Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to p... | You could probably get an easier calculation by going over the logarithmic derivative,
$$
\log(y(x))=\frac12(\log(3x)-\log(2x+3c))
\\~\\
\implies
\frac{y'(x)}{y(x)}=\frac12\left(\frac1x-\frac2{2x+3c}\right)
=\frac1{2x}-\frac1{3x}y(x)^2
$$
The last step is obtained by doing the minimum to eliminate the constant $c$.
As ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that the inequality $\left|\int_{0}^{1} f(x)\,dx\right| \leq \frac{1}{12}$ holds for certain initial conditions
Given that a function $f$ has a continuous second derivative on the interval $[0,1]$, $f(0)=f(1)=0$, and $|f''(x)|\leq 1$, show that $$\left|\int_{0}^{1}f(x)\,dx\right|\leq \frac{1}{12}\,.$$
My attemp... | Consider the following integral:
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx. $$
By integrating by parts twice, you get
$$\int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f^{\prime\prime}(x)\, dx = \underbrace{\left(\frac{x^{2}}{2}-\frac{x}{2}\right)f'(x)\bigg|_0^1}_{0} - \int_0^1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Finding the area enclosed by three curves I need to find the area that is enclosed between three curves. I was trying to look for the smartest solution visually, but haven't gotten anywhere.
Curves:
$$y_1= 2x^2$$
$$y_2 = x^2$$
$$y_3 = \frac{1}{x}$$
Please help
| I'm not sure how to solve this visually, but we can solve it with integrals.
First, we need to find where the different equations intersect each other.
*
*$y_1=y_2\rightarrow2x^2=x^2\rightarrow x^2=0\rightarrow x=0$
*$y_1=y_3\rightarrow2x^2=\frac{1}{x}\rightarrow x^3=\frac{1}{2}\rightarrow x=\frac{1}{\sqrt[\leftroot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part Here, $\{-3.4\}=0.6$.
The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number;
after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$
... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2... | I think I found the solution.
Lebesgue's identity says $(k^2 + l^2 + m^2 + n^2)^2 = (2kn + 2lm)^2 + (2ln - 2km)^2 + (k^2 + l^2 - m^2 - n^2)^2$, so if every odd prime $p$ can be written as the sum of four primes such that non of the right-hand-side terms is equal to $p^2$ the question is solved.
Lagrange's four square t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove: $\int_0^2 \frac{dx}{\sqrt{1+x^3}}=\frac{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{1}{3}\right)}{6\Gamma\left(\frac{1}{2}\right)}$ Prove:
$$
\int_{0}^{2}\frac{\mathrm{d}x}{\,\sqrt{\,{1 + x^{3}}\,}\,} =
\frac{\Gamma\left(\,{1/6}\,\right)
\Gamma\left(\,{1/3}\,\right)}{6\,\Gamma\left(\,{1/2}\,\right)}
$$
First... | Consider
$$
y^2=4x^3+4,g_2=0,g_3=-4
$$
The real period of $\wp(z)$ is
$$
\omega_1=\int_{-1}^{\infty} \frac{1}{\sqrt{1+x^3}} \text{d}x
=\frac{\Gamma\left ( \frac{1}{3} \right )\Gamma\left ( \frac{1}{6} \right ) }{2\sqrt{\pi} }
$$
And
$$
\wp\left ( \frac{\omega_1}{2} \right ) =-1
$$
We have the addition formula:
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 3
} |
Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $... | What about the ratio test?
\begin{align*}
\limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\to\infty}\frac{n3^{n}|x-5|^{n}}{(n+1)3^{n+1}|x-5|^{n+1}}\\\\
& = \limsup_{n\to\infty}\left(\frac{n}{n+1}\right)\frac{1}{3|x-5|} = \frac{1}{3|x-5|} < 1\\\\
& \Rightarrow |x - 5| > \frac{1}{3} \Rightarrow \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3782557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \... | According to the Matrix Determinant Lemma, $\det(A+uv^T)= \det(A)(1+v^T A^{-1}u)$, hence
$$\det(A) = \det(4I + 2\cdot\mathbf 1 \mathbf 1^T) = \det(4I)(1+2\cdot \mathbf 1^T(\tfrac{1}{4}I)\mathbf 1) = 4^n (1 +\tfrac{1}{2}n) $$
For $n=5$, that's $4^5\cdot \frac{7}{2} = 4^4 \cdot 14$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$p\equiv 1\pmod 4\Rightarrow p=a^2+b^2$ and $p\equiv 1\pmod 8\Rightarrow p=a^2+2b^2$, what about for $p\equiv 1\pmod {2^n}$ in general Primes $p$ with $p\equiv 1\pmod 4$ can be written as $p=a^2+b^2$ for some integers $a,b$. For $p\equiv 1\pmod 8$ we have $p=a^2+2b^2$. Can primes that satisfy $p\equiv 1\pmod{2^n}$ for ... | it gets harder, and we cannot just impose congruence conditions.
Added: one thing I've not seen in print is this: as soon as $p \equiv 1 \pmod 8,$ we find that $-1$ is a fourth power mod $p,$ and
$$ z^4 + 1 \equiv 0 \pmod p $$
has four distinct roots.
A prime can be expressed as $p = x^2 + 32 y^2$ if and only if $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
... | My first SOS is same as Nguyen Huyen. Here is my second and third.
$$\displaystyle a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$\displaystyle=\frac34 \left( b-c \right) ^{2} \left( b+c \right) +\frac34 \left( a-b
\right) ^{2}a+\frac14\, \left( a+b-2c \right) ^{2}a+\frac12\, \left(b+c-2a \right) ^{2}a$$
$$\disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$ How to evaluate:
$$\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ$$
Can somebody help me handle it?
I have no idea what to do.
This is my attempt:
$$\cos24^\circ-\cos(60^\circ+24^\circ)-\cos12^\circ+\sin (12^\circ+30^\circ)$$
| $$\begin{split}
\cos24^\circ-\cos84^\circ-\cos12^\circ+\sin42^\circ &= \cos24^\circ-\cos84^\circ-\cos12^\circ+\cos48^\circ\\
&= \cos24^\circ +\cos48^\circ-\cos84^\circ-\cos12^\circ \\
&= 2\cos36^\circ\cos12^\circ-2\cos48^\circ\cos36^\circ\\
&=2\cos36^\circ(\cos12^\circ-\cos48^\circ)\\
&=2\cos36^\circ(2\sin18^\circ\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c... | Another way.
We need to prove that:
$$\sum_{cyc}a^{n+1}\geq\sum_{cyc}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}},$$ which is true by Muirhead because $$(n+1,0,0)\succ\left(n+\frac{1}{3},\frac{1}{3},\frac{1}{3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
find the complex integral: $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$. Problem with integral formula.... Question I am trying to find the complex integral: $\displaystyle\int_0^\infty \frac{z^6}{(z^4+1)^2}dz$.
My Attempt (and eventual question): $\int_0^\infty \frac{z^6}{(z^4+1)^2}dz=\frac{1}{2}\int_\infty^\infty\frac{z^6... | Note
\begin{align}
\int_0^\infty \frac{z^6}{(z^4+1)^2}dz
&\overset{z\to\frac1z}
= \int_0^\infty \frac{dz }{(z^4+1)^2}
= \frac14\int_0^\infty \frac1{z^3}d \left(\frac{z^4}{z^4+1}\right)\\
&=\frac34 \int_0^\infty \frac{dz }{z^4+1}
\overset{z\to\frac1z} = \frac38\int_0^\infty \frac{1+z^2 }{z^4+1}dz\\
& = \frac38\int_0^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$ is equal to 1? $\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
By maths calculator it results 1.
I calculate and results $\sqrt{-\frac{1}{2}}$.
$\sqrt{\left(\frac{-\sqrt3}2\right)^2+{(\frac12)}^2}$
$\sqrt{\frac{-{(3)}^{{\displaystyle\frac12}\times2}}{2^2}... | The square of a negative number is positive.
$$\left(-\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that ... | Let $z^3=8$.
Thus, $$(z-2)(z^2+2z+4)=0,$$ which gives
$$\{2,-1+\sqrt3i,-1-\sqrt3i\}$$ or
$$\left\{2(\cos0+i\sin0),2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right), 2\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How were amplitudes of the $\cos$ and $\sin$ chosen? I don't understand why we use $\displaystyle\sqrt{1^2+\left(\frac{1}{2}\right)^2}$ in the below transformation. Can someone help to explain?
from
$$f(x)=\frac{3}{5}-\frac{3}{5}e^t\left(\cos(2t)+\frac{1}{2}\sin(2t)\right)$$
transform to
$$f(x)=\frac{3}{5}-\frac{3}{5}\... | Let’s concentrate on the important part, which is of the form
$$
f(x)=a\cos x+b\sin x
$$
which we want to express as
$$
f(x)=A(\cos\varphi\cos x+\sin\varphi\sin x)
$$
A necessary (and sufficient) condition is that
$$
A\cos\varphi=a,\qquad A\sin\varphi=b
$$
and therefore $a^2=A^2\cos^2\varphi$, $b^2=A^2\sin^2\varphi$. H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}... | You should recall that $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$, since $(c-d)(c+d)=c^2-d^2$. This should help you simplify the expression with the square root.
Though I don't understand how you obtained what you wrote, from what I can see you should get:
$$ x+ \frac{2x^3}{3}-\frac{2(x^2+1)^{\frac{3}{2}}}{3}=x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How to prove that $\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$ The sum $$\sum_{n=0}^{\infty} (-1)^n \ln \frac{3n+2}{3n+1}=\frac{1}{2} \ln 3$$ has been encountered in the post below:
How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$
I would like to know as to how this... | We can obtain something analogous to the Wallis product using$$\prod_{n\ge1}(1-z^2/n^2)=\frac{\sin\pi z}{\pi z}\implies\prod_{n\ge1}(1-z^2/(2n+1)^2)=\frac{\frac{\sin\pi z}{\pi z}}{\frac{\sin\pi z/2}{\pi z/2}}=\cos\frac{\pi z}{2},$$so$$\prod_{n\ge0}\frac{(6n+2)(6n+4)}{(6n+1)(6n+5)}=\prod_{n\ge0}\frac{1-1/(6n+3)^2}{1-4/(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that there are no integer solutions to $x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$ I have struggled on this problem for quite a bit of time now, asked some of my peers and teachers, and I am yet to find the solution. Here is the problem:
Prove that there are no integer solutions to the equation $$x\left(... | We have
$$xy^2-x = 2y+\frac{y}{x}$$
$$x^2y^2-x^2=2xy+y$$
$$x^2y^2-x^2-y-2xy = 0$$
Solve as a quadratic in $x$
$$(y^2-1)x^2-(2y)x-y = 0$$
Use quadratic formula
$$x = \frac{2y\pm \sqrt{4y^2+(4y^3-4y)}}{2(y^2-1)}$$
$$x = \frac{2y\pm \sqrt{4y^3+4y^2-4y}}{2y^2-2}$$
We can factor out a $2$ to get
$$x = \frac{y\pm \sqrt{y^3+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Find the stronger version of $9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0$ For $a,b,c \geqslant 0.$ Then $$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0.$$
I use computer and found that the following stronger inequality holds for all reals of $a,b,... | There is the following stronger version.
Let $a$, $b$ and $c$ be non-negatives. Prove that:
$$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3}\geq$$
$$\geq4(3\sqrt3-4)(a^3+b^3+c^3-3abc)abc.$$
The equality occurs for $a=b=c$ and for $(a,b,c)=t(6+4\sqrt3,1,1)$, where $t\geq0$ and for any cyclic perm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the set of all possible values of a function similar to Nesbitt's Inequality
Let $x,$ $y,$ $z$ be positive real numbers. Find the set of all possible values of
$$f(x,y,z) = \frac{x}{x + y} + \frac{y}{y + z} + \frac{z}{z + x}.$$
This seems extremely similar to Nesbitt's inequality, in which I did some reasearc... | I think $1 < f(x,y,z) < 2.$ Indeed, because
$$\frac{x}{x+y} \geqslant \frac{x}{x+y+z}.$$
Equality occur when $x = 0$ or $z = 0.$
Therefore
$$f(x,y,z) \geqslant \frac{x+y+z}{x+y+z} = 1.$$
But $x,y,z$ are positive real numbers, so $f(x,y,z) > 1.$
Another
$$\frac{x}{x+y} < \frac{x+z}{x+y+z},$$
equivalent to
$$\frac{yz}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving limit using the definition of a limit Heres the question (provided for context) about proving a limit by using the definition of a limit:
$\mathop {\lim }\limits_{x \to 4} \left({x^2} + x - 11\right) = 9$
So, let’s get started. Let $\varepsilon > 0$ be any number then we need to find a number $\delta > 0$ so... | You have
$$
|x+5|=|x-4+9|\leq|x-4|+9\leq\delta+9
$$
so
$$
|x-4||x+5|\leq\delta(\delta+9)<\varepsilon
$$
and you can see that you can choose
$$
0<\delta<\frac{9}{2} \left(\sqrt{1+\frac{4\varepsilon}{81}}-1\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$
Calculate the volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$.
My approach:
For $z = 0$:
\begin{equation*}
2x + y = 4 \implies y = -2x + 4
\end{equation*}
and... | Your approach is correct and you can calculate it also as
$$\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}\int\limits_{0}^{4-2x-y}\,dy\,dx\,dz=\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}(4-2x-y)\,dy\,dx =\int\limits_{0}^{2}\int\limits_{0}^{4-2x}(4-2x-y)\,dx\,dy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the longest chord of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passing through $(0,-b)$? Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be an ellipse. How to find the longest chord in such ellipse which passes through the point $(0,-b)$ when
*
*$\ a> \sqrt{2}b>0$,
*$\ 0<a\leq \sqrt{2}b$
Frankly I do no... | Let $(x_1,y_1)$ be the coordinates of the other end of the chord, with $x_1>0$ (the solution is symmetric with respect to $y$-axis, so this is not a limitation).
The condition to impose is: the tangent in $(x_1,y_1)$ should be orthogonal to the chord. The tangent is
$$
t:\ \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1
$$
A vecto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da$
Evaluate
$$
\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da
$$
My try: rearranging original integral
$$
\int \frac{{1+\tan^{3/2}}(a)}{\sqrt{\sin^3(a)\sin(a+b)}}\,da
= \int \frac{{(1+\tan^{3/2}}(a... | Hint:
$\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da$ = $\int \frac{1}{\sqrt{\sin^3(a)\sin(a+b)}}\,da + \int \frac{1}{\sqrt{\cos^3(a)\sin(a+b)}}\,da$
Take first part now.
$\sqrt{\sin^3(a)\sin(a+b)} = \sqrt{\sin^4(a)(\cos b + \cot a \sin b)}$
say, $t = \cos b + \cot a \sin b$
$dt = - \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
subsets where consecutive difference is never $2$
Show that the number of subsets of $\{1,\cdots, n\}$ with size $k$ and where the difference between consecutive pairs of elements is never equal to $2$ is $\sum_{j=0}^{k-1} {k-1\choose j}{n-k-j+1\choose n-2j-k}.$
I got that the desired number is equal to $\sum_{j=0}^{... | The generating function approach is sound. We show the following is valid for $M:=\min\{k-1,\lfloor(n-k)/2\rfloor\}$.
\begin{align*}
\color{blue}{[x^{n-k}]\frac{\left(1-x+x^2\right)^{k-1}}{(1-x)^{k+1}}=\sum_{j=0}^{M}\binom{k-1}{j}\binom{n-k-j+1}{n-2j-k}}\tag{1}
\end{align*}
In (1) we use the upper limit $M$ to assure t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is $\mathbb{E}\!\left(X^2 \mid X + 2Y \right)$ if $X,Y \sim \mathcal{U}\left(\left[-1,1\right]\right)$ independently? $X$ and $Y$ are independent random variables of the uniform distribution on $\left[-1,1\right]$, $X,Y \sim \mathcal{U}\!\left(\left[-1,1\right]\right)$.
What is $\mathbb{E}\!\left(X^2 \mid X + 2Y ... | Define $h:\mathbb{R} \to \mathbb{R}$ such that $h(x)=\frac 1 3 \chi_{[-3,3]}$.
The claim is that $\mathbb{E}\left(X^2 | X + 2Y \right)=h(X+2Y)$.
Suppose $B \in \mathscr{B}([-3,3])$.
Let us prove $\int \chi_{B}(x+2y)x^2dx dy=\int \chi_{B}(x+2y)h(x+2y)dxdy$.
If $g(x,y)=(x+2y,x)$ then $|\det g (x,y)|=2$ and thus we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfr... | Instead of dividing by $\cos^2\alpha$, you could do the following,
\begin{align}&\sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\sin^2\alpha\cos^2\alpha + \cos^4\alpha\\ &= \sin^2 \alpha + \tan^2 \alpha +\sin^2\alpha +\cos^2\alpha(\sin^2\alpha + \cos^2\alpha) \\
&= \sin^2 \alpha + \tan^2 \alpha + \cos^2\alpha
\\ &=1+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Number of digits of $xy$ is either $m+n$ or $m+n-1$ Given positive integers $x$ and $y$ with $m$ and $n$ digits respectively, must $xy$ have either $m+n$ or $m+n-1$ digits?
If $x$ has $m$ digits, then $10^{m-1} \le x < 10^m$. Likewise, if $y$ has $n$ digits, then $10^{n-1} \le y < 10^n$. Multiplying those two inequalit... | Yes, you are right.
From $10^{m-1} \le x$ and $10^{n-1} \le y$ it follows that $10^{m+n-2} \le xy$, so $xy$ has at least $m+n-1$ digits. Also, from $x<10^m$ and $y<10^n$ it follows that $xy < 10^{m+n}$. Since $10^{m+n}$ is the least integer having $m+n+1$ digits, $xy$ has less than $m+n+1$ digits and thus $xy$ has at m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inverse of a matrix and matrix relation Let $A$ be the matrix $$ A= \begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix} $$ I am trying to find $A^{-1}$ as a relation of $I_{3}, A$ and $A^{2}$ and also to prove that $A^{2006}-2A^{2005}=A^{2}-2A$. For the first one I noticed that $$A^{n}= \begin{pma... | In this case it is not necessary to calculate the characteristic polynomial of $\;A\;$ and apply the Cayley-Hamilton theorem.
Since $\;A=\begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix}\;$ and $\;A^2=\begin{pmatrix}
4 & -3 & -3\\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix}\;,\;$ the first column of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number of ordered pairs $(a,b)$ of real numbers satisfying certain conditions with a polynomial How many ordered pairs $(a,b)$ of real numbers are there such that $$(2a^2+1)+(2a^2-1)i$$ is a solution to the equation $$x^2-10x+b^2+4b+20=0$$ There should be $6$.
Here is what I have so far:
$$\begin{align}f(x)&=x^2-10x+b^... | Case 1. $2a^2-1=0$. Then $a=\pm 1/\sqrt{2}$. For each of these $a$, plug $2a^2+1=2$ into the equation and find $b=-2$ -we get $2$ solutions $(a,b)$: $(\frac{1}{\sqrt{2}},-2)$ and $(\frac{-1}{\sqrt{2}}, -2)$.
Case 2. $2a^2-1\ne 0$, then the number $(2a^2+1)+(2a^2-1)i$ is not real. So the number $(2a^2+1)-(2a^2-1)i$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the domain of $f(x)=\sqrt\frac{x^2+6}{x-4}$
Given $f(x)=\sqrt{\dfrac{x^2+6}{x-4}}$, find the domain for the function $f(x)$.
My attempt failed:
So the first constraint is $\frac{x^2+6}{x-4}\geq0$ because of the radical.
Now, multiplying both sides by $x-4$ I get $x^2+6\geq0, x\neq4$.
And now $x^2\geq-6$, so $... | $x^{2}+6 >0$ for all $x$ so $\frac {x^{2}+6} {x-4}$ exists and is $ \geq 0$ iff $x>4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Diophantine equation: $x^4+4=py^4$
find all primes p for which $x^4+4=py^4$ is solvable for integers.
My try: I started out with $p=2$ . LHS must be even, thus $(x^4+4)\mod(16)=4$ similarly
$(2y^4)\mod (16)=${$0,2$}. Thus for $p=2$ there are no solutions.
Now I tried factoring i.e $x^4+4=(x^2+2x+2)(x^2-2x+2)=py^4.$ ... | It is easy to see that $x$ is odd (otherwise $4$ divides $py^4$ and $x, y$ are both even, which gives contradiction mod $16$). Then the $\gcd$ of $x^2 + 2x + 2$ and $x^2 - 2x + 2$ is $1$. Therefore we must have \begin{eqnarray}x^2 + 2x + 2 &=& u^4\\x^2 - 2x + 2 &=& pv^4\end{eqnarray} or \begin{eqnarray}x^2 + 2x + 2 &=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Simi... | $x^2 = 6[x] - 6= 6([x]-1)$. Forget about the greatest aspect of GIF and concentrate on the integer aspect of GIF. $x^2$ is multiple of $6$ and $x = \pm\sqrt {6k}$ where $k=[x]-1$.
$x^2 \ge 0$ so $[x]-1\ge 0$ so $x\ge [x] \ge 1$ so $k \ge 1$.
And if $k \ge 6$ then $x =\sqrt{6k} \le \sqrt {k^2} = k$ so $[x]\le x \le k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfra... | By AM-GM we have
$$
\frac{(a+b+c) + (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}{2} \geq \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}.
$$
Note that by the assumption, we have
$$
3 + \frac{1}{3} \geq a + \frac{1}{a}
$$
and similarly for the other variables. Therefore
$$
3 \cdot \frac{10}{3} \cdot \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative num... | Now, let $x^2+y^2=2uxy.$
Thus, since $xy\geq0$ and for $xy=0$ our inequality is true, we can assume that $xy>0$, which gives $u\geq1$ and we need to prove that:
$$8(x^2+xy+y^2)^3\geq6(-3xy(x+y))^2$$ or
$$2(2u+1)^3\geq27(u+1)$$ or
$$16u^3+24u^2-15u-25\geq0,$$ which is obvious for $u\geq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\f... | Another way.
By C-S $$\sum_{cyc}\frac{1}{a^3+b+c+d}=\sum_{cyc}\frac{\frac{1}{a}+b+c+d}{(a^3+b+c+d)\left(\frac{1}{a}+b+c+d\right)}\leq$$
$$\leq\frac{\sum\limits_{cyc}\left(\frac{1}{a}+b+c+d\right)}{(a+b+c+d)^2}=\frac{\sum\limits_{cyc}\left(abc+3a\right)}{(a+b+c+d)^2}.$$
Thus, it's enough to prove that:
$$(a+b+c+d)^3\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| We need to prove that $$c^2+ab\leq(a+b)c$$ or
$$a^2+b^2\leq(a+b)\sqrt{a^2-ab+b^2}$$ or
$$a^2+b^2\leq\sqrt{(a+b)(a^3+b^3)},$$ which is true by C-S.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + ... | $\left(x-1/x\right)^2=x^2+1/x^2-2$. If you substitute $y=x-1/x$, your equation becomes $y^2+2-2y-1=0$, so $y^2-2y+1=0$, so $y=1$. Then you find $x$ from $x-1/x=1$: $x^2-x-1=0$, $x=(1\pm \sqrt{5})/2$. Each of these is a double root of the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why is it enough to imply that $x^3+x+1|x^6+x^5+x^4+x^3+x^2+x+1$? I tried to understand a solution but I am stuck at how did the solution concluded that $x^3+x+1\mid x^6+x^5+x^4+x^3+x^2+x+1.$
Consider the field of element $8$ given by $\mathbb{F}=\frac{\mathbb{Z}_2[X]}{\langle X^3+X+1\rangle}.$ Now I can tell that each... | Well, the general result is that the elements of $\Bbb F_q$, $q=p^n$ with $p$ prime, satisfy $x^q-x=0$. Moreover, $x^q-x$ is the product of all irreducible polynomials of degree $d\geq 1$ over $\Bbb F_p$ with $d\mid n$.
In your case, $q=2^3$ and $x^8-x = x(x-1)(x^3+x+1)(x^3+x^2+1)$ over $\Bbb F_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z... | Since $x+y+z \geqslant 3 \sqrt[3]{xyz}$, we have $xyz (x+y+z)\geqslant 3 (xyz)^{4/3}$.
Using the given condition, we have $1 \geqslant 3 (xyz)^{4/3}$. This is $xyz \leqslant \frac{1}{3^{3/4}} $
Also, we have $(x+y)(y+z)(z+x) = (x+y+z)(xy+ yz + zx) - xyz $
Now $ -xyz \geqslant - \frac{1}{3^{3/4}} $ and
$$ (xy+ yz + zx) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the e... | Much simpler: $z^4 =(2+3i)^4= 1\cdot (2+3i)^4$
and $z = 1^{\frac 14} (2+3i)$, where $1^{\frac 14}$ is understood to mean the four complex fourth roots of $1$, namely $\pm 1, \pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
How to solve system of equations using inverse matrix? System of equations is the following:
$$x + 4y + 2z = 10$$
$$4x - 3y+0z = 6$$
$$2x + 2y + 2z = 14$$
Here is my solution:
$$det(A) = 1 *(-3 * 2 - 0 * 2) -4 * (4 * 2 - 0 * 2) + 2 * (4 * 2 - (-3) * 2)$$
$$= -6 -32 + 28$$
$$= -10$$
$$
+\begin{pmatrix}
-3 & 0 \\... | $$A=
\begin{matrix}
1 & 4 & 2 \\
4 & -3 & 0 \\
2 & 2 & 2 \\
\end{matrix}
$$
$$b=
\begin{matrix}
10 \\
6 \\
14 \\
\end{matrix}
$$
To find the inverse of A:
1.Find the Determinant: $\Delta A$ =-10
2.Find the matrix of cofactors:
$$\begin{matrix}
-6 & -8 & 14\\
-4 & -2 & 6\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. le... | You have
$$
x_n = \prod_{k=1}^n \left(1+\frac{1}{2^k}\right)
= e^{\sum_{k=1}^n\ln \left(1+\frac{1}{2^k}\right)}
\leq e^{\sum_{k=1}^n\frac{1}{2^k}}
\leq e^{\sum_{k=1}^\infty\frac{1}{2^k}}
= e
$$
where for the first inequality we used that $\ln(1+x)\leq x$ for all $x>-1$. This shows the sequence is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing the Change of Basis Matrix between two vector spaces For the two bases, $A$ and $B$ whose columns represent the basis vectors respectively, I was asked to find the transformation matrix from $A$ to $B$
$$ A=
\begin{Bmatrix}
1 & -1 & 1 \\
2 & 2 & 1 \\
1 & -1 & 3 \\
\end{Bmatrix} $$
$$
B= ... | The equation $
\left(\begin{matrix}
1 \\
2 \\
1 \\
\end{matrix}\right) = x_{11}\left(\begin{matrix}
-3 \\
2 \\
-3 \\
\end{matrix}\right) + x_{21}\left(\begin{matrix}
1 \\
-1 \\
-1 \\
\end{matrix}\right) + x_{31}\left(\begin{matrix}
5 \\
4 \\
9 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of sequence involving reciprocal of the sum of the first squares Let $(x_n)$ a sequence defined by $x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}.$ What is $\lim_{n\to \infty}x_n$?
My idea: $1^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}<\frac{k(k+1)(k+2)}{3}$. Thus
$$x_n>3\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=3\left(\frac 12 -... | Hint: Note that
$$
\frac{1}{k(k+1)(2k+1)} = \frac 1{k+1} + \frac 1k - \frac 4{2k+1}.
$$
So for instance, with $n = 10$, we end up with the sum
$$
\left(1 + 2\cdot \frac12 + \cdots + 2\cdot \frac 1{10} + \frac 1{11} \right) - 4\left(\frac 13 + \frac 15 + \cdots + \frac 1{21} \right) = \\
1 + 2 \left( \frac 12 - \frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | Use the high-school formula
$$a^3+b^3=(a+b)(a^2-ab+b^2).$$
You'll obtain
\begin{align}
\cos^6x+\sin^6x&=(\cos^2+\sin^2x)(\cos^4x-\cos^2x\sin^2x+\sin^4x)\cr
&=\cos^4x-\cos^2x\sin^2x+\sin^4x
\end{align}
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 3
} |
$6(m + 1)(n − 1), 6 + (m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
Find all pairs of integers (m, n) such that the integers $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ are simultaneously perfect cubes.
I assumed $6(m + 1)(n − 1), 6 +
(m − 1)(n + 1)$ and $(m − 2)(n + 2)$ to be ... | Suppose $m$ and $n$ such that the expressions are perfect cubes. Let $a$, $b$ and $c$ be integers such that
\begin{eqnarray*}
a^3&=&6(m+1)(n-1)&=&6mn-6m+6n-6,\\
b^3&=&6+(m-1)(n+1)&=&mn+m-n-6,\\
c^3&=&(m-2)(n+2)&=&mn+2m-2n-4.
\end{eqnarray*}
Then comparing coefficients shows that
$$a^3-18b^3+12c^3=90,$$
so in particular... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across... | Define a new operation $p \otimes q = \frac{{p + q}}{{1 - pq}}$.
And it's easy to find that it's associative.
$p \otimes q \otimes r = \frac{{p + q + r - pqr}}{{1 - pq - pr - qr}}$.
It means $p + q + r = pqr \Leftrightarrow p \otimes q \otimes r = 0$.
And then $\begin{array}{l}
\frac{{2x}}{{1 - {x^2}}} \times \frac{{2y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 4
} |
Inverse formulas of rotations using the substitution method
Given the direct rotations
$$
\mathcal R \begin{cases}
x=X\cos\vartheta-Y\sin\vartheta\equiv x'\cos\vartheta-y'\sin\vartheta\\
y=X\sin\vartheta+Y\cos\vartheta\equiv x'\sin\vartheta+y'\cos\vartheta
\end{cases} \tag 1 $$
to find the inverse rotations,
$$\mathca... | Let $c=\cos\vartheta$ and $s=\sin\vartheta$. So, your system is$$\left\{\begin{array}{l}x=cX-sY\\y=sX+cY\end{array}\right.$$and we have\begin{align}\left\{\begin{array}{l}x=cX-sY\\y=sX+cY\end{array}\right.&\iff\left\{\begin{array}{l}X=\frac1c(x+sY)\\y=\frac sc(x+sY)+cY=\frac scx+\left(\frac{s^2}c+c\right)Y=\frac scx+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a^{\lambda b} + b^{\lambda a} + a^{\lambda b^2} + b^{\lambda a^2} \le 2$ for positive reals $a+b=1$
Problem 1: Let $\lambda = \frac{\ln (7 + 3\sqrt{5})}{\ln 2} - 1 \approx 2.77697$. Let $a, b$ be positive real numbers with $a + b = 1$. Prove (or disprove) that
$$a^{\lambda b} + b^{\lambda a} + a^{\lambda b... | working the weaker case and using $b=1-a$, we have
$$f(a)=a^{\frac{25}{9} (1-a)}+(1-a)^{\frac{25}{9}a }+a^{\frac{25}{9} (1-a)^2}+(1-a)^{\frac{25 }{9}a^2}$$
The derivative cancels close to $a=\frac 14$. In fact, the first iterate of Newton method is extremely close to $\frac 6 {25}$
$$f\left(\frac{1}{4}\right)\sim 1.94... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $10x^3-6x^2-12x-8=0$ Solve the equation $10x^3-6x^2-12x-8=0$
I was doing this question but got stuck and hence looked at the solution, which went as follows $(x+2)^3=x^3+6x^2+12x+8$ and hence the equation is written as $11x^3=(x+2)^3$ from which we have that $\frac{x+2}{x}=\sqrt[3]{11}$ and hence $x=... | I imagine the first thing is to try to factor. That will fail as this has not rational roots but with $8 = 2^3$ as the last term of a third degree polynomial they think in terms of $(ax \pm 2)^3 = a^3 \pm 6a^2 + 12 a \pm 3$. That fails as there is no such possible $a$ but the coefficients of $6,12, 8$ are just too co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Please check my proof for $x^2+|x-2|>1$ Let $f(x)=x^2+|x-2|-1$.
If $x <2,$ then
$f(x)=x^2-x+1 \implies f'(x)=2x-1 \implies f_{min}=f(1/2)=\frac{3}{4} >1.$
If $x>2$, then $f(x)=x^2+x-3 \implies f'(x)=2x+1>0$. So the function is increasing for $x>2$
The function $f(x)$ has just one min, so $f(x)>f(1/3)=3/4 >0$, hence fo... | Ehen $x>2$ the inequation is $$x*2+x-3 >0\implies x \in (-\infty, \frac{-1-\sqrt{13}}{2}) \cup (\frac{\sqrt{13}-1}{2}, \infty)$$
So the solution is $x \in (2, \infty).$
When $x \le 2$, the inequation becomes $x^2-x+1>0$ which is true as $B^2< 4AC$. Then the solutions are $(-\infty, 2]$.
Finally, the solutions of this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3848989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proving $ 1+2f'(x)+\frac{2}{x(1+x^2)}\left(\frac{3x}{2}+f(x) \right)\ge \frac{6x^2}{1+8x^2} $.
Put
\begin{align*}
f(x)=\left( -\frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}-\left( \frac{x}{2} +\sqrt{\frac{1}{27}+\frac{x^2}{4}} \right)^{1/3}
\end{align*}
Prove that
$$
g(x):=1+2f'(x)+\frac{2}{x(1+x^2)}\... | The hint.
Let $f(x)=y$.
Thus, $$y^3+y+x=0,$$ which gives $$3y^2y'+y'+1=0$$ or $$y'=-\frac{1}{1+3y^2}$$ and we need to prove a polynomial inequality of one variable $y$.
I got that finally we need to prove that:
$$y^2(6y^{14}+16y^{12}-10y^{10}+y^8+94y^6+94y^4+26y^2+1)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Equation of Conics The general equation of the conic section is : $ax^2+2hxy+by^2+2gx+2fy+c=0$,
where
$$\Delta=\begin{vmatrix}a&h&g\\h&b&f \\g&f &c\\
\end{vmatrix}$$
This equation can also be analysed to distinguish whether it is an equation of pair of straight lines, parabola, ellipse or hyperbola.
*
*If $\Delta=0$ ... | Let's go the other way:
Any pair of lines has equation $k(a_1 x+b_1 y+c_1)(a_2 x+b_2 y+c_2)=0,$ which corresponds to the matrix being $\begin{pmatrix}a_1 \\b_1\\c_1\end{pmatrix}\begin{pmatrix}a_2&b_2&c_2\end{pmatrix},$ which is of rank $1$ and hence has zero determinant. Also all the 2 by 2 minors are zero in particula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Bisector of the acute angle formed between the lines $4x-3y+7=0$ and $3x-4y+14=0$
The bisector of the acute angle formed between the line $4x-3y+7=0$ and $3x-4y+14=0$ has the equation...
By calculating the intersection point, we get is as $(2,5)$.
But I couldn't proceed because I don't know how to find the equation ... | Equation of angle bisector lines are given by normalized equations of the two lines.
So, $\frac{L1}{|L1|} \pm \frac{L2}{|L2|} = 0$
So, $\frac {4x-3y+7} {\sqrt{4^2+3^2}} \pm \frac {3x-4y+14} {\sqrt{4^2 + 3^2}} = 0$
That gives us both angle bisectors: $x - y + 3 = 0, \, x + y - 7 = 0$
Now the slopes of original lines ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Verify if the following limit exists (by the formal definition): $\lim_{(x,y)\to(0,0)}\frac{x^3-y^3}{x^2+y^2}$ First, I tried approaching the point through a few curves: ($x=0,\,y=0,\, y=m\cdot x,\, y=x^2,\, x=y^2)$ and to all of those I got $0$ for my limit. Since this isn't enough to prove that the limit actually exi... | Theorem: If $\lim f(x) = 0$ and $g(x)$ is bounded, then $\lim f(x)g(x) = 0$.
$$\frac{x^3-y^3}{x^2+y^2} = \frac{(x-y)(x^2+xy+y^2)}{x^2+y^2} = \frac{(x-y)xy + (x-y)(x^2+y^2)}{x^2+y^2} = (x-y)\frac{xy}{x^2+y^2}+x-y$$
Because $\begin{aligned}\lim_{(x, y)\to(0, 0)} x-y = 0\end{aligned}$ we have $\begin{aligned}\lim_{(x, y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | If you replace with $t:= 2-x$, the question becomes
$$\lim _{t\to 0}\frac{\sqrt{t+4}-2}{\sqrt{t+1}-1}$$
From wolframalpha, reading the diagram for $\sqrt{t+1}-1$, at the neighbour of $0$ it's something like $t$, or $$\sqrt{t+1}-1=t+O(t^2)$$; similarly, $$\sqrt{t+4}-2=\frac12 t+O(t^2)$$
, here we are using the big O n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 3
} |
Proving that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$ I have to show that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$.
What I know is that $\sum_{k=0}^{k=n} \binom{n}{k} \cdot k = 2^{n -1} \cdot n$.
How do I proceed from there?
| $$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{2n-1}+\dbinom{2n}{2n}=2^{2n}$$
$$\dbinom{2n}{0}+\dbinom{2n}{1}+\dbinom{2n}{2}+...+\dbinom{2n}{n-1}+\dbinom{2n}{n}=2^{2n-1}$$
$$\dbinom{n}{k}=\dbinom{n}{n-k}$$
$$k\dbinom{n}{k}+(n-k)\dbinom{n}{n-k}=n\dbinom{n}{k}$$
$$0\dbinom{2n}{0}+1\dbinom{2n}{1}+2\dbinom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculating probability that $P(X\geq Y+1)$, where $X$ and $Y$ are independent random variables
Let $X$ and $Y$ be independent random variables with probability functions given by
$$
p_{X}(x)=\left\{\begin{array}{cc}
\frac{2^{x}}{15}, & x \in\{0,1,2,3\} \\
0, & \text { otherwise }
\end{array}\right. \text { e } p_{Y}(... | We have $$\{X\geqslant Y+1\} = \bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}, $$
and the union is of disjoint events. By independence, then
\begin{align}
\mathbb P(X\geqslant Y+1) &= \mathbb P\left(\bigcup_{i=1}^3\bigcup_{j=0}^{i-1} \{X=i\}\cap\{Y=j\}\right)\\
&= \sum_{i=1}^3\sum_{j=0}^{i-1}\mathbb P(X=i)\mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help with solving an elliptic integral I have the elliptic integral below, and currently I am stuck on how I can solve it, could someone please help me out:
$$\int^{\frac{L}{15}}_{0}2\pi\sqrt{\frac{a^{2}+b^{2}}{2}}dx$$
where: $a = W-\frac{0.608W^{2}}{L^{2}}x^{2}$ and $b = H-\frac{1.68H^{2}}{L^{2}}x^{2}$ ($H$, $W$ and $... | Using whole numbers
$$\frac{a^2+b^2}2=A-B x^2+C x^4$$
$$A=\frac{H^2+W^2}{2} \qquad B=\frac{2 \left(21 H^3+76 W\right)}{25 L^2} \qquad C=\frac{2 \left(441 H^4+5776\right)}{625 L^4}$$ and the antiderivative
$$I=\int \sqrt{\frac{a^2+b^2}2}\,dx$$ contains, as you expected , the elliptic integrals of the first and secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $x^4-3x^2+5\ge0$
Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$
$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x... | Hint :
First solve for $\sqrt{z} + \sqrt{z+7} = 7 $
Next substitute $z=x^4-3x^2+5$ and solve for $x$.
Actually easy to see $\sqrt{9} + \sqrt{9+7} = 7 $.
$9$ is only solution as monotonic function has unique root (when it does).
So only need to solve for $x^4-3x^2+5=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$. (COMC) 1997 Part B Question 4 Find all real values of $x, y$ and $z$ such that $x-\sqrt{yz}=42$, $y-\sqrt{xz}=6,z-\sqrt{xy}=-30$.
I was preparing for my COMC Contest so I was doing past years exams. This question really ... | Using your substitution and subtracting the equations pairwise, we end up with:
$72 = 42 - (-30) = a^2 - bc - (c^2 - ab ) = (a-c) ( a+b+c)$
$36 = 6 - (-30) = b^2 - ca - (c^2 - ab) = (b-c) (a+b+c )$
$ 36 = 42 - 6 = a^2 - bc - (b^2 - ca) = (a-b) ( a + b + c) $
Hence $ a-c = 2 (b-c) = 2(a-b)$, or that $ b = \frac{ a+c}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$ Let $a,b,c,d$ be positive real numbers that fulfill two conditions:
$$a+b+c+d \leqslant 2$$$$ab+bc+cd+ad \geqslant 1$$
Prove that $|a-b+c-d|\leqslant \frac{1}{16}$
Let:
$a+c=x$ and $b+d=y$
Both $x$ and $y$ are positive.
$$x+y \leqslant 2$$$$xy \geqslant 1$$
$$-4xy \le... | By AM-GM $$1\leq ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+c+b+d}{2}\right)^2\leq1,$$
which gives $$a+c=b+d$$ and $$a+b+c+d=2$$ or
$$a+c=b+d=1,$$ which gives $$|a-b+c-d|=0<\frac{1}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = ... | Here is a different approach: Consider the change of variable $b_n=a_{n+4},$ so that the only initial condition is given by $b_0=0.$ Notice that the equation becomes
$$(n+5)a_{n+5}-(n+4)a_{n+4}+3a_{n+4}-12(n+5)H_{n+5}=0,$$
$$(n+5)b_{n+1}-(n+4)b_{n}+3b_{n}-12(n+5)H_{n+5}=0,$$
take $f(x)=(x+4)b_x$ so that $$\Delta (f)=f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$
By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.
Hence it suffices to prove $$\sum {\le... | We need to prove that:
$$\sum_{cyc}\frac{1}{(x+y)^2}\geq\frac{9}{4(xy+xz+yz)}$$ or
$$\sum_{cyc}\left(\frac{1}{(x+y)^2}-\frac{3}{4(xy+xz+yz)}\right)\geq0$$ or
$$\sum_{cyc}\frac{4xz+4yz-2xy-3x^2-3y^2}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}\frac{(z-x)(3x+y)-(y-z)(3y+x)}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{3y+z}{(y+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Stronger than Nesbitt's inequality using convexity and functions Hi it's a refinement of Nesbitt's inequality and for that, we introduce the function :
$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$
With $a,b,x>0$
Due to homogeneity we assume $a+b=1$ and we introduce the function :
$$g(a)=\frac{a}{1-a+x}$$
Showing... | Stronger than Nesbitts inequality for non-cyclic style
*
*If $a,\,b,\,c$ are positive real numbers, then
$$ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$
*If $a,\,b,\,c$ are positive real numbers, then
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}+\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How I can solve this limit only with algebra I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.
$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$
first all I use the logarithms properties and rewrite l... | Let $a = \lim_{x\to 0} \log({\frac{1 + x}{1 - x}})^\frac{1}{2x}$. Then:
$$\exp(a) = \lim_{x\to 0} \exp \left(\log({\frac{1 + x}{1 - x}})^{\frac{1}{2x}} \right) = \lim_{x\to 0} \left(\frac{1 + x}{1 - x} \right)^{\frac{1}{2x}} = \lim_{x\to 0} \left(\frac{1/x + 1}{1/x - 1} \right)^{\frac{1}{2x}}$$
Now let $u = \frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that $$x-... | EDIT: The question was updated, so here's my new answer.
Notice the LHS is
and since $x>0$, it's clear that it is $<0$ for all $x$.
Old answer: This is false. Indeed, since $x>0$, we have
\begin{align*}
&x-4x \frac{1}{\sqrt{x}+x} +3x \left( \frac{1}{\sqrt{x}+x} \right)^2 -3 \left(\frac{1}{\sqrt{x}+x} \right)^2 >0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
indefinite integral of $\int \frac{dx}{\sqrt{(x)(x+1)(x+2)}}$ I tried integration by parts considering $dx=du$
$\frac{1}{\sqrt{(x)(x+1)(x+2)}}=v$ but im not getting the answer.
My attempt....
$uv=\sqrt{\frac{x}{(x+1)(x+2)}}$ and-$\int{dv u}$=$\frac{I}{2}$+$\int{\frac{\sqrt{x}}{2\sqrt{(x+1)}(x+2)^{\frac{3}{2}}}}$+$\int{... | In general, an integral with square-root of a polynomial of degree $3$ or $4$ is an elliptic integral.
Maple evaluates this in terms of the elliptic integral of the first kind $F$:
$$
\int \!{\frac {1}{\sqrt {x \left( x+1 \right) \left( x+2 \right) }}}
\,{\rm d}x=2\;{ F} \left( \frac{\sqrt {x+2}}{\sqrt{2}},\sqrt {2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
Find all solutions to the diophantine equation $7^x=3^y+4$ in positive integers.
I couldn't have much progress.
Clearly $(x,y)=(1,1)$ is a solution. And there's no solution for $y=2$.
Assume $y \ge 3$ and $x \ge 1$.
By $\mod 9$, we get ... | It's $3(3^a-1)=7(7^b-1)$ with $a=x-1$ and $b=y-1$.
Therefore $7\mid3^a-1$, so $a$ is a multiple of (what?).
Therefore, $3^a-1$ is a multiple of $13$.
Therefore, $7^b-1$ is a multiple of $13$.
Therefore, $b$ is a multiple of (what?).
Therefore, $7^b-1$ is a multiple of $9$.
Therefore, $3(3^a-1)$ is a multiple o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving Riordan's identity that $\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1$ In a book titled "Advances in Problem Solving," authored by Sailesh Shirali, Riordan's identity is mentioned which can also be written as
$$S=\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1~~~~(1)$$
Here, we prove it by the integral ... | We seek to show that
$$S= \sum_{k=0}^{n-1} {n-1\choose k} \frac{(k+1)!}{n^{k+1}} = 1.$$
The sum is
$$\frac{(n-1)!}{n}
\sum_{k=0}^{n-1} \frac{k+1}{(n-1-k)!} \frac{1}{n^k}.$$
We get without the factor in front
$$\sum_{k=0}^{n-1} \frac{-n+k+1}{(n-1-k)!} \frac{1}{n^k}
+ \sum_{k=0}^{n-1} \frac{n}{(n-1-k)!} \frac{1}{n^k}
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Showing $\alpha(\beta+1)\leq\frac{5}{3}\alpha^2+\frac{1}{3}\beta^2$ for nonnegative integers $\alpha$ and $\beta$. I found this lemma in a paper I was reading and it was not proved. There doesn't seem to be any obvious factorized form so how would one go about proving the inequality holds?
$$\alpha(\beta+1)\leq\frac{5}... | Alternative proof inspired by Michael Hardy:
$$5\alpha^2+\beta^2-3\alpha\beta-3\alpha \geqslant 0 \iff 20\alpha^2+4\beta^2-12\alpha
\beta-12\alpha \geqslant 0$$
$$20\alpha^2+4\beta^2-12\alpha \beta-12\alpha
= (2\beta-3\alpha )^2+11\alpha^2-12\alpha \\
=\begin{cases}
(2\beta-3\alpha )^2+\alpha (11\alpha-12)>0, & \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$? If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?
I tried this problem for a while but made no p... | Hint: Note that
$$
\frac ac + \frac ca + \frac bd + \frac db = 20 \implies\\
\frac ab \cdot \frac bc + \frac cd \cdot \frac da + \frac bc \cdot \frac cd + \frac da \cdot \frac ab = 20.
$$
With that in mind, compare the expanded sums
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\right)^2, \quad
\left(\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequal... | Hints:
Put Let $a=\frac{x}{3\sqrt2},$ $b=\frac{y}{3\sqrt2}$ and $c=\frac{z}{3\sqrt2}.$
And $x=\tan(u)$,$y=\tan(v)$,$z=\tan(w)$ with $u+v+w=\pi$
Then the function $f(u)=\sqrt{3+\frac{18}{\tan(u)^2}}$ is convex on $(0,\pi)$ see here
Remains to apply Jensen's inequality .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Proof: not a perfect square Let $y$ be an integer.
Prove that
$$(2y-1)^2 -4$$
is not a perfect square.
I Found this question in a discrete math book and tried solving it by dividing the question into two parts:
$$y = 2k , y = 2k + 1$$
But that got me nowhere.
| $(2y-1)^2-4=4(y^2-y)-3$ If it were a perfect square it would be $=c^2$, where c is an integer. Solve for $y$ in $4(y^2-y)-3-c^2=0$ and get $y=\frac{4\pm \sqrt{16+16(3+c^2)}}{8}=\frac{1\pm \sqrt{4+c^2}}{2}$.
However $c^2+4$ cannot be a square, unless $c=0$ (where $y$ is not an integer). Assume $c^2+4=b^2$ so $b=c+a$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone ou... | We are going to replace $4^n=12^n-3^n+1-6M$
$\begin{align}12^{n+1}-4^{n+1}-3^{n+1}+1
&=12.12^n-4.4^n-3.3^n+1\\
&=12.12^n-4(12^n-3^n+1-6M)-3.3^n+1\\
&=8.12^n+3^n-3+24M
\end{align}$
$12^n$ and $24M$ are obviously divisible by $6$
Notice $3^n-3=3\times\underbrace{(3^{n-1}-1)}_{\text{even}}$ is also divisible by $6$ so you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| For me it's just Vieta's formulas (or comparing coefficients as if $x^2+5x$ were a single variable):
$(x^2 + 5x + \color{red}4)(x^2 + 5x + \color{red}6) \color{red}{-48} = (x^2 + 5x + \color{red}{12})(x^2 + 5x \color{red}{-2})$?
$4+6=12+(-2)$, and $4\cdot 6 - 48 = 12 \cdot (-2)$, so yes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Cross section of cylinder $5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$ Prove that
$$5x^2 + 5y^2 + 8z^2 − 2xy + 8yz + 8zx + 12x − 12y + 6 = 0$$
represents a cylinder whose cross-section is an ellipse of eccentricity $\frac{1}{\sqrt 2}$.
I know how to find the plane of a cross section if I know the directio... | The equation can be written $5(x+1)^2+5(y-1)^2+8z^2-2(x+1)(y-1)+8(y-1)z+8(x+1)z-6=0,$ and we can translate.
From @WillJagy's comment look at the plane $x+y-z=0.$ We're interested in the intersection $\langle 5x^2+5y^2+8z^2-2xy+8yz+8xz-6,x+y-z\rangle.$
Now rotate so that the normal vector $[\frac1{\sqrt{3}},\frac1{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate... | Start with $x-x^3/6<\sin(x)<x$; there are many nice proofs here.
Then we have
$$
\sum_{k=1}^{n}(k/n^2)-(k/n^2)^3/6 <\sum_{k=1}^{n}\sin(k/n^2) < \sum_{k=1}^{n}k/n^2
$$Using the fact that $\sum_{k=1}^{n}k^p = \frac{n^{p+1}}{p+1}+\text{ lower order terms}$, we have
$$
\frac{n^2+\cdots}{2 n^2}- \frac{n^4+\cdots}{24 n^6} <\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$? What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
I know that $289 = 17 \times 17$
so would it be $2^8\equiv 256\bmod17 =1$
and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this
| $256 \equiv 1 \pmod {17}$ but $256\not \equiv 1 \pmod {289}$ which we need.
But not $289 = 17\times 17$ so $\phi (289) = 17\cdot16$ so $2^{17\cdot 16}\equiv 1\pmod {289}$ by Eulers theorem.
But the order might be something smaller that divides $17\cdot 16$.
We can figure that $2^8 = 17*15 + 1 \equiv 17*(-2) + 1\pmod{17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Trying to find the sum of cosines I have been trying to calculate this sum, could someone confirm if all my working is correct please :) :
$$S_k(x)=\sum_{n=0}^k\cos(nx)=\Re\sum_{n=0}^ke^{inx}=\Re\sum_{n=0}^k\left(e^{ix}\right)^n=\Re\left(\frac{1-e^{i(k+1)x}}{1-e^{ix}}\right)=\Re\left(\frac{1-\cos((k+1)x)-i\sin((k+1)x)}... | You could easily use $$\cos(nx)=\frac{e^{inx}+e^{-inx}}{2}$$ to make calculations bit easier. Let me propose you an alternative less messy solution, which uses telescoping method. Note that $$\sin((n+1)x)-\sin((n-1)x)=2\cos(nx)\sin(x).$$ Now telescoping gives $$\sin((n+1)x)+\sin(nx)-\sin(x)=2\sin(x)\sum_{k=1}^n\cos(kx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$ for natural numbers $a,b,c$ Find natural numbers $a<b<c$ such that:
$$\sum_{k=a}^{b-1} k=\sum_{k=b+1}^c k$$
I found infinitely many solutions when $a=1$, where $b$ and $c$ can be found in the following way:
Calculate, in the simplest form, the continued fractions of the se... | Not an answer, but rather a long comment.
Hint. Use the well-known formula $$\sum_{i=1}^n i=\frac{n\cdot (n+1)}{2}$$ in order to deduce $$\sum_{k=a}^{b-1} k=\sum_{k=1}^{b-1}k-\sum_{k=1}^{a-1}k=\ldots=\frac{(b-a)(a+b-1)}2$$ Similarly $$\sum_{k=b+1}^c k=\frac{(c-b)(c+b+1)}2$$Thus, your problem comes down to solving the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
understanding the solution to a particular differential equation? so I ran across this problem:
$(3x^2+1)y' -2xy = 6x$
I solved it by integrating factor $ e^{\int\frac{-2x}{3x^2+1} dx}$
.
.
.
$\int D[(3x^2+1)^{\frac{-1}{3}}y] = \int 6x(3x^2+1)^{\frac{-4}{3}} dx$
.
.
.
$(3x^2+1)^{\frac{-1}{3}}y = -3(3x^2+1)^{\frac{-1}... | You could transform the equation simply as
$$
(3x^2+1)y'=2x(y+3)
\\~\\
y'=\frac{2x}{3x^2+1}(y+3)
$$
This is a separable equation and the case of the constant solution $y=-3$ appears directly as the one to exclude for performing the separation of variables. Then continue
$$
\int\frac{dy}{y+3}=\int\frac{2x\,dx}{3x^2+1}.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing that if $3x$ is even then $3x+5$ is odd I'm learning the absolute basics of how to do proofs, and am really struggling.
If 3x is even then 3x+5 is odd.
This is the solution:
I get that even numbers are 2n and odd numbers are 2n+1. For the life of me, I CANNOT get it into that form shown below. I feel so dumb... | Your problem is that $3x + 5 = 2k +1$ is not an assumption. It is the conclusion you need to prove.
Your one and only assumption is that $3x = 2n$ for some integer $n$.
so you start with
$3x = 2n$.
.... then you do a bunch of steps ....
.... steps .....
.... and get in the end ........
Conclusion: $3x + 5 = 2(????????... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Finding Exponents of Generating Function $A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $ I have the following generating function:
$$A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $$
Among multiple of $2$, we will get a multiple of 6 whenever we take a multiple that is a ... | We obtain
\begin{align*}
\color{blue}{A(x)}&=\prod_{k=1}^\infty \frac{1-x^{6k}}{\left(1-x^{2k}\right)\left(1-x^{3k}\right)}\\
&=\prod_{k=1}^\infty \frac{1+x^{3k}}{1-x^{2k}}\tag{1}\\
&=\prod_{k=1}^\infty \frac{1-x^k+x^{2k}}{1-x^k}\tag{2}\\
&\,\,\color{blue}{=\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\tag{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$ So there exist a lot of similar questions which ask the $n$th term of
$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider e... | OEIS has an entry for this sequence. And they give a formula:
a(n) = A000290(m+1), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2)
The sequence named A000290 is just the perfect squares, so A000290(m+1) means $(m+1)^2$. And $m$ is given as $n-\frac{t(t+1)}2$, where $t=\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A problem in limits to show that $\lim_{n\to\infty}$ $p_n$/$q_n$ = $π/4$ Define two sequences of real numbers $p_n$ and $q_n$ with $n = −1, 0, 1, 2, \ldots$ as follows:
$$p_n = 2p_{n−1} + (2n−1)^2p_{n−2}$$ and
$$q_n = 2q_{n−1} + (2n−1)^2q_{n−2}$$
for every $n \geq 1$, and starting with $p_{−1}= 0, q_{−1}= 1, p_0 = q_0 ... | Let $r_n := p_n/q_n$. We can compute the first few times to observe that: $r_0 = 1$, $r_1 = \frac{2}{3} = 1-\frac{1}{3}$, $r_2 = \frac{13}{15} =1-\frac{1}{3} + \frac{1}{5}$, $r_3 = \frac{76}{105} = 1-\frac{1}{3} + \frac{1}{5} -\frac{1}{7}$. So this gives us enough motivation to conjecture that: $$\frac{p_n}{q_n} = 1 -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Question related to the following function $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$ Let $f:\left( {0,1} \right) \to R$ be defined by $f\left( x \right) = \frac{{b - x}}{{1 - bx}}$
, where 'b' is a constant such that $0 < b < 1$. Then
(A) $f$ is not invertible on (0,1)
(B) $f\ne f^{-1}$ on (0,1)and $f'(b)=\frac{1}... | You also may show that $f(f(x))=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factor $x^5-1$ for binary codes Currently, I have that I can do:
$$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1) $$
Then I thought I could do:
$$ x^5-1 = (x-1)(x(x+1)(x^2+1)+1) $$
However I want to use the factorization to find all cyclic codes in $[5, k]$. For that, I need to find all irreducible factors of $ x^5-1 $. I am n... | Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already
$$
x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ .
$$
To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\iint _S (y^2+x^2)\, dA$ by change of variables.
Let $S$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Compute $$\iint _S (y^2+x^2)\, dA.$$
I have tried to use the change of variables $u=xy$ and $v=x+y$. Then $$\begin{align}x^2+y^2=v^2-2u \\ x^2-y^2=v... | EDIT
The substitution $u=xy$, $v=x^2-y^2$ does the job. The Jacobian with respect to $x$ and $y$ is
$$
J=\left(\begin{matrix}
y & x\\
2x & -2y
\end{matrix}\right)
$$
so $|\det(J)|=2(x^2+y^2)$, which means $dudv=2(x^2+y^2)dxdy$ and your integral becomes
$$\begin{align} \iint _S (y^2+x^2) dxdy= \int _1^3 \int_1^4 2 \;dud... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equati... | HINT
\begin{align*}
y'' + y = \frac{1}{y^{3}} & \Longleftrightarrow y''y' + yy' = \frac{y'}{y^{3}}\\\\
& \Longleftrightarrow (y')^{2} + y^{2} = -\frac{1}{y^{2}} + c\\\\
& \Longleftrightarrow (y')^{2} = \frac{cy^{2} - y^{4} - 1}{y^{2}}\\\\
& \Longleftrightarrow y' = \pm\sqrt{\frac{cy^{2} - y^{4} - 1}{y^{2}}}
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
General term of a sequence and its limit. Having a sequence $\;\sqrt[3]5\;,\;\sqrt[3]{5\sqrt[3]5}\;,\;\sqrt[3]{5\sqrt[3]{5\sqrt[3]5}}\;,\;\ldots\;,\;$ what could be the general term and its limit ?
Note that the second term is cube root of $5$ and inside that there is cube root of $5$ again and the third term is cube r... | Observe the pattern that, if the $n$th term of the sequence is denoted $c_n$, then
$$c_n = 5^{1/3 + (1/3)^2 + (1/3)^3 + \cdots + (1/3)^n}$$
To see this, notice that, for instance:
\begin{align*}
c_2 &= \sqrt[3]{ 5 \sqrt[3] 5} = \left( 5 \cdot 5^{1/3} \right)^{1/3} = 5^{1/3} \cdot 5^{1/3^2} = 5^{1/3 + 1/3^2} \\
c_3 &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\fra... | *Note: * For $k=1$ to $3$ we have to evaulate the sum for $k\in\{1,2,3\}$:
\begin{align*}
\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}\tag{1}
\end{align*}
Hint: Evaluation at $n=2$ is not feasible, since in that case we have division by zero, which is not admissible. In order... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does $\int \tan^3x\sec^2x \space dx$ have 2 solutions? So normally, you evaluate $\int\tan^3x\sec^2x \space dx$ by substituting $u = \tan x$ and $du = \sec^2x\space dx$ right? So,
$$\begin{equation}\begin{aligned}
\int\tan^3x\sec^2x \space dx &= \int u^3 \space du \\
&= \frac{u^4}{4} + C\\
&= \frac{tan^4x}{4} + C \... | There's no mistake. Both methods resulted in the same solution, up to a constant of integration. Indeed, notice that:
\begin{align*}
\frac{1}{4}\tan^4 x + C
&= \frac{1}{4}(\tan^2 x)^2 + C \\
&= \frac{1}{4}(\sec^2 x - 1)^2 + C \\
&= \frac{1}{4}(\sec^4 x - 2\sec^2 x + 1) + C \\
&= \frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What's wrong with this proof of $3=0$ starting from $x^2+x+1=0$? What am I missing?
$$ x^2 + x + 1 = 0$$
Then
$$ x^2 + x = -1$$
$$ x(x+1) = -1$$
$$ x(-x^2) = -1$$
$$x^3 = 1 $$
$$x = 1 $$
but
$$(1)^2 + (1) + 1 = 3 $$
So
$$ 3 = 0$$
| Nothing wrong: it is true that if $x^2+x+1=0$ and $x\in\Bbb R$, then $x=1$. Keeping all the pieces together $$x^2+x+1=0\Leftrightarrow\begin{cases}x^2+x=-1\\ x+1=-x^2\end{cases}\Leftrightarrow \begin{cases}-x^3=-1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, u... | Since every term but the last is divisible by $a$, so we only need to deal with $(2x)^n \bmod \left(x^2-x+1\right)$. Let's start from small ones, and try to find patterns:
$$
\begin{aligned}
((2 x)^1 &\bmod \left(x^2-x+1\right))=\color{gray}{2x} \\
((2 x)^2&\bmod \left(x^2-x+1\right)) =\color{red}{-4+4 x} \\
((2 x)^3 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
I can't understand the result of this limit $$ \lim_{n\to\infty} \frac{n^4-n^3+1}{\sqrt{n}+n^2-n^3}
$$
I think the answer is $-\infty$ because I take the highest degree coefficients $$n^4 (numerator)$$ and $$-n ^ 3(denominator)$$
and I was thinking $+/- = -$, but the actual answer is $+\infty$, this is probably a dumb... | \begin{align} \lim_{n\to\infty} \left(\frac{n^4-n^3+1}{-n^3+n^2+\sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \frac{n^3-n^2+\frac1n}{-n^3 + n^2 + \sqrt{n}}\right) \\
\\
=\lim_{n\to\infty} \left(n \ . \ \left(-\ \frac{n^3-n^2+\frac1n}{n^3 - n^2 -\sqrt{n}}\right)\right) \\
\\
=-\ \lim_{n\to\infty} \left(n \ . \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Mistake when substituting constraint $4x^2+y^2=1$ into a function $f(x,y)=x^2+y^2$ in extrema problem Consider the function $f(x,y)=x^2+y^2$ under the constraint $4x^2+y^2=1$. The extrema of $f$ under that constraint can be easily found with Lagrange multipliers, and they are attained for $(0,1)$, $(0,-1)$ for maximum ... | Solving
$$
\min_{x,y}(\max_{x,y})(x^2+y^2)\ \ \text{s. t.}\ \ \ 4x^2+y^2=1
$$
using the Lagrange multipliers method, reduces to determine the stationary points for
$$
\nabla(x^2+y^2)+\lambda\nabla(4x^2+y^2-1)=0
$$
by solving for $x,y,\lambda$
$$
\cases{
2x+8x\lambda=0\\
2y+2y\lambda=0\\
4x^2+y^2-1=0
}
$$
giving as solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| Rewriting the inequality $$2\sin^2(\frac{x}{2})\ge\frac{x^2}3 \iff (\sqrt2 \sin(\frac{x}{2}) - \frac{x}{\sqrt3})(\sqrt2 \sin(\frac{x}{2}) + \frac{x}{\sqrt3}) \ge 0 \tag{1}$$
Assume $0\le x \le 1$. Obviously we have $$\sqrt2 \sin(\frac{x}{2}) + \frac{x}{\sqrt3}\ge 0$$ So we should analyze $$f(x,a) = \sin(\frac{x}{2})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
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.