Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Computing problems and generating functions
Q. Find the generating function for the sequence $\{a_n\}$, where $a_n$ is the number of solutions to the equation: $a+b+c=k$ when $a, b, c$ are non-negative integers such that $a\ge2, 0\le b\le3$ and $2\le c\le5.$
As far as I have understood, here, $a_n$ must be the coeffi... | Let's do a few simplifications:
*
*$(x^{2} + x^{3} + \cdots) = x^{2}(1 + x + x^{2} + \cdots) = x^{2} \cdots \dfrac{1}{1-x}$
*$1 + x + x^{2} + x^{3} = \dfrac{1-x^{4}}{1-x}$ (by the sum of a geometric series)
*$x^{2} + x^{3} + x^{4} + x^{5} = x^{2}(1 + x + x^{2} + x^{3}) = x^{2} \cdot \dfrac{1-x^{4}}{1-x}$
So reall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$
I am aware of using ... | Hint:
For $n>1$, we can write
$$\frac{a_n}2=2^{-2^{-n}}>2^{-\frac1n}>1-\frac1n$$ because $2^n>n$ and because by the binomial theorem
$$\left(1-\frac1n\right)^{-n}>\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\cdots+1>2.$$
Now it suffices to prove convergence of $1-\dfrac1n$ to $1$, which is easy, and will al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.
Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.
My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^... | First solution of the problem
We prove that for each $x \in(0,\pi)$, there exists $n,y\in \mathbb{N}$ such that
$$\pi y + \frac{\pi}{k^2} \le x k^n\le \pi y +\pi - \frac{\pi}{k^2} \iff y +\frac{1}{k^2}\le k^n\frac{x}{\pi} \le y+1-\frac{1}{k^2} \tag{1}$$
because $$(1) \iff |\sin(xk^n)| \ge \sin \left(\frac{\pi}{k^2} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$
In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$
(note: we... | Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below
\begin{align}
& \frac{s^2-(2R+r)^2}{4R^2}\\
= &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\
=& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\
=& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the integral $ \iint_R\sqrt{y/x}\,e^{\sqrt{xy}}\,dA$ The $R$ is the region of integration, described in the following image.
To solve this integral we make a change of coordinates with $u=\sqrt{xy}$ and $v=\sqrt{\frac{y}{x}}$. Furthermore, in the region of integration we have $x>0$ and $y>0$. Then $u^2=xy$ y... | Your solution is correct. Still we can do without changing variables:
\begin{align*}
[\ldots]&=\int_1^2\int_{1/y}^y\sqrt\frac{y}{x}e^{\sqrt{xy}}\,dx\,dy
\\&=\int_1^2 2e^{\sqrt{xy}}\Bigg|_{x=1/y}^{x=y}\,dy
\\&=2\int_1^2(e^y-e)\,dy
\\&=2(e^y-ey)\Bigg|_{y=1}^{y=2}=2e(e-2).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$ Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$
I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.
| Let $f(x)=\frac{\cos(3x)}{x^4+x^2+1}$. Inasmuch as $f$ is an even function, we can write
$$\begin{align}
\int_0^\infty \frac{\cos(3x)}{x^4+x^2+1}\,dx&=\frac12 \int_{-\infty}^\infty \frac{e^{i3x}}{x^4+x^2+1}\,dx\\\\
&=\pi i \left(\text{Res}\left(\frac{e^{i3z}}{z^4+z^2+1}, z=e^{i\pi/3}\right)+\text{Res}\left(\frac{e^{i3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$
Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$... | You have
$$\begin{aligned}
a_{n+1} &=\sqrt{na_n +2n+1}\\
&= \sqrt{n^2+2n+1+(n a_n -n^2)}\\
&=(n+1) \sqrt{1+\left(\frac{n}{n+1}\right)^2\left(\frac{a_n}{n}-1\right)}
\end{aligned}$$
Hence denoting $b_n =\frac{a_n}{n}$
$$\begin{aligned}
b_{n+1} -1
&=\sqrt{1+\left(\frac{n}{n+1}\right)^2\left(b_n-1\right)}-1
\end{aligned}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$ $x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$... | Probably the following is shorter and easier.
Firstly, note that for positive reals $a$ and $b$ the inequality
$$
\frac{a^2}{b}\ge 2a-b
$$
holds (since it's equivalent to $\frac{(a-b)^2}{b}\ge 0$). Applying this observation six times gives
$$
\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\ge(2x-z+2y-z)+(2x-y+2z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this us... | Knowing the answer from taking the derivative, I'd cheat by "observing"
$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac 2x\right)+\left(\frac 1x+\frac {9}{2y}\right)$$
$$\ge 4+2\sqrt{\frac x2 \cdot \frac 2x}+\frac{(x+2y)(\frac 1x + \frac{9}{2y})}{x+2y} \tag{AM-GM}$$
$$\ge 4+2+\frac{(1+3)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Question on solving one algebra relation to obtain another If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:
If $(1-2x)(1-2y)(1-2z)... | Let's call $\begin{cases} U=\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\\V=\frac b{a+b}+\frac c{b+c}+\frac a{c+a}\end{cases}$
We have trivially $U+V=3$
Now $U-V=\frac {a-b}{a+b}+\frac {b-c}{b+c}+\frac {c-a}{c+a}=\cdots=-\frac {a-b}{a+b} \times\frac {b-c}{b+c}\times\frac {c-a}{c+a}=27$
Show the identity $\frac{a-b}{a+b}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrating $\frac1{2\sigma\sqrt{2\pi}}\int_{-\infty}^\infty(1+erf(\frac{z-\mu_i}{\sigma_i\sqrt2}))\exp( -\frac12(\frac{z-\mu}\sigma)^2)dz$ I am interested in the integral of
$$\frac{1}{2\sigma \sqrt{2\pi}}\int_{-\infty}^\infty \left( 1+\operatorname{erf}\left(\frac{z-\mu_i}{\sigma_i\sqrt{2}}\right)\right)\exp\left( -\... | Integral of one error function and Gaussian
Mathematica does not evaluate the integral. Surprisingly, there is an exact result. By changing variables, $z-\mu=x$, your integral of $\operatorname{erf} \times \text{Gaussian}$ may be written as
$$
I(b)=\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(ax+b) \exp \left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \i... | From
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2 \iff\\
\frac{a^2+bc-bc}{a^2+bc}+\frac{b^2+ac-ac}{b^2+ac}+\frac{c^2+ab-ab}{c^2+ab}<2 \iff\\
3-\frac{bc}{a^2+bc}-\frac{ac}{b^2+ac}-\frac{ab}{c^2+ab}<2 \iff\\
\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}>1$$
and applying Titu's Lemma
$$\frac{bc}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
How to show that this $2n \times n^2$ matrix has rank $2n-1$? The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 &... | Things will perhaps be clearer if we write these $2n$ rows from $\Bbb{R}^{n^2}$ as
$n \times n$ matrices.
For $n=3$ we are looking at
$$\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits... | Note
$$I=\int\limits_{0}^{1} \frac{\ln(1-x+x^2)}{\sqrt{x}(1+x)}{d}x
\overset{t=\sqrt x} = 2\int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)}{1+t^2}dt
$$
Let $J(a) = \int_0^1 \frac{\ln(\frac14(1+t^2)^2 \sec^2a -t^2)}{1+t^2}dt
$.
Then, $J(0) = \int_0^1 \frac{2\ln\frac{1-t^2}2}{1+t^2}dt=
-2G$
$$J’(a)= \int_0^1 \left(\frac{\tan a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$? How to evaluate
$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$
The integral arises in the computation of
$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as
$$ \scripts... | This may be counterproductive but we can obtain another form from @yarchik's solution.
$$
\begin{aligned}
S%
&=-\sum _{n=0}^{\infty } \frac{(-1)^n }{4n+3}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\sum _{n=0}^{\infty } \frac{(-1)^n }{n+3/4}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\int_0^1x^{-1/4}\sum _{n=0}^{\infty } (-x)^nH_{n+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac... | Let $ n\in\mathbb{N}^{*} $ be an odd integer, we have : \begin{aligned}\frac{1}{1\times 2}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\left(n+1\right)}=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k-1}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ &=\sum_{k=1}^{n}{\frac{1}{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculating this series integral $\int (\frac{x^{2n-2}}{(2n-2)!})\cdot \frac{x^{2n-1}}{(2n-1)!}$ I've calculated this integral and would like the communities feedback and support towards my solution:
$$\int \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\cdot \left(x \cdot \frac{... | Please keep in mind: the derivative of a product is not the product of the derivatives!
In this case you should simply multiply both factors obtaining
$$\int\prod_{k=0}^{2n-1}\frac {x^k}{k!}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{\sum_{k=0}^{2n-1}k}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{n (2n-1)}dx=
\frac{x^{n (2n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $ I have the conjecture for the integral
$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$
where $G$ is the Catalan constant, followin... | Using the third of these equations we can express your integrals in terms of the Clausen function:
\begin{align}
I &\equiv \int \limits_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \log (\cot(t))\, \mathrm{d} t + \int \limits_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \log (\cot(t))\, \mathrm{d} t \\
&= \frac{1}{2} \left[- \operatornam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
How to simplify $\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$? The problem is as follows:
Simplify the following expression:
$B=\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$
The alternati... | The denominator is $\sin(55^\circ) \sin(57.5^\circ) \sin(67.5^\circ)$ and note that the angles add up to $180^\circ$.
Using the identity $\sin \alpha \cdot \sin \beta \cdot \sin \gamma = \frac14(\sin 2\alpha + \sin 2 \beta + \sin 2\gamma)$ listed in Wikipedia Further "conditional" identities for the case α + β + γ = 18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & ... | for D_3 case;I think LU decomposition can help to rewrite as $AB$ and use $|AB|=|A||B|$
$$\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{matrix}\right)=\\
\left(\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
There is a rectangle in 3-dimensional space. If I am only given the $x$ and $y$ coordinates, how do I find the $z$ coordinates? We are given the $x$ and $y$ coordinates of the four vertices of a rectangle.
How do I find the $z$-coordinates?
I know that this 3 dimensional rectangle must obey all the usual conditions of ... | (Rewritten after missing an entire degree of freedom!)
Let's say the vertices are in (clockwise or counterclockwise order) $\vec{A} = (x_A, y_A, z_A)$, $\vec{B} = (x_B, y_B, z_B)$, $\vec{C} = (x_C, y_C, z_C)$, and $\vec{D} = (x_D, y_D, z_D)$.
In order for the quadrilateral to be a parallelogram, pairs of opposing sides... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If we place $1$ to $n^2$ in an $n\times n$ table, what is the smallest $s$ where $s$ is the max of $a+b$ where $a,b$ are numbers in adjacent cells? Primary question : Let $n$ be an integer greater than 1. Numbers $1,2,...,n^2$ are placed in an $n\times n$ table, one number per cell. For each such configuration, let $... | Indeed, a minimum of $n^2 + \lfloor \frac n 2\rfloor + 1$ can be arranged for any $n \geq 3$, as follows, where $m = \lfloor \frac n 2 \rfloor$:
\begin{matrix}
n^2, & 1, & n^2 - 1, & 2, & n^2 - 2, & \dots \\
m + 1, & n^2 - n + m, & m + 2, & n^2 - n + m + 1, & m + 3, &\dots \\
n^2 - n, & n + 1, & n^2 - n - 1, & n + 2, &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4046654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and... |
The point, $Z$ on the dark green line has coordinates
$x = x(P)$ and $y=area(\triangle PQR)$
So the ratio you are looking for is
$\dfrac{6-2}{6-1.5}=\dfrac{4}{4.5}=\dfrac 89$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
studying the function $f(x)= \sqrt{x^2+x}-x$ I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.
In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.
$ \frac{2x+... | Note that the domain of the derivative
$$f^\prime(x)\:=\: \frac{2x+1}{2 \sqrt{x^2+x}}-1$$
is not equal to the domain of $f$ since the denominator zeros $\{-1,0\}\,$ have to be excluded.
The sign analysis is quickly achieved using the AGM inequality:
If $0<x$ then
$$\sqrt{x(x+1)}\:<\:\frac{2x+1}{2}\quad\implies\; 0\;<\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$? $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$
$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$
$3x+5=x-1$
$x=-3$
But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$
So I tried to do other things... | The equation:
$$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}
$$
obviously holds if $x^2-9=0$ provided that the exponents are non-negative. This is the case for $x=3$.
If $x^2-9\ne0$ we can divide both sides by $(x^2-9)^{(x-1)}$ obtaining:
$$(x^2-9)^{(2x+6)}=1.$$
This equation holds if $2x+6=0$ or $x^2-9=1$. The roots of the equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\s... | Problem like this is fun to solve for sure. Let's do the backward direction: If $m+n > 2$, then either $m \ge 2$ or $n \ge 2$. Let's say $m \ge 2$, then rewrite it as $\dfrac{x^2y}{x^2+y^2}\cdot x^{m-2}y^{n-1}$. Observe that $x^{m-2}y^{n-1} \to 0$ or $1$, and $0 \le \left|\dfrac{x^2y}{x^2+y^2}\right| \le |y|$, thus it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$ I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$
Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$
Where $H_n$ is the n-th ha... | Maple gives a rather unpleasant expression involving complex dilogarithm functions:
$$-\frac{19 \pi^{2}}{144}+\frac{7 \ln \! \left(2\right)^{2}}{12}-\frac{\ln \! \left(2-\sqrt{3}\right)^{2}}{6}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Algebra question with sequences. I dont know how to solve it. Can somebody help but also tell me how to approach these kind of questions.
Terms of the sequence $a_1, a_2, \ldots, a_{2020}$ are obtained by the rule
$$
a_2 = \frac{1+a_1}{1-a_1}, \qquad
a_3 = \frac{1+a_2}{1-a_2}, \qquad \ldots, \qquad
a_{2020} = \frac{1+... | Given equatiobn is:
$$A_{n+1}=\frac{1+A_n}{1-A_n} \implies A_{n+1} (1-A_n)=1+A_n~~~(1)$$
Let $$1-A_n=\frac{B_n}{B_{n-1}} \implies A_{n+1}=1-\frac{B_{n+1}}{B_n}$$Then (1) becomes
$$\frac{B_n-B_{n+1}}{B_n}\frac{B_n}{B_{n-1}}=2-\frac{B_n}{B_{n-1}}$$
$$\implies B_n-B_{n+1}=2B_{n-1}-B_n \implies B_{n+1}+2B_{n-1}-2B_n=0$$
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
B... | Well if $a = 2$ you have $z=2; y=-\frac 19$ and $x = whatever$.
But if $a-2\ne 0$ keep at it.
Multiply row 2 by $\frac 9{a-2}$ and subtract getting rows.
$\begin{array}\\
1&2&2&|\,\,1\\
0&a-2&1&|\,\,2\\
0&0&(a-2)-\frac 9{a-2}&|-\frac 9{a-2}\\
\end{array}$
Now if $(a-2)-\frac 9{a-2} = 0$ we have a case with not solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Conclude integral is zero without calculation I have the following two integrals:
$$\int_0^{2\pi} \frac{\cos{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
and I know both are zero. I can argue why the seco... | Hint: Use the substitution $\phi = \pi/2 - \theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$
where $s$ is the semi-perimeter.
I started by applying the rule of sines:
*
*$a=k\sin A$
*$b=k\sin B$
*$c=k \sin C$
where $k$... | What is true is the relation $a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$ where $r$ is the inradius, $R$ is the circumradius and $s$ is the semi perimeter of the triangle.
By the sine rule, we can write the length of the sides as $a=2R\sin A,b=2R\sin B,c=2R\sin C$ therefore we have $$a\cos A+b \cos B+c\cos C=R(\sin 2A+\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A question about limit Can anyone help this questions?
Find the limit of
$$1+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\sqrt[5]{5+....+\sqrt[n]{n}}}}}$$
I can only solve that this formula is less than 3 but can not find the exact answer for this.
| It can be shown that $1.9<a_0<2$:
We use following inequalities:
$\sqrt[k]{k+1}>1; \sqrt[k]{k}<\sqrt[k]{k+2}<2$
Seris of $a_n$ is increasing and from first inequality we have:
$a_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n]n}}<\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n-1]{n-1+2}}}<\cdot\cdot\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality consisting of sum using mathematical induction I wish to seek help for a question from the Finnish matriculation exam in mathematics. There are multiple methods to solve the exercise, but I am ONLY interested in solving it using mathematical induction. The question is as follows:
$$\text{let}\ n\ \text... | Thanks to the help of @maxmilgram, I've managed to figure out the solution.
SOLUTION:
*
*Test with n = 1, which gives:
$$0\le1\le\frac{127}{7}, \text{which is true}$$
*Assumption:
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
*Induction step:
$$\frac{127}{7}\cdot\ n^7\le\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
... | For the system $\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases}\quad$ just offset the variables $\begin{cases}x=1+X\\y=1+Y\\z=1+Z\end{cases}$
Then you get $\begin{cases}3+X+Y+Z=3\\3+2X+2Y+2Z+X^2+Y^2+Z^2=3\end{cases}\implies X^2+Y^2+Z^2=0$
Which in the real domain has only the $(0,0,0)$ solution.
For $x+y+z=-3$ proceed s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What is the following limit? How can I calculate the following limit?
What do I miss here? What am I doing wrong?
$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightar... | Using the limit $\lim_{x\to 0}(e^x-1)/x=1$ the denominator of the expression under limit can be safely replaced by $x^6$. Next we add and subtract $6x^2$ in numerator and express it as $$6(x^2-\sin x^2)+6x\sin x-6x^2+x^4$$ This allows us to split the desired limit as a sum of two limits the first one of which is $$6\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right)... | Hint:$$\frac{x^{2x}}{(x+1)^{x+1}}=x^{2x-(x+1)}\frac{1}{\left(1+ \frac{1}{x} \right)^{x+1}} \sim\frac{x^{x-1}}{e} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. than... | $$(2z+3)^3=\frac{1}{64} \implies 2z+3=\frac{1}{4}, \frac{\omega}{4}, \frac{\omega^2}{4}$$
$$\implies z=\frac{-11}{8}, \frac{\omega -12}{8}, \frac{\omega^2-12}{8}.$$ Here $\omega$ is cube-root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$... | Knowing the form of the answer, it is natural to apply the substitution $t= \frac{\sqrt2 x}{1-x^2}$, which leads to
$$1+t^2 = \frac{1+x^4}{(1-x^2)^2},\>\>\>\>\>
dt = \frac{\sqrt2 (1+x^2)}{(1-x^2)^2}dx$$
Then
$$\int\frac{1+x^2}{1-x^2}\frac{1}{\sqrt{1+x^4}}dx
=\int \frac{1+x^2}{1-x^2}\frac{\frac{(1-x^2)^2}{\sqrt2(1+x^2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$?
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$
Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdo... | $$3^x+2^x=7\cdot3^x-7\cdot2^x$$
$$8\cdot2^x=6\cdot3^x$$
$$\left(\frac32\right)^x=\frac 43$$
and
$$x=\log_{3/2}\frac43.$$
As every steps are equivalent, this is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Shorthand limits of matrices? Is there any relatively easy way to solve limits with matrices?
$$
\lim_{h\rightarrow \infty}
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}^h
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
\frac{1}{rc} & 0 & 1-\frac{1}{rc}
\... | The problem here is the power $h$. But if the matrix is diagonalizable, there is an easy way to handle it. Let
$$H :=\begin{pmatrix}1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}$$
And suppose there is some matrix $Q$ with $Q^{-1}DQ = H$ where $$D =\begin{pmatrix}d_1&0&0\\0&d_2&0\\0&0&d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
F... | Actually, the limit
$\lim\limits_{x\to 0}\frac{(5x+8)^{1/3}-2}{x}$ (of the form $\frac{0}{0}$)
$=\lim\limits_{x\to 0}\frac{5.\frac{1}{3}(5x+8)^{-2/3}}{1}$, by L' Hospital's rule (since the limiting value of both numerator and denominator is 0 and the first derivative of the denominator w.r.t. $x$ is non-zero, the rule ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Areas in triangle
In the above isosceles right triangle ABC, with its two sides $AB = AC = 1$ unit, we take a random point D on the hypotenuse and draw perpendicular lines to the sides AB and AC, which intersect them at points E and F respectively.
Show that the maximum of the three areas AFDE, EBD and CDF is always ... | Start with D as the mid point of BC. WLOG, imagine D moves up by a vertical distance of $\Delta x$ .
\begin{align}S_{AEDF}=\frac{1}{4}-\Delta x^2, S_{\triangle DFC}=\frac{1}{2}(\frac{1}{2}+\Delta x)^2\end{align}
\begin{align}S_{AEDF}\geq \frac{2}{9}(\Delta x \leq\frac{1}{6})\\
S_{\triangle DFC}\geq \frac{2}{9}(\Delta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Compute $\int \frac {dx} {(a + b \cos x)^2} $ for $a > b.$ Let $a>b$ be real numbers. As the title suggests, I would like to compute $$\int \frac 1 {(a + b \cos x)^2} \, dx.$$
My attempt consisted of converting the cosine to half-angle tangents, substituting the half-angle tangent for $t$, and simplifying to $$\int \fr... | Proceed as follows after half-angle substitution $t=\tan\frac{x}2$
\begin{align}
& \int \frac{2(1+t^2)}{[(a+b)+(a-b)t^2]^2} \, dt\\
=& \frac2{a^2-b^2} \int \frac{a[(a-b)t^2+(a+b)]+ b[(a-b)t^2-(a+b)] }{[(a+b)+(a-b)t^2]^2} \, dt \\
=& \frac2{a^2-b^2} \int \frac{a[(a-b)+\frac{a+b}{t^2} ]+ [b[(a-b)-\frac{a+b}{t^2}]}{[(a-b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
Calculate
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$
where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2... | In polar coordinates, the integration region is enclosed by $\theta \in (0,\frac\pi4)$ and $ r\in (\frac1{ \sin \theta + \cos\theta},1)$. Then, the integral becomes
$$I=\int_0^{\pi4}\int_{\frac1{\sin \theta + \cos\theta}}^1 \frac1{r^2}drd\theta= \int_0^{\pi4}(\sin \theta +\cos \theta -1 )d\theta=1-\frac\pi4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Trapezium from two diagonals and two non-parallel sides Is it possible to construct (and to calculate) a trapezium from it's two non-parallel sides and it's two diagonals, with other words $b,d,e,f$ are given:
I read out the equations system
$\begin{array}{|l l}
(1) & a =c+p+q \\
(2) & h^2 =e^2 -(a-q)^2 \\
(3) & h^2 ... | Per the cosine rule
$$a^2 = b^2+e^2 -2be \cos C= d^2 +f^2 -2df \cos D\tag1
$$
and note that the triangles ADB and ACB are equal in area
$$\frac12 be \sin C =\frac12df \sin D\implies b^2e^2 (1-\cos^2 C )= d^2f^2(1-\cos^2 D)\tag2
$$
Substitute (1) into (2) to eliminate $\cos C$ and $\cos D$, yielding
$$a^2= \frac{(b^2-e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4094466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Hypergeometric functions for $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c\geq 0$ The sum is: $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c \geq 0$.
The equivalent hypergeometric function can be found in page 214, Concrete Mathemtic... | The dumbest mistake is in this $(-1)^n$ in:
$$\frac{(a-k-n)(b-k-n)(c-k-n)(-1)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n $$
Clearly, it should be canceled as well! And the idea in the comment works!
Let we assume $a \leq b \leq c$. So, we have $n = -c$!
Now, $$\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Uncertain how to solve this trigonometric equation I am currently attempting to discover how to find the general solutions to $$\sqrt 3\tan^2x=2\tan x+\sqrt 3$$
The given solutions are $x= \frac{\pi}{3}+ \pi k$ , $\frac{5\pi}{6} + \pi k$
To solve this equation I removed the square root from $\sqrt 3\tan^2x$ leaving me ... | Elaborating on player3236's comment: Let $u = \tan x$. Then your equation is $\sqrt{3} u^2 = 2u + \sqrt{3}$. Solve this quadratic equation, and then use $u = \tan x$ to solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can we prove $\gcd(3n^2+1, 2n-3)\mid 31$ I want to show that $\gcd(3n^2+1, 2n-3)$ divides 31 $\forall n\in\mathbb{N}$.
I have tried to begin by eliminating the $3n^2$ factor on the left by adding and subtracting multiples and powers of $2n-3$ but I still haven't figured out how. Can someone give me some hint?
| Note if $\gcd(k,a) = 1$ then $\gcd(a, b) = \gcd(a, bk)$. THis is because $k$ having no factors in common with $a$ then multiplying $b$ by $k$ will not add any common factors of $a$.
And as $3|3n^2$ then $3\not \mid 3n^2 + 1$. And as $2|2n$ then $2\not \mid 2n-3$. And as $3,2$ are prime $\gcd(3n^w+ 1, 3) = 1$ and $\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4098541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Converting a radical to a fractional exponent I want to understand how to convert a radical to a fractional exponent. Given the following equation:
$\sqrt[3]{(x)^6\cdot x^9}=\sqrt[3]{x^{24}\cdot x^9}=\sqrt[3]{x^{33}}=x^{\frac{33}3}=x^{11}$
How does: $\sqrt[3]{(x)^6\cdot x^9} = \sqrt[3]{x^{24}\cdot x^9}\;\;$??
| Power rules of exponent :
*
*$(a^m)^n=a^{mn}$ $\to \sqrt[3]{(x^4)^6\cdot x^9} = \sqrt[3]{x^{24}\cdot x^9}$
*$a^ma^n=a^{m+n}$ $\to \sqrt[3]{x^{24}\cdot x^9}$$=\sqrt[3]{x^{33}}$
*$\sqrt[n]{a}=a^{\frac{1}{n}},\sqrt[n]{a^m}=a^{\frac{m}{n}} \to \sqrt[3]{x^{33}}$$=x^{\frac{33}{3}}=x^{11}\;$
Or,
$\sqrt[3]{(x^4)^6\cdot x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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The functional equation $ f \left( x ^ 2 y + f ( y ) \right) = x ^ 2 f ( y ) + f \big( f ( y ) \big) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \left( x ^ 2 y + f ( y ) \right) = x ^ 2 f ( y ) + f \big( f ( y ) \big) $$
for all $ x , y \in \mathbb R $.
It's straightforward to check that for a... | In fact, it's not true that linear functions are the only solutions. For any $ a , b \in \mathbb R _ { 0 + } $, the function $ f : \mathbb R \to \mathbb R $ defined with
$$ f ( x ) =
\begin {cases}
a x & x \ge 0 \\
b x & x < 0
\end {cases} $$
is another solution. This can be verified easily, using the fact that $ y $, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| $$x^2+xy+y^2=1\implies y_\pm=-\frac{1}{2} \left(x \pm\sqrt{4-3 x^2}\right)$$
Using $y_-$ gives
$$f=x^2-3xy_m-2y_m^2=\frac{7 x^2}{2}+\frac{1}{2} \sqrt{4-3 x^2} x-2$$ Compute the derivative, set it equal to $0$ and solve for $x$. You should get
$$x_1=\sqrt{\frac{1}{39} \left(26+7 \sqrt{13}\right)} \qquad x_2=-\sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4105835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\... | If you like the gaussian hypergeometric function
$$J_n= \int \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}\,dx=\frac{x^{4 n-1} }{4 n-1}\,\,_2F_1\left(n-\frac{1}{4},n;n+\frac{3}{4};-x^4\right)$$ Assuming that $n \gt \frac 14$, for $x=0$ the rhs tends to $0$ and
$$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}=\frac{\Gamma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$ and $a^2 + b^2 = c^2$ I have found this in an Romanian magazine. We have to solve for natural numbers:
$$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$$
$$a ^ 2 + b ^ 2 = c ^ 2$$
After some elementary calculations, we get that $bc + 1 = a(b + c)$. I tr... | (This is not a solution. I made an error.)
Show that (If you get stuck, show your work.)
*
*$(a, b, c)$ is a primitive pythagorean triplet.
*If $(a, b, c ) = (m^2 - n^2, 2mn, m^2+n^2)$ with $m > n > 0$, then there are no solutions.
Substituting in, we get $n^4 = m^4 - 4mn^3$.
View this as a quartic in $ x = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ .
What I Tried: I wrote :-
$$\rightarrow \frac{7}{2^{1/2} + 2^{1/4} + 1} = \frac{2^2 + 2 + 1}{2^{1/2} + 2^{1/4} + 1}$$
Then I ration... | Let $2^{1/4}=p$
So, we need
$$\dfrac{1+p^4+p^8}{1+p+p^2}$$
Now $1+p^4+p^8=(1+p^4)^2-(p^2)^2=?$
and $1+p^2+p^4=(1+p^2)^2-(p)^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the limit of the following sum as it approaches a divergence I am interested in the behaviour of summations like the following:
$$
\lim_{\alpha\to 0}\sum_{n=1}^{\infty}\frac{e^{in\alpha}}{\sqrt{\beta+n^{2}}}
$$
where $\beta$ is a constant. Just putting $\alpha=0$ doesn't help because the sum doesn't co... | I will assume that $\alpha>0$ and $\beta \geq 0$. Let us write
$$
\sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{{\sqrt {\beta + n^2 } }}} = \sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{n}} + \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)e^{in\alpha } } .... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4113365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Length of cardiod $r=1+\sin\left(\theta\right)$ I'm pretty sure this is correct since I'm able to justify each of the steps but I'm just wondering if there's a more efficient way that isn't just skipping steps? Because I had to do 2 substitutions, one of which converted sine into cosine.
\begin{align*}
\text{Length }= ... | Instead
\begin{align*}
\text{L }= & \ 2 \sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 +\sin\theta}\> d\theta
= \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\left(\sin\frac{\theta}2+ \cos\frac{\theta}2 \right)^2} \>d\theta\\
= &\ 2\sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sin\frac{\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this quest... | $$\begin{align}{\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}&=T(x)>0&\end{align}$$
$$T^2(x)=2x+2 \sqrt{x^2-4x+4}$$
$$T^2(x)=2x+2|x-2|$$
$$\begin{align}\color {gold}{\boxed {\color{black}{T(x)=\sqrt{2x+2|x-2|}}}}\end{align}$$
*
*If $x≥2$, then $\begin{align}T(x)&=\sqrt{2x+2x-4}\\
&=2\sqrt{x-1}\end{align}$
*If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx$ So, I need to integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx, n \in \mathbb{N}$.
I've tried partial fractions, but I get nowhere.
Then I tried using substitution, using the substitution $t = 1+x^2$, but I get:
$$
\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}}dx = \... | Denote
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx.
$$
Using integration by parts it's easy to see that
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx = \begin{bmatrix} u = \frac{1}{(1+x^2)^{n+3}} & dv = x^{2n-1} dx \\ du = - \frac{2(n+3)x}{(1+x^2)^{n+4}} dx & v =\frac{x^{2n}}{2n}\end{bmatrix} =
$$
$$
= \frac{x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not th... |
Most likely that's not the author's intention
It could very well be. If we replace $y = x + 2$ we have
$\frac 1{y-2} + \frac 1{y-1} +\frac 1y + \frac 1{y+1}+\frac 1{y+2} = 0$ so
$(y-1)y(y+1)(y+2) + (y-2)y(y+1)(y+2) + (y-2)(y-1)(y+1)(y+2) + (y-2)(y-1)y(y+2) + (y-2)(y-1)y(y-1) =0$
$(y^2-1)(y^2+2y) + (y^2-4)(y^2+ y) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
A little bit of trouble computing the characteristic polynomial of a matrix? I'm reading Smirnov's Linear Algebra and Group Theory, here:
He says that is the characteristic equation but when I compute it on Mathematica, I obtained: $-(\lambda -2)^2 \left(\lambda ^3-27 \lambda -26\right)$. For reference, I computed i... | I assume there is a typo in the matrix $A$ and the issue is that $a_{41}$ should be negative, so lets use this and find the Jordan canonical form. We have
$$A = \begin{pmatrix}
-2 & -1 & -1 & 3 & 2 \\
-4 & 1 & -1 & 3 & 2 \\
1 & 1 & 0 & -3 & -2 \\
-4 & -2 & -1 & 5 & 1 \\
4 & 1 & 1 & -3 & 0 \\
\end{pmatrix}$$
The e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise i... | Symmetrize the integrand with $y= t-\frac x2$
\begin{align}
& \int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt\\
= & \int_{-\infty}^\infty\frac{1}{(a^2+(y+\frac x2)^2)(a^2+(y-\frac x2)^2)}dy\\
=&\
\frac2{x(4a^2+x^2)}\int_{-\infty}^\infty
\frac{{\frac x2 +(y+\frac x2)}}{a^2+(y+\frac x2)^2 }
+ \frac{{\frac x2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
how to prove that $1^x+2^x+3^x+4^x+\cdots+N^x$ will never sum to a prime number except $1^x+2^x$? I am a a web developer programming in PhP which is limited to large calculations, but running a quick script shows that $1^x+2^x+3^x+4^x+\cdots+N^x$ can never sum to a prime number unless in the case of $1^x+2^x$, such as ... | This is also only a partial solution that may provide more insight.
We can split into two cases, $x$ even and odd.
If $x$ is odd, then note that $i^x \equiv -(n-i)^x \mod n$. Then, consider $1^x+ 2^x...+n^x$ If $n$ is odd, we can pair up $1$ and $n-1$, etc. Each pair’s sum is divisible by $n$, $n^n$ is divisible by $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Are these the only "intersections" between the following series, and why all of them are multiples of $10$ As a web developer that programs in PhP, I enjoy running some scripts to see some of math wonders, however, PhP is limited to large calculations.
I wanted to see if there are any intersections for the following $2... | I think you are trying to find tuples $(a,b) $that satisfy the equation
$$\sum_{i = 1}^{a} i = \sum_{i = 1}^{b}\sum_{j =1}^{i}j$$
The left hand side is
$$\sum_{i = 1}^{a}i = \frac{a(a + 1)}{2}$$
and the right-hand side is
$$\sum_{i = 1}^{b}\sum_{j = 1}^{i}j = \frac{b(b + 1)(b + 2)}{6}.$$
Then, the equation becomes \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Binary $[15,5,d]$ cyclic code My goal is to find a generator polynomial for a binary $[15,5,d]$ cyclic code that corrects all errors with weight less than or equal to $3$.
I have to find a degree $10$ monic polynomial that divides $x^{15}-1$ ($\bmod 2$) and has weight $7$. If $15$ and $5$ were coprime I would know how ... | If you try systematically dividing by quartics you will see that more polynomials than $x^4+x^3+x^2+x+1$ factor into $x^{15}+1$ over $\mathbb F_2$:
$$x^{15}+1=(x+1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$$
Then
$$(x^2+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1)=x^{10}+x^9+x^8+x^6+x^5+x^2+1$$
is the desired weight-$7$ polyn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the determinant of nth order $$
\begin{vmatrix}
2 & 2 & ... & 2 & 2 & 1 \\
2 & 2 & ... & 2 & 2 & 2 \\
2 & 2 & ... & 3 & 2 & 2 \\
... & ... & ... & ... & ... & ... \\
2 & n-1 & ... & 2 & 2 & 2 \\
n & 2 & ... & 2 & 2 & 2
\end{vmatrix}
$$
I got this in my linear algebra homework. In th... | Let's do some elementary operations on rows and columns:
$$\begin{align}
\begin{vmatrix}
2 & 2 & ... & 2 & 2 & 1 \\
2 & 2 & ... & 2 & 2 & 2 \\
2 & 2 & ... & 3 & 2 & 2 \\
... & ... & ... & ... & ... & ... \\
2 & n-1 & ... & 2 & 2 & 2 \\
n & 2 & ... & 2 & 2 & 2
\end{vmatrix}
&\overset{1}=
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $... | $\lfloor{\log_{10}(2^n)}\rfloor = \lfloor{\log_{10}(10^n) - \log_{10}(5^n)}\rfloor = n + \lfloor{ - \log_{10}(5^n)}\rfloor = n - \lfloor{\log_{10}(5^n)}\rfloor - 1$
So:
$\lfloor{\log_{10}(2^n)}\rfloor + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n - \lfloor{\log_{10}(5^n)}\rfloor - 1 + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and Jack are in the same group.
b-)Dennis , John and Jack are in different groups.
c... | Part b should have been $\frac{60 \times C(22,4) \times C(18,4) \times C(14,4) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
part c should have been $\frac{20 \times C(22,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove convexity of $f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$
Prove convexity of $$f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$
I am looking for an easy (or relatively easy) way to show $f$ is convex.
I tried to split the function into:
$... | $g$ is not convex, as $(2,2)$-entry of $\nabla^2 g(x)$ is negative when $x_1$ is negative.
We have $\nabla f(x) = \begin{bmatrix} \frac{2x_1+1}{x_2+x_3} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2}\end{bmatrix}$.
$$\nabla^2 f(x) = \begin{bmatrix} \frac{2}{x_2+x_3} & - \frac{2x_1+1}{(x_2+x_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)}$ is a decreasing function How do I prove that $f(x)$ is a decrasing function
$$f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)},x>3$$
I did:
$$f{'(x)}=-\frac{4}{(x-3)^2}+\frac{2}{(1-2x)^2}$$
$$f^{'}(x)<0$$
then I got
$$-\frac{7}{9}<x<1$$
So where do I go from here
| \begin{align}f(x) &= \frac{1+11x-6x^2}{(x-3)(1-2x)}\\&=\frac{3(-2x^2+7x-3)+11x-21x+1+9}{(x-3)(1-2x)}\\
&=3 + \frac{10(1-x)}{(x-3)(1-2x)}\\
&=3 + \frac{4}{x-3}- \frac{2}{1-2x}\\
&= 3 + \frac4{x-3}+\frac{1}{x-0.5}\end{align}
For $x>3$, as $x$ increases, we can see that $f(x)$ decreases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find and solve a recurrence relation In a casino, there are $k(\geq 2)$ colored poker chips, one of which is blue. Find and solve a recurrence relation for the number of ways Joseph can stack n of these chips without consecutive blue chips.
Attempt
Let $a_n$ be the number of such arrangements of $n$ poker chips. If we ... | If the reccurence relation is
$$a_n=(k-1)(a_{n-1}+a_{n-2})$$ the solution is given by
$$a_n=\frac {2^{-(n+1)}} K \Big[(K-k-1) (k-K-1)^n+(K+k+1) (k+K-1)^n\big]$$ where $K=\sqrt{(k-1)(k+3)}$.
This generates the sequence
$$\left(
\begin{array}{cc}
0 & 1 \\
1 & k \\
2 & (k-1) (k+1) \\
3 & (k-1) \left(k^2+k-1\right) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find Two Rank 1 Matrices I'm looking for rank one matrices B and C such that A = B + C and BC = 0. Where
$$A = \begin{bmatrix}
0 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 0
\end{bmatrix}$$
with eigenvalues 0, -2, 6 and eigenvectors \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \begin{bmatrix... | What about $B=\left(\begin{array} & 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{array}\right)$ and $C = \left(\begin{array} & -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{array}\right)$?
Both matrices have rank 1, their sum is $A = B+C$ and
$BC = \left(\begin{array} & 0 & 0 & 0 \\ 0 & 0 & 0 \\0 & 0 & 0\end{array}\right)$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $\int e^{x^2+x}(4x^3+4x^2+5x+1)dx$ One of the solutions tried to write the integral a
$$\int e^{x^2+x}((2x+1)p(x) + p’(x))dx$$
Where $p(x)=2x^2+bx+c$
I have no idea why they chose the lead coefficient to be $2$.
Going with this, $b=1$ and $c=0$
So
$$\int e^{x^2+x} ((2x+1)(2x^2+x)+4x+1)dx $$
Now even after substit... | $$\begin{align}
\int e^{x^2+x}(4x^3+4x^2+5x+1)dx &= \int e^{x^2+x}(4x^3 + 2x^2 + 2x^2 + x + 4x + 1)dx\\
&= \int e^{x^2+x}(2x+1)(2x^2 + x) + e^{x^2+x}(4x+1)dx\\
&= \int e^{f(x)}(f'(x)g(x) + g'(x)) dx
\end{align}$$
where $f(x) = x^2 + x$ and $g(x) = 2x^2 + x$. You should be able to see (integration by parts is one way... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Condition that satisfies inequality involving fractions and the floor function I want to solve the following inequality for $a$ :
$$\left \lfloor \frac{a}{b} \right \rfloor > \frac{a}{b+1}$$
Where both $a$ and $b$ are strictly positive integers.
My end goal with this is, for a given $b$, to be able to find the smallest... | *
*The floor function obeys$$\mathrm{ bn\le a<b(n+1),y=b, n\in\Bbb N}$$ Graph,
*Using the identity above: $$\mathrm{y_1=b=\left\lfloor\frac ab\right\rfloor>y_2=\frac a{b+1}}$$
*Therefore when we let a vary:$$\mathrm{a>b(b+1)=b^2+b, bn\le a<b(n+1),y=b, a,b,n\in\Bbb N}$$
*Graphical proof
But this only gives the boun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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$\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx$ converges or diverges Determine if the integral $\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx$ converges / diverges:
My attempt:
Notice that $\lim_{x \to \infty}(1+\frac{-1}{x})^x=e^{-1}$, so for every $\varepsilon>0$ there exists $M>0$ such that for every $x>M$ we have $|(1+\frac{-1}... | A continuation of @Gary's observation:
As you have showed, we know that for every $x \ge M$ (for some $M\geq 1$) we have $(1+\frac{-1}{x})^{x^2} \le \frac{1}{2^x}$. Therefore
$$\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx =\int_{1}^M(1-\frac{1}{x})^{x^2}dx+\int_{M}^\infty(1-\frac{1}{x})^{x^2}dx\le \int_{1}^M(1-\frac{1}{x})^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Estimate of $\int_y^{+∞}\frac{[2-({\rm erf}(x+a)+{\rm erf}(x-a))]^2}{e^{-(x+a)^2}+e^{-(x-a)^2}}\,\mathrm dx$ $\DeclareMathOperator{\erf}{erf}\DeclareMathOperator{\erfc}{erfc}$I am trying to find either a closed form expression or an $y$-dependent analytical estimate for
$$\int_y^{+\infty}\frac{[2-(\erf(x+a)+\erf(x-a))]... | $\DeclareMathOperator{\erf}{erf}\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\paren#1{\left(#1\right)}$Assume $a > 0$. For any $ν \in \mathbb{R}$, by l'Hospital's rule,\begin{gather*}
\lim_{x → +∞} \frac{\displaystyle\int_x^{+∞} t^ν \exp(-t^2) \,\d t}{x^{ν - 1} \exp(-x^2)} = \lim_{x → +∞} \frac{-x^ν \exp(-x^2)}{-(2x^ν - (ν ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I ... | We can use your idea of finding the antiderivative. We have:
\begin{align}
e^{a\sin^{-1}(x)}& =1+a\sin^{-1}x+\dfrac{(a\sin^{-1}x)^2}{2!}+...+\dfrac{(a\sin^{-1}x)^5}{5!}+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x+\dfrac{x^3}{6}+...\right)^2+\dfrac{a^3}{3!}\left(x+\dfrac{x^3}{6}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
Generating functions in discrete mathematics - How to find coefficient from polynomial with multiplication So I encountered a certain problem when studying generating functions that I seem to be unable to fully solve.
The problem is as follows:
Carol is collecting money from her cousins to have a party for her aunt. If... | It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. With $A(x)$ a generating function we can write for instance
\begin{align*}
[x^{14}]\left(1+2x^5+x^{10}\right)A(x)=\left([x^{14}]+2[x^9]+[x^{4}]\right)A(x)\tag{1}
\end{align*}
In (1) we apply the rule $[x^{p-q}]A(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Line integral , potential function and Gradient theorem
Given the field $F=(\frac {x+y}{x^2+y^2},\frac {-x+y}{x^2+y^2}$ , calculate $\int_C Pdx+Qdy$ where $C$ is the line $y=\frac{x+25}{7}$ that begins at point $(3,4)$ and ends at $(-4,3)$
What I did was:
First I checked if $P_y=Q_x$ and got $P_y=\frac {x^2-y^2-2xy}{... | Your working is correct but you have some calculation mistakes in the end.
$\phi(-4,3)=\frac{\ln(9+16)}{2}+\arctan(\frac{-4}{3})$
$\phi(3,4)=\frac{\ln(9+16)}{2}+\arctan(\frac{3}{4})$
Line integral is $\phi(-4,3) - \phi(3,4) = \arctan(-\frac{4}{3}) - \arctan(\frac{3}{4})$
$ = - \arctan(\frac{4}{3}) - \arctan(\frac{3}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$? Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. Finally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)... | Too long to comment :
One can prove that
$$D(m^2)\lt D(p^k)\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}$$
Proof :
We have
$$\begin{align}&D(p^k)-D(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\sigma(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\frac{2p^km^2}{\sigma(p^k)}
\\\\&=2p^k-\frac{p^{k+1}-1}{p-1}-2m^2+\frac{2p^km^2(p-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom... | Since you have taken $3/2$ while doing synthetic division, we can write
$$ 2x^3 +7x^2-13x -3 = (x- \frac32)(2x^2 + 10x + 2) $$
Now rearranging terms in RHS we get
$$2x^3 +7x^2-13x -3 = (2x- 3)(x^2 + 5x + 1)$$ which is the same answer you got when you did the long division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a - b = 3$, $a + b + x = 2$, then find the value of $(a - b)[x^3 - 2ax^2 + a^2x - (a + b)b^2]$ After solving the question I got to the conclusion that it has something to do with $(a-b)$ being a factor of the whole polynomial. And since the value of $(a-b)$ is already given we can further solve the equations to get... | Let $f(x)=(a-b)[x^3-2ax^2+a^2x-(a+b)b^2]$. Now using the Remainder Theorem, which tells us that if you divide a polynomial $f(x)$ by $(x-y)$ then the remainder is $f(y)$. Now, we know that $(a-b)$ is a factor of $f(x)$ so $x=3$ is a root. Substituting it in $a+b+x=2$ we get $$a+b+3=2$$ $$\implies a+b=-1$$ Now add $a-b=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x ... | Just for your information since you will need it quite often with differential equation problem
$$\large\frac{d^2 t}{dx^2}=-\frac {\frac{d^2 x}{dt^2} } {\Big[\frac{d x}{dt} \Big]^3}$$
Just checking
$$x=\tan(t) \implies \frac{d x}{dt}= \sec ^2(t)\implies \frac{d^2 x}{dt^2}=2 \tan (t) \sec ^2(t)\implies \frac{d^2 t}{dx^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lower bound on probability of choosing odd integer out of $2m$ integers
The probability of choosing an integer ‘$k$’ out of $2m$ integers
$1,2,3,…2m$ is inversely proportional to $k^4$.If $P_1$ is the probability that chosen number is odd and $P_2$
is the probability that chosen number is even, then
A. $P_1 < \frac{2}... | For $m=1$ you have $P_1 = \frac{\frac1{1^4}}{\frac1{1^4}+\frac1{2^4}} = \frac{16}{17}> \frac23$
Now consider the position as $m \to \infty$.
Let $S_\infty= \sum\limits_{k=1}^\infty \frac{1}{k^4}$ so $\frac{S_\infty}{2^4} = \sum\limits_{k=1}^\infty \frac{1}{(2k)^4}$ and $P_1 \to \frac{S_\infty-\frac{S_\infty}{2^4}}{S_\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a$ and $b$ such that $x^3-4ax^2-8bx-3+a+b$ has a triple root $$P(x)=x^3-4ax^2-8bx-3+a+b$$
NOTE: a & b are both real and the solution is not necessarily unique
So far i've done this:
If $P(x)$ is a $3$rd degree polynomial and has a triple root, then it has the form:
$$
P(x)=(x-n)^3=x^3-3x^2 n+3xn^2-n^3.
$$
Then... | Doing the same as @Soumya Patel, the cubic equation
$$8k^3−3k^2+6k−24=0$$ has only one real root (the discriminant being negative).
Using the hyperbolic method for one real root, we have
$$k=\frac{1}{8} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)$$
Now,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$ is divided by $(x^2 +1)$, then find the remainder
If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$
is divided by $(x^
2
+1)$, then the remainder is:
How Do I solve this question without the tedious long division?
Using remainder theorem , we can ta... | With $y:=x^2$,$$P(x)=x^{19} +x^{17} +x^ {13} +x^ {11} +x^ 7 +x^ 5 +x^ 3=(y^9+y^8+y^6+y^5+y^3+y^2+y)x$$ and you are asked the remainder of the division by the polynomial $x^2+1=y+1$.
You can write
$$P(x)=(Q(y)(y+1)+R(y))x$$ where $R$ is of degree $0$, i.e. a constant (which you know how to compute by division by a linea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.
If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.
My attempt to proof:
Suppose ($ACBD$) $= -1$ then $\frac{AC}{CB} = \frac{AD}{BD}$ then $AC \cdot BD = CB \cdo... | Rephrasing of problem statement
Let $ A C = a, CB = b, BD = c$.
We are given that $ac = (a+b+c)b$.
WTS $(a+b+c)c = (b+c)(c+ (a+b)/2)$.
Taking the difference of the 2 sides, and multiplying by 2
WTS $ -ab + ac - b^2 - bc = 0$.
Notice that this is exactly the condition $ac = (a+b+c)b$. Hence we are done.
Note:... | {
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"url": "https://math.stackexchange.com/questions/4184226",
"timestamp": "2023-03-29T00:00:00",
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$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$?
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$
Since $\frac{1-a^3}{a}=\frac{1-b^3}{b}$, $a^3=b^2a-\frac{a}{b}-1$.
$\therefore a^3+b^3+c^3=ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1$
Since $a^3+b^3+c... | It looks like you don't need to do any calculations.
We can construct a new cubic equation with roots $a^3,b^3,c^3.$
$$a^3=x_1,~ b^3=x_2,~ c^3=x_3$$
$$\begin{align}r=\frac{1-a^3}{a}=\frac{1-b^3}{
b}=\frac{1-c^3}{c}\end{align}$$
$$\begin{align}&\implies1-a^3=ar \\
&\implies(1-a^3)^3=a^3r^3 \\
&\implies(1-x)^3=xr^3 \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Polynomial Inequality Holds for All Real Numbers x Problem: The quadratic polynomial $P(x),$ with real coefficients, satisfies
$$P(x^3 + x) \ge P(x^2 + 1)$$ for all real numbers $x.$ Find the sum of the roots of $P(x).$
Work: Since $P(x)$ is a quadratic polynomial, let $P(x)=ax^2+bx+c$. Then, we have $$a(x^3+x)^2+b(x^3... | You have
$$(x^3 - x^2 + x - 1) (a(x^3 + x^2 + x + 1) +b) \ge 0.$$
So, for any $x > 1$, $a(x^3+x^2+x+1)+b \ge 0$ and for any $x < 1$, $a(x^3+x^2+x+1)+b \le 0$.
It follows from continuity that $a(1^3+1^2+1+1) + b = 0$, that is, $-b/a = 4$.
You can further check that $x^2-4x$ satisfies the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $a$ and $b$ such that the graph of $y=-2\sin3x+1$ passes through $(-\frac\pi{12},a)$ and $(b,1)$ Please look at the attached image through the link. I already figured out $b$, which I'm pretty sure is $2\pi/3$. However, I'm confused on figuring out $a$. This is what I tried:
\begin{align*}
y & = -2\sin(3x)+1\\
... | To find b simply let $1 = -2\sin(3x) +1$ and it is when $\sin3x = 0$ for the some $\frac{\pi}{4} < x < \frac{\pi}{2}$
Since $\arcsin(0) \in (0, \pi, 2\pi...) $ you may obtain $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}...$ from which clearly $\frac{\pi}{3}$ is the desired value
To find a is a similiar method, let $a = 1 - 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing $\int_{0}^{2\pi}\cos(x)\cos(2x)\cos(3x)\,dx = \frac\pi2$ An integral from MIT Integration Bee:
Show that
$$I = \int_{0}^{2\pi}\cos(x)\cos(2x)
\cos(3x)\,dx = \frac\pi2$$
This integral appeared in the 2019 paper. Below is my own solution:
$$\begin{align}
I &= \int_{0}^{2\pi}\cos(x)\cos(2x)(\cos x\cos2x-\sin x\s... | Another method:
$$I=\int_0^{2\pi} \cos x\cos 2x(\cos 3x+\cos x-\cos x)\, dx=\\
\int_0^{2\pi} \cos x \cos 2x (2\cos 2x\cos x-\cos x)\, dx=\\
\int_0^{2\pi} \cos^2x\cos 2x(2\cos2x-1)\,dx=\\
\int_0^{2\pi} 2\cdot \frac{1+\cos 2x}{2}\cdot \frac{1+\cos 4x}{2}\,dx- \frac{1+\cos 2x}{2}\cdot \cos 2x\,dx=\\
\int_0^{2\pi} \frac12+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 6
} |
Question on Series Transformations from "The Harmonic Series and Logarithmic Intragrals" This is from an example in Section 1.7 of the book, "The Harmonic Series and Logarithmic Integrals" by, Ali Shadhar Olaikhan
$$\sum_{n=0}^\infty \frac{-1^n}{(2n+1)^3} = \mathcal{J}\sum_{n=1}^\infty \frac{i^n}{n^3}$$
I don't underst... | \begin{align*}
\Im \sum_{n=1}^\infty \frac{\mathrm{i}^n}{n^3}
&= \Im \left( \frac{\mathrm{i}^1}{1^3} + \frac{\mathrm{i}^2}{2^3} + \frac{\mathrm{i}^3}{3^3} + \frac{\mathrm{i}^4}{4^3} + \cdots \right) \\
&= \Im \frac{\mathrm{i}^1}{1^3} + \Im \frac{\mathrm{i}^2}{2^3} + \Im \frac{\mathrm{i}^3}{3^3} + \Im \frac{\mathrm{i}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x... | My second solution:
It suffices to prove that, for all $x \in (0, 1/2]$,
$$\ln x \ln (1 - x) \le \ln^2 2.$$
Let $f(x) = \ln x \ln (1 - x) - \ln^2 2$. We have
$$f'(x) = \frac{(1 - x)\ln (1 - x) - x\ln x}{x(1 - x)}.$$
Since $u\mapsto \ln u$ is concave on $(0, \infty)$, we have (see the remarks)
$$\frac{1 - 2x}{1 - x}\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac... | HINT
Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation:
\begin{align*}
\left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\
& \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\
& \Longleftrightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If A is the 4 by 4 matrix of ones, find the eigenvalues and the determinant of A−I So I want to find the eigen values and eigen vectors of a matrix with all 1's
\begin{bmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix}
Only 1 independent would be left,
\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}... | HINT
To begin with, obtain the characteristic polynomial:
\begin{align*}
\det(A - \lambda I) & =
\begin{vmatrix}
1 - \lambda & 1 & 1 & 1\\
1 & 1 - \lambda & 1 & 1\\
1 & 1 & 1 - \lambda & 1\\
1 & 1 & 1 & 1 - \lambda
\end{vmatrix}\\\\
& =
\begin{vmatrix}
4 - \lambda & 1 & 1 & 1\\
4 - \lambda & 1 - \lambda & 1 & 1\\
4 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| There are indeed not enough restrictions. We can write $\cos(\theta) = \frac a h$, $\sin(\theta) = \frac b h$, hence $h\cos(\theta)+h\sin(\theta)+h=30$, or $\cos(\theta)+\sin(\theta)+1 = \frac{30}h$.
$\cos(\theta)+\sin(\theta)$ is equal to $\sqrt 2\sin(\theta +\frac{\pi}4)$, so we get $\sqrt 2 \sin(\theta +\frac{\pi}4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality... | Multiplying the numerators of the fractions by $a^2$, $b^2$ and $c^2$ gives a wrong inequality:
$$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{a^2}{a^2(b^2+c^2+bc)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}$$ and we need to prove that:
$$(a+b+c)^4\geq9\sum\limits_{cyc}(2a^2b^2+a^2bc),$$ which is wrong fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Determine whether the series absolutely converge, conditionally converge or diverge The series is $\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$.
In my book I just see examples and exercises for determining whether the series absolutely converge, conditional converge or diverge in alternating series.
This series is not ... | First off, your series isn't alternating, so you cannot use the Alternating Series Test. The terms "absolutely\conditionally converge" ONLY apply to alternating series. If your series is not alternating, it either just converges or diverges. These terms don't apply. You could say it's absolutely convergent, because the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given a function, find values of when it is equal to $100$. Question: Let $f(n) = 2n - [\frac{(1 + \sqrt{8n – 7})}{2}]$ for each positive integer, where $[x]$ is denoted for the largest integer less than or equal to $x$ for any real number $x$. Find the sum of all possible value(s) $n$ such that $f(n) = 100$.
My soluti... | $$2n- \left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor= 100$$
$$2n-100 =\left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor $$
$$ \frac{\sqrt{8n-7}-1}2\le 2n-100 \le \frac{1+\sqrt{8n-7}}{2}$$
$$4n-200-1 \le \sqrt{8n-7} \le 4n-200 + 1$$
$$(4n-201)^2 \le 8n-7 \le (4n-199)^2$$
The first inequality is satisfied for $$46 \le n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- N... | Here is another approach.
By AM-GM,
$(a+b+c)(a^2+b^2+c^2) \geq 3 (abc)^{2/3} \cdot 3 (abc)^{1/3} = 9abc$
$4 (a^2+b^2+c^2) \geq \cfrac{18abc}{s}$
$13 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + \cfrac{18abc}{s}$
Using $a^2+b^2+c^2 \geq ab+bc+ca$,
$13 (a^2+b^2+c^2) \geq 9(ab+bc+ca) + \cfrac{18abc}{s}$
$26 (a^2+b^2+c^2) \geq 18(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
What is the most complete binary quadratic form table? Is there a free, easily accessible quadratic form resource — maybe even just a big web-accessible table — which shows “all known results” [reasonably speaking] about numbers of the form $mx^2+ny^2$ for various $m,n$?
Immediate motivation: Given integers $a$ and $x... | Particular case where $(x, 3)=1$
For $3x^2+5$ we may write:
$x^2\equiv 1\bmod 3\Rightarrow 3x^2+5\equiv 8\equiv 2\bmod 3$
We may conclude that primes of the form $p=3m+2$ can divide $3x^2+5$ , for example:
$x=1\rightarrow 3x^2+5=8, k=0\rightarrow p=2; 2|8$,
$x=6\rightarrow 3x^2+5=113=3\times 37+2$
$x=4\rightarrow 3x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4205723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.