Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving an inequality(with a term in cosine) using the mean value theorem Given that $x\in \mathbb{R}$ and
$-2\leq x\leq 2$
prove:
$|cos(x) -1 +\frac{x^2}{2}| \leq\frac{2}{3} $
What I have so far:
for all x in between -2 and 2, $cos(x)\in[0, 1] $ and $-1 +\frac{x^2}{2}\in[-1, 1] $
I've tried to solve for this inquality... | It's easier.
$$
\cos(x)-1+\frac{x^2}{2} = \sum_{k = 0}^\infty \frac{(-1)^k x^{2k}}{(2k)!} -1+\frac{x^2}{2} = \sum_{k = 2}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So:
$$
\left \lvert \cos(x)-1+\frac{x^2}{2} \right \rvert \leq \left \lvert \sum_{k=2}^\infty \frac{ x^{2k}}{(2k)!} \right \rvert = \left \lvert \frac{x^4}{4!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the side length of an inscribed tilted cube in a spherical octant
A cube of unknown side length is to be inscribed in the unit spherical octant where all coordinates are nonnegative. The cube is tilted, and has four vertices belonging to the same cube face on the curved surface of the unit sphere, and two vertic... | Attaching a coordinate reference frame to the cube with its $z'$ axis pointing from the origin towards the center of the cube, and its $x'$ axis horizontal (parallel to the $xy$ plane), we can express the coordinates of the vertices.
We have
$\hat{z'} = (\cos(\theta) \cos(\frac{\pi}{4}) , \cos(\theta) \sin(\frac{\pi}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{\pi/2} \frac{y \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$ This integral:
$$\int_{0}^{\pi/2} \frac{y \sqrt{\cos(y)}}{\sqrt{\cos(y)} + \sqrt{\sin(y)}} dy$$
came up when I was playing around with trying to solve this similar problem:
Is there a closed form for $\int_0^{\pi/2} \dfrac{e^{-x}\sq... | Disclaimer : this is not a full answer.
Let
$$I_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\cos y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy$$
and
$$J_{n} = \int_{0}^{\frac{\pi}{2}} \frac{y^{n} \sqrt{\sin y}}{\sqrt{\cos y} + \sqrt{\sin y}} dy.$$
Substitute $y' = \frac{\pi}{2} - y$ then
$$I_{n} = \int_{0}^{\frac{\pi}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find the closed form of $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}$, where $n\in N$? In my post, I found the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi\left(a^{2}+b^{2}\right)}{4 a^{3} b^{3}}$$
Then I w... | Let $a=p+r$, $b=q+r$ and, for sure, as you did $x=\tan^{-1}(t)$
$$I_n=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} (x)+q \sin ^{2} (x)+r\right)^{n}}=\int_0^\infty \frac{(1+ t^2)^{n-1} }{(a+b\,t^2)^n }\,dt$$There is an antiderivative
$$\int\frac{(1+ t^2)^{n-1} }{(a+b\,t^2)^n }\,dt=\frac t {a^{n}}\, F_1\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there positive integer solution of $a^b - c^d = 6$? Question: is there any integer solution of
$$a^b - c^d = e$$
where $e=6$, and $a,b,c,d \geq 2$? Any idea to attack this type of problem? Thanks.
Adam Rubinson comments that $a$ and $c$ are both odd by considering on $\mathbb Z/2 \mathbb Z$. We have $b \neq d$ since... | THIS IS NOT AN ANSWER, BUT IS TOO LONG FOR THE COMMENTS SECTION
Does $a^b-c^d=6$ have solutions in the integers, $a,b,c,d \ge 2$?
$\{a,c,6\}$ are pairwise coprime, because any prime factor of two of them is a factor of all three of them, and $6$ has only $2,3$ as prime factors. For $b,d \ge 2$, any prime factor of $a,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is this interval notation for the solution of an inequality problem correct? Is this notationally correct?
$$
\begin{align*}
5 - x^2 &< -2, \\[1ex]
5 - x^2 + (-5) &< -2 + (-5), \\[1ex]
-x^2 &< -7, \\[1ex]
(-x^2)(-1) &> (-7)(-1), \\[1ex]
... | I suggest either spelling out that logical symbol (for readability and because some readers may not know what it means) $$x < -\sqrt{7} \quad\text{or}\quad x > \sqrt{7},$$ or actually adopting interval notation $$(-\infty,-\sqrt{7})\cup (\sqrt{7},\infty)$$ as suggested by paw88789.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the solutions of the PDE $x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$ How to find all solutions of the PDE
$x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}=(x+y)u$?
My way was:
The Lagrange-Charpit equations are: $\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{(x+y)u}$.
... | $\frac{\frac{1}{x}dx}{x}=\frac{\frac{1}{y}dy}{y}=\frac{\frac{1}{u}du}{(x+y)}$
$\implies\frac{\frac{1}{x}dx+\frac{1}{y}dy}{x+y}=\frac{\frac{1}{u}du}{x+y} $
$\implies \frac{1}{x}dx+\frac{1}{y}dy=\frac{1}{u}du$
Solve and get second integral curve to form the solution surface.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ Calculate double integral $\int_Se^{\frac{x}{y}}dxdy$ for the region $1 \le y \le2$ and $y \le x \le y^3$
What I have tried:
$y = 1, y=2 \\ x = y, x=y^3$
$1 \le x \le 2, \text{when }x=y \\
1 \le x \le 2^{\frac{1}{3}}, \text{when }x=y^3 \\
2 \le x \le 8, \text{ when}... | Integrating wrt $x$ first (the inner integral in equation below),
\begin{equation}
\int_1^2 dy \int_y^{y^3} dx \, e^{x/y} = \int_1^2 dy \, y \, \left[ e^{x/y} \right]_{y}^{y^3} = \int_{1}^2 dy \, y \, [e^{y^2} - e] = \frac{e^4 - e }{2} -\frac{3}{2}e = \frac{1}{2} e^4 - 2 e \,.
\end{equation}
We can see that $x\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
True or False: There is a $6\times 6$ matrix $A$ with $\text{Rank}(A)=4$ and $A^3 =0$ I understand how to do it if the question changed $A^3$ to $A^2$, because then you can just use the rank–nullity theorem. $\text{Rank}+\text{Nullity}=6$, $\text{Rank}=4$ so $\text{Nullity}=2$ so of the $6$ column vectors $v_1,v_2,\dot... | If $A = \begin{bmatrix} 0 && 0 && 1 && 0 && 0 && 0 \\
0 && 0 && 0 && 1 && 0 && 0 \\
0 && 0 && 0 && 0 && 1 && 0 \\
0 && 0 && 0 && 0 && 0 && 1 \\
0 && 0 && 0 && 0 && 0 && 0 \\
0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix}$
Then
$A^2 = \begin{bmatrix} 0 && 0 && 0 && 0 && 1 && 0 \\
0 && 0 && 0 && 0 && 0 && 1 \\
0 && 0 && 0 && 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$ Is there a closed form integral for $$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$$ for $-1 < x < 1$?
This integral is related to Legendre polynomials as
$$\frac{-1}{z}\ln\left(\frac{1-x z +... | Here's an approach using differentiation under the integral sign:
$$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\
&=\left.\frac{(x+1)\ln\left(\sqrt{z^2-2xz+1}+x-z\right)-2x\ln\left(\sqrt{z^2-2xz+1}-z+1\right)}{x^2-1}\right]_{z=0}^{z=1}\\
&=\frac{x\ln(2)-x\ln(1-x)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdo... | A solution (in Chinese) is as follows. Is it correct?
\begin{align*} \sum_{k=1}^n\frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \sum_{k=1}^n\frac{k^{p+1}}{k\sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}}\\ &=\sum_{k=1}^n\left(\frac{k}{\sqrt[k]{k!}}\right)^p\frac{1}{\sqrt[k]{a_1a_2\cdots a_k}}\\& \le e^p\sum_{k=1}^n\frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)... | It should be mentioned that this is one of the Schur polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the equation of the straight line that contains a point, is perpendicular to a line , and is parallel to a plane. Find the equation of the straight line that contains the point $(3,7,5)$, is perpendicular to line $x-1=\frac{y+7}{-2}=\frac{z-2}{3}$, and is parallel to plane $2x+y-z-2=0$.
To make this possible, shou... | If your searched for line contains the point $(3,7,5)$, we can write it as
$$f=\begin{pmatrix}3\\7\\5\end{pmatrix}+r\begin{pmatrix}a\\b\\c\end{pmatrix}$$
With a free variable $t$ the given line can be rewritten to
$$g=\begin{pmatrix}t\\-2t-5\\3t+1\end{pmatrix}=\begin{pmatrix}0\\-5\\1\end{pmatrix}+t\begin{pmatrix}1\\-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Are $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime to each other, $n \in \mathbb{N}$? Let $p$ be a prime integer greater than $2$.
Then I want to prove the followings:
$(1)$ $\frac{p^2+1}{2}$ and $\frac{p^5-1}{2}$ are coprime to each other.
$(2)$ $\frac{p^2+1}{2}$ and $\frac{p^{5n}(p^5-1)}{2}$ are coprime ... | This is for the question in the title. Let $q$ be a prime dividing both numbers. Then $q\mid p^2+1$, say $p^2+1 = Kq.$ Also note that $q\neq p$, so we must have $$q\mid p^5 -1 = p^5 + p^3 - p^3 - p +p -1 $$ $$= p^3(p^2+1) - p(p^2+1) + p-1$$
$$=p^3Kq -pKq + p-1.$$
Therefore $q\mid p-1$. And so $q\mid p^2-1$. Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving that $a_i,b_i \to 0$ where $a_{i+1} := |b_i - a_i|$ and $b_{i+1} := |a_i - a_{i+1}|$? Let $a_0 := 1$ and $b_0 := \sqrt{2}$, and define
\begin{align*}
a_{i+1} &:= |b_i - a_i| \\
b_{i+1} &:= |a_i - a_{i+1}|
\end{align*}
Prove that $\lim_{i \to \infty}{a_i} = 0$ and $\lim_{i \to \infty}{b_i} = 0$.
Context: I am gi... | Prove, by induction that
$$a_n = \left(\sqrt{2} - 1\right)^n, ~~b_n = \left(\sqrt{2} - 1\right)^{n-1} - \left(\sqrt{2} - 1\right)^n. \tag1 $$
Empirically true for $n=1$.
$$a_{n+1} = |b_n - a_n| = |\left(\sqrt{2} - 1\right)^{n-1} \times [ 1 - 2(\sqrt{2} - 1) ]|$$
$$ = |\left(\sqrt{2} - 1\right)^{n-1} \times (3 - 2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality cas... | Partial hint :
We have a two first not hard to show :
Let $x>0$ then define :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}$$
Then :
$$f''(x)\leq 0$$
Second fact :
Let $x>0$ then define :
$$(f(e^x))''<0$$
So we use Jensen's inequality with the constraint $c\geq d\ge a\geq 1$ we need to show for $x,y\leq 1$ :
$$2f\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
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Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$ EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:
We do this using the epsilon-N definition of the limit of :
$$\forall \v... | Do
$\frac{n^2+n-1}{n^2+2n+2}=\frac{1+\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}+\frac{2}{n^2}}$
Make use of the fact that this is each $1$ plus something that is getting smaller and smaller the bigger $n$ gets.
$\frac{1}{1+x}\approx 1-x$
in first order for small $x$.
So the division goes over into a simple binomial multi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find residue of $f(z) = \frac{\sin z}{(z^2+1)^2}$ at $z = \infty$ Find residue of $f(z) = \dfrac{\sin z}{(z^2+1)^2}$ at $z = \infty$. Then this is the same as finding the residue at $z=0$ for $\dfrac{-1}{z^2}f(1/z)= \dfrac{-z^2 \sin 1/z}{(z^2+1)^2}$
$z = 0$ is a pole. But $z^n \sin 1/z \to \infty$ when $z \to 0, n = 1,... | The function $\frac{-z^2}{(z^2+1)^2}$ is holomorphic near $z=0$. So you can expand it as a power series at $z=0$:
$$
\frac{-z^2}{(z^2+1)^2}=a_0+a_1z+a_2z^2+\cdots\tag{1}
$$
where you can work out the coefficients by looking at the Taylor expansion of $\frac{z}{(z^2+1)^2}$ at $z=0$ and then multiply by $-z$.
Using the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$ Find the locus of a point, sum of whose distances from $(c, 0)$ and $(-c, 0)$ is constant and greater than $2c$.
Try
Let the point be $(h, k)$. Then by given condition
$\sqrt {(h-c) ^2+k^2} + \sqrt {(h+c) ^2 ... | Yeah, basically it s ellipse.. Let $2a=2c+\alpha$ then we can derive the standatd ellipse equation from it. The trick to deal with the sqrt is to square twice.
$$
\sqrt {(x-c) ^2+y^2} + \sqrt {(x+c) ^2 +y^2} = 2c + \alpha := 2a\\
\sqrt {x^2+y^2+c^2-2cx} + \sqrt {x^2+y^2+c^2+2cx} = 2a
$$
Let $A=x^2+y^2+c^2$ Square once
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$
My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \... | Hint :
Using :
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
And $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$
We got :
$$8-6(ab+bc+ca)+5abc\leq 2$$
You can conclude using uvw's method .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$ in $\Delta ABC$
In $\triangle ABC$, $\sin B=-\cos C$. Find the minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$.
According to the law of sines, $\begin{aligned}\frac{a^2+b^2}{c^2}=\frac{\sin^2A+\sin^2B}{\sin^2C}\end{aligned}$.
Solution $1$
Let $... | Everything should be correct in solution $2$ because AM-GM can be used as $\sin^2 C$ is always non-negative.
For solution $1$ however, after $\sin A = 1 - 2k^2$ your steps don't follow: something must have been gone catastrophically wrong there. Using the values that we obtained for $\sin A, \sin B$ and that $\sin^2 C ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
Aft... | We have, $$A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$$
According to Caley-Hamilton Theorem,
$$|A - \lambda I| = 0$$
So we have,
$$A - \lambda I = \begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix} - \lambda \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = \begin{bmatrix}2 - \lambda&1&1\\0&1 - \lambda &0\\1&1&2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$ I just happened to find a problem and an elegant solution.
The question asks us to solve the following equation
$$\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I am answering this question below but I would love if you can also share a differ... | Suppose $$A=\sqrt[3]{10-x}+\sqrt[3]{30-x}=\sqrt[3]{15-x}+\sqrt[3]{25-x}$$
I will use the following
\begin{align}
A^3=&(p+q)^3 \\
=&p^3+q^3+3pq(p+q) \\
=&p^3+q^3+3pq(A)
\end{align}
Then
\begin{align}
A^3=&(10-x)+(30-x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A) \\
=&(40-2x)+3(\sqrt[3]{10-x})(\sqrt[3]{30-x})(A)
\end{align}
\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
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find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2... | Note: The following answer is incorrect, see comments. I don't have a correct version yet. However since the other answer mentions this, I've marked it as community wiki and made it visible.
Let
$$
L
=
\lambda (x^3+y^3+z^3-3)+{x^3\over 3y+1}+{y^3\over 3z+1} +{z^3\over 3x+1}\\
$$
Differentiating with respect to $\lambda... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove decay function concaved-up Below decay function were used for solving TVM, for interest rate.
$\displaystyle f_n(x) = \frac{n\;x}{(1+x)^n-1}\qquad$, where $n>1,\;x>-1$
I wanted to show decay function is concaved-up.
In other words, I wanted to show $f_n^{''}(x) > 0$
This is what I get for second derivative, confi... |
For $x≠0$, all we need is to show
$(f_n(x) + ({n-1 \over 2})x-1) > 0$
Let $R=1+x>0$
$\displaystyle f_n(x) = \frac{n\;(R-1)}{R^n-1} \stackrel{?}{>}\; \left(1 - \left(\frac{n-1}{2}\right)\;(R-1)\right)$
Or, $\displaystyle \; n \stackrel{?}{>}
\left( \frac{R^n-1}{R-1} \right)
\left(1 - \left(\frac{n-1}{2}\right)\;(R-1)... | {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Explaining symmetry in sequences defined by $\frac{a_{n+1}}{n+1}=\frac{a_n}{n}-\frac12$. (Eg, for $a_1=5$, we get $5,9,12,14,15,15,14,12,9,5$) I've noticed recently the following fact: consider ratios $\frac{a_n}{n}$ with $n = 1,\dots,10$. If we require that
$$
\frac{a_{n+1}}{n+1} = \frac{a_n}{n} - \frac12
$$
that is, ... | $b_n = \dfrac{a_n}{n}\,$ is an arithmetic progression with starting value $\,a_1\,$ and common difference $\,-\dfrac{1}{2}\,$, so:
$$
b_n = \frac{2a_1 - n + 1}{2} \quad\implies\quad a_n = n \cdot \frac{2a_1 - n + 1}{2}
$$
For $\,2a_1 \in \mathbb N\,$, it follows that:
$$
\require{cancel}
\begin{align}
a_{2a_1+1 - n} &=... | {
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"answer_count": 2,
"answer_id": 0
} |
Volume of $y=x^2+1; y=-x^2+2x+5; x=0; x=3$ about $x$ axis (Shell Method). I was working on this exercise for an assignment. However, I get stuck in the following part.
$
y=-x^2+2x+5
$
Complete the square
$y=-(x^2-2x)+5$
$(b/2)^2=(-2/2)^2=1$
$y=-(x^2-2x+1-1)+5$
$y=-(x^2-2x+1)+5+1$
$y=-(x-1)^2+6$
$y-6=-(x-1)^2$
... | Yes, it can be solved through the shell method. Consider the picture below:
Let us first see what is the volume of the region for which $0\leqslant x\leqslant2$. The shell method tells us that it is equal to\begin{multline}2\pi\left(\int_1^5y\sqrt{y-1}\,\mathrm dy+\int_5^6y\left(\left(1+\sqrt{6-y}\right)-\left(1-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Flux of $\mathbf{F}=(xz,yz,x^2+y^2)$ across paraboloid $z=4-x^2-y^2,\ z\geq 0$ I have calculated the flux of the vector field $\mathbf{F}=\begin{bmatrix}xz\\ yz\\ x^2+y^2\end{bmatrix}$ outward across the surface of the paraboloid $S$ given by $z=4-x^2-y^2,\ z\geq 0$ (with outward-pointing normal having positive $\mathb... | I got a different result,
$$\begin{align}
\iint_S \mathbf{F}\cdot d \mathbf{S}&=\iint_{x^2+y^2\leq 4}
\langle (xz,yz,x^2+y^2),(2x,2y,1)\rangle dx dy\\
&=\int_{0}^{2\pi}\int_{r=0}^2(2r^2(4-r^2)+r^2)r drd\theta\\
&=2\pi\int_{r=0}^2(9r^3-2r^5)\,dr=2\pi\left[9r^4/4-r^6/3\right]_0^2=\frac{88\pi}{3}.
\end{align}$$
The result... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \ma... | The answers are actually the same.
$\begin{align*}
\theta = 2\arcsin \sqrt x &\implies \sin \frac \theta 2 = \sqrt x \\
&\implies \sin^2 \frac \theta 2 = x \\
&\implies \frac{1-\cos \theta}{2} = x \\
&\implies \cos \theta = 1-2x \\
&\implies \sin(\theta+\pi/2) = 2x-1 \\
&\implies \theta = \arcsin(2x-1) - \pi/2
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$? Let $A$ be a diagonal matrix and $B$ be a Hermitian and Idempotent matrix, then is $(I + ABA^H)^{-1} = (I - \frac{1}{2}ABA^H)$, where $I$ is an identity matrix?
I am reading a paper, which sta... | This can be shown by a direct computation as such:
\begin{align}
(I + ABA^H)(I - \frac{1}{2}ABA^H) &= I - \frac{1}{2}ABA^H + \frac{1}{2}(ABA^H)^2 \\
&\stackrel{\spadesuit}{=} I - \frac{1}{2}ABA^* + \frac{1}{2}AB^2A^* \\
&= I - \frac{1}{2}ABA^H + \frac{1}{2}ABA^H = I
\end{align}
Assuming that $A^HA = I$ at $\spadesuit$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Smith Normal forms in a polynomial ring
Let $M$ be a $\mathbb{C[x]}$ module with generators $m_1,m_2$ and relations : $$(x^2+ix)m_1+(x+i)m_2=0 \\ (-2x+2i)m_1+ (x^2+1)m_2=0 $$
Find integers $t,n_1,...,n_s \in \mathbb{N_0}$ and $\lambda_1,\lambda_2,...,\lambda_s\in \mathbb{C}$ such that the following holds: $$\mathbb{C[... | You're very close to finishing it. Once you have a unit, like $4i$, you can rescale a row or column to make it $1$ and then begin canceling all the other entries in its row and column.
\begin{align*}
\begin {pmatrix} x+i & 4i \\0 & -x^3-3x+2i\end {pmatrix} &\leadsto
\left(\begin{array}{rr}
4 i & x + i \\
-x^{3} - 3 x +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$ if $|z_{1}| < 1$ and $|z_{2}| < 1$. Given $z_{1},z_{2} \in \mathbb{C}$
Prove that $|z_{1}+z_{2}| < |1+\overline{z_{1}}\cdot z_{2}|\;$, if $|z_{1}| < 1$ and $|z_{2}| < 1$.
My Professor gave my classmates and I, the following answer:
$|z_{1} + z_{2}| \leq |z_... | Square both sides:
\begin{align*}
|z + w| < |1 + \overline{z}w| & \Longleftrightarrow |z + w|^{2} < |1 + \overline{z}w|^{2}\\\\
& \Longleftrightarrow (z + w)(\overline{z} + \overline{w}) < (1 + \overline{z}w)(1 + z\overline{w})\\\\
& \Longleftrightarrow z\overline{z} + z\overline{w} + \overline{z}w + w\overline{w} < 1 ... | {
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"url": "https://math.stackexchange.com/questions/4483122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$y''+y=x^2+1, y(\pi)=\pi^2, y'(\pi)=2\pi$ - By Laplace Transform I am solving the following IVP by Laplace Transform:
$$y''+y=x^2+1,\qquad y(\pi)=\pi^2, \qquad y'(\pi)=2\pi$$
Let $f(x)=u_{\pi}(x)y(x-\pi).$ Then,
$$f''(x)+f'(x)=u_{\pi}(x)(x-\pi)^2+u_{\pi}(x), \qquad f(0)=\pi ^2, \qquad f'(0)=2\pi.$$
Using the Laplace Tr... | After Laplace transformation we have
$$
Y(s) = \frac{s^2+y_0s^4+s^3y'_0+2}{s^3(s^2+1)}
$$
with inverse
$$
y(x) = x^2+\sin (x) y'_0+y_0 \cos (x)+\cos (x)-1
$$
here $y_0$ and $y'_0$ are generic constants so imposing the initial conditions we have
$$
\cases{
y(\pi) = \pi^2-y_0-2=\pi^2\\
y'(\pi) = 2\pi-y'_0=2\pi
}
$$
and s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
When not to use $ \lim\limits_{x\rightarrow 0}\frac{\log\left( 1+x\right) }{x}=1 $ My Teacher has told to use this as a standard result so when question is in this I directly apply this result as in this
$$\lim\limits_{x\rightarrow 0}\dfrac{\log\left( 5+x\right) }{x}-\dfrac{\log\left( 5-x\right) }{x}=2/5$$
but here
$$\... | The proposed result is not valid.
Having said that, you may be also interested in the power series method:
\begin{align*}
\ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \ldots
\end{align*}
whenever $|x| < 1$. If we restrict the values of $x$ to $(-1/2,1/2)$, we may claim as well that
\begin{align*}
\ln(1 + 2x) = ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit question involving the the greatest integer function and fractional part The value of $$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\sin \left\{ {\frac{2}{n}} \right\}}}{{\left[ {2n\tan \frac{1}{n}} \right]\left( {\tan \frac{1}{n}} \right)}} + \frac{1}{{{n^2} + \cos n}}} \right)^{{n^2}}},$$ where $[.]$... | Recall the Maclaurin series of $\sin x$ and $\tan x$:
$$\sin x= x-\frac{x^3}6+O(x^5),\qquad x\to0,$$
$$\tan x= x+\frac{x^3}3+O(x^5),\qquad x\to0.$$
As a result,
$$\sin\left\{\frac 2n\right\}=\sin\left(\frac2n\right)=\frac2n-\frac4{3n^3}+O\left(\frac1{n^5}\right),\qquad n\to\infty,$$
$$\tan\left(\frac1n\right)=\frac1n+\... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What am I doing wrong in integrating $\frac1{\sqrt{x} + \sqrt[4]{x}}$ $\displaystyle\int \dfrac{1}{\sqrt{x} + \sqrt[4]{x}}\mathrm dx$
Putting $x = t^4$ and $dx=4t^3\mathrm dt$,
$$ = \int \dfrac{4 t^3 \mathrm dt}{t^2 + 1}\\ = 4\int \dfrac{t^3 + t - t \mathrm dt}{t^2 + 1} \\= 4\int \dfrac{t^3 + t}{t^2 + 1}\mathrm dt - 4\... | The problem lies in the first step. After the substitution, you should have obtained$$\int\frac{4t^3}{t^2+t}\,\mathrm dt\left(=4\int t-1+\frac1{t+1}\,\mathrm dt\right).$$
| {
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"url": "https://math.stackexchange.com/questions/4485266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
when ring with roots is a field I encountered the following problem in abstract algebra textbook. To determine if the set
$$\{a+b\times 3^{1/3}+c \times 3^{2/3},\quad a,b,c \in \mathbb{Q}\}$$
is a ring, and, if so, a field as well.
I think it's clearly a ring because the product of two such element is also in the set,... | Hint: $\,$ let $\,\omega\,$ be a primitive cube root of unity so that $\,\omega^3=1\,$ and $\,\omega^2+\omega+1=0\,$, then:
$$
(a + b t + ct^2)(a + b\omega t + c\omega^2 t^2)(a + b \omega^2 t + c\omega t^2) = c^3\, t^6 + (b^3 - 3 a b c)\, t^3 + a^3
$$
With $\,t = \sqrt[3]{3}\,$:
$$
\frac{1}{a + b \sqrt[3]{3} + c \sqrt[... | {
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Evaluate $\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}$ . Problem:
Evaluate
$$
\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}.
$$
My question:
This is a question from a previous complex analysis qualifying exam that I am trying to work through. I know that the straightforward... | A variation: It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain
\begin{align*}
\color{blue}{\underset{z=0}{\text{Res}}}&\color{blue}{ \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}}
=[z^{-1}]\frac{\left(z^6-1\right)^2}{z^5\left(2z^4-5z^2+2\right)}\\
&=[z^{4}... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tr... | Hint
If you solve
$$f(n) = \frac{n-4}{n}f(n-1)+1$$ the solution is given by
$$f(n)=\frac {n+1} 5+\frac C{(n-3) (n-2) (n-1) n}$$ Then, ..., try to find the mistake.
| {
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"url": "https://math.stackexchange.com/questions/4488489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is i... | Here's an approach that involves checking considerably fewer than $24$ cases, with it also having the advantage that it helps to provide a somewhat intuitive explanation of why the relation holds.
First, note that $\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{n-\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Linear upper bound for function involving Lambert W I have a function
$$f(x)=\frac{2-2x}{\ln x}W_{-1}\left(\frac{\ln x}{(2-2x)\sqrt x}\right)$$
Looking at the plot, it appears somewhat linear in shape and empirically, it seems that we have $\lim_{x\to\infty}f(x)/x=3$. Can this limit be proven, and if so, is there a sim... | Fact 1: It holds that, for all $x > 2$,
$$f(x) \le 3x \cdot \frac{(x - 1)(3\ln x + 2\ln 3 - 2)}{3x\ln x - 2x + 2}.$$
Fact 2: It holds that $f(x) \ge 3x$ for all $x > 2$.
By Facts 1-2, using the squeeze theorem, we have
$$\lim_{x\to \infty} \frac{f(x)}{3x} = 1.$$
Proof of Fact 1:
Let
$$y := \frac{1}{3x} \cdot \frac{2-2... | {
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"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Polynomial division by polynomial without remainder I have two polynomials:
$$
Q(x)=x^{624} + x^{524} + x^{424} + x^{324} + x^{224} + x^{124} + x^{n}
$$
$$
P(x) = x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1
$$
And the question: what is the largest natural value of the number $n \leq 100 $ at which the polynomial $Q(x... | First, using the geometric series formula, it is easy to see that
$$
P(x) = \sum_{i=0}^6 x^i = \frac{x^7-1}{x-1}
$$
If we let $H(x) = Q(x)/P(x)$ for some polynomial $H(x)$, then rearranging the equation gives the following:
$$
(x-1)Q(x) = (x^7-1)H(x)\\
x^{625}-x^{624} + x^{525}-x^{524} + \dots+ x^{125}-x^{124} + x^{n+1... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff ... | I guess this is a typo, they meant to say
$$ 1-a \ge b-a,$$
which is true from
$$ (1-a)(1+a) \ge b(b-a)$$
since $1+a \ge 1 \ge b$.
(Of course, $1-a\ge b-a$ is the same as $1\ge b$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Asymptotic expansion of solutions of $x = \tan\left(x\right)$ I'm working on having an asymptotic expansion of the solution of $\tan\left(x\right)=x$ for $x \in I_n$ where $I_n = \left]-\frac{\pi}{2}+n\pi ; \frac{\pi}{2} + n\pi\right[$. Let's note $x_n$ this solution.
I've used that
$$
\tan\left(x_n\right)=x_n \Leftrig... | Say that you are looking for the $n^{\text{th}}$ zero of function
$$f(x)=x-\tan(x)$$ which is discontinuous at any $x=(2 n+1)\frac{\pi }{2}$. Since this cannot be a root, consider instead
$$g(x)=x\cos(x)-\sin(x)$$ and expand as series around $x=(2 n+1)\frac{\pi }{2}$.
This gives
$$g(x)=-(-1)^n-\frac{1}{2} \left(\pi ... | {
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Understanding the alternate method for evaluating the number of ways for below question Number of positive ($0$ excluded) integral solutions which contains only $1,2,3$ as variable values to the equation $a_1 +a_2 + a_3 + \ldots + a_6 = 12$ are?
My method was this: It's equivalent to finding coefficient of $x^{12}$ in... | As JMoravitz indicated in the comments, the correct expression should be
$$\binom{11}{5} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}\binom{5}{5}$$
A particular solution of the equation
$$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 12 \tag{1}$$
corresponds to placing $6 - 1 = 5$ addition signs in the $12 - 1 = 11$ spaces between ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $... | Hint: You can also use Hölder's Inequality, like so:
$$\left(8x^4+27y^4+64z^4 \right) \cdot \left(\frac12+\frac13+\frac14 \right)^3 \geqslant \left(|x|+|y|+|z|\right)^4 \geqslant (x+y+z)^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove $f(-\frac12) \le \frac{3}{16}$ if all roots of $f(x) = x^4 - x^3 + a x + b$ are real
Let $a, b$ be real numbers such that all roots of $f(x) := x^4 - x^3 + ax + b$ are real. Prove that $f(-1/2) \le 3/16$.
The question was posted recently which was closed,
due to missing of contexts etc.
My attempt: $f(-1/2) \le... | Suppose $b= a/2+c$ where $c>0$. Then we have
$$\underbrace{x^4-x^3+ax+{a\over 2}}_{p(x)} =-c$$
This means that graph of $p$ and line $y=-c<0$ have at least two different common points (if it has only one then $p(x)=(x-d)^4-c$ for some real $d$, this case is not possible) so $p$
has exactly two minimums with negative va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \thet... | Given $I(a)=\int_{0}^{\pi} \ln (a+b \cos x)dx
=\pi \ln\frac{a+\sqrt{a^2-b^2}}2$
$$\int_{0}^{\pi} \frac{1}{a+b \cos \theta}d\theta
=\frac{dI(a)}{da}=\frac{\pi}{\sqrt{a^{2}-b^{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$
My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
& \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
& \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -... | I give here another parametrization, a little harder and less practical but somewhat enlightening for the beginners. The focus is $F=\left(\dfrac{-7}{40},\dfrac{11}{40}\right)$ and the directrice is $4x+12y-1=0$ then if $(x,y)$ is any point of the parabola then its distance to the focus and to the directrice are equal ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
| We can do
$$\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q} \times \frac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+q^2}+q} = \frac{(\sqrt{x^2+p^2}-p)(\sqrt{x^2 + q^2}+q}{x^2}$$
First $$\frac{\frac{\sqrt{x^2+p^2}-p}{1}\times \sqrt{x^2 + q^2} +q}{x^2} \tag{1}$$
and
$$\frac{\frac{\sqrt{x^2 + p^2}+p}{\sqrt{x^2 + p^2}+p}\times \frac{\sqrt{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to $a+b+c=1$ and $a,b,c>0$
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to
$a+b+c=1$ and $a,b,c>0$
My attempt:
Since $f(a,b,c) \ge 0$, then we can minimizing the addends individually. Furthermore, minimizing fractions is the same... | This solution was posted by Arkady Alt on LinkedIn. Let $t:=ab+bc+ca.$ Since $3abc=3abc(a+b+c)\leq(ab+bc+ca)^2=t^2$,
$a^2+b^2+c^2=1-2t$ and $3t\leq(a+b+c)^2=1$ then
$$(1/(a^2+b^2+c^2))+(1/(9abc))-6≥(1/(1-2t))+(1/(3t^2))-6=$$
$$((4t+1)(1-3t)^2)/(3t^2(1-2t))≥0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1... | Even more elementary:
Note that $x_{n+1} = {1 \over n+1} x_n +1$. Note that
$x_{n+1} \le x_n$ iff $x_n \ge {n+1 \over n}$. Also note that $x_1 = 2$.
If $x_n \ge {n+1 \over n}$ then $x_{n+1} \ge {n+1 \over n} \ge {n+2 \over n+1}$. Hence $x_n$ is non increasing and bounded below by $0$. Hence it has a limit $L$ and from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Real solutions to the depressed cubic equation How can I find the allowed domain to this depressed cubic inequality
$$x^3 - 3 x + 2 \cos(\frac{3 \sqrt{3} n}{2}) \geq 0$$
where $n$ is a real non-negative number. Using Cardano's method, I can obtain some solutions for $x$ ($x_1$, $x_2$, and $x_3$) that should be real and... | Let $\alpha = \frac{n\sqrt 3}{2}$.
Using $\cos 3\alpha = 4\cos^3 \alpha - 3 \cos \alpha$, the equation is written as
$$x^3 - 3x + 2(4\cos^3 \alpha - 3 \cos \alpha) = 0$$
or
$$(x^3 + (2\cos \alpha)^3) - 3(x + 2\cos \alpha) = 0$$
or
$$(x + 2\cos\alpha)(x^2 - 2x\cos\alpha + 4\cos^2\alpha - 3) = 0$$
or
$$(x + 2\cos\alpha)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ..... | Possibly adding to the confusion, but
$$
x+{1\over 3!}x^3+{1\over 5!}x^5+...
$$
is the Taylor series for $\sinh x$ at $x=0$ and
$$
1+{1\over 2!}x^2+{1\over 4!}x^4+...
$$
is the Taylor series for $\cosh x$ at $x=0$ so your sum is
$$
\cosh\frac12-\sinh\frac12
$$
which is of course $\exp(-1/2)$ as noted in the other answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\li... | From
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
we have $$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}-ax-b}{x}) = 0$$
hence
$$\lim_{x \to \infty}(\frac{\sqrt{4x^2+2x+1}}{x})=a=2$$
$$b=\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-2x) +\frac{1}{2}$$
and
$$\sqrt{4x^2+2x+1}-2x=\frac{2x+1}{\sqrt{4x^2+2x+1}+2x}$$
u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove the following equality in a triangle
If $A,B,C$ are angles of a triangle then prove that $$\frac{\operatorname{tan}A}{\operatorname{tan}B}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}$$ where $\angle A$ is opposite to side $a$ , $\angle B$ to side $b$ and $\angle C$ to side $c$.
I wasn't having a clue of proceeding. So, I a... | Use the sine and cosine rules: $$\tan A =\frac{\sin A}{\cos A}, \tan B=\frac{\sin B}{\cos B}$$
$$\text{Sine Rule:}\quad \sin A=\frac{a}{2R}, \sin B=\frac{b}{2R}$$
$$\text{Cosine Rule:}\quad \cos A=\frac{c^2+b^2-a^2}{2bc}, \cos B=\frac{c^2+a^2-b^2}{2ac}$$
Thus, $$\tan A=\frac{abc}{R}\frac{1}{c^2+b^2-a^2}, \tan B= \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible ... | (Too long for a comment.)
While the other answers addressed how to express the sextic in radicals, I was also curious about what other contexts it may arise. When faced with an equation, one trick is to look at its discriminant $d$ since it is an important invariant and may give clues. The OP's original equation is
$$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substitut... | First note that we can get
$$2xy = (x+y)^2 - x^2-y^2 = 1 - 2 = -1$$
We can then apply binomial theorem again to obtain $x^3+y^3$ and $x^5+y^5$
$$
\begin{align*}
x^3 + y^3 &= (x+y)^3 - 3x^2y - 3xy^2 \\
&= 1 + \frac{3}{2}(x+y) = \frac{5}{2} \\
x^5 + y^5 &= (x+y)^5 - 5x^4y - 10x^3y^2 - 10x^2y^3 - 5xy^4 \\
&= 1 + \frac{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x... | Fact 1: Let $n \ge 2$. Let $x_1, x_2, \cdots, x_n \ge 0$ and $x_1 > 0$. Then
$$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j} \ge \frac{2n - 1}{n^2}(x_1 + x_2 + \cdots + x_n).$$
(The proof is given at the end.)
By Fact 1, we are done.
Proof of Fact 1:
We use Mathematical Induction.
It is easy to verify the case $n = 2$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4537840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Limit of 2 variable function I'm trying to determine the limit of this function:
$$\lim_{(x,y)\to (1,2)} \frac{xy^2-4xy-y^2+4x+4y-4}{x^2+y^2-2x-4y+5}$$
I tried to approach in many different ways, such as
$$\lim_{t\to 1} f(t,2t) \quad, \quad\lim_{t\to 1} f(t,2) \quad, \quad \lim_{t\to 2} f(1,t) $$
But i got that the lim... | Let $f(x,y)=xy^2-4xy-y^2+4x+4y-4$ and $g(x,y)=x^2+y^2-2x-4y+5$. Then, for each $(x,y)\in\Bbb R^2$,\begin{align}f(x,y)&=f\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)(y-2)^2\end{align}and\begin{align}g(x,y)&=g\bigl((x-1)+1,(y-2)+2\bigr)\\&=(x-1)^2+(y-2)^2.\end{align}Therefore\begin{align}\lim_{(x,y)\to(1,2)}\frac{xy^2-4xy-y^2+4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact ... | It's easier to use the chain rule, that is, if $h=g\circ f$, then $$\frac{dh}{dx} = \frac{dg}{df}\cdot \frac{df}{dx}$$ or, written in another way, $$(g\circ f)' = g'\circ f\cdot f'$$
In this case, you can take $g$ as the square root funcion that is $g(x)=\sqrt{x}$, and $f$ as $f(x)=\frac{1-x}{1+x}$.
This quickly leads ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:
Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$
I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{... | Observe that, the given polynomial is exactly quadratic with respect to the variable $x$. You have:
$$
\begin{align}
&3x^2y+y-x=0\\
\iff &3yx^2-x+y=0\\
&\Delta_x=1-12y^2\ge 0\\
\iff &-\frac {1}{2\sqrt 3}\leq y\leq \frac {1}{2\sqrt 3}
\end{align}
$$
This result tells us, $\forall x\in\mathbb R$ if $3x^2+y-x=0$, then $-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations... | Since OP said "I want integer solutions for this," meaning $b, c,$ and $d$ should be integers. In that case, $c$ can only be $1$ or $-1$ since the second eqiation implies:
$$c(b−d) = −1 \tag{i}$$
Also, in either case of $c = \pm 1$, the first eqiation implies:
$$b + d = (\pm 1)^2 - 6 = -5 \tag{ii}$$
From the equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Check if this series is convergent or not I've been going crazy for a long time in determining if this series converges or diverges. Most likely it converges.
$\displaystyle \sum_{n=1}^\infty \left(\frac{n}{2}\, \sin\frac{1}{n}\right)^\frac{n^2+1}{n+2}$
I am stuck in the necessary condition.
I tried to solve in this wa... | Letting
$$
a_n
=
\left(
\frac{n}{2}
\cdot
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}
$$
we have $a_n > 0$ for all positive integers $n$, so we get
$$
\lim_{n\to\infty}
|a_n|
^
\frac{1}{n}
=
\lim_{n\to\infty}
(a_n)
^
\frac{1}{n}
=
\lim_{n\to\infty}
\left(
\frac{1}{2}
\cdot
\frac
{\sin\Bigl({\large{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How do I show that $\sin(x^2+3x+2)$ is not periodic? I know that if $g(x)$ is periodic then $f(g(x))$ is periodic.
This is a sufficient condition but not necessary as $\sin(x)=f(g(x))$ is periodic where $g(x)=x$(non periodic) and $f(x)=\sin(x)$.
Can the above condition be made into a necessary and sufficient condition ... | Suppose it is periodic with period $2\pi T$ for some $T>0$. This means that for all $x$, $\sin(x^2+3x+2)=\sin((x+2\pi T)^2+3(x+2\pi T)+2)$. Expanding the RHS gives $\sin(x^2+3x+2)=\sin(x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2)$.
If $\sin(u)=\sin(v)$, then either $u=v-2\pi n$ or $u=\pi-v+2\pi n$ for some constant integer $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15... | If $n$ is even then $n^2+15n$ is even. If $n$ is odd, then $n^2+15n$ is still even. So $n^2+15n+122$ is even for every $n$.
So we just need to determine when the expression is divisible by $3$.
$$n^2+15n+122 \equiv n^2 +2 \pmod{3}.$$
It's easy to check that $n=1$ and $n=2$ are the only solutions. So the answer is "a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ \sum_{n=0}^{\infty} 2^{-n(n+1)/2} = \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}$ Prove the following equality:
$$ \sum_{n=0}^{\infty} 2^{-n(n+1)/2} = \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}$$
| We can write
\begin{align*}
\sum_{n=0}^{\infty} 2^{-n(n+1)/2} &= \sum_{n=0}^{\infty} 2^{-n(n+1)/2} \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k} \sum_{n=0}^{\infty} 2^{-n(n+1)/2} \prod_{k=1}^{\infty} (1 - \frac{1}{2^k})^{(-1)^k}\\
&= \prod_{k=1}^{\infty} (1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determination of complex logarithm: $\mathrm{Log}(z) = i$ I read that the determination of complex logarythm $\mathrm{Log}(z)$ is defined as:
$$\mathrm{Log}(z) := \ln|z| + i\mathrm{Arg}(z)$$
with $\mathrm{Arg}(z)$ the main argument of the complex number. I need to solve:
$$\mathrm{Log}(z) = i$$
Assuming $z =: x + iy$, ... |
Assuming $x + y \cdot \mathrm{i} := z$, the right answer is $\cos(1) + \sin(1) \cdot \mathrm{i} $. I found: [...]
Where is the problem in my calculations?
The problems are the step from $\arctan\left( \frac{y}{x} \right) = 1$ to the next one and $x = y$.
The argument is not always the arctangent of $\frac{y}{x}$, e.g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a clos... | Let $u$ be a noncubic integer with $\lfloor\sqrt[3]{u}\rfloor = n \in \mathbb{Z}^+$. Assuming $\sqrt[3]{u} \le x < \sqrt[3]{u+1}$, we can reformulate the equation as follows.
$$ \sqrt{nu} + \sqrt{(x-n)(x^3-u)} = x^2 $$
Squaring both sides, we have
$$(\sqrt{n} x - \sqrt{u})^2 = 0$$
Hence the equation holds for $x = \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{... | You have a good start!
I'd suggest trying to reduce the number of variables. You can solve the first equation for $B=1-A$. Then try multiplying the second equation by 3 to get $12A-3B+3C=0$. Add this to the 3rd equation. Combine with the substitution for $B$ and solve for $A$. Then you can work backwards through the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
\int_a^bf(x)dx = (b-a)\int_... | Let $x=2t+3$, then $dx=2dt$. The bounds of the integral change like this: $x=3=2t+3\implies t=0$ and $x=2t+3=5\implies t=1$. Hence,
$\begin{align}
\int_3^5\frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}}&=\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{2t(2-2t)}}2dt\\
&=\sqrt[3]{2}\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{t(1-t)}}dt\\
&=\sqrt[3]{2}\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A not straightforward limit The original limit is:
$$\lim_{x \to +\infty} \frac{e^{\sin \frac{1}{x}}-1-\frac{1}{x}}{\ln \left(1+\frac{x^2}{(1+x)^3} \right )-\frac{x^2}{(1+x)^3}}$$
I performed the substitution $y=\frac{1}{x}$ which leads to
$$\lim_{y \to 0^{+}} \frac{e^{\sin y}-1-y}{\ln \left ( 1+\frac{y}{(1+y)^3} \righ... | That is not what I have.
$\lim_\limits{y\to 0} \frac {e^{\sin y} -1 - y}{\ln (1+ f(y)) - f(y)}$
The Talor expansion of the numerator is
$1 + \sin y + \frac 12 \sin ^2 y+\cdots - 1 - y$
Expanding each of those sine functions
$1+ y - \frac 16y^3 +\cdots + \frac 12 y^2 - \frac 16 y^4+\cdots -1-y$
Which will ultimately be
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute the limit of a determinant of a given $n \times n$ matrix Compute:
$\displaystyle \lim_{n \rightarrow \infty} \begin{vmatrix}
1+x & -x & 0 & 0 & \cdots & 0 & 0 \\
-\frac{1}{2} & 1+\frac{x}{2} & -x & 0 & \cdots & 0 & 0 \\
0 & -\frac{1}{3} & 1+\frac{x}{3} & -x & \cdots & 0 & 0 \\
0 & 0 & -\frac{1}{4} & 1+\frac{x... | Hint : $D_{n+2} = \left( 1+\dfrac{x}{n+2} \right) D_{n+1} - \left( \dfrac{x}{n+2} \right) D_{n}$ rewrites as $$D_{n+2} -D_{n+1} = \frac{x}{n+2} (D_{n+1} - D_{n})$$
Let $A_n = D_{n+1}-D_n$ : you get $$A_{n+1} = \dfrac{x}{n+2}A_n$$
Can you proceed from here ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why can't we multiply both sides of $x-1=0$ by $x$ to create another solution $x=0$? Consider the equation $x-1=0$. The obvious solution is $x=1$. However, what is stopping us from creating more solutions by multiplying both sides by arbitrary values? For example:
$$ \begin{align}
x-1 & = 0 \\
x\cdot(x-1) &= 0\cdot x \... | Taking your question, every one of your steps is valid.
$$\begin{equation}x - 1 = 0\end{equation}$$
has one solution, $x = 1$, and
$$x(x-1) = 0$$
has two solutions, $x = 0$ and $x = 1$.
Where this goes wrong is in assuming that this holds for the original equation: you are implicitly saying
$$\begin{aligned}0(0 - 1) &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Find a complex number that satisfies both $|z-3+i|=3$ and $|z+2-i|=|z-3+2i|$. What I did:
Let $z=x+yi$, Then:
$$|(x-3)+(y+1)i|=3$$
$$|(x+2)+(y-1)i|=|(x-3)+(y+2)i|$$
Since $|x+iy|=\sqrt{x^2+y^2}$:
$$(x-3)^2+(y+1)^2=9$$
$$y=\frac{5x-4}{3}$$ Hence:
$$(x-3)^2+\bigg(\frac{5x-1}{3}\bigg)^2=9$$
Or
$$34x^2-64x+1=0$$
So
$$x_{1,... | Geometrically u are looking for the intersection of the circle centered at $(3,-1)$ with radius 3 and the perpendicular bisector of the line segment ralating the points $(-2,1)$ and $(3,-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(r, \theta) = (\cos\theta, \sin\theta)$ is continuous I am trying to prove that $f(r, \theta)= (\cos\theta, \sin\theta)$ is continuous using an $\varepsilon - \delta$ definition. Here is my work so far.
Let $f(r, \theta) = (\cos\theta, \sin\theta)$ and $(r_0, \theta_0) \in \mathbb{R}^2$. We will show that $f_r$... | $\sqrt {(\cos\theta - \cos\theta_0)^2 + (\sin\theta - \sin\theta_0)^2}\\
\sqrt{\cos^2\theta + \cos^2\theta_0 - 2\cos\theta\cos\theta_0 + \sin^2\theta + \sin^2\theta_0 - 2\sin\theta\sin\theta_0}\\
\sqrt {2 - 2\cos(\theta - \theta_0)}\\
2\sin\frac {\theta - \theta_0}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Arrangements of $a,a,b,b,b,c,c,c,c$ in which no two consecutive letters are the same I have the following question:
We are going to generate permutations from a,a,b,b,b,c,c,c,c. Please compute the number of permutations such that:
(a) for any consecutive 4 elements, they are not all the same;
(b) for any consecutive 3... | Such problems become intractable very rapidly, I prefer to use a form of the generalized Laguerre polynomial as described by Jair Taylor
Define polynomials for $k\geq 1$ by $q_k(x) =
\sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$.
The number of permutations will be given by
$$\int_0^\infty \prod_j q_{k_j}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Generalise a function, $f\left(x\right)=x^4+ax^3+bx^2$ such that $f$ and $f'$ only has integer roots.
Consider a function, $f:\mathbb{R}\rightarrow \mathbb{R},\:f\left(x\right)=x^4+ax^3+bx^2$ where $a,b\in \mathbb{Z}\setminus \left\{0\right\}$ and $b\ne \left(\frac{a}{2}\right)^2$ with three distinct real roots whose ... | Since $$f(x)=x^2(x^2+ax+b),\qquad f'(x)=x(4x^2+3ax+2b)$$
it is necessary that there are integers $c,d$ such that
$$a^2-4b=c^2,\qquad 9a^2-32b=d^2$$
Eliminating $b$ gives
$$a^2+2(2c)^2=d^2$$
which is of the form $x^2+2y^2=z^2$, so
$$a= \pm k|B^2-2A^2|,\quad 2c=\pm 2ABk,\quad d=\pm k(B^2+2A^2)$$
(where $A,B$ are positive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Showing that $\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$ is the square of a Fibonacci number I was experimenting with products of the form
$$\prod_{k=1}^{n} \left( a + b\cos(ck) \right)$$
when I found that the expression
$$\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \righ... | Changing notation slightly (my $n$ is your $n + 1,$ and my $F_n$ is your $F_{n + 1},$ if that's not too confusing a way to put it!), let $\omega$ be a primitive $n$th root of unity. Then
$$
\sqrt{5}F_n = \varphi^n - \frac{(-1)^n}{\varphi^n} = \frac{(\varphi^2)^n - (-1)^n}{\varphi^n} = \prod_{k=0}^{n-1}\frac{\varphi^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 2
} |
Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y... | We can obtain the homogenious equation as follows.
$$\begin{align}
&\Longleftrightarrow \frac{x}{3}\cdot \frac{d(y^2)}{dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{3x^2dx} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
&\Longleftrightarrow x^3\cdot \frac{d(y^2)}{d(x^3)} = \sqrt{(x^3)^2-y(^2)^2}+y^2\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x... | This approach is somewhat similar to yours. It is more of a comment than an answer, but it is too long to fit in the comments section.
Let the integral in question be $I$. Letting $x \to \dfrac{1-x}{1+x}$ at the beginning yields
$$I = 2\int_{0}^{1}\frac{\ln\left(1-\left(\frac{1-x}{1+x}\right)^{4}\right)}{1+\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ ... | Since, $|5-x|\geqslant 0$, then you have:
$$\begin{align}&\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}\geqslant 0\\
\iff &x^2-9x+24\geqslant 6x^2-59x+149\geqslant 0\\
\iff &-5(x-5)^2\geqslant 0\\
\iff &x=5. \end{align}$$
This implies that, if there's a solution, then it is $5$. Indeed, $x=5$ is a solution.
Remember that, some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How many solutions are there to the equation? I need to find the number of solutions in positive integers to the following equation:
$$x_1 + x_2 + x_3 + x_4 = 19$$
where $x_2 \neq 2x_3$ and $x_1 \neq x_2$
I know to solve the cases like $x_1 \ge 2x_3$, but I don't understand how to do the jump from this inequality to th... | A generating function approach. We start with the number of solutions of
\begin{align*}
&\color{blue}{x_1+x_2+x_3+x_4=19}\tag{1}\\
&\color{blue}{x_1,x_2,x_3,x_4\geq 1}
\end{align*}
We represent the solutions of a variable $x_j\geq 1, 1\leq j\leq 4$ as generating function for $x_j$ which is
\begin{align*}
z+z^2+z^3+\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\s... | We have,
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\left(\frac{1}{\sqrt{n+1}}\cdot \frac{1}{\sqrt{n+1}-\sqrt{n}}\cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\right)$$
To rationalize the denominator $\sqrt {n+1}-\sqrt n$, we need to use the conjugate based o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5... | When $z\neq 0,$ that means we want all rational solutions to:
$$m^2+n^2-x^2-y^2=1\tag 1$$
We pick one solution, $\mathbf s_0=(m_0,n_0,x_0,y_0)=(-1,0,0,0)$ and any directional vector $\mathbf p=(p_1,p_2,p_3,p_4)$ with the $p_i$ integers with $\gcd(p_1,p_2,p_3,p_4)=1.$
Then $\mathbf s_0+t\mathbf p$ yields a solution for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove $\cos(2\pi/7)$, $\cos(4\pi/7)$, and $\cos(6\pi/7)$ are roots of the equation $8x^3 + 4x^2 - 4x - 1 = 0$. The first part of the question asked you to show that $\theta = 0, 2\pi/7, 4\pi/7, 6\pi/7$ when $\cos(4\theta) = \cos(3\theta)$. By using $\cos(\theta) = \frac{1}{2} \left(z + \frac{1}{z} \right)$, I have mana... | Using your work, note that $\cos(4 \theta) = \cos(3 \theta)$ gives:
$$(8u^4 - 8u^2 + 1) - (4u^3 - 3u) = 0 \implies 8u^4 - 4u^3 - 8u^2 + 3u + 1 = 0$$
where $u = \cos \theta$. $u = 1$ is a root, and hence dividing through by $u - 1$ we get:
$$\begin{array}{c|rrrr} 1 & 8 & -4 & -8 & 3 & 1 \\ & \downarrow & 8 & 4 & -4 & -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4587537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that the equation $X^2=J_n$has no solution if $n\geq2$, if $J_n$ is the jordan block of size $n$ with zeros on diagonal. More clearly:
$$
J_n=
\begin{pmatrix}
0 & 1 & 0 &\cdots & 0 & 0 \\
0 & 0 & 1 &\cdots & 0 & 0 \\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0 & 0 & 0 &\... | Let $y$ be any $n$-dimensional vector not in Ker$(X)$, such that $X^2y$ is nonzero. [That $X^2y$ is nonzero for some $y$ is what would have to be for $X^2$ to be $J$, and furthermore, for Dim Ker$(X^2)$ to be only $1$.] Then the fact that $X$ is nilpotent gives $X^ny=0$ for some positive integer $n$. Thus, let $n$ now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4587913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Without calculus, what is the maximum value of the rational function : $f(t)= \frac {30t} {t^2 + 2}$ Source : Stewart, Precalculus.
The original question is to graph the function : $f(t)= \frac {30t} {t^2 + 2}$.
Desmos construction : https://www.desmos.com/calculator/19kybcl3hc
It can be shown that $f(x)$ tends to $0$ ... | We can also work out a generalization in this way. We can establish by the "zero-discriminant" quadratic polynomial method you used or by application of the AM-GM inequality (Feng's approach) that $ \ x + \frac{1}{x} \ $ has its relative minimum of $ \ +2 \ $ at $ \ x \ = \ 1 \ $ for $ \ x \ > \ 0 \ $ and its relative... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which i... | I wanted to provide the trigonometric approach as a $3$rd answer, thinking it might be useful to some readers.
Substitute $x=\sin\theta$, where $0\leqslant \theta\leqslant \frac {\pi}{2}$ based on the fact $x\geqslant 0$, then we have:
$$\begin{align}&\sin\theta+\sin \theta\cos \theta=\cos \theta\\
\implies &\cos \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4593991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead... | A novel way to solve the problem is to change the question, with equvalent answer.
$\cos(A-B) \;=\; \sin(A)\sin(B) \;+\; \cos(A)\cos(B)$
$\cos(\quad C \quad) \;=\; \sin(A)\sin(B) \;-\; \cos(A)\cos(B)$
Only difference between the two are the sign of cosine product term.
We change the question! $\; a=5,\; b=4,\;\cos(C)=7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prove $a^x+a^{1/x}\le 2a$ ($\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$
Prove $a^x+a^{1/x}\le 2a$ (with $\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$.
Is it true? I can't figure out how to prove it.
It may use the conclusion $\ln x+\frac{1-x}{x}\ge 0, \forall x\in (0,+\infty) $ as this question is followed.
... | Yes, it is true. Fix $a$ in the range $1/2 \le a < 1$. It suffices to prove
$$
a^x + a^{1/x} \le 2a
$$
for $0 < x \le 1$, because the left-hand side is invariant under the substitution $x \leftrightarrow 1/x$. This inequality can be rearranged to
$$
g(x) := a^{-1/x} (2a - a^x) \ge 1 \, .
$$
We have $g(1) = 1$, and w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$ I ask for an inequality which is a follow up of this question: Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ :
$$\int_{0}^{\infty}x^{-x}dx<\pi-\ln\pi$$
You can find a nice proof @RiverLi among others and also an attempt of mine.
My try
We have:
$$\int_{1}^{\phi}\phi^{\lef... | I put here my progress so far we have :
$$\int_{2}^{\pi}k\left(x\right)dx+\int_{1}^{2}f\left(x\right)dx+\int_{0}^{1}h\left(x\right)dx+\int_{\pi}^{4}\frac{3+\frac{3.5}{1000}}{\ln3}x^{\frac{12}{1000}+\frac{2}{5}}\left(\frac{3+\frac{3.5}{1000}}{\ln3}\right)^{-x^{1+\frac{2}{5}+\frac{12}{1000}}}dx+\int_{4}^{5}ex^{\frac{12}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integ... | Another standard approach, set $t = x + \sqrt{x^2 +x -2}$, then
\begin{align*}
& (t -x)^2 = x^2 +x -2 \\
& t^2 -2tx = x -2 \\
& x = \frac{t^2 +2}{1 +2t} \\
& dx = 2\frac{t^2 +t -2}{(1 +2t)^2}dt \\
\end{align*}
moreover, express the square root in terms of $t$
\begin{align*}
\sqrt{x^2 +x -2} &= t - x = \\
& = t - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (... | I think that I got a different way
Another identity
$$\tan(x+y+z) = \frac{\tan x + \tan y + \tan z - \tan x \tan y \tan z}{1 - \tan y \tan z - \tan x \tan y - \tan x \tan z}$$
Now let $x = 20^{\circ}, y = -40^{\circ}, z=80^{\circ}$, so that $x+y+z = 60^{\circ}$
Let's do some math, using your identities in your post
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $\log_x(9x^2)\cdot\log_3^2(x)=4$ Solve the equation $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
The answers are $x_1=\dfrac19$ and $x_2=3$.
For the range we have: $$D_x:\begin{cases}x>0\\x\ne1\\9x^2>0\iff x\ne0\end{cases}\iff x\in(0;1)\cup(1;+\infty)$$
I wrote the equation as follows using $\log_a(b)=\dfra... | $$\log_x(9x^2)\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{\log_3(9)+\log_3(x^2)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow \dfrac{2+2\log_3(x)}{\log_3(x)}\cdot\log_3^2(x)=4$$
$$\Rightarrow 2\log_3 (x)\left[1+\log_3(x)\right]=4$$
$$\Rightarrow \left[\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{\sqrt{1+x^4}}{1-x^4}dx$ How to evaluate
$$\int \frac{\sqrt{1+x^4}}{1-x^4}dx.$$
If we substitute $x=\sqrt{\tan{\theta}}$, it becomes
$$\int \frac{1}{2\cos{\theta}\cos{2\theta}\sqrt{\tan{\theta}}}d\theta.$$
What can I do next ?
Edit
Are these steps correct ?
$$=\int \frac{1}{2\cos{\theta}\cos{2\the... | Maple also gives a non-elementary answer to the original integral in terms of elliptic integral functions. However, for the integral over $\theta$ it does give an elementary answer, which translated back to the original integral becomes
$$ \frac{\sqrt{2}\, \mathrm{arctanh}\! \left(\sqrt{\sin\left(2 \arctan \! \left(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the symmetric matrix given its eigenvalues and eigenvector.
$A$ is a $3 \times 3$ symmetric matrix. It has the eigenvalue $\lambda_1 = 3$ with the eigenvector
$$
\begin{bmatrix}
1 \\ 0 \\ -1
\end{bmatrix}
$$
and a double eigenvalue $\lambda_2 = -1$.
Find the matrix $A$.
Since $A$ is symmetric, it can be diagon... | (I will use row vectors because they are easier to type and use up less space; interpret them as transposes of the vectors you want)
Symmetric matrices are always orthogonally diagonalizable. So we know that the eigenspace of $-1$ will be the orthogonal complement of the eigenspace of $3$; you know the eigenspace of $3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure ... | \begin{align}
\sum_{i=1}^n i\binom{2n}{n-i}
&= \sum_{i=0}^n i\binom{2n}{n-i} \\
&= \sum_{i=0}^n (n-i)\binom{2n}{i} \\
&= n\sum_{i=0}^n \binom{2n}{i} - \sum_{i=1}^n i\binom{2n}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+\sum_{i=0}^{2n} \binom{2n}{i}\right) - 2n\sum_{i=1}^n \binom{2n-1}{i-1} \\
&= \frac{n}{2}\left(\binom{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt... | *
*No.
*Since it is false, this is irrelevant.
*I would write the function like this $y=x\arcsin x(1-x^2)^{-1/2}$. By triple product rule,
$$y'=\arcsin x(1-x^2)^{-1/2}+x\frac{1}{\sqrt{1-x^2}}(1-x^2)^{-1/2}+\\x\arcsin x(-1/2)(1-x^2)^{-3/2}(-2x)$$
$$y'=\frac{\arcsin x}{(1-x^2)^{1/2}}+\frac{x}{{1-x^2}}+\frac{x^2\arcsin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrating $\frac{x^3}{\sqrt{x^2 + 4x + 6}}$ Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.
My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2... | Undoing the substitution is exactly how you would match the two results.
By the Pythagorean theorem, a right triangle with a reference angle $t$ such that $\tan(t)=\frac{x+2}{\sqrt2}$ has its sides occurring in a ratio of $\sqrt2$ (leg adjacent to $t$) to $x+2$ (leg opposite $t$) to $\sqrt{(x+2)^2+2}=\sqrt{x^2+4x+6}$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"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.