.` Sqrt[(4.` (1.` + 1.` b^2)^2)/sal^4 - (
16.` (0.25` - 0.5` b^2 + 0.25` b^4 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Finding equilibrium points of system of equations I'm asked to solve system of ODE for finding equilibrium points of this
$$\frac{dx}{dt} = a_1x(1-x) + a_2x\frac{z}{1+z}+a_3xy$$
$$\frac{dy}{dt} = b_1y(1-y) + b_2y\frac{z}{1+z}+b_3xy$$
$$\frac{dz}{dt} = c_1z(1-z) + c_2z\frac{x}{1+x}+c_3z\frac{y}{1+y}$$
I checked some rel... | Hint
An equilibrium point is a triple $\left(x , y , z\right)$ satisfying the non-linear system
\begin{equation}
\renewcommand{\arraystretch}{1.5} \left\{\begin{array}{rcl}{a}_{1} x \left(1-x\right)+{a}_{2} x \displaystyle \frac{z}{1+z}+{a}_{3} x y&=&0\\
{b}_{1} y \left(1-y\right)+{b}_{2} y \displaystyle \frac{z}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An Elegant New Proof of Herons Formula? A triangle with side lengths $a, b, c$ an altitude($h$), where the height($h_a$) intercepts the hypotenuse($a$) such that it is the sum of two side lengths, $a = u +v$ and height($h_b$) intercepts hypotenuse($b$) such that it is also the sum of two side lengths $b = x + y$, we ca... | In my post, I had shown that
$$
r^{2}(x+y+z)=x y z \tag*{(result 1)}
$$
and
$$
K=r s \tag*{(result 3)}
$$
where $s=x+y+z$ is the semi-perimeter of the triangle ABC whose area is $K$ and sides $a=y+z, b=z+x $ and $c=x+y$.
From (1), $r^{2} s=x y z$ implies that
$$
K^{2}=r^2s^2 =sxy z=s(s-a)(s-b)(s-c)
$$
Taking square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the function $f(x)$. Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?
I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.
| You stated, that $f$ is a polynomial function, hence we can write $f$ as $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ plugging that into your given equation we get
$$f(x+2)-f(x)=a_n(x+2)^n+a_{n-1}(x+2)^{n-1}+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$
using the binomial formula, we can extend this to
$$a_n(x^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to remove roots from an equation? The question here is how (if it is even possible) to remove the square root terms and transform the following equation to a polynomial with one unknown $x$.
The coefficients $a$, $b$, $c$, and $d$ are known and also $r$.
$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} +... | Well the $r-x^2$ screams that they want a trig substitution as lone student's answer suggest.
But you can always remove roots but bringing terms over and squaring.
$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$
$\sqrt{x}(a + b\sqrt{r^2-x^2}) = - d - c\sqrt{r^2-x^2}$
$x(a^2 + b(r^2-x^2) + 2ab\sqrt{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate from the number of solutions for $\sin 2x +\cos 3x - \sin 4x = 0$? The problem is as follows:
First find the number of solutions for the equation from below:
Assume $x \in [0,2\pi]$
$\sin 2x +\cos 3x - \sin 4x = 0$
Let $n$ be the number of solutions. Using this $n$ find the sum for:
$5\tan^2\left(\frac... | I Think $n=6$.
Algebraically:
$$\sin 2x +\cos 3x -\sin 4x = \\ 2\sin x \cos x +\cos 2x \cos x-2\sin x \cos x \sin x -4 \sin x
\cos x \cos 2x= \\ 2 \sin x \cos x +(1-2 \sin^2 x)\cos x - 2 \sin^2 x \cos x -2 \sin x \cos x(2-4 \sin^2 x) = \\ 2 \sin x \cos x (1-2+4 \sin^2 x) + \cos x (1- 2 \sin^2 x - 2 \sin^2 x)= \\ 2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution for approximating square roots? I just saw a post on instagram which said that
$$\sqrt{x}\approx \frac{x+y}{2\sqrt{y}}$$
I tried it out on a few values, and surprisingly, it came within 1 decimal point of the actual answer. Is there a reason for this or is it coincidental?
y is the closest perfect square to x
| Google AM–GM inequality.
If we square any real number $z$, which can be expressed as a difference of two other real numbers $x$ and $y$, the result is always greater or equal $0$, this means $\color{blue}{z^2 \geq 0}$, assume $z = x-y$, it follows $$\color{blue}{z^2} = (x-y)^2 = x^2 \color{red}{-2xy} + y^2 = x^2 \color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4045847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find the last two digit of $5^{121}*3^{312}$ The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$
and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of $5^{143}*3^{312}$ are $25$.
Don't know why $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies th... | I have no idea what the book is trying to do.
I guess Id do it this way.
$5^2\cdot 3^4 = 25 *81 = 2025 \equiv 25\equiv 5^2 \pmod {100}$.
So $(5^2 \cdot 3^4)^{78} \equiv 25^{78} \pmod {100}$.
SO $5^{156}\cdot 3^{312} \equiv 5^{156} \pmod {100}$.
Now $5^2 \equiv 25 \pmod {100}$ so by induction if $5^k \equiv 25 \pmod {10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Unique solutions for $\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text{and}\quad a+b+c=1.$
Let $A,B,C$ be strictly positive real numbers satisfying $A+B+C=1$ and let $a, b, c $ be real variables. Suppose $a, b, c$ satisfy the following system of equations:
$$\frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}=1,\quad\text... | The solution is unique only if $a,b,c > 0$. This is because by Titu's lemma: $1 = \dfrac{A^2}{a}+\dfrac{B^2}{b}+\dfrac{C^2}{c} \ge \dfrac{(A+B+C)^2}{a+b+c}= \dfrac{1^2}{1} = 1$, and you have equality which means the $=$ must hold, and this occurs when $A = ak, B = bk, C = ck \implies k =1 \implies a = A, b = B, c = C$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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For two odd primes $pLet us fix two primes $p,q$ with $2<p<q$. How can we find positive integers $a,b$ which solve the equation $a^2+b^4=pq$ without brute force?
Interestingly there exist sometimes two solutions:
*
*$5\cdot13=7^2+2^4=8^2+1^4$
*$5\cdot821=3^2+8^4=53^2+6^4$
*$17\cdot113=25^2+6^4=36^2+5^4$
In more rar... | With kind assistance I have received a useful hint for a direction to investigate, wich I would like to share with you all:
In the case that $p\equiv3\pmod4$ or $q\equiv3\pmod4$ no solution exist.
Let us set $p=r^2+s^2$ and $q=u^2+v^2$. If $p\equiv1\pmod4$ and $q\equiv1\pmod4$ we exactly obtain one solution $r>s>0$ for... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find constant $c$ such that all intersection points of two spheres have perpendicular tangent planes I have come to a problem in a multivariable calculus book that I'm having trouble with.
The problem statement is :
"Find a constant $c$ such that for any point of intersection of the two spheres
$(x-c)^{2} + y^{2} + z^{... | Math is beautiful in that a good problem gives you everything that you need to solve it, and nothing more.
Taking this into consideration, let's see what we're given.
We have:
$$ (x-c)^2 + y^2 + z^2 = 3 $$
$$ x^2 + (y-1)^2 + z^2 = 1 $$
So, we can define two functions:
$$
f(x,y,z) = (x-c)^2 + y^2 + z^2 - 3
$$
$$
g(x,y,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof regarding convergence I am supposed to show that the sequence converges and to find the limit.
We've just started going over sequences and induction so I'm very limited in what I know/can use and I was wondering if the general pattern of the proof is okay and if it's correct.. Also it took me quiet a bit of time ... | Your proof is honestly very tedious and should be replaced ! You must make use of the recursive property of the sequence directly. Here is how induction works for your problem. Base case: $a_2 = 4 \ge 3 = a_1$ is true. Assume $a_n \ge a_{n-1}$ for $n \ge 2$, then using the step $a_n \ge a_{n-1} \implies a_{n+1} = 1+\s... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).
By and large these things require considerations o... | 1875 book by Reuschle. The technique was initiated by Gauss, some 30 years before Galois theory.
Indeed, on page 249 in Cox, Galois Theory, we find
$$ \sum_{w=1}^{16} \; (w|17) \; \zeta_{17}^w \; = \sqrt{17}. $$
Letting $\eta_0$ refer to the quantity of interest, calling the sum of the others $\eta_1$ as in Reuschle b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove $\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$ for $a^{2}+b^{2}+c^{2}+abc=4$
Prove:
$$\frac{a^{2}}{b^{2}} +\frac{b^{2}}{c^{2}} +\frac{c^{2}}{a^{2}} +\frac{15abc}{4}\geq \frac{27}{4}$$ for $a,b,c>0$ such that $$a^{2}+b^{2}+c^{2}+abc=4.$$
I tried to note $a=\cos ... | With the following link $a^2+b^2+c^2+abc=4 \Rightarrow abc \leq 1$ thus the maximum value is $1$ and we assume that $a=1,b=1,c=1$. $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{15abc}{4}$, we substitut $1$ we get exactly $\frac{27}{4}$, this is the maximum value that we could get. If $a,b,c$ is less that $1$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Collatz Conjecture: Does it follow that since there is no $1$-cycle, if $ab > 1$, then $\frac{3^a - 2^a}{2^{a+b} - 3^a}$ cannot be an integer? For me, one of the most interesting results of the Collatz Conjecture is that if an $m$-cycle exists, then $m > 68$. So, there are no non-trivial $1$-cycles, $2$-cycles, ..., u... | Say you start with $n_0=2^kt-1$, apply $a$ times the Collatz function $f(n)=\frac{3n+1}{2}$, than $b$ times the Collatz function $f(n)=\frac{n}{2}$, you can easily see that you end up with $$n=\frac{3^a}{2^{a+b}}n_0+\frac{3^a-2^a}{2^{a+b}}$$
or
$$n\cdot{2^{a+b}}-3^a\cdot n_0=3^a-2^a$$
and in the case of a cycle $n=n_0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067889",
"timestamp": "2023-03-29T00:00:00",
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} |
A limit of a given sequence Let $(a_n)_{n\ge1}$ a sequence of real positive numbers such that $a_n+\frac{n}{a_{n+1}^2} \le a_{n+1} \le \frac{a_{n-1}^2}{a_n}+\frac{n+1}{a_n^2}$ for any $n \ge 1$. Prove that $b_n = \frac{a_n}{\sqrt[3]{n^2}}$ is a convergent sequence and find its limit.
First of all, it's clear that $a_n$... | We have $a_n$ increasing and positive, and
$$
a_{n} \leq a_{n+1} - \frac{n}{a_{n+1}^2},
$$
Thus $a_{n+1}^3 \geq n$, and
$$
\frac{a_{n-1}^2}{a_n} + \frac{n+1}{a_n^2} \leq a_n + \frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2},
$$
and using $a_{n}^3 \geq (n-1)$
$$
\frac{(n-1)^2}{a_n^5} + \frac{3-n}{a_n^2} \leq \frac{(n-1)^2}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Two mysterious missing angles in the sine values of acute angle list? There is a well known list of 5 for trig values of special angles between 0$^o$ to $90^o$:
$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$
, $\sin 60^o =\... | The missing angles are:
$$\sin 22.5º = \frac{\sqrt{2 - \sqrt2}}{2}$$
$$\sin 67.5º = \frac{\sqrt{2 + \sqrt2}}{2}$$
There are other values listed here on Wikipedia, but the pattern is not obvious. For instance, $\sin 18º = \frac{\sqrt5 - 1}{4}$, but $\arcsin \frac{\sqrt5 + 1}{4}$ is not $90º - 72º = 36º$ but $54º$. In fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the sum of a sequence of reciprocal of factorial I wonder how to calculate the sum of the below sequence
\begin{align}
S & = \frac{1}{2!} - \frac{2}{3!} + \frac{3}{4!} - \frac{4}{5!} + \frac{5}{6!} - ... \\
& = \frac{2 - 1}{2!} - \frac{3 - 1}{3!} + \frac{4 - 1}{4!} - \frac{5 - 1}{5!} + ...\\
& = 1 - \f... | $$\sum_{n!=2}^\infty (-1)^{n-1}\frac1{n!} = 1 - \frac1e.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to derive the sum-to-product formula for cos(a-b)-cos(a+b)? I know how to derive the sum-to-product formulas for cos+cos, sin+cos, and cos+sin. But I cannot seem to do it for cos-cos. My problem is the negative out front. I cannot product it from any of my proofs.
Here is my attempt:
I know that cos(a-b) = cosacosb... | Develop the right side:
$$-2\sin\frac{a+b}2\,\sin\frac{a-b}2=-2\left[\sin\frac a2\cos\frac b2+\sin\frac b2\cos\frac a2\right]\left[\sin\frac a2\cos\frac b2-\sin\frac b2\cos\frac a2\right]=$$
$$\stackrel{\text{squares difference}}=-2\left[\sin^2\frac a2\,\cos^2\frac b2-\sin^2\frac b2\,\cos^2\frac a2\right]=$$
$$=-2\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solving $|2x-5| + |7-2x| =2 $ $$|2x-5| + |7-2x| =2 $$
I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-
*
*1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.
*2nd case, where $2.5<x\le3.5... | The short cut way: let $y=2x$, what are the values that $y$ can take such that the distance from $5$ and the distance from $7$ are exactly $2$. If $y$ is less than $5$, then the distance from $7$ is beyond $2$. SImilarly if $y$ is more than $7$, then the distance from $5$ is beyond $2$. We can easily check that any val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with $\int \frac{1}{(1-t^2)t^2} \, dt$ $$\int \frac{1}{(1-t^2)t^2} \, dt$$
By using partial fractions I get:
$$\frac{1}{(1-t)(1+t)t^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+1} + \frac{D}{1-t}$$
$$1 = t^3 (-A-C+D) + t^2 (-B+C+D) + At+ B$$
$$-A-C+D = 0$$
$$-B+C+D=0$$
$$A=0$$
$$B=1$$
So, $A=0, B=1, C=\frac{1}{2}... | We may utilize $$\dfrac1{t^2(1-t^2)}=\dfrac{(1-t^2)+t^2}{t^2(1-t^2)}=\dfrac1{t^2}+\dfrac1{1-t^2}$$
Finally $$\dfrac1{1-t^2}=\dfrac{1+t+1-t}{2(1-t)(1+t)}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding the orthogonal trajectories of $y^2=x^2(1-cx)$ Here is my initial attempt:
$c=\frac{x^2-y^2}{x^3}$
$2y\frac{dy}{dx}=2x-3cx^2$
$-2y\frac{dx}{dy}=2x-\frac{3x^2-3y^2}{x}$
$-2xydx=(2x^2-3x^2-3y^2)dy$
$-2xydx=(-x^2-3y^2)dy$
$2xydx=(x^2+3y^2)dy$
$\frac{dy}{dx}=\frac{2xy}{x^2+3y^2}$
Let $y=ux$; $\frac{dy}{dx}=u+x\frac... | I find always a little sad that such issues aren't associated with a graphical representation in order to have a geometrical feeling of what really happens.
In particular here, I wondered how a family of conical curves (a family of ellipses tangent to the $x$-axis, in red) could have an orthogonal family with third deg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve definite integral $\int_{\frac{\sqrt2}2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$ by variable substitution I have this integral
$$\int_{\sqrt{2}/2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$$
I have to solve it by substituting x.
I believe that the easiest way would be to substitute with $t = \sqrt{x^2+1}$. The other way... | Note that, with the substitution $t=\sqrt{x^2+1}$
$$ x = \sqrt{t^2-1}, \>\>\>\>\>dx = \frac t{\sqrt{t^2-1}}dt$$
Then, the integral becomes
$$\int_{\frac{\sqrt{2}}2}^{\sqrt{3}} \dfrac{1}{x\sqrt{x^2+1}}dx
= \int_{\sqrt{\frac32}}^{2} \dfrac{1}{t^2-1}dt
= \frac12 \ln\frac{t-1}{t+1}\bigg|_{\sqrt{\frac32}}^{2}\\
= \frac12\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to f... | Let $x=\sqrt{6 + \sqrt{6 + \sqrt{6 + …}}}$ Next, square both sides: $x^2=6 + \sqrt{6 + \sqrt{6 + …}}$ Then, subtract 6 from each side: $x^2-6=\sqrt{6 + \sqrt{6 + …}}$ Wait a minute..... the right side of the equation is x! Substituting in, we get $x^2-6=x$, and $x^2-x-6=0$. This is a simple quadratic we can solve, yiel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I... | $3^2=9.$
$3.5^2 = \frac{35}{10} \times \frac{35}{10} = \frac{35 \times 35}{100} = \frac{1225}{100} = 12.25.$
$9 < 12 < 12.25.$
$\sqrt{9} < \sqrt{12} < \sqrt{12.25}.$
$3< \sqrt{12} < 3.5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3... | Use trig identity: $\sin^2\theta+\cos^2\theta=1$ $$\implies\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\left(\frac{3}{\sqrt{10}}\right)^2}=\frac{1}{\sqrt{10}}\quad \forall \quad 0\le\alpha\le \frac{\pi}{2}$$
$$\implies \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^2}=\frac{1}{\sqrt{5}}\quad \foral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, i... | If $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$ then $a_3$$=0$
So $a_1$ = $a_{3}/2$ $=0 .$ But you can’t divide by $0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Proving Proposition 2.4.3. in Artin I am trying to prove Proposition 2.4.3. in Artin. He leaves this fact unproved. It states:
Let $x$ be an element of finite order $n$ in a group, and let $k$ be an integer that is written as $k = nq + r$ where $q$ and $r$ are integers and $r$ is in the range $0 \leq r < n$. Then:
(a)... | You proofs of parts a) and b) are correct.
For part c), note first that $$\left(x^k\right)^{n/d}=\left(x^n\right)^{k/d}=e^{k/d}=e$$ so that the order of $x^k$ divides $\frac nd$.
On the other hand, let $m$ be a positive integer such that $\left(x^k\right)^m=e.$ Then $x^{km}=e$ so that $n\mid km$, and the $$\frac nd \mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets... | Well the very least $x$ can be is $0$ and if so, you can have $z$ be any value from $0$ to $20$ inclusive (by letting $y = 20-z$). SO all (integer) values from $-20$ to $0$ are possible but no smaller are possible.
Now we just have to find all possible positive values.
Every positive integer is a multiple of $4$ minus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to find reduction formula for $I_n=\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$ I have to find the reduction formula for the following :
$$\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$$
I took this integral from wikipedia. By using parts this is what I got.
${2a^{-1}(px+q)^{n}\sqrt{ax+b}-2pna^{-1}\int\{px+q}^{n-1}\sqrt{ax+b}dx$
The a... | We first perform integration by parts on $\sqrt{ax+b}$.
\begin{aligned}
I_{n} &=\int \frac{(p x+q)^{n}}{\sqrt{a x+b}} d x \\
&=\frac{2}{a} \int(p x+q)^{n} d(\sqrt{a x+b}) \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int(p x+q)^{n-1} \sqrt{a x+b} d x
\end{aligned}
Rationalization gives back our integrals. $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the matrix totally unimodular? Let $A=\begin{bmatrix}1&0&1&1&0&0&1&0&0\\0&1&0&1&0&0&0&0&0& \\0&0&1&1&1&0&0&0&0\\0&0&-1&-1&0&1&-1&0&0\\0&0&0&-1&0&0&0&1&0\\0&0&-1&-1&0&0&0&0&1\end{bmatrix}$, is the matrix totally unimodular?
I realised that if I change the order of columns in A I could get change the matrix I would g... | Successively applying column-swapping, $C_3 \leftrightarrow C_5$, $C_4 \leftrightarrow C_6$, $C_7 \leftrightarrow C_8$ and $C_8 \leftrightarrow C_9$, we get,
$A^{\prime}=\begin{bmatrix}1&0&0&0&0&0&1&1&1\\0&1&0&0&0&0&0&1&0& \\0&0&1&0&0&0&1&1&0\\0&0&0&1&0&0&-1&-1&-1\\0&0&0&0&1&0&0&-1&0\\0&0&0&0&0&1&-1&-1&0\end{bmatrix}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$u_{n+1}=u_n^{u_n}$
*
*$u_0>1$
*$u_{n+1}=u_n^{u_n}$
We want to show that :
*
*the sequence $u_n$ diverges
*$ \sum \dfrac{1}{u_n}$ converges
*$ \exists N \in \mathbb{N} , \forall k : u_{N+k} \geq k+2$
*$\exists C >0, \forall n \geq N : \sum_{k=n+1}^{ \infty} \dfrac{1}{u_k} \leq \dfrac{C}{u_{n+1}} $
*$\forall... | We can show by induction that $u_n >u_0^{u_0^{n}}$ for all $n$. Since $u_0 >1$ this will certainly imply converegence of $\sum \frac 1 {u_n}$.
[$u_0^{n}=(1+(u_0-1))^{n}> (u_0-1)n$ by Binomial expansion so $u_0^{u_0^{n}}>u_0^{(u_0-1)n}$. Apply ratio test].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find integral $\int_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$ Find $$\int\limits_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$$
$$
\lim_{a\rightarrow 0^+}\int\limits_{a}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx=\lim_{a\rightarrow 0^+}-\int\limits_{0}^{1}\left ( 1-x \r... | A simpler way to evaluate it is to use the following identity:
$$\boxed{\int _0^1\frac{x\ln ^n\left(t\right)}{1-xt}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(x\right)}$$
Applying it to the desired integral yields:
$$\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=\int _0^1\ln \left(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In an acute angled triangle, prove that $\sin A + \sin B + \sin C > 2$, where $0Is there any way to prove this? When I check intuitively by taking $A=0$ and $B=C=\frac{\pi}{2}$, the value of the expression becomes $2$ and as I changed the angles the value kept increasing in $0$ to $\frac{\pi}{2}$. I tried using Jensen'... | $$\sin A + \sin B + \sin C=\sin A +\sin B + \sin(\pi-A-B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+\sin(A+B)$$
$$=2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})+2\sin(\frac{A+B}{2})\cos(\frac{A+B}{2})$$
$$=2\sin(\frac{A+B}{2})\Bigl(\cos(\frac{A-B}{2})+\cos(\frac{A+B}{2})\Bigr)$$
$$=4\sin(\frac{A+B}{2})\Bigl(\cos\frac{A}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respective... | Let
\begin{align}
a^2+9&=k
\tag{1}\label{1}
,\\
b^2+16&=k
\tag{2}\label{2}
,\\
1+(a+b)^2&=k
\tag{3}\label{3}
\end{align}
for some $k>0$.
Then \eqref{3}$-$\eqref{1}$-$\eqref{2} gives
\begin{align}
ab &= 12-\tfrac12\,k
\tag{4}\label{4}
,\\
a^2b^2 &= (12-\tfrac12\,k)^2
\tag{5}\label{5}
.
\end{align}
On the other hand, fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometr... | Writing $\sin^2A=a,$ we have $a+a^2=1$
We need $$1+a=\dfrac a{1-a}$$
As $1-a\ne0,1+1^2\ne1$ $$\iff a=(1-a)(1+a)\iff a=1-a^2\iff a+a^2=1$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Volume inside paraboloid and sphere
Find the volume inside the paraboloid $az=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2a^2$
Since we have $x^2+y^2$ I thought about polar coordinates , but before that I showed the functions as $z_1=\frac {x^2+y^2}{a}$ for the paraboloid and $z_2=\sqrt{2a^2-x^2-y^2}$ and I am not s... | The intersection between the sphere and the paraboloid takes place when $2a^2-z^2=az$, that is, when $z=a$. So, if $(x,y,z)$ belongs to the region whose volume we are trying to compute, then:
*
*if $z\leqslant a$, then $0\leqslant x^2+y^2\leqslant az$;
*if $z\geqslant a$, then $0\leqslant x^2+y^2\leqslant 2a^2-z^2$.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$4k^2+n$ is prime Find all $n$ positive integers such that $4k^2+n$ is prime for every nonnegative integer $k$ less than $n$
My progress $k=0$ we get $n$ is a prime number
let $k=n-l$ we get $n+4n^2+4l^2-8nl$ is a prime number
which means $l$ doesn't divide $n(4n+1)$ since $n$ is prime
$l$ doesn't divide $(4n+1)$ for ... | Note $n$ must be an odd integer so $n \equiv 1, 3 \pmod{4}$. As you already stated, for primes $n \equiv 1 \pmod{4} \implies n = 4j + 1, \; j \in \mathbb{N}$, then $k = j$ gives
$$4j^2 + 4j + 1 = (2j + 1)^2 \tag{1}\label{eq1A}$$
As $n \gt 1 \implies j \gt 0$, this is not a prime. Thus, this only leaves $n \equiv 3 \pmo... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
Find a lower bound for a Riemann integral. Let $a,b$ be strictly positive constants and let $p\in(1/4,3/4)$. Is there $c>0$ such that
$$\int_0^1(1-t^a)^{1/2+p}[2t-(2+b)t^{b+1}]dt>c \quad ?$$
*The constant $c>0$ may depend on $a,b$ or $p$.
I tried several values for $(a,b,p)$, and always observed a positive integral. S... | Firstly, note that
\begin{align}
\int_0^1\lvert 1-t^a\rvert^{1/2+p}\lvert 2t-(2+b)t^{b+1}\rvert dt\leq\int_0^12t-(2+b)t^{b+1}=2<\infty.
\end{align}
Suppose that $1/4<p<1/2$. Now, using the Fubini's Theorem and the Binomial Theorem for the fractional exponent $q:=1/2+p$ ,
\begin{align}
&\int_0^1(1-t^a)^q[2t-(2+b)t^{b+1}... | {
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"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | This is a partial answer which leaves one integral unevaluated. I'm not sure if it really helps or if it is just a reformulation replacing an unevaluated sum by an unevaluated integral.
We are looking for the sum
$$s = \sum_{k=3}^{\infty} \frac{\log(k)}{k^2-4}$$
Inserting
$$\log(k) = \int_0^1 \frac{x^{k-1}-1}{\log (x)}... | {
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"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 5
} |
Discretization of second order ODE, apply Z transform and inverse I have the following ODE:
$\frac{\mathrm{d^2y(t)} }{\mathrm{d} t} + 2\frac{\mathrm{dy(t)} }{\mathrm{d} t}+4y(t)=e^{-2(t-2)}u(t-2)$
With $u(t)$ being the unit step function. I am than asked to discretize the ODE (with sampling T=1), which gave me:
$y(n+2)... | You can perform partial fractions and simplify first.
Let,
$${Y(z)} = \frac{1}{z(z-e^{-2})(z+\sqrt3i)(z-\sqrt3i)} =\frac{a}{z}+\frac{b}{z-e^{-2}}+\frac{c}{z+\sqrt3i}+\frac{c^*}{z-\sqrt3i}$$
$z = 0 \Rightarrow a = -\dfrac{e^2}{3}$
$z= e^{-2} \Rightarrow b = \dfrac{e^6}{1+3e^4}$
$z = -\sqrt3i \Rightarrow c =\dfrac{1}{-\s... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$
Prove by induction that for $m, n ∈ N$, $m^{2n+1} − m$ is divisible by $6.$
What I have thus far:
Base case: $m=n=0$;
$0^{0+1}-0 = 0$, which is divisible by $6$.
Base case(2): $m=n=1$;
$1^{2+1}-1 = 0$, which is divisible by $6$.
Proposition: ... | Your induction step isn't right
$m^{2(n+1) + 1} - m = m^{2^n+1}m^2 - m=$
$([m^{2n+1} -m] + m)m^2 - m = $
$(6k + m)m^2 - m =$
$6km^2 + m^3- m$.
So we must prove $m^3 -m$ is always a multiple of $6$.
Hint: $m^3 - m = m(m^2-1) = m(m-1)(m+1)$
=====
In general you can do two proofs by induction, one where one variable is i... | {
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"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\sum_{n=1}^{\infty}\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n}$ Proving the convergence of a series by Weierstrass M-test.
$$\sum_{n=1}^{\infty}\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n}.$$
$$\frac{(2^n + 3^n)\sin(n)}{2^n + n^2\cdot3^n} \leq \frac{2^n + 3^n}{2^n + n^2 \cdot 3^n} = \frac{1 + \left(\frac{3}{2}\right)... | \begin{align*}
\left|\dfrac{(2^{n}+3^{n})\sin n}{2^{n}+n^{2}3^{n}}\right|\leq\dfrac{2^{n}+3^{n}}{2^{n}+n^{2}3^{n}}\leq\dfrac{3^{n}+3^{n}}{n^{2}3^{n}}=\dfrac{2}{n^{2}},
\end{align*}
now $\sum n^{-2}<\infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
| General solution:
I will solve a specific quartic equation that is a specific case of a general quartic equation.
Let, $a≠0,~ b≠0$, then we have
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x^2+\frac e{ax^2}+\frac ba x+\frac d{ax}+\frac ca=0$$
$$\begin{align}&x^2+\frac ea \times \frac 1{x^2}+\frac ba \left(x+\frac {d}{bx}\right)+\fra... | {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
} |
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
... | Consider
$$\left( \begin{matrix}
n \\
k \\
\end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=R}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$
so
$$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix}
n+r \\
r \\
\end{matrix} \right)}=\frac{1}{2\pi i}\int\limits_{\left| z \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 8,
"answer_id": 4
} |
A system of equations with degree 2 Let $a,b,c,d \in \mathbb R$. Suppose the following holds
\begin{align}
a^2-c^2 &=1 \\
b^2-d^2 &=-1 \\
ad-bc &= \pm1 \\
ab-cd &=0
\end{align}
How can I find
$a,b,c$ and $d$. I'm trying to show something regarding differential geometry
and arrive with these equations. I tried to solve... | Just two by two matrices,
$$
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{array}
\right)
\left(
\begin{array}{cc}
a&-b \\
-c&d \\
\end{array}
\right) =
\left(
\begin{array}{cc}
1&0 \\
0&1 \\
\end{array}
\right)
$$
so that the right hand factor must be the inverse of $ M=
\left(
\begin{array}{cc}
a&c \\
b&d \\
\end{arr... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Set of all real values of $a$ for which the equation $(a-4)\sec^4x+(a-3)\sec^2x+1=0, (a\ne4)$ has real solutions
Find set of all real values of $a$ for which the equation $(a-4)\sec^4x+(a-3)\sec^2x+1=0, (a\ne4)$ has real solutions.
Let $\sec^2x=t$. So, the equation becomes $(a-4)t^2+(a-3)t+1=0$.
Since $\sec^2x\ge1\im... | I think you overdid it. It is quite neat
$$\sec^2x=\frac{-(a-3)\pm\sqrt{(a-3)^2-4(a-4)}}{2(a-4)}=-1,\frac{1}{4-a}$$
So, obviously $\sec^2x\geq 1$
So $$\frac{1}{4-a}\geq 1$$
which gives $\frac{a-3}{a-4}\leq 0$
Thus $a\in[3,4)$
| {
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"answer_count": 1,
"answer_id": 0
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Find the sum of $\sum\limits_{n=-\infty}^{\infty}\frac{2^n}{1+7^{2^n}}$ My attempt:
$\sum\limits_{n=-\infty}^{\infty}\frac{2^n}{1+7^{2^n}}=\sum\limits_{n=-\infty}^{1}\frac{2^n}{1+7^{2^n}}+\sum\limits_{1}^{\infty}\frac{2^n}{1+7^{2^n}}+\frac 18\\
=\sum\limits_{1}^{\infty}\frac1{2^n(1+7^{1/2^n})}+\sum\limits_{1}^{\infty}\... | You can write this as a telescoping sum. Note that
$$
\frac{2^n}{7^{2^n}-1} - \frac{2^n}{7^{2^n}+1}=\frac{2^n(7^{2^n}+1)-2^n(7^{2^n}-1)}{7^{2^{n+1}}-1}=\frac{2^{n+1}}{7^{2^{n+1}}-1}.
$$
So
$$
\begin{eqnarray}
\sum_{n=-\infty}^{\infty}\frac{2^{n}}{7^{2^n}+1} &=& \sum_{n=-\infty}^{\infty}\left(\frac{2^n}{7^{2^n}-1} - \f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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A Fibonacci conjecture: $\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}$ Given the Fibonacci sequence $F_n$, that is $$F_1=F_2=1,F_{n+2}=F_{n+1}+F_n.$$
I find that
$$
\log_21<\log_32<\log_53<\log_85<\cdots,
$$
i.e. we have $$
\log_{F_{n+1}}{F_n}<\log_{F_{n+2}}{F_{n+1}}, \quad \text{for} \; n\geqslant1\tag{1}
$$
If $n=2k$,... | Conjecture: $ \log_{F_{n+1} } F_n < \log_{F_{n+2} }F_{n+1}.$
This is equivalent to:
$$\dfrac{\log F_{n}}{\log F_{n+1}} < \dfrac{\log F_{n+1}}{\log F_{n+2}}$$
Multiply both sides by $\log F_{n+1}\log F_{n+2}$ and you'll get $\log(F_n )\log (F_{n+2}) < (\log (F_{n+1}))^2$ ($\log$ here is the natural logarithm).
One c... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.... | Reduce to this:
$(a-b)(a^3-b^3)\ge 0$
Do you see why it is always satisfied?
| {
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"answer_count": 3,
"answer_id": 1
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Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right... | There is no need for so much work as you have done.
Just note that the expression under limit is of the form $$x^2(A^{1/7}-B)$$ where both $A, B$ tend to $1$. Then we can write the above expression as $$x^2\left(\frac{A^{1/7}-1}{A-1}\cdot(A-1)+1-B\right)$$ Now $x^2(1-B)\to 1/2$ and $$x^2(A-1)=\frac{x^2(x-1)}{x^3+1}\to ... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$ Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$
This looks like a problem that would do well with a conversion to polar coordinates ($r^2 = x^2 + y^2$ and $x = r\cos(\theta)$ $y = r\sin(\theta)$). However, I am confused on how to change $y' =... | $x = r \cos\theta, y = r \sin\theta$
$dx = \cos\theta \ dr - r \sin\theta \ d\theta$
$dy = \sin\theta \ dr + r \cos\theta \ d\theta$
Given equation is $dy = \frac{\sqrt{x^2+y^2}-x}{y} dx$
So we get,
$r \sin\theta \ (\sin\theta \ dr + r \cos\theta \ d\theta) = r \ (1 - \cos\theta) \ (\cos\theta \ dr - r \sin\theta \ d\t... | {
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"answer_id": 0
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Prove: $\cos^3 \frac x3 + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x$ I tried to solve the following identity:
$$\cos^3 \frac x3 + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x$$ I applied formulas, $\cos (a+b)$ and $\cos^3 x$ and I arrived at
$$\cos 3x + 2 \cos x - ... | The formula you arrived at is wrong (put $x=\frac {\pi}{2}$ to see that). A correct method would be this:
$$\cos 3x=4\cos^3 x-3\cos x$$
So, $$\cos^3 x=\frac {\cos 3x+3\cos x}{4}$$
Now you just have to apply this formula to the given three angles.
| {
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"question_score": "2",
"answer_count": 1,
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Prove that the number $1^k + 2^k + ... + 12^k$ is divisible by $13$ Prove that the number $1^k + 2^k + ... + 12^k$ is divisible by $13$ for $k = 1, 2, ..., 11$.
My attempt:
If k is odd number, then $1^k +...+6^k +7^k ... + 12^k \equiv 1^k + ... + 6^k - 6^k - 5^k - ... -1^k \equiv 0 (mod 13)$.
But if k is even number th... | To be specific: $2$ is a primitive root modulo $13$. Hence any number from $1$ to $12$ can be written uniquely as $2^n$, for some $n$ between $1$ and $12$. Hence after a rearrangement your sum becomes:
$$2^k+ 2^{2k}+ \dots + 2^{12k} \pmod{13}.$$
If we denote your sum by $S$, then
$$ S \equiv \sum\limits_{n=1}^{12} 2^... | {
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Prove that $\sqrt[n]{a^n+x}=a+\frac{x}{na^{n-1}}-r$ ,$a>0$,$x>0$ Prove that
$\sqrt[n]{a^n+x}=a+\frac{x}{na^{n-1}}-r$ ,$a>0$,$x>0$
given $0$$<$$r$$<$$\frac{n-1*x^2}{2n^2*a^{2n-1}}$.
What I did.
$\sqrt[n]{a^n+x}=a(1+\frac{x}{a^n})^\frac{1}{n}=a(1+\frac{1}{n}*\frac{x}{a^n}+\frac{1}{2}*\frac{1}{n}*\frac{1-n}{n}*(\frac{x}{a... | The second term should be
$\frac{1}{2}\frac{1}{n}(\frac1{n}-1)(\frac{x}{a^n})^2
=-\frac{n-1}{2n^2}(\frac{x}{a^n})^2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \... | The parabolas represented by the quadratic functions $ \ P(x) \ = \ ax^2 + bx + c \ $ and $ \ Q(x) \ = \ – ax^2 + dx + c \ \ , \ \ ac \ne 0 \ \ , $ "open" in opposite "vertical" directions and have a common $ \ y-$intercept. The absolutely minimal number of $ \ x-$intercepts that could be arranged is one, by placing ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find all integers n such that $(\frac{n^3-1}{5})$ is prime
Find all integers n such that $(\frac{n^3-1}{5})$ is prime?
My Approach:
I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.
$n^3-1=10,15,25,35,55,...,215$
which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$
My doubt is that h... | If we split all integer $n$ into $[5m+1,5m+2,5m+3,5m+4,5m+5]$ you can show that only numbers $(5m+1)^3-1$ are divisible by $5$ as $[2^3-1,3^3-1,4^3-1,5^3-1]$ are all not divisible by $5$
Now $$\frac{(5m+1)^3-1}{5}=3m+15m^2+25m^3=m(3+15m+25m^2)$$
and the product $m(3+15m+25m^2)$, can only be a candidate prime if $m=\pm1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$? Between which two consecutive integer numbers is $\sqrt{2}+\sqrt{3}$?
My thoughts: $\sqrt{2}$ is $\approx1,4$ and $\sqrt{3}$ is $\approx{1,7}$ so their sum must be of the interval $(3;4)$. Any more strict approaches? Thank you in advance!
| First we have $\sqrt{2} + \sqrt{3} < \sqrt{4} + \sqrt{4} = 4$. To show that $\sqrt{2} + \sqrt{3} > 3$, we square both sides, and see that it suffices to show that $5 + 2\sqrt{6} > 9$. But this follows from $\sqrt{6} > 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $ I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double series which I have asked here. How can I solv... | Here is a magical solution where no advanced results are used.
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$
$$=-\underbrace{\int _0^{\infty }\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx}_{J}+\underbrace{\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$ Find all $x,y,z>0$ such that
$$3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$$ $$xy+yz+zx = 1$$
The only solution should be $x=\frac{1}{3}$, $y = \frac{1}{2}$, $z=1$.
There is a way to do it with $x = \tan \a... | Rewrite the second equation $xy+yz+zx = 1$ as $z=\frac{1-xy}{x+y}$ and evaluate
$$z+\frac1z= \frac{(x^2+1)(y^2+1)}{(x+y)(1-xy)}$$
Substitute above $z+\frac1z = \frac{z^2+1}z$ into the first equation
$$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y = \frac{5(z^2+1)}z \tag1$$
to get
\begin{align}
&(x^2+1)\left(\frac3{5x}- \frac{y^2+1... | {
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Let $c$ be a positive real number for which the equation $x^4-x^3+x^2-(c+1)x-(c^2+c)=0$ has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$ Let $c$ be a positive real number for which the equation
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$
I tried to to solve usin... | Suppose that $\alpha$ is a root to the equation:
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
Okay, then we have:
$\alpha^4-\alpha^3+\alpha^2-(c+1)\alpha-(c^2+c)=0$
$\begin{align}
\text{You can} & \text{put it in standard form as:} \\
& \\
& c^2+(1+\alpha)c^1 + (-\alpha^4+\alpha^3-\alpha^2 + \alpha)c^0=0 \\
\end{align}$
$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find a linear transformation $T\begin{pmatrix}-1\\-2\end{pmatrix}$ $T:\mathbb{R}^2 \Rightarrow\mathbb{R}^3, T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\4\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}4\\-1\\1\end{pmatrix}$. Find $T\begin{pmatrix}-1\\-2\end{pmatrix}$ and $T\begin{pmatri... | It's easy to see that the matrix representation of $T$ is $A=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1}$
so applying this transformation to a vector $(x,y)$ you get $$T_A(x,y)=\pmatrix{-1 & 4 \\ 4 &-1 \\ 3 &1} \pmatrix{x \\ y}=\pmatrix{-x+4y \\ 4x-y \\ 3x+y }$$
so now just plug the values you want $$T_A(-1,-2)=\pmatrix{-7 \\-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4160043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of an AP. If these are three consecutive terms of a GP, then $\cfrac{a}{c}$ is
Let $a, b$ and $c$ be the $7$th, $11$th and $13$th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\cfrac{a}{c}$ is ... | It is unfortunate that the complete question was not posted, though one might suspect this was a multiple-choice question; it would be helpful, but not essential (as we'll see), to know whether we are to arrive at a numerical value or just an expression in terms of the arithmetic difference and/or the geometrical facto... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$ to function I want to calculate $3-\frac{5}{2}+\frac{7}{24}-\frac{9}{720}+...$, while using $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$.
Here is my attempt:
Firs I proved that the radius of converges for this series is $R= \infty$.
$$\sum_{n=1}^\... | $$S=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-2)!}(2n+1)x^{2n}$$
Let $k=n-1$, then
$$S=\sum_{k=0}^{\infty} (-1)^k (2k+3)\frac{x^{2k+2}}{(2k)!}$$
We konw that $$x^3\cos x=\sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+3}}{(2k)!}$$
D.w.r.t. $x$, then
$$-x^3\sin x+3x^2 \cos x=S$$
OP's summation is verified. The sum of the numerical... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding the coordinate of a point after multiple transformations
Find the new coordinates of $(2,3)$ if following transformations take place:
(i) Origin is shifted to $(1,1)$
(ii) Axes are rotated by an angle $45^{\circ}$ in anticlockwise sense
(iii) Origin is shifted to $(1,1)$ and then axes are rotated by angle $45^... | (i) Origin is shifted to $(1, 1)$, so coordinates become $(1, 2)$.
(ii) If the axes are rotated by $45^\circ$ anticlockwise then the relation between
the original coordinates $(x,y)$ and the new coordinates $(x', y')$ is
$ x = \cos 45^\circ x' - \sin 45^\circ y' $
$ y = \sin 45^\circ x' + \cos 45^\circ y' $
which impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$ The following exercise is from the book of Churchill of complex analysis. Solve the integral
$$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$$
where $-1<a<3$ the book of complex analysis of Churchill suggests the indented countour:
And also the book gives you the answer $\frac{\pi(1-a)}... | $$ \left | \int_{C_r}\frac{z^a}{(z^2+1)^2} dz \right |\le \int_{C_r} \left | \frac{z^a}{(z^2+1)^2} dz \right | \le \int_{C_r}\frac{r^a}{(1-r^2)^2} |dz|= \frac{\pi r^{a+1}}{(1-r^2)^2} \to 0$$
as $a+1 \gt 0$.
| {
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"url": "https://math.stackexchange.com/questions/4170309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $... | Just compute
$$
(u-v)^2=(b+b^4-b^2-b^3)^2=-(b^4+b^3+b^2+b+1)+5=5.
$$
We have used $b^5=1$ in the second step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{ab... | Another approach, as you requested.
I write your question as (L)$\iff $(R). Since (L) is clearly
valid if $a+b=0,$ it suffices to prove (L) implies (R).
I now assume that (L) is valid. By scaling, we can assume that $c=1.$
From
$$\frac{1}{a} +\frac{1}{b} +1=\frac{1}{a+b+1},$$
we get
$$\frac{a+b}{ab}=\frac{1}{a+b+1}-1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neithe... | $(2\sin\theta+\cos\theta)^2 = 3$
$4\sin^2\theta + \cos^2\theta + 4 \sin\theta \cos\theta = 3 \sin^2\theta + 3\cos^2\theta$
$\sin^2\theta + 4\sin\theta \cos\theta = 2\cos^2\theta$
Now dividing both sides by $\cos^2\theta$,
$\tan^2\theta + 4\tan\theta = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My a... | Let the polynomial be $P(x)$. It is given that for some polynomials $Q(x),Q_1(x)$
$$P(x)=(x^2+x+1)Q(x)+1-x$$
$$P(x)=(x^2-x+1)Q_1(x)+3x+5$$
Now it is well known that $x^2+x+1$ has the zeros as $\omega,\omega^2$ and $x^2-x+1$ has the zeroes $-\omega,-\omega^2$, where $\omega=\frac{-1+i\sqrt{3}}{2}$
So We get
$$P(\omega)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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How Can I calculate $\int_{0}^{1}\frac{x^{2n}}{x^2+3}dx$ While simplifying a big integral I had to calculate the following two integrals
$$\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}, \int_{0}^{1}\frac{x\space \tan^{-1}(x)}{3x^2+1}$$
$$I=\int_{0}^{1}\frac{x\space \tan^{-1}(x)}{x^2+3}$$
$$\implies I=\int_{0}^{1}\fr... | I think we can get a closed form. For that we simply solve
$$\Delta I_n +4 I_n = \frac{1}{2n+1},\ I_0 = \frac{\pi}{3\sqrt{3}}$$ (credit of @Martund)
with which we obtain
$$I_n = (-3)^n\left(C-\frac{1}{3}\sum\left(-\frac{1}{3}\right)^{n}\frac{1}{2n+1}\right)=(-3)^n\left(C-\frac{1}{3}\int^{-\frac{1}{3}} \frac{t^{2n}}{t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$(\frac{x}{x+1})^x$ decreasing I want to show that $f(x) = \left(\dfrac{x}{x+1}\right)^x$ is decreasing for $x > 0$; this is clear from plotting its graph. Taking its derivative,
$$f'(x) = \frac{x^x}{(x+1)^{x+1}}\left(1+(x+1)\log\left(\frac{x}{x+1}\right)\right).$$
So we need to show that $1+(x+1)\log\left(\frac{x}{x+1... | By taking a derivative, you can easily show that
$$\log(1+y) \le y$$
for all $y > -1$, with equality iff $y=0$ (this is a standard inequality). In particular, since $-1<-\frac{1}{x+1}<0$,
$$\log\left(\frac{x}{x+1}\right) = \log\left(1 - \frac{1}{x+1}\right) < - \frac{1}{x+1}.$$
Therefore
$$1+(x+1)\log\left(\frac{x}{x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum of $\cos^2(x) (\csc^2(\frac{\pi}{2n} - \frac{x}{n}) + \csc^2(\frac{\pi}{2n} + \frac{x}{n}))$ for $0 \leq x < \pi/2$ and $n \geq 3$ I am trying to proove that $x=0$ is the minimum point of the function
$$f(x) = \cos^2(x)\left( \csc^2(x_+) + \csc^2(x_-) \right)$$
in the interval $0 \leq x < \pi/2$ where $x_\pm = ... | Trying to proove that $x=0$ corresponds to $\large \color{red}{a}$ minimum point is not so difficult if you make a Taylor expansion of
$$f_n(x)=\cos ^2(x) \left(\csc ^2\left(\frac{\pi -2 x}{2 n}\right)+\csc ^2\left(\frac{\pi+2
x }{2 n}\right)\right)$$
$$f_n(x)=2 \csc ^2\left(\frac{\pi }{2 n}\right)\Bigg[1+ \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How do you prove the Cauchy product (multiplication) of two infinite power series (generating functions) which have different exponents/indices? I'm trying to multiply 2 generating functions ( (1/(1-x) ) and ( 1/(1-x^5) ) which have different denominations so that I can find the coefficient. When browsing math.stackexc... | \begin{align*}
\frac{1}{1-x} \frac{1}{1-x^r}
&= (1+x+x^2+\cdots)(1+x^r+x^{2r}+\cdots) \\
&= 1+x+x^2+\cdots \\
&+ x^r+x^{r+1}+x^{r+2}+\cdots \\
&+ x^{2r} + x^{2r+1} + x^{2r+2} + \cdots \\
&+ \cdots
\end{align*}
Every exponent from $x^0, \ldots, x^{r-1}$ is included once, every exponent from $x^r, \ldots, x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Computing the series $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}$ I want to know how to compute the closed form for the series
$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}, \quad |A| < 1,$$
with or without special functions. When I plugged this into Mathematica, it gave... | Just for reference, I guess, I finally did go back and take the suggestion of @samario28 and use the generalized binomial theorem,
$$ (1 + z)^\alpha = \sum_{n=0}^\infty \frac{(\alpha)_n}{n!} z^n = \sum_{n=0}^\infty \binom{\alpha}{n} z^n,$$
where $(\alpha)_n$ denotes the falling factorial. Just for good measure, a ratio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Derivative of an integral with change of variable. I have this problem.
I want to compute the following integral in terms of $z$, where $x = az$
$$
\frac{d}{da}\int a^2x^2dx
$$
Computing the integral first, the derivative as the last passage, I obtain what I believe to be the correct solution, i.e.,
$$
\frac{d}{da}\int... | We substitute $x=az$ right from the beginning and obtain
\begin{align*}
\color{blue}{\frac{d}{da}\int a^2x^2dx}&=\frac{d}{da}\int a^2(az)^2d(az)\\
&=\frac{d}{da}\left(a^5\int z^2 dz\right)\tag{1}\\
&=\left(\int z^2 dz\right)\left(\frac{d}{da} a^5\right)\tag{2}\\
&\,\,\color{blue}{=\left(\frac{1}{3}z^3+C\right)\left(5a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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What is the origin and meaing of the term "Telescoping Series"? I looked into Carl. B. Boyer and Morris Kline books of math history, some calculus books like Apostol and Swokowski, many pages on the internet and even the Tractatus de Seriebus Infinitis of Jacobi Bernoulli with no sucess to find out the origin and meani... | Picture an old telescope (a.k.a. spyglass) that is retractable.
When the terms of a series contain differences, internal terms can be canceled, much like the segments of the telescope overlapping as it is contracted. For example, using the identity
$$
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1},
$$
\begin{alignat*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$. The following question is taken from the practice set of JEE exam.
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$.
$$\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x\\\implies {^6}C_... | $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$
$\implies\alpha^3(\alpha^2+\alpha+1)+(\alpha^2+\alpha+1)=0$
$\implies(\alpha^3+1)(\alpha^2+\alpha+1)=0$
$\implies\alpha^3=-1, \alpha^2+\alpha+1=0$
Now from your equation, $(1+\alpha)^6=1+x\alpha$
$(1+\alpha)^6-1=x\alpha$
$\alpha(\alpha+2)(\alpha^2+3\alpha+3)(\alpha^2+\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to evaluate $\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}$?
Evaluate $$\lim_{x\to y}\frac{\tan x-\tan y}{x-y}.$$
I used elementary methods to evaluate the limit. Here is my solution:
Let $d=x-y$. Then as $x\to y$, we have $d\to 0$. Now,
$\lim\limits_{x\to y}\frac{\tan x-\tan y}{x-y}\\ =\lim\limits_{d\to 0}\frac{... | You can use the definition of the derivative to evaluate such a limit :
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$
But here's an approach if you want to exercise more, by the end of this answer I'll mention the important properties you have to know :
\begin{align}
\lim_{x\to y} \frac{\tan x -\tan y}{x-y}&=\lim_{x\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Given $3$ points on a unit circle, figure out something about them. Question: Given three points $(a, b), (c, d)$ and $(x, y)$ on the unit circle in a rectangular coordinate plane, find the maximum possible value of the expression $(ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy + a)^2 + (dx - cy - b)^2. $
Answer: We will... | It is correct. Here is a way to see it by complex numbers and geometry. By abuse of notations, using complex numbers to represent points, one writes $$A=(a,b)=a+bi,B=(c,d)=c+di,C=(x,y)=x+iy,$$ three complex numbers on the unit circle, so $\overline{A}A=1,$ etc. Then it is easy to see using multiplication of complex num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculate surface of a triangle It's quite a simple question solvable with many methods but the $2$ of them I picked don't agree.
First of I have chosen the triangle with corners: $(3,0,0) \quad (0,3,0) \quad (0,0,6)$
Thus, this one here:
First Method is by evaluating the surface Integral: $A = \int_{S}1\,\mathrm{dS} ... | You must have:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\left(\frac{1}{2}\sqrt{18}\right)^\color{red}{2}}\right)^2} =13.5.$$
Or you can note that the triangles at the base are isosceles:
So $AC=BC=\frac{\sqrt{18}}{2}$ is what you need:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Selecting identical and distinguishable balls when the order matters I made up a question such that
There are $3$ identical blue balls ,$2$ identical red balls , $5$ distinguishable yellow balls ,$4$ distinguishable green balls. We want to make a mixture consisting of $8$ balls by using these balls. How many ways are ... | Your work is all correct.
Here is a way to simplify your work. Give each of the yellow and green balls its own generating functions, since $5$ distinguishable yellow balls is the same as 5 different colors each with one ball. The g.f. for each of these colors with a single ball is $1+x$, for both the ordered and unor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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For any $ a \in \mathbb{N}$ ; $f(x)=x^4+(4a+2)x^2+1$ is irreducible in $\mathbb{Z}[X]$ & reducible mod $p$ for all $p$. A.1 First we show $f(x)$ is reducible mod $p$ for all $p$.
Taking $y=x^2$ ; $f(x)$ becomes $y^2+(4a+2)y+1$. Solving the equation $y^2+(4a+2)y+1=0$ we get
$$
y=-(2a+1)\pm2\sqrt{a(a+1)}
$$
Now replacing... | $f$ has no real roots, so the only way to factor it over $\mathbb Z[x]$ is as two second degree polynomials of the form $(x-r)(x-\bar r)$, where $r$ is a complex root. You already found the roots, so the two factors would have to be
$$
(x-x_1)(x-x_3) = x^2+2a+1+2\sqrt{a(a+1)}
$$
and
$$
(x-x_2)(x-x_4) = x^2+2a+1-2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality: $\frac{a+b+c}{2} \geq \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}$ I am asking this question on behalf of another student, who sought my help through a school mentoring scheme, and claims that the question is similar to that found in Question $1$ of the British Maths Olympiad.
Prove that for all positi... | Looking at part a), I think the intention of part b) is
\begin{eqnarray*}\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}
& \stackrel{a)}{\leq} & \frac{(a+b)^2}{4(a+b)} +\frac{(b+c)^2}{4(b+c)} + \frac{(a+c)^2}{4(a+c)} \\
& = & \frac 14(a+b + b+c + a+c) \\
& = & \frac{a+b+c}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How can I establish the convergence (or divergence) of $ \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x $ I want to find out wether the following integral converges
$ \displaystyle \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x \tag{1} $
The integral above converges iff the foll... | Note that $x\to \ln(\sin^2(x))\leq 0$ and it is periodic of period $\pi$. Notice that
$$\int_0^{\pi}\ln(\sin^2(x))\,dx=-2\pi\ln(2).$$
(see for example Complex integration with trigonometric and logarithm? ).
Moreover $x\to\frac{x^2 }{(1+x^2)^2}\geq 0$, it is increasing in $[0,1]$ and decreasing in $[1,+\infty)$. Hence
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is the order of magnitude of $\sum\limits_{n=1}^kn\binom{k}{n}\frac{(2k-2n-1)!!}{(2k-1)!!}$ as $k\to\infty$? What is the order of magnitude of the function $f(k)$ below in the limit as $k \rightarrow \infty$? Does it diverge, converge to a positive limit, or converge to zero?
$$
f(k) = \sum\limits_{n=1}^{k} n \bin... | We are interested in the asymptotics of
$$g(n) = \sum_{k=1}^{n-1} k {n\choose k}
\frac{(2n-2k-1)!!}{(2n-1)!!}
= n \sum_{k=1}^{n-1} {n-1\choose k-1}
\frac{(2n-2k-1)!!}{(2n-1)!!}.$$
Now we have
$$(2n-1)!! = \frac{(2n-1)!}{2^{n-1} \times (n-1)!}$$
so we get for our sum
$$\frac{n! \times 2^{n-1}}{(2n-1)!}
\sum_{k=1}^{n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Show that $\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$
Show that $$\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$$
My book wrote that
$$=\frac{\Gamma... | Your method was correct, you just made a small mistake in the simplification. It is true that
\begin{align}
\int_0^{\dfrac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\dfrac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta
\end{align}
equals
\begin{align}
& \beta(\frac{3}{4}, \frac{1}{2})\; \beta(\frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Angle chasing problem: Find $\angle Q_{2024}Q_{2025}P_{2025}$ in the quadrilateral.
$ABCD$ is a convex quadrilateral where $BC = CD$, $AC = AD$, $\angle BCD = 96^\circ$ and $\angle ACD = 69^\circ$. Set $P_0 = A, Q_0 = B$ respectively. We inductively define $P_{n+1}$ to be the center of the incircle of $\triangle CDP_n... | Let us start with the triangle $P_{n-1}CD$ with angles $2x$, $2y$, $2z$ in $P_{n-1},C,D$ respectively. Let $I=P_n$ be its incenter. Then:
$$
\hat P_n=
\hat I
:=
\widehat{CID}=180^\circ-y-z=90^\circ+\frac12(180^\circ-2y-2z)=90^\circ+\frac 12\cdot 2x=90^\circ+\frac 12\hat P_{n-1}\ .
$$
The same relation applies for $\hat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of exponential within a region Are there methods to compute the following integral for $a \leq b$? Here $x\in\mathbb{R}$
$$
\int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
Substitution
The error function is
$$
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt
... | You can not compute it directly. But you can use normal distribution in other to have a very accurate estimate. In fact by supposing that $b\leq0$, the region $a\leq -\frac{x^2}{2}\leq b$ is the same as $\{\sqrt{-2b}\leq x\leq \sqrt{-2a}\}\cup\{-\sqrt{-2a}\leq x\leq -\sqrt{-2b}\}$.
So, $\begin{align*}
\int_{a\leq -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate the integral with divergence theorem
Given $F=(xz,y,z^2)$ and $E=\{(x,y,z)\;|\;y\geq0,x^2+y^2+z^2\leq4\}$ calculate the integral $\int_{\partial E} \langle F,N\rangle$ in two ways using the divergence theorem.
First $\text{div}(F)=3z+1$ .
Second $\int_E \text{div}(F)=\int_{\partial E} \langle F,N\rangle$ , ... | As already pointed out by Math Lover, the left part is
$$\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}(3z+1)\, dV=\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}1\, dV=\frac{1}{2}\left(\frac{4 \pi}{3}\cdot 2^3\right)=\frac{16 \pi}{3}$$
where the integral of $3z$ is zero by symmetry.
For the right part, the integral over the disc is
$$\iint_{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find Value of $(3\tan^2\alpha \;+ 4\tan^2\beta)$ Let $\vec V_1$ and $\vec V_2$ are two vectors such that $\vec V_1= 2(\sin\alpha+\cos\alpha) \hat i+\hat j$ and $\vec V_2=\sin\beta \; \hat i +\cos\beta \; \hat j$, where $\alpha$ and $\beta$ satisfy the relation $2 (\sin \alpha \; + \cos\alpha)\sin\beta=3-\cos\beta, $ Fi... | Trick question! Doesn't really have anything to do with vectors.
Note that the maximum value of $a \sin \theta+b \cos \theta$ is $\sqrt {a^2+b^2}$, which occurs when $\tan \theta =\frac ab$.
Here, we've been given that:
$$2(\sin \alpha+ \cos \alpha)\sin \beta+\cos \beta=3$$
Now, maximum value of LHS is $$\sqrt{4(\sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$ The question is
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$.
My attempt:
This reduces to finding primes $p,q$ such that p divides $q^3+1=(q+1)(q^2-q+1)$ and q divides $p^3+1=(p+1)(p^2-p+1)$.
Now we have 4 cases: If p divides $q+1$ and q d... | $Solution\ :$
\begin{gather}
So,\ given \notag\\
p^{3} +q^{3} +1\ \equiv 0\bmod pq \notag\\
And\ we\ know\ that, \notag\\
( a+b+c)^{3} -a^{3} -b^{3} -c^{3} =3\cdotp ( a+b)( b+c)( c+a) \notag\\
\notag\\
So,\ a^{3} +b^{3} +c^{3} =( a+b+c)^{3} -3\ ( a+b) \ ( b+c) \ ( c+a) \notag\\
\notag\\
\therefore p^{3} +q^{3} +1\ \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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.