Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
| From
$$
2 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = (x+y) \frac{(x-y)^2+x^2+y^2}2
$$
it follows that $x+y> 0$. On the other hand, for arbitrary $t > 0$
$$
x =-t \quad , \quad y = \sqrt[3]{2+t^3}
$$
satisfies $x^3+y^3 =2$ and
$$
x + y = \sqrt[3]{2+t^3} - t = \frac{2}{(\sqrt[3]{2+t^3})^2 + t \sqrt[3]{2+t^3} + t^2} \to 0
$$
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$ The inequality:
$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$
Conditions:... | By Holder inequality,we have
$$\left(\sum\dfrac{a^2}{(a+b+c)^2}\right)\left(\sum a(a+b+c)\right)^2\ge \left(\sum_{cyc} a^{\frac{4}{3}}\right)^3$$
therefore
$$\sum\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{(a^{4/3}+b^{4/3}+c^{4/3}+d^{4/3})^3}{[(a+c)^2+(b+d)^2+(a+c)(b+d)]^2}$$
use Holder we have
$$a^{\frac{4}{3}}+c^{\frac{4}{3}}\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Solving Linear Equations with Modulo I'm having trouble approaching the following problem:
Solve the following system of equation -
$$3x+5y\equiv 14\mod 17$$
$$7x+3y\equiv 6\mod 17$$
Multiplying the top by $3$ and the bottom by $5$, I get $x = -3/5$ and $y = 51/15$, but I'm unsure how to apply (mod $17$) with $2$ vari... | So what you do is elimination of variables, and then deal with it differently:
start by eliminating $y$: multiply the first equation by three, second equation by five, then you get:
$$
9x+15y \equiv 42 (\equiv 8) \mod 17 ; 35x+15y \equiv 30 (\equiv 13) \mod 17
$$
Now, subtract the first equation from the second, and y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.
Can anyone provide a hint about what should be done?
| Notice that
$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$
Thus,
$$9x-3 = \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$
$$9x-3 = 0 \implies x=1/3$$ or
$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$
But... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
the value of $\frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_{k}|$ is Let $S_{k},$ where $k=1,2,3,\cdots \cdots ,100$ denote the sum of the infinite geometric series whose first term is $\displaystyle \frac{k-1}{k!}$ and the common ratio is $\displaystyle \frac{1}{k},$ then the value of $\displaystyle \frac{100^2}{100... | $$\frac{k-1}{(k-2)!}-\frac{k}{k!}=\frac{(k - 2) k^2}{k!}>0 \implies \bigg|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}\bigg|=\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}$$
Then as @Archis said use telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$?
If $a,b,c$ are the sides of a triangle then show that-
$\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$
I got this probl... | Your inequality is wrong! Try $a=b=c$.
It should be
$$\frac{1}{b+c-a} +\frac{1}{c+a-b} +\frac{1}{a+b-c} \geq \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$$
For the proof we can use the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since our inequality is fifth degree, we see that $f(w^3)\geq0,$ where $f$ is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to derive this formula and provide a proof? I have a formula that "filters" out values of x where x mod 3 is 0:
$$f(x) := 3\cdot\left\lfloor \frac x2\right\rfloor\, +\,(x\mathrm{\,mod\,} 2)\,+\,1$$
e.g.
f(0) = 1
f(1) = 2
f(2) = 4
f(3) = 5
f(4) = 7
f(5) = 8
f(6) = 10
etc.
I've come up with this formula through intui... | A recursive definition would be $f(0)=1$ and
$$
f(x+1)=\begin{cases}
f(x)+1 & \text{if $3\nmid f(x)+1$}\\[4px]
f(x)+2 & \text{if $3\mid f(x)+1$}
\end{cases}
$$
Let's examine your function $f$.
We should have
$$
0<f(x+1)-f(x)\le2
$$
and $f(x+1)-f(x)=2$ if and only if $f(x)+1$ is a multiple of $3$.
We have
$$
f(x+1)-f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all reals $x$,$y$ satisfying the following equation:
Find all positive reals $x,y \in \mathbb{R}^+$ satisfying:
$$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$
Since this involves higher exponents I am unable to tackle this problem. Please help me.
| Just to give a non-AM-GM answer, let $u=xy$, which must be positive in order for $x^{10}+y^{10}=10xy-8$ to have a solution, and note that $x^{10}+y^{10}=(x^5-y^5)^2+2(xy)^5$. Thus
$$\begin{align}x^{10}+y^{10}-10xy+8
&=(x^5-y^5)^2+2(u^5-5u+4)\\
&=(x^5-y^5)^2+2(u-1)(u^4+u^3+u^2+u-4)\\
&=(x^5-y^5)^2+2(u-1)^2(u^3+2u^2+3u+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty... | Hint
$$A=\left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}\implies \log(A)={\sqrt{ax^2 +bx+c}}\log\left(1+\dfrac{1}{x} \right)$$ Now, use Taylor series $$\log\left(1+\dfrac{1}{x} \right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ and you will arrive to the result (not only the limit but also how it is ap... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the limit $\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$ I have been trying to find the limit,
$$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$
and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says... | Another approach (apart from the one mentioned in my comments) is to use Binomial Theorem. Let the expression be denoted by $$\frac{u - v}{w}$$ and then multiply numerator and denominator by $$t = u^{11} + u^{10}v + \cdots + v^{11}$$ to get $$\frac{u^{12} - v^{12}}{wt}$$ and note that since $u, v$ both tend to $1$ as $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim_{x \to 1}\frac{1-x^2}{\sin\pi x}$ without L'Hopital I'm trying to evaluate the given limit without using L'hopitals rule. $$\lim_{x \to 1}\frac{1-x^2}{\sin(\pi x)}$$
Replacing $x$ by $1$ leads to $\frac{0}{0}$.
I have tried multiplying by $\frac{1+x^2}{1+x^2}$ and resoving the $\sin x$ factor by doing:
$... | Write
$$\frac{1-x^2}{\sin\pi x}=\frac{1-x^2}{\sin(\pi-\pi x)}=\frac{1-x^2}{\sin\pi(1- x)}=\frac{(1-x)(1+x)}{2\sin\frac{\pi(1- x)}{2}\cos\frac{\pi(1- x)}{2}}=\frac12\times\frac{2}{\pi}\frac{\frac{\pi(1- x)}{2}}{\sin\frac{\pi(1- x)}{2}}\times\frac{1+x}{\cos\frac{\pi(1- x)}{2}}$$
so $t=1-x$ we have
$$\lim_{x\to1}\frac{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
| $(2^x+2^y)^2=2^{2x}+2^{x+y+1}+2^{2y}$ so we can take $y=1007$ and $x=1009$ to conclude $2^{2018}+2^{2017}+2^{2014}$ is a square.
how to obtain all $n$:
$9(2^{2014})=2^{2017}+2^{2014}=k^2-2^n$.
Notice that we must have $k^2\equiv 1 \bmod 3$ implying $2^n\equiv 1\bmod 3$.
So $n=2m$.
From here we have $2^{2014}\times 9= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
| Trial and error can be avoided. Note that minimizing $x+y+z$ where the positive real numbers $x,y,z$ have a specified product $1001$ is achieved by making $x,y,z$ as nearly equal as possible.
Here we minimize $(1+a)+(1+b)+(1+c)$ when we minimize $a+b+c$. Since $a,b,c$ are required to be (positive?) integers, the mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trigonometric Inequalities-Proof Verification. Prove the following inequalities:
If $x\in\Bbb R$, $x\ge0$ then,
(1) $-x\le \sin x\le x$. Moreover, does this inequality hold if $x<0$
(2) $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
(3) $x-\displaystyle\frac{x^3}{6}\le \sin x\le x$
(4) $1-\displaystyle\frac{x^2}{2}\le \... | Everything looks fine. For (5) use (1) and the fact that $|a| \le |b|$ equivalent $-|b| \le a \le |b|$.
For $x \ge 0$ then, using (1) $|sin(x)|\le x$, therefore $|sin(x)|\le |x|$.
For $x \lt 0$ then, using (1) $x \le sin(-x) \le -x$ equivalent $x \le -sin(x) \le -x$ equivalent $x \le sin(x) \le -x$ equivalent $-|x| \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of convergence $\frac{n^2+3n+2}{6n^3+5}$
I need to prove that $a_n = \frac{n^2+3n+2}{6n^3+5}$ converges as $n\to\infty$. I think it will converge to $0$, so I want to prove it.
Let $\epsilon >0$. Then
side-work
$$\left|\frac{n^2+3n+2}{6n^3+5}\right|< \left| \frac{n^2+3n+2n}{6n^3+5}\right| < \left| \frac{n^2+5n... | Unless you have some particular reason for an epsilon-delta proof:
$$\frac{n^2+3n+2}{6n^3+5} = \frac{n^2}{n^3} \cdot \frac{1 + 3/n + 2/n^2}{6+5/n^3} \;\;\to\;\; 0 \cdot \frac{1}{6} \;=\; 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
| We have $$(a+1)(b+1)(c+1)=9+a+b+c+ab+bc+ca,$$
since $$abc=8.$$ We get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}=2$$ and $$\frac{ab+bc+ca}{3}\geq \sqrt[3]{(abc)^2}=4$$ Multplying by $3$ and adding up, we get the statement above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Quadratic equation with parameters If $a$, $b$, $c$ are real numbers such that $2a + 3b + 6c = 0$, prove that $ax^2+bx +c=0$ has a solution in the interval $[0, 1]$.
This should use high school maths or a little bit more then that.
| \begin{align*}
0 &= 2a+3b+6c \\
f(x) &= ax^2+bx+c \\
f(0) &= c \\
f\left( \frac{1}{2} \right) &= \frac{a}{4}+\frac{b}{2}+c \\
&= -\frac{a}{12} \\
f(1) &= a+b+c \\
&= \frac{a}{3}-c
\end{align*}
Case I: $ac \ge 0$
$$f(0) f\left( \frac{1}{2} \right) = -\frac{ac}{12} \le 0$$
*
*$\exists x\in \left[ 0, \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Converting a power series to a function
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$
I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?
| Just another way to do it.
$$A=\sum_{n=0}^\infty (n+1)^2x^n =\sum_{n=0}^\infty (n^2+2n+1)x^n $$ Now, write $$n^2+2n+1=(n(n-1)+n)+2n+1=n(n-1)+3n+1$$ $$A=\sum_{n=0}^\infty n(n-1)x^n+3\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n$$ $$A=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+3x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\ta... | Quite long but it's correct to me.
To try with something else, you could:
1. Use Taylor Series
$$\tan(x)\approx x$$
$$\cos(x) \approx 1 - \frac{x^2}{2}$$
2. Use a different approach
For example the exponential representation
$$(\cos(x)-1)^{\tan(x)}
= \text{exp}\Big( \tan(x)\log\left(-2\sin^2\frac{x}{2}\right)\Big)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Contour integration with semicircular arcs. I am trying to compute
$$\int_0^\infty \frac{\ln^2 x}{1+x^2} dx $$
using a complex contour integration of $f(z) = \ln^2 z / (1+z^2)$ around a closed contour with the segments:
$$C_1 : [-R,-\rho]\\C_2: \{z: |z| = \rho, 0 \leq \arg z \leq \pi\}\\ C_3: [\rho , R] \\ C_4: \{z:... | By the residue theorem
$$\oint_{C_1 \cup C_2 \cup C_3 \cup C_4}f(z) \, dz = 2\pi i Res(f,i) = 2\pi i (i \pi^2/8) = -\frac{\pi^3}{4}.$$
It is straightforward to show that the integrals over $C_2$ and $C_4$ vanish in the limit as $R \to \infty$ and $\rho \to 0$.
Hence,
$$\tag{1} -\frac{\pi^3}{4} = \int_0^\infty \frac{\ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt... | The solution set of the second equation is the cubic parabola
$$\gamma:\quad y=x^3-x^2$$
which intersects the symmetry line $y=x$ in the three points
$$\left({1-\sqrt{5}\over2},{1-\sqrt{5}\over2}\right),\quad(0,0),\quad\left({1+\sqrt{5}\over2},{1+\sqrt{5}\over2}\right)\ .\tag{1}$$
Otherwise $\gamma$ lies in the interio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Flipping a coin - expected value We have seven coins, two of them have tails on both sides. We choose (randomly) a coin and we flip a coin untill two tails appear (not necessarly in row). Calculate expected number of throws.
My attempt:
$\mathbf{P}\left( X=0\right) =\mathbf{P}\left( X=1\right) =0 \\
\mathbf{P}\left( X=... | You have $\frac 27$ chance of being guaranteed two tails in two throws. You have $\frac 57$ chance of expecting to take four throws, so the expected number of throws is $2 \cdot \frac 27+ 4\cdot \frac 57=\frac {24}7$ It should definitely be under $4$ as it would be four without the two two-tailed coins. Those will l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\co... | $2(4 \cos^3 \frac π9 - 3 \cos \frac π9)$
As $\cos 3\theta = 4 \cos^3 \theta - 3\cos \theta$
= $2(\cos (3 × \frac π9))$
=$2 × \cos(\frac π3)$
=$2 × \frac 12 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the integral of $\int \frac{x^4}{x-1}$ I was not happy with the responses so I went ahead and solved the integral to get feedback on the answer:
Dividing as the book shows us:
we get $x^4$ divided by $x-1$ which gives us the quotient plus the remainder over the divisor which is: $x^3 + \frac{x^3}{x-1}$
$\int x^3+\... | $$\int\frac{x^4-1+1}{x-1}dx=\int\frac{x^4-1}{x-1}dx+\int\frac{1}{x-1}dx$$
note
$$x^4-1=(x-1)(x^3+x^2+x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,\lambda^{625}$ I have the following question :
$K=\frac{\mathbb{Z}_5[x]}{(f)}$ is a field with 125 elements.
$\lambda=x+(f)\in K$
$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,$ and $\lambda^{625}$ express $\lambda^{625}\i... | We have $\frac{\mathbb Z_5[x]}{(f)}\cong \mathbb Z_5(a)$ then number of the element in $\mathbb Z_5(a)=125=5^3$ then $[\mathbb Z_5(a):\mathbb Z_5]=3$ then $f(x)$ is a polynomial of degree $3$ and donot have roots in $\mathbb Z_5$, then $\beta=0$.
we have $\alpha\in \mathbb Z_5 $,then $\alpha^2=0,1,4 $
if $\alpha^2=0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$ Let $\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand.
| You can write the whole thing as $100a + 10b + c - a^3 - b^3 - c^3$.
Rearranging the terms gives $(100a - a^3) + (10b - b^3) + (c - c^3)$, which you can solve one summand at a time:
You maximize $100a - a^3$ for $a = 6$.
You maximize $10b - b^3$ for $b = 2$.
You maximize $c - c^3$ for $c = 0$ or $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving trigonometrical identity from given identity.
Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$
Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.
| You are almost finished. From where you left off, let us say $$b \sin^2 A =a \cos^2 A \iff \frac{\sin^2 A}{a}=\frac{\cos^2 A}{b}=c$$
For some $c$. Thus $$\sin^2 A+\cos^2 A=1 \iff ac+bc=1 \iff c=\frac{1}{a+b} \tag{1}$$
Now, note $$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\frac{a^4c^4}{a^3}+\frac{b^4c^4}{b^3}=ac^4+bc^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
An expression for $f(2x)$ as a rational function of $f(x)$ A real-valued function $f$ defined on the $\mathbb{R}\setminus\{1\}$ by
\begin{equation*}
f(x) = \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1}
\end{equation*}
is invertible, and
\begin{equation*}
f^{-1}(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3} .
\end{equatio... | Rational functions of the form $x\mapsto \frac{ax+b}{cx+d}$ (with $ad\ne bc$) form a group under composition (Möbius transformations). You are given such an $f$ (and its inverse $f^{-1}$) and want some $g$ such that $g\circ f=f\circ 2$ (where the duplication function $2(x)=\frac{2x+0}{0x+1}$ is a Möbius transformation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to solve this problem using Jensen's inequality? This is the task to find the minimum value of $x^5 + y^5 + z^5 - 5xyz$
where $x,y$ and $z$ are any positive numbers.
I know that I may use this inequality:
$$(t_1\cdot t_2\cdot t_3\cdots t_n)^{\frac{1}{n}} \leq \frac{t_1+t_2+t_3+\cdots +t_n}{n}$$
| By the inequality we have that $\frac{x^5 + y^5 + z^5 + 1 + 1}{5} \ge \sqrt[5]{x^5y^5z^5\cdot1 \cdot 1}$. Hence $x^5 + y^5 + z^5 + 2 \ge 5xyz$
$$x^5 + y^5 + z^5 - 5xyz = x^5 + y^5 + z^5 + 2 - 5xyz - 2 \ge 5xyz - 5xyz - 2 = -2$$
It's obtained for $x=y=z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $A^n$ for all $n \in \mathbb{N}$. I'm learning linear algebra and need help with the following problem:
Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.
My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a patte... | The method descirbed in this answer still works even when $A$ is not diagonalizable. The upside of this method is that you are not required to calculate eigenvectors or inverses of matrices.
The characteristic polynomial is $P_A(X)=(2-X)(X-1)^2$. The Cayley-Hamilton theorem says that $P_A(A)=(2Id-A)(A-Id)^2=0$. Conside... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simultaneous equation that turns into quadratic Solve for $x$ and $y$:
\begin{cases}
y &= 4x^2 - x - 6 \\
y &= 2 - x.
\end{cases}
I have tried rearranging to get $x + y = 2$ and then substituting $y$ into it but hit a dead end. I'm pretty sure it turns into a quadratic equation. Any help would be appreciated thanks.
| You can substitute $y$ into the first equation to get:
$4x^2-x-6=2-x$
$4x^2-8=0$
Plus this into the quadratic formula where $a=4$, $b=0$, and $c=-8$
$x = \dfrac{-(0)\pm \sqrt{(0)^{2} - 4 (4)(-8)}}{2(4)}$
$x=\pm \sqrt {2}$
$y=2\pm \sqrt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
All pair of $m,n$ satisfying $lcm(m,n)=600$
Find the number of pairs of positive integers $(m,n)$, with $m \le n$, such that
the ‘least common multiple’ (LCM) of $m$ and $n$ equals $600$.
My tries:
It's very clear that $n\le600$, always.
Case when $n=600=2^3\cdot 3\cdot 5^2$, and let $m=2^{k_1}\cdot 3^{k_2}\cdot 5^... | As you have very nicely written, you have the factorisation
$$600=2^3\cdot 3\cdot 5^2.$$
Now recall that the lcm of two numbers, given their prime factorisation, is the product of their prime factors to the highest power. For example
$$lcm(2\cdot 3\cdot 5^7,2^2\cdot 5^6)=2^2\cdot3\cdot5^7$$
So you have to find all th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
values of $a$ in inequality If $\sqrt{xy}+\sqrt[3]{xyz}<a(x+4y+4z)$ and $x,y,z>0$ then $a$ is
with the help of am gm inequality
$\displaystyle \sqrt{xy}\leq \left(\frac{x+y}{2}\right)$ and $\displaystyle \sqrt[3]{xyz}\leq \left(\frac{x+y+z}{3}\right)$
so $\displaystyle \sqrt{xy}+\sqrt[3]{xyz}\leq \frac{x+y}{2}+\frac{x... | We'll prove that $$x+4y+4z\geq3\left(\sqrt{xy}+\sqrt[3]{xyz}\right).$$
Indeed, by AM-GM
$$4z+x+4y-3\sqrt{xy}=4z+2\cdot\frac{x+4y-3\sqrt{xy}}{2}\geq$$
$$\geq3\sqrt[3]{4z\left(\frac{x+4y-3\sqrt{xy}}{2}\right)^2}=3\sqrt[3]{z\left(x+4y-3\sqrt{xy}\right)^2}.$$
Thus, it remains to prove that $$\left(x+4y-3\sqrt{xy}\right)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Power series expansion of a holomorphic function One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.
I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$ is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{... | We could also recall the binomial series expansion valid for $|z|<1$
\begin{align*}
\frac{1}{(1-z)^3}&=\sum_{n=0}^\infty\binom{-3}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+2}{2}z^n\\
&=\frac{1}{2}\sum_{n=0}^\infty(n+2)(n+1)z^n \tag{1}
\end{align*}
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I ... | $$sin(x)sin(3x) = sin(2x-x) sin(2x+x) < \frac{1}{4}$$
$$ \text{or, } \frac{cos(x) - cos(2x)}{2} < \frac{1}{4}$$
$$ \text{or, } {cos(x) - cos(2x)} - \frac{1}{2} < 0$$
$$ \text{or, } {cos(x) - 2cos^2(x) + 1} - \frac{1}{2} < 0$$
Let $y = cos(x)$
$$ \text{or, } - 2y^2 + y + \frac{1}{2} < 0$$
Solving,
$$ \frac{1-\sqrt{5}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" s... | Note that
$$n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$$
Multiplying both sides by $\dfrac {n+1}6$
$$\frac {n(n+1)(2n+1)}6+(n+1)^2=\frac {(n+1)(n+2)(2n+3)}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
To prove $\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$ To prove:
$$\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$$
I tried taking lcm but could not get to anything.
| \begin{align*}
\frac {1}{\sin 10^\circ} - \frac {\sqrt 3}{\cos 10^\circ} &= \frac {\cos 10^\circ - \sqrt 3\sin 10^\circ}{\sin 10^\circ \cos 10^\circ} \\
&= 2\left(\frac {\frac 12\cos 10^\circ - \frac {\sqrt 3}{2}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 2\left(\frac {\sin 30^\circ\cos 10^\circ - \cos 30^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Sum of the squares of the sides of a triangle
I have solved this problem, but I had to use a calculator, how do I solve it without using a calculator(wont be allowed to use a calculator in the examination).
Here is my attempt:
| $$a^2+b^2+c^2=4R^2(\sin^2\frac{\pi}{7}+\sin^2\frac{2\pi}{7}+\sin^2\frac{4\pi}{7})=$$
$$=2R^2\left(3-\cos\frac{2\pi}{7}-\cos\frac{4\pi}{7}-\cos\frac{8\pi}{7}\right)=2R^2\left(3+\frac{1}{2}\right)=7R^2$$
Because $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is the product of two consecutive integers $1$ $\pmod n$? Is there a simple test for $n$ to determine if there exists an integer $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. For example, $n$ $=$ $3$ and $n$ $=$ $7$, there are no integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. But for $n$ $=$ $5$ and $n$ $=$ $11$, t... | You are asking when the polynomial $x^2+x-1$ has a root mod $n$. Since $x(x+1)$ is always even, clearly $n$ must be odd for such an $x$ to exist. In that case $2$ is invertible mod $n$, and so by the quadratic formula $x^2+x-1$ has a root mod $n$ iff the discriminant $5$ has a square root mod $n$.
So you are asking f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Proving functions are linearly independent I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:
Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove th... | Your case 2 and case 3 are wrong.
For case 2,
just because
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
does not imply that
the $a_i$ are all zero.
Actually,
for any $x \ne 0$,
and any $a_2, a_3$,
$a_1$ can always be chosen
so that
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
by setting
$a_{1}=-\dfrac{a_{2}sin(x) + a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
| We have that
$$
c=\frac{(a+b)^2}{a^2+b^2}=\frac{a^2+2ab+b^2}{a^2+b^2}=1+\frac{2ab}{a^2+b^2},
$$
so $\dfrac{2ab}{a^2+b^2}$ must be an integer. But by Cauchy inequality $2ab \leq a^2+b^2$ and an integer can be only if $2ab=a^2+b^2$. It implies that that $a=b$ and $c=2$.
From the first equation we have $2 a^2=c^3=8 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $7$ marbles in a box, $3$ are white and numbered from $1$ to $3$ and other $4$ are grey and numbered from $4$ to $7$. Calculate Probabilities Question: You have $7$ marbles in a box, $3$ are white and are numbered from $1$ to $3$ and the other $4$ are grey and are numbered from $4$ to $7$. You extract them one by... | From the first part, there are $3!$ sequences of having chosen all white balls, which are as follows: $$\begin{array}{ccc} 1 & 2 & 3\\ 1 & 3 & 2\\ 2 & 1 & 3\\ 2 & 3 & 1\\ 3 & 1 & 2\\ 3 & 2 & 1\end{array}$$
Out of this, only one sequence is in the order $1 2 3$, so, the probability is $$\frac{1}{6} \times P(\text{First ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to deduce $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$ from $A+B+C=\pi$?
If $A+B+C=\pi$,$$\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2\tag1$$$$\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos \dfrac C2\tag2$$$$\cos A+\cos B+\cos C=4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2+1\tag3... | hint for all of them: $\sin(C) = 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$, and also the little identities such as $\sin\left(\frac{A+B}{2}\right) = \cos\left(\frac{C}{2}\right)$. And the for $3)$, use $\cos(C) = 1- 2\sin^2\left(\frac{C}{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Functional equation $f\left(\frac{x+y}{2}\right)+f\left(\frac{2xy}{x+y}\right)=f(x)+f(y)$ implies $2f(\sqrt{xy})=f(x)+f(y)$ Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation
$$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =... | Solution modified from here
By repeated application of the function property for some positive reals $a,b,c,d$ as you suggested:
$$\begin{align*}f(a)+f(b)+f(c)+f(d)&= f\left({a+b\over2}\right)+f\left({2ab\over a+b}\right)+f\left({c+d\over2}\right)+f\left({2cd\over c+d}\right) \\
&=f\left({{a+b\over2}+{c+d\over2}\over2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to find parameterization if 2 points are given? It is given $a=\begin{pmatrix} y-z \\ 3x \\ 2x \end{pmatrix} $ and points $A=(3,1,-2)$ and $D=(4,2,1)$. Give parameterization $t\in \left[0,1\right]$
$x=?$ Solution: $t+3$
$y=?$ Solution: $t+1$
$z=?$ Solution: $3t-2$
| If you're trying to find a parametrization between $A$ and $D$, we can create ${\bf v} = \vec{D} - \vec{A}= (4,2,1) - (3,1,-2) = (1,1,3)$. So, if we want to make a line in $3D$ passing through $A$ and $D$, we need the vector parallel to the line and an initial point. Let's take point $A$.
Then, $$\begin{align}\vec{r}(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite integration with variable substitution gives wrong result I have the following definite integral:
$$\int^{5}_{-5}{\sqrt{25-x^2} dx}$$
I do a variable substitution:
$$y = 25 - x^2$$
$$x = \sqrt{25 - y}$$
$$dy = -2x~dx$$
$$dx = \frac{dy}{-2x} = -\frac{1}{2\sqrt{25-y}}$$
I get the new integral:
$$-\frac{1}{2}\int... | $\sqrt{25-x^2}$ is an even function.
So $$\boxed{\int_{-5}^5 \sqrt{25-x^2} dx=2\cdot \int_{0}^5 \sqrt{25-x^2} dx}$$
Now proceed the way you have, by substitution with $y$ and so on. You will get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | Hints
*
*For $7$, what are the possible values of $n^3 \pmod 7$? (It suffices to check $-3, \ldots 3$.)
*For $9$, what are the values of $n^3 - 1, n^3, n^3 + 1 \pmod 9$? For example, for $n = 4 \pmod 9$ we have $n^3 - 1 = 0 \pmod 9$, so the given product is also divisible by $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align... | HINT:
$$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$$
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} =... | Quadratic equation $ax^x+bx+c=0$ has two roots, $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$.
We can investigate their behavior when $a \to 0$ by calculating their limits. We assume $b>0$ (we can always mupltiply the equation by -1):
\begin{split}
\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".
Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$
My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\... | $$\begin{align}
\sum_{k=0}^n(-1)^k\binom nk 4^{n-k}
&=4^n\sum_{k=0}^n \binom nk \left(-\frac14\right)^k\\
&=4^n\left(1-\frac 14\right)^n\\
&=4^n\left(\frac 34\right)^n\\
&=3^n\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recurs... | If $k\in[1,n]$ then the difference between $\frac{1}{\sqrt{n^2+k}}$ and $\frac{1}{n}$ is rather small:
$$ 0\leq \frac{1}{n}-\frac{1}{\sqrt{n^2+k}} = \frac{k}{n\sqrt{n^2+k}(n+\sqrt{n^2+k})}\leq \frac{1}{2n^2} $$
hence $\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$ tends to $1$ as $n\to +\infty$, since $\sum_{k=1}^{n}\frac{1}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence m... | Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 5
} |
Calculate :$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}$ Find the:
$$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}:a>0$$
My Try:
$$\frac{1}{a^6}\int_{0}^{\infty}{ }\frac{dx}{1+(\frac{x-\frac{1}{x}}{a})^6}$$
$$\int_0^\infty{dx\over1+u^6}={1\over2}\left({\pi\over2}+{\pi\over6}+0\right){\pi\over3}$$
Is it ... | Let $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}dx$$
substitute $\displaystyle x= \frac{1}{t}$ Then $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(t-\frac{1}{t}\bigg)^6}\cdot \frac{1}{t^2}dt = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}\cdot \frac{1}{x^2}dx$$
So $$2I = \int^{\infty}_{0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Most efficient way of calculating primitive cube roots of unity I understand the definition of a primitive cube root of unity in a finite field $\mathbb{F}_p$ to be all those numbers $x$ such that $x^3=1$ but $x\neq 1$ and $x^2 \neq 1$
When we have a small $p$, say $p=7$, we can compute these through 'brute force' - th... | Perhaps I miss something here, but it all sums up to calculate the roots of $\;x^2+x+1=0\,\pmod p\;$ , and thus we have to know whether $\;-3\;$ is a quadratic residue, and then
$$\binom{-3}p=1\iff p=1\pmod3$$
Thus, for example in $\;\Bbb F_{17}\;$ there are no primitive roots cube roots of unity, but in $\;\Bbb F_{19}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=... | Generally we look for a zero on the right side of the equation. This is standard form and allows everyone to work in the same way, consistently.
If we do not have an equation but just a quadratic polynomial, put it in normal descending order.
$5x^2−10x+7 = 0 \ \ $ is the standard equation form.
$P(x) = 5x^2−10x+7 \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
$\int \sec^3x dx$ in disguise I found this integral,
$$\int \sqrt{x^2+1}dx$$
on a problem list and I think it is a sneaky way of hiding a $\int \sec^3xdx$ problem but I am not sure if what I did was correct though, because of what happens at the end.
So what I did was use trig-substitution and let $u=\tan\theta$ and ... | From your trig sub $x=\tan\theta$,
\begin{align}
\int \sqrt{x^2+1}dx&=\int \sec\theta(\sec^2\theta d\theta)
\end{align}
That is, $\sqrt{x^2+1}=\sec\theta$ and $dx=\sec^2\theta d\theta$. The integrand only changed to $sec^3\theta$ after both of these pieces are considered.
Reassuringly, $\sqrt{x^2+1}\ne\sec^3\theta$, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$
Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$.
I have no idea even how to begin approaching this problem.
| Suppose $p$ is a prime with the property that the only prime factors of $p-1$ and $p+1$ are $2$ and $3$
We assume $p>17$
Then, one of the numbers $p-1$ and $p+1$ contain only the prime factor $2$, hence is a power of $2$.
Case $1$ : $p-1$ is a power of $2$. Then $p=2^n+1$ and $p+1=2^n+2=2\cdot (2^{n-1}+1)$.
Case $2$ :... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determine value of $m$ Consider $m = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ . Now determine $m$ so that expression has solution .
I used Wolfram to find range of $f(x) = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ but it was unable to find it!
| $m = \frac{\sin x}{\cos x} + \frac{\sin x + \cos x}{\sin x - \cos x}$
$m = \frac{\sin x(\sin x - \cos x) +\cos x(\sin x + \cos z)}{\cos x(\sin x - \cos x)}$
$m = \frac{\sin^{2}x + \cos^{2}x}{\sin x \cos x - \cos^{2}x}$
$(\sin x \cos x - \cos^{2}x)m - 1 = 0$
$(\tan x - 1)m - (1+\tan^{2}x) = 0$
$tan^{2}x - m\tan x +m+1 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$
Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.
I am very confused about this problem. I tried to find the upper and l... | This is all about providing an accurate approximation for the involved generalized harmonic sum. I will use a technique (creative telescoping) clearly outlined in the first chapter of these course notes.
We may notice that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n-\frac{1}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares. Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.
| If we can find $y$ such that $y^2-1$ is a sum of two squares, say $y^2- 1 = a^2+b^2$, then take $n = y^2-1$ and we'll have $n = y^2-1 = a^2+b^2$, $n+1 = y^2-1 +1 = y^2 +0$, $n+2 = y^2 + 1$. So, we want to show the existence of infinitely many $y$ such that $y^2-1$ is a sum of two squares.
Look at $3^{2^k}$ for integers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Determine whether $\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$ converges. Determine whether $$\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$$ converges.
Doing some rough work, I realize that this function near $\infty$, behaves like $$\frac{1}{x\sqrt{x^2}} = \frac{1}{x^2}$$
I know this function converges, but I am hav... | Easier and more accurate to evaluate directly. I took a look at that integral and it reminded me of something to do with secant or was that inverse secant? Tried an integral table quick search but failed to find the one needed. Paper and pen a couple of minutes gives:
By trig substitution
Let $x/2 = \sec u;\ \ x = 2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$
Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$
I think I can solve the problem by putting all the variables on one side and then $0$ on the other side, then factoring the side with the variables to become a square. This proves the inequali... | Let $f:[1,\infty) \to \mathbb{R}$ be defined by
$$f(a) = a^2 - a - 1 + \frac{1}{a^a - a + 1}$$
The goal is to show $f(a) \ge 0$ for all $a \in [1,\infty)$.
First suppose $a \ge 2$. Then
\begin{align*}
f(a) &= a^2 - a - 1 + \frac{1}{a^a - a + 1}\\[6pt]
&> a^2 - a - 1\\[6pt]
&\ge 2^2 - 2 - 1 = 1\\[6pt]
&> 0\\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfra... | Ok, so we have that $a+b=1$.
From AM-GM, this immediately tells us that $ab \leq 0.25$. Furthermore, since you said that $\frac{1}{a} + \frac{1}{b} = \frac{1}{ab}$, this shows that $\frac{1}{a} + \frac{1}{b} \geq 4$.
Now, we are given two numbers $a + \frac 1a$ and $b+ \frac 1b$. Their sum is greater than or equal to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+... | That method will work fine, but you can simplify it by noting that $ f(n) = n^2(n+1)^2$ therefore $\,n,$ or $n+1$ is even, so $4$ divides its square, so also $f(n)$ (provable directly or by induction if need be).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is this elementary proof of FLT correct? Consider the classic FLT for $n=3$, $x^3+y^3=z^3 $
Without loss of generality,we can rewrite as $a^3+(b-a)^3=c^3…Equation (1)$
We can also assume there is no common factor between $a,b-a$ and $c …Assumption (1)$
Since $m+n$ is a factor of $m^3+n^3$, $b$ divides $c^3$.
Let $p$ b... | No. What makes you think $b$ and $c$ have to have a common prime factor?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square? Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square?
Context for this question: For pedagogical purposes I was trying to create an example of a cubic polynomial with both its critical values and its zeros all at integers. Aft... | Here is an alternative solution that I came up with shortly after Ahmed S. Attaalla posted his answer.
If $a^2 - a + 1 = b^2$
for some $b \in \mathbb{Q}$ then $a^2 - a + (1-b^2) = 0$, whence by the quadratic formula we have
$$a = \frac{1 \pm \sqrt{1 - 4\left(1-b^2\right)}}{2} = \frac{1 \pm \sqrt{4b^2-3}}{2}$$
If this i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Infinite series of formula How do I compute this infinite series
$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$
I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:
${\frac{(7n+32)(3^n)}{n(n+... | I would do it as follows.
$$ \sum_{n=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}=
7\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n} +
32\sum_{n=0}^{\infty}{\frac{1}{n(n+2)}{\left(\frac34\right)}^n}$$
Now, we have that
$$\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n}=
{\left(\frac34\right)}^{-2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenva... | The $x$ shouldn't be outside the vector. The solution to equations I,II, and III is
\begin{pmatrix}
x\\
-2x\\
z
\end{pmatrix}
where $x$ and $z$ are arbitrary. Every vector of this form is an eigenvector for $A$. You can write each such vector as a linear combination of two vectors $e_1$ and $e_2$ defined by
$$e_1:=
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Is this matrix diagonalisable or not? We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).
We have matrix $A=\begin{pmatrix}
3 & -1 & 0\\
2 & 0 & 0\\
-2 & 2 & -1
\end{pmatr... | From I and II, we have $y = 4x$ and $y = -2x$. The only way these can simultaneously hold is if $x = y = 0$. However, $z$ is indeed a free variable. We find that
$$
E_A(-1) = \{(0,0,z)^T : z \in \Bbb R\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$p^{p+1}+(p+1)^p-1$ a perfect square For which primes $p$ is $$p^{p+1}+(p+1)^p-1$$ a perfect square?
Context:
Once again a modified problem, this time from $p^{p+1}+(p+1)^p$ a perfect square, for which the answer is no such $p$ exist. For the modified form above, however, $p=2,3$ work, the main trouble I've had being... | This is my experience :
$A=P^{p+1}+(p+1)^p -1$
$p^{p+1}=(p^2)^\frac{p+1}{2}=k_1 p^2$
$(p+1)^p=∑ C^p_r p^r +1= k_2 p^2 +1$
$⇒A=P^{p+1}+(p+1)^p -1=(k_1+k_2)p^2 $
Also:
$P^{p+1}=(p+1-1)^{p+1}=∑ C^{p+1}_r (p+1)^r +1=t_1 (p+1)^2+1$
$(p+1)^p = (p+1)^2 (p+1)^{p-2}= t_2(p+1)^2$
$⇒A=P^{p+1}+(p+1)^p -1=(t_1+t_2)(p+1)^2 $
$⇒A=P^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^... | Also we can use C-S twice:
$x^4+y^4=\frac{1}{2}(1+1)(x^4+y^4)\geq\frac{1}{2}(x^2+y^2)^2=\frac{1}{8}((1+1)(x^2+y^2))^2\geq\frac{1}{8}\left((x+y)^2\right)^2=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Noob question about $\int \frac{1}{\sin(x)}dx$ I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$
After replacing $u = \cos(x)$,
$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} ... | Where did I make a mistake?
First, you rather have
$$
\frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left|\frac {u-1} {u+1}\right| + C
$$then observe that
$$
\left|\frac {\cos(x)-1}{\cos(x)+1}\right|=\frac {1-\cos(x)}{1+\cos(x)}
$$ since
$$
1 - \cos(x)\ge 0,\quad 1+ \cos(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to solve this limit? $ \lim_{x \to +\infty} \frac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$ There's a limit which confuses me:
$$ \lim_{x \to +\infty} \dfrac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$$
Is it possible to use L'Hôpital's rule here?
| An asymptotic way:
Let $\displaystyle F:x\mapsto \int_0^x e^{t^2} dt$.
Note that $$\frac{(\int^x_0e^{t^2}dt)^2}{\int_0^xe^{2t^2}dt} = \sqrt 2\frac{(F(x))^2}{F(\sqrt 2 x)}$$
Moreover, $$\begin{align}
F(x)&=\int_0^1 e^{t^2} dt + \int_1^x \frac{1}{2t}\cdot 2te^{t^2}dt \\
&= \int_0^1 e^{t^2} dt + \frac{e^{x^2}}{2x}-\frac e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 ... | We have
$$
\det(B) = 1 \times \det(E) + 1 \times \det(D) = (1 - 4) + (2 - 1) = -2
$$
where $ E $ is
$$
\begin{bmatrix}
1 & 2 \\ 2 & 1
\end{bmatrix}
$$
Then
$$
\det(B^4) = \det(B)^4 = (-2)^4 = 16
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating limits using taylor expansion $$\lim_{x\to 0^{+}} (\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2$$
I have this limit in my sheet and the answer is $\frac 13$.
But I don't know how to approach any step using taylor expansion near zero (this is our lesson by the wa... | As $x \to 0^+$, one has $\ln x<0$, giving
$$
\arctan\left(\ln\left(x+x^2\right)\right)=-\frac \pi2-\arctan\left(\frac1{\ln\left(x+x^2\right)}\right),
$$ and using $\dfrac1{\ln\left(x+x^2\right)} \to 0$ one gets, by applying a standard taylor series expansion,
$$
\arctan\left(\frac1{\ln\left(x+x^2\right)}\right)=\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Menelaus theorem & collinear points From vertex C of the right triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves.
So far I have:
Construct point M on AC such that AM=MC. WE wa... | You have to check that by choosing $M$ as the midpoint of $AC$ then
$$ \frac{AM}{CM}\cdot\frac{CF}{KF}\cdot\frac{KE}{AE} = 1 $$
holds, but $\frac{AM}{CM}=1$ and by the bisector theorem $\frac{KE}{AE}=\frac{CK}{CA}$, so it is enough to prove that
$$ CF\cdot CK = AC\cdot KF $$
or
$$ \frac{CA}{CK} = \frac{FC}{FK} = 1+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that$ k^3n-kn^3$ is divisible by $6 $ for all n∈N. Hello I have problem with solution of task.
Prove that $k^{3}n-kn^3$ is divisible by $6$ for all $n∈N$, $k∈N$ .
Help me, please.
I know, when $n^3-n$ is divisible by 6.
$n^3-n= (n-1)(n)(n+1)$ and is divisible.
I having similar idea $(kn)^3-(kn)= (kn-1)(kn)(kn+1)... | $k^{3}n-kn^3 = kn(k+n)(k-n)$
If this is divisible by both $2$ and $3$, then it is divisible by $6$.
Assuming $k$ is an integer, if either $k$ or $n$ is even, then the expression is even, and if they are both odd, then $k+n$ is even, so the expression is still even.
If either $k$ or $n$ are divisible by $3$, then the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Relations similar to $\sin(\pi/18) + \sin(5\pi/18) = \sin(7\pi/18)$ This I checked numerically (and can be proven analytically easily). I guess there are many similar relations between the numbers
$$\{ \sin\frac{m\pi}{q}, 1\leq m \leq q \} , $$
where $q$ is some integer.
Can anyone give some reference?
| I have studied this problem and found two kinds of solutions.
[Solution 1]
$ \cos\left(\frac{\pi k}{3 n}\right)=\cos \left(\frac{\pi (n-k)}{3 n}\right)+\cos \left(\frac{\pi (k+n)}{3 n}\right) $
Example:
$ \cos\left(\frac{\pi }{15}\right)=\cos\left(\frac{4 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
$ \cos \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}... | Write
$$\frac{x\cos x-\sin x}{x^3}=\frac{\cos x-1}{x^2}+\frac{x-\sin x}{x^3} $$
The first fraction goes to $-\frac{1}{2}$ (it follows from $\frac{\sin x}{x} \to 1$, no de l'Hopital or Taylor needed), while the second goes to $\frac{1}{6}$ (see Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a... | We can see the series $x_{n+1}$ is monotone increasing as $n\rightarrow \infty$ using the Monotone Convergence Theorem:
Monotone Increasing $\Rightarrow x_{n+1}>x_{n}, \forall n$
$\Rightarrow x_n+\frac{1}{x_n}>x_n$
$\Rightarrow \frac{1}{x_n} > 0$
Which is true since $x_1=a$ and $a>0$.
So now we know we have an increasi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Expectation of a random variable Suppose $Y$ is normal random variable.
Show that for any integer $n \ge 1$,
$$
E[(Y- \mu)^{2n}] = \sigma^{2n} \frac{2n!}{2^{n}n!}.
$$
I've mangaged to show that that $f(\mu+y)$ = $f(\mu-y)$ and that therefore $(Y-\mu)$ and $(\mu - Y)$ have the same distribution. I think I have to s... | Assume $Y\sim N(\mu,\sigma)$ is a random variable with a normal distribution. Then for any $n\geq 1$,
$$
E[(Y-\mu)^{2n}] = E\left[\left(\sigma\cdot \dfrac{Y-\mu}{\sigma}\right)^{2n}\right] =\sigma^{2n}E\left[\left( \dfrac{Y-\mu}{\sigma}\right)^{2n}\right].
$$
Let $X=\dfrac{Y-\mu}{\sigma}$. Then $X\sim N(0,1)$ has the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $x^{-1/2}\leq x^{n}y\leq x^{1/2}$ Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}\leq x^{n}y\leq x^{1/2}$.
I have no idea how to do this.
| For $x,y>0$ the required inequality is equivalent to write
$$1\le x^{n+\frac{1}{2}}y\le x.$$
The right hand side:
Let $f(x,y)=x-x^{n+\frac{1}{2}}y$ we want to show that $f(x,y)\ge 0$. So that
\begin{align}
x-x^{n+\frac{1}{2}}y = x \left(1-x^{n-\frac{1}{2}}y\right) \ge 0 \qquad(?)
\end{align}
But since $x>0$ so we need... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Surds question grade 9 I am a student and I need help answering this question.
Simplify:
$\frac{6}{\sqrt{28}}$-$\frac{9}{\sqrt{63}}$
What I did:
$\frac{6}{\sqrt{28}}$- $\frac{9}{\sqrt{63}}$
=$\frac{6}{\sqrt{7×4}}$ - $\frac{9}{\sqrt{7×4}}$
= $\frac{6}{2\sqrt{7}}$ - $\frac{9}{3\sqrt{7}} $
$\frac{6}{2\sqrt{7}}$ × $\frac... | Simply note that
$$\frac{6}{\sqrt{28}}-\frac{9}{\sqrt{63}}=\frac{6}{2\sqrt7}-\frac{9}{3\sqrt7}=\frac{3}{\sqrt7}-\frac{3}{\sqrt7}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
... | The sequence $a_n = 3^{2n-1}+2^{2n-1}$ fulfills the recurrence relation
$$ a_{n+2} = 13 a_{n+1} - 36 a_n $$
and since both $a_1$ and $a_2$ are $\equiv 0\pmod{5}$, by the previous recurrence relation every term of the sequence is $\equiv 0\pmod{5}$. Anyway, it is probably easier to notice that if $d$ is a positive odd i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach:
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underb... | Introduce the Euler substitution:
Let $u=\dfrac{\sqrt{x^2+5x}}{x}$ ,
Then $x=\dfrac{5}{u^2-1}$
$dx=-\dfrac{10u}{(u^2-1)^2}~du$
$\therefore\int\dfrac{x+2}{\sqrt{x^2+5x}+6}~dx$
$=-\int\dfrac{\dfrac{5}{u^2-1}+2}{\dfrac{5u}{u^2-1}+6}\dfrac{10u}{(u^2-1)^2}~du$
$=-\int\dfrac{(2u^2+3)10u}{(6u^2+5u-6)(u^2-1)^2}~du$
$=-\int\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
What is the dimension of the following function space? Construct a function space out of the following:
$v_1 = x^4+x^3+x^2+x+1$
$v_2 = x^4+x^2+1$
$v_3 = x^3+x$
$v_4 = x+1$
Where the space S is defined as linear product of the $v$'s:
$S=lin(v_1,v_2,v_3,v_4)$
What is the dimension of S? Since $v_1$ is the linear product... | Pass from polynomials to coordinate vectors:
$$v_1\to(1,1,1,1,0)\;,\;\;v_2\to(1,0,1,0,1)\;,\;\;v_3\to(0,1,0,1,0)\;,\;\;v_4\to(0,0,0,1,1)$$
Now form the corresponding matrix with the above as rows, and reduce the matrix (Gauss elementary operations):
$$\begin{pmatrix}
1&1&1&1&0\\
1&0&1&0&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ I wish to find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ as part of a larger problem. I can see that the solution will involve De Moivre's theorem somehow, but I cannot see how to ap... | Misha beat me on time, so I will give a slightly different point of view. It is well-known that $\cos\frac{2\pi}{7}$, $\cos\frac{4\pi}{7}$ and $\cos\frac{6\pi}{7}$ are algebraic conjugates, roots of the Chebyshev polynomial
$$ p(x)=8x^3+4x^2-4x-1.$$
On the other hand
$$ \tan^6\frac{\pi}{7}=\left(\frac{1}{\cos^2\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine the acceleration given the position function of a particle
The position function of a particle in a test laboratory is $s(t) = \frac{10t}
{t^2+3}$.
Determine the acceleration of the particle of particle after 4 seconds.
The speed $v(t)$ of the particle is the derivative of the position $\frac{ds(t)}{dt}$.... | $$s(t) = \frac{10t}{t^2+3}$$
$$\implies v(t) = \frac{ds}{dt} = \frac{(t^2+3)\cdot 10 - 10t\cdot 2t}{(t^2+3)^2} $$
$$v(t) = \frac{10t^2+30-20t^2}{(t^2+3)^2} = \frac{30-10t^2}{(t^2+3)^2}$$
$$\implies a(t) = \frac{dv}{dt} = \frac{(t^2+3)^2\cdot -20t -(30-10t^2)\cdot 2(t^2+3)\cdot 2t}{(t^2+3)^4}$$
$$a(t) = \frac{-20t(t^2+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How t... |
Here's a way to prove it without using trigonometry. For sake of discussion, in the pictured triangle assume $∠B$ is lower left corner and side $a$ is the bottom (side $c$ is left and $b$ is right). Now look at the three cases where $∠C$ is $<90°$, $=90°$, and $>90°$.
The easy case is $∠C =90°$. In this case $c=2a$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Using 3 equations, find $2a-b+c$. From the problem, I extract the following 3 equations.
*
*$a+b+c=10$
*$ab+bc+ca=31$
*$abc=30$
The question is to find $(2a-b+c)$.
Using the equation $(a+b+c)^2 = a^2 + b^2 + c^2 +2 (ab+bc+ca)$, I found $a^2+b^2+c^2=38$.
Using the equation $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2... | the first equation gives:
$$a+b=10-c$$
the second one:
$$c(a+b)=31-ab$$ with the first
$$c(10-c)=31-ab$$ and with $$c=\frac{30}{ab}$$ we obtain
$$c(10-c)=31-\frac{30}{c}$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find det(A) for matrix, find C and adj(A) I am asked to find the determinant of matrix A by using cofactors method. I understand how to do that portion, but then it asks to also find matrix C, and adj(A). How can I go about that?
for det(A) I got -20.
$ A = \begin{bmatrix}
1 & 3 & -3 \\
-3 & -3 & 2 \\
... | $A = \begin{bmatrix}
1 & 3 & -3 \\
-3 & -3 & 2 \\
-4 & 4 & -6 \\
\end{bmatrix}$
$\det A = 4$.
We know $A^{-1}=\frac{1}{\det A} adj (A)$, where $adj$ is adjoint of matrix $A$.
$adj(A)= \left( \begin{array}{ccc} -26 & -30 & -3 \\ 10 & -6 & 7 \\ -24 & -16 & 6 \end{array} \right)$.
$\therefo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \l... | HINT:
$$\dfrac1{n+1}>\dfrac1{n+r}$$ for $2\le r\le n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Roundest ellipse with specified tangents Suppose you are given two points $\mathbf p_1$ and $\mathbf p_2$ in the plane with associated vectors $\mathbf v_1$ and $\mathbf v_2$. You want to find an ellipse passing through $\mathbf p_1$ tangent to $\mathbf v_1$ and through $\mathbf p_2$ tangent to $\mathbf v_2$. There are... | At the moment I can offer an analytical solution, which is quite cumbersome but manageable. I hope someone will be able to give a simpler solution.
To reduce the number of parameters, we can set up a coordinate system such that $\mathbf{p}_1=(\alpha,0)$, $\mathbf{p}_2=(0,\beta)$, $\mathbf{v}_1=(0,1)$, and define then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
Let $n$ be an even perfect number greater than $ 6$. Prove that $n \equiv 4 \pmod 6$. Things I've gotten so far:
Proven that $2^p-1\equiv \pmod 3$ (for some odd prime)
$n$ is in the form $n =( 2^p-1)(2^{p - 1})$ where $(2^p - 1)$ is prime.$ n > 6$, so $p > 2$.
So I have the congruence $2^p-1 \equiv 1 \pmod 3$,
and I ... | Observe that $$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+\cdots+x^a+1).$$ Hence a number of this form could be prime if and only if $x=2$ and $a=1$ and $b$ is a prime. Here it has shown that an even perfect number is of the form $2^{p-1}(2^p-1),$ where $2^p-1$ is a prime. Hence we only need to consider about prime powers!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $\displaystyle\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$ using squeeze theorem
Find $$\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$$ using squeeze theorem
Tried using binomial expansion, but have no idea on how to continue.
| Your idea of a use of the binomial theorem was good, writing$$A=x - \sqrt{x+1}\times\sqrt{x+2}=x-\sqrt x \times\sqrt{1+\frac{1}{x}}\times\sqrt x \times\sqrt{1+\frac{2}{x}}$$ that is to say $$A=x\left(1-\sqrt{1+\frac{1}{x}}\times \sqrt{1+\frac{2}{x}}\right)$$ Now consider, using the binomial theorem with $t=\frac 1x$$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Identify $\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$
Identify $$\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$$
My Try :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$
Now : $$\frac{1}{x^3}+\frac{1}... | Write
\begin{align}
\sqrt{x^4+x+1}-\sqrt{x^4+x+5}
=\frac{(\sqrt{x^4+x+1}-\sqrt{x^4+x+5})(\sqrt{x^4+x+1}+\sqrt{x^4+x+5})}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}
\end{align}
and use
\begin{align}
\frac{1}{2\sqrt{x^4+x+5}}\leq\frac{1}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\leq \frac{1}{2\sqrt{x^4+x+1}}
\end{align}
and
\begin{align}
\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
| Let $\tan\frac{x}{2}=t$.
Hence, we need to solve
$$\frac{2}{\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}}=1+\frac{6t}{1-t^2}$$ or
$$t^4+t^3-4t^2-t+1=0$$ or
$$t^4-t^3-t^2+2t^3-2t^2-2t-t^2-t+1=0$$ or
$$(t^2-t-1)(t^2+2t-1)=0,$$
which gives
*
*$\tan\frac{x}{2}=\sqrt2-1$ or $\frac{x}{2}=22.5^{\circ}+180^{\circ}k$, where $k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_{0}^1 (a+bx+cx^2)^2dx$. Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.
What I did is
$$
\int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \ri... | Just substitute $a=1-b-c$ and minimize the function of two variables
$$g(b,c)=f(1-b-c,b,c)=\int_0^1(1-b-c+bx+cx^2)^2 dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding solution of trigonometric equation How to find the solutions of this trigonometric equation $$\sum_{m=1}^6 cosec(\theta + \frac{(m-1)\pi}{4}) cosec(\theta + \frac{m\pi}{4}) =4\sqrt{2}$$ if $0<\theta<\pi/2$.
| HINT:
$\csc\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\csc\left(\theta+\dfrac{m\pi}4\right)=\dfrac1{\sin\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\sin\left(\theta+\dfrac{m\pi}4\right)}$
$\dfrac{\sin(A-B)}{\sin A\sin B}=\dfrac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}=\cot B-\cot A$
Here $A=\theta+\dfrac{m\pi}4$ and $B=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a+b+c=ab+ac+bc$ then $\sum\limits_{cyc}\frac{a}{a^2+2b}\leq1$.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=ab+ac+bc$. Prove that:
$$\frac{a}{a^2+2b}+\frac{b}{b^2+2c}+\frac{c}{c^2+2a}\leq1$$
We can try the following.
Replace $a\rightarrow\frac{1}{a}$, $b\rightarrow\frac{1}{b}$ and $c\rightarrow\fra... | The BW helps.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that
$$\sum_{cyc}\frac{a}{a^2+\frac{2b(ab+ac+bc)}{a+b+c}}\leq\frac{a+b+c}{ab+ac+bc}$$ or
$$\sum_{cyc}\frac{a}{a^2(ab+ac+bc)+2b(a+b+c)}\leq\frac{1}{ab+ac+bc}$$ or
$$63(u^2-uv+v^2)a^7+(149u^3-11u^2v+105uv^2+149v^3)a^6+$$
$$+(145u^4-59u^3v-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.