Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
let $f(x)=(3(x+x^2))/14$ and $x$ between $0$ and $2$ , zero otherwise be the pdf for a random variable $X$ ,Find the median and the mode? let f(x)=(3(x+x^2))/14 and x between 0 and 2 ,
zero otherwise be the pdf for a random variable X
Find the median and the mode
`
Could you please help me Is it correct or not?
| If $f(x)$ represents a legal probability density function (which you should check based on the support of the distribution), then for the median we find an $m$ such that
$$\int_0^m f(x) dx = \frac{1}{2}$$
$$\iff \int_0^m (3(x+x^2))/14 dx = \frac{1}{2}$$
$$\iff \frac{3}{14}\int_0^m x+x^2 dx = \frac{1}{2}$$
$$\iff \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,... | PARTIAL SOLUTION
$$f(x,y,z)=\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}$$
$$g(x,y,z)= x^2+y^2+z^2+xyz-4=0$$
$$ \begin{align}\mathcal{L}(x,y,z,\lambda) =& \;f(x,y,z) + \lambda g(x,y,z)\\
=&\;\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}+\lambda(x^2+y^2+z^2+xyz-4) \end{align}$$
$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Inverse tangents in a cyclic order
If
$$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)$$then find $\tan\theta$
I tried to use these as sides of a triangle and use their properties, but other than that I am clueless. I cannot thin... |
If
$$\theta= \tan^{-1}\left(\color{red}{\sqrt{\frac{a(a+b+c)}{bc}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{b(a+b+c)}{ac}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{c(a+b+c)}{ab}}}\right)$$then $$\tan\theta=\color{red}{0}$$
Proof:
$$\tan^{-1}\left( \sqrt {\frac {x (x+y+z)}{yz}} \right)+\tan^{-1}\left( \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$ over the unit circle? How can I integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$?
$(z^2 + 1/2)^2$ isn't in a form where I can use the Cauchy Integral Formula, and we haven't covered using residues yet in my class so I'm assuming that's not the way to go (it's also fairly ugly t... | You can apply Cauchy's integral theorem by expanding the factor $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ in partial fractions. You can obtain the partial fraction expansion without having to solve any equations as follows. The singularities are at $z = \alpha^{\pm} = \pm \frac{i}{2}\sqrt{2}$. The partial fraction e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$ Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then
*
*$L < 1$
*$L > 1$
*$L < \frac{\pi}{4}$
*$L > \frac{\pi}{4}$
I got some idea from this video link. But got stuck while evaluating the second integral.
Please help!!
Thanks in advance!
| There is a "closed-form" expression for this integral.
Using partial fractions,
$$ \dfrac{1}{1+x^8} = - \dfrac{1}{8} \sum_{\omega} \dfrac{\omega}{x - \omega}$$
where the sum is over the roots of the polynomial $x^8 + 1$. Then integrate:
$$ \int_0^1 \dfrac{dx}{1+x^8} = -\dfrac{1}{8} \sum_\omega \omega (\log(1-\omega)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
| Just to be a bit different, note that a binomial expansion tells us
$$7^7=(1+6)^7=1+7\cdot6+(\text{stuff})\cdot6^2\equiv43\equiv-2\mod 9$$
since $6^2\equiv0$ mod $9$, and thus, since $13\equiv4$, $19\equiv1$, and $23\equiv-4$ mod $9$, we have
$$1^7+7^7+13^7+19^7+23^7\equiv1^7-2+4^7+1^7+(-4)^7=1-2+4^7+1-4^7=0\mod 9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove $(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$ Let $a,b,c,d>0$ and such $a+b+c+d=1$, show that
$$(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$$
since
$$a+b\ge 2\sqrt{ab}$$
I think this will not hold.because we have $256abcd\le 1$
| We need to prove that $\prod\limits_{cyc}(a+b)^3\geq16(a+b+c+d)^4a^2b^2c^2d^2$.
Let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Since $a$, $b$, $c$ and $d$ are positive roots of the equation
$x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$, we see that (by the Rolle's theorem) the equations
$v^2x^2-2w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Method to find smallest value of $x$ for which $x^2-x+C$ is composite. Problem statement: Given the function $f(x)=x^2-x+C$, where $x$ is a positive integer $>1$ and $C$ is a positive integer ($C=0$ is also allowed), find some method and/or set of rules to always find the smallest value of $x$ that will give a non-prim... | If $C$ is even then $f(2)$ is not prime, and if $C=1$ then $f(4)$ is not prime. So suppose $C>2$.
Let $p$ be the smallest prime dividing $C$. Then $p$ also divides $f(p)$, so $f(p)$ is not prime unless $f(p)=p$. But this is the case if and only if $C=p(2-p)$, which is impossible as $C$ is positive. Hence $2\leq x\leq p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$ Integrate
$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$
Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$
$x=1\rightarrow u={\pi\over 4}$
$x=0\rightarrow u=0$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-... | Let $y = x^2$ and $z = \sqrt{y^{-1} - y}$, we have
$$\begin{align}
\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx
= & \int_0^1 \frac{x\sqrt{x^{-2} - x^2}}{x^2(x^{-2}+x^2)}dx
= \int_0^1 \frac{\sqrt{x^{-2} - x^2}}{(x^{-2}+x^2)}\frac{dx}{x}\\
= & \frac12 \int_0^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{dy}{y}
= -\frac12 \int_{y=0}^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Finding the degree of $\sqrt[3]{2} + \sqrt[3]{3}$ over $\mathbb Q$ I am practicing writing down random algebraic numbers and finding their degrees over $\mathbb Q$ and have fumbled when coming to $\sqrt[3]{2} + \sqrt[3]{3}$. Mathematica tells me the minimal polynomial is $x^9-27 x^6-135 x^3-729$, which is a polynomial ... | First, Mathematica is wrong. The minimal polynomial is $-125-87X^3-15X^6+X^9$.
The irreducible equation for $\sqrt[3]2$ over $\Bbb Q$ is $X^3-2$, and the irreducible equation for $\sqrt[3]2+\sqrt[3]3$ over $\Bbb Q(\sqrt[3]3\,)$would apparently be $F(X)=(X-\sqrt[3]3\,)^3-2=X^3-3\sqrt[3]3X^2+3\sqrt[3]9X-5$. Now, to get a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration methods: euler's substitution I've looked it up on the internet but I'm having trouble as to how to proceed using Euler's substitution.
For example, how does one solve the following integrals using Euler's substitution:
*
*$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$
and
*
*$\int_0^{\infty} \dfrac{xdx}{\sqr... | For the first problem:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$$
$$x+\sqrt{x^2+1} = t \implies \sqrt{x^2+1} = -x+t \implies x= \dfrac{t^2-1}{2t}$$
Calculate the derivative $dx$:
$$dx=\dfrac{t^2+1}{2t^2}dt$$
Substitution of $dx$ and $\int_1^2\frac{dx}{x+\sqrt{x^2+1}}$:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}} = \int_1^2 \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x... | When computing $1+x^2+(x^2)^2+(x^2)^3+(x^2)^4$, the series is in powers of $x^2$ not $x$. So the proper expression is $$1+x^2+(x^2)^2+(x^2)^3+(x^2)^4=\frac{1-(x^2)^5}{1-(x^2)}=\frac{1-x^{10}}{1-x^2}.$$ By contrast, $$\frac{1-x^{9}}{1-x}=1+x^1+x^2+\cdots +x^8$$ which differs from the above series in it has both odd and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all the primitive roots of $13$
Find all the primitive roots of $13$
My attempt:
Since that $13$ is a prime I need to look for $g$ such that $g^{13-1}\equiv 1\pmod{13}$
There are $\phi(12)=4$ classes modulo $12$
how can I find the classes?
| My quick & dirty method was to note that $13-1=12$ has prime factors $2,3$ & quickly find squares and cubes of a numbers, since those cannot be primitive roots, and not checking those which have already turned up. Any indication of a "short cycle" (like $5$) also gets discarded immediately.
$2: 2^2 \equiv 4, 2^3 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Write the index table for the primitive root $3$ of $25$
Write the index table for the primitive root $3$ of $25$
My attempt:
$$
\begin{array}{c|lcr}
k & 0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\
\hline
3^k & 1&3&9&2&6&18&4&12&11&8&24&22&16&23&19&7&21&13&14&17 \\
\end{array}
$$
so the index table will be
$... | By saying $3$ is a primitive root of $25$ it only means that the positive integers less than 25 and relatively prime to it are all of the form $3^k$ for an appropriate exponent $k$ so your table is essentially right, only just exclude the $5,10,15,20$ because they are not coprime with 25.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculation of area union of 3 overlapping circles Please help calculate the total un-overlapped area of union inside a boundary consisting of perimeters of 3 different circles as shown. The two larger circles are 4.25 mm in diameter and the center circle is 3.28 mm.
All the circles are centered on x-axis, the distance... | Let's start by finding the combined area of the two larger circles. The area of one of the larger circles, which I will call $A_O$, is:
$$
\begin{align*}
A_O &= \frac14\pi d^2\\
&= \frac14\pi(4.25)^2\\
&\approx14.186
\end{align*}
$$
Thus the area of both circles, ignoring their intersection for a moment, is $2A_O$. To ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the equation of a parabola when only given the x-intercepts and the Axis of symmetry? i have been given the following problem.
A tennis ball is lobbed from ground level and must cover a horizontal
distance of 22m if it is to land just inside the opposite end of the
court. If the opponent is standing ... | The general equation of a parabola is
$$
f(x) = a x^2 + bx + c
$$
it has three parameters.
We know $f(11) = 0$, so we have
$$
f(11) = 0
$$
we also know
$$
3 = f(11-4) = f(7)
$$
further we lobbed from ground level, such that we are 22m away, so this means
$$
0 = f(-11) = a 11^2 - 11 b + c
$$
This gives the linear system... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Equation involving the Jacobi symbol: $\left( \frac {-6} p \right) = 1$? I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.
I have that:
$$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$
and I know that $\left( \frac 2 p \right) ... | We have $(-6/p)=1$ if (i) $(2/p)=1$ and $(3/p)=1$ or (ii) $(2/p)=-1$ and $(3/p)=-1$.
You seem to have taken care of (i). We get the solutions $p\equiv 1\pmod{24}$ and $p\equiv 7\pmod{24}$.
For (ii) we want a) $p\equiv 3\pmod{8}$ and $p\equiv 2\pmod{3}$ or b) $p\equiv 5\pmod{8}$ and $p\equiv 2\pmod{3}$.
a) has the solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many ways can all six numbers in the set $\{4, 3, 2, 12, 1, 6\}$ be ordered Is there an easy way to solve the problem?
How many ways can all six numbers in the set $S = \{4, 3, 2, 12, 1, 6\}$ be ordered so
that $a$ comes before $b$ whenever $a$ is a divisor of $b$?
By analyzing each number in $S$, I get the ans... | I would solve this problem by looking at the divisor lattice for $12$ and then doing casework. Here is the divisor lattice:
You're asking us to put any number that is below another on a tree before in the list. Therefore, $1$ is the first number since it's at the bottom. The next number is either $2$ or $3$, since tho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding certain coefficients in Taylor expansion of $ \log(1 +qx^2 + rx^3)$ This exercise is part of the STEP $3$ paper from $2014$.
At a certain point in the problem, we 're supposed find $a_n$ for $n = {2,5,7,9}$
where $a_n$ is the coefficient of $x^n$ in the series expansion around $0$ of $ \log(1+qx^2 + rx^3)$.
I ... | I think the suggested solution is straightforward enough not to be dismissed out-of-hand.
We know
$$
\log(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} \pm\dotsm
$$
Let $u = qx^2 + rx^3$. Then
$$
\log(1+qx^2 + rx^3) = (qx^2 + rx^3) - \frac{1}{2}(qx^2 + rx^3)^2 + \frac{1}{3}(qx^2 + rx^3)^3 + \dotsm
$$
Only t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type... | $$(xy-7)^2=x^2+y^2 $$
$$x^2y^2-14xy+49 = x^2+y^2 $$
$$x^2y^2-12xy+49 = x^2+2xy+y^2 $$
$$(xy-6)^2+13=(x+y)^2 $$
$$13 = (x+y)^2-(xy-6)^2 $$
$$13 = (x+y+xy-6)(x+y-xy+6) $$
The factors on the right must be factors of the prime $13$.
We conclude $x+y=7$ and easily enumerate all solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after ge... | You can assume $x>0$ and substitute $x=1/t^2$, with $t>0$, so the limit becomes
$$
\lim_{t\to0^+}\frac{a\sqrt{1+t^2}+b\sqrt{1+2t^2}+c\sqrt{1+3t^2}}{t}
$$
The limit is zero if $a+b+c=0$ and infinite otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Simplify $\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$ I am given this expression to simplify:
$\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$
and I know the correct answer is $\sin^2(x)$
I was able to reduce the second fraction to a bit nicer
$\frac{\sin(2x)}{2}=\frac{2\sin(x)\cos(x)}{2\tan(x)\cot(x)}=\frac{2\sin(x)... | $$\frac { \cos 2x }{ \cot x-1 } -\frac { \sin 2x }{ 2 } =\frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } }{ \frac { \cos { x } -\sin { x } }{ \sin { x } } } -\sin { x } \cos { x } =$$
$$=\frac { \sin { x } \left( \cos { x } +\sin { x } \right) \left( \cos { x } -\sin { x } \right) }{ \cos { x } -\sin { x } } -\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convex integral inequality I cannot prove that if $f(x)$ is convex on $[a,b]$ then
$f\Big(\frac{a+b}2\Big) \le \frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 .$
| For convenience, we can take $b=1,a=-1$.
For the first inequality, use
$$f(0) \le \dfrac{f(x) + f(-x)}{2}$$
and integrate both sides from $x=-1$ to $1$.
For the second, use
$$ f(x) = f\left( \dfrac{1-x}{2} (-1) + \dfrac{1+x}{2}(1)\right) \le
\dfrac{1-x}{2} f(-1) + \dfrac{1+x}{2} f(1) $$
and integrate both sides from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | Note that $$\frac{1}{x^{4}+x^{2}+1}=\frac{1}{2}\left(\frac{x+1}{x^{2}+x+1}-\frac{x-1}{x^{2}-x+1}\right)
$$ and $$\begin{align}
\int\frac{x+1}{x^{2}+x+1}dx= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{x^{2}+x+1}dx \\
= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 0
} |
The solution of equation $4+6+8+10+\cdots +x =270$ is 15. The solution of equation
$4+6+8+10+\cdots +x =270$
is $x=15$.
How can I prove it?
I ve tried the geometric sequence but I cannot figure out the pattern.
| Since $$1+2+3+4+.....+(n-1)+n = \frac{n(n+1)}{2}$$
$$2+4+6+8+.....+2(n-1)+2n = n(n+1)$$
That is $$4+6+8+.....+2(n-1)+2n = n(n+1)-2$$
Thus $$n(n+1)-2 \leq 270$$
$$n^2+n-272 \leq 0$$
$$(n+17)(n-16) \leq 0$$
So $$n=16$$
Thus $$4+6+8+10+12+14+16+18+20+22+24+26+28+30+32=270$$
I think the question should be $$(4+6+8+...)\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
With the linear approx. of $f(x)= sin(x)$ around $0$ Calculate $\lim_{\theta\to 0} \frac{sin\theta}{\theta}$ With the linear approximation of $f(x)= sin(x)$ around $0$, calculate:
$$ \lim_{\theta\to 0} \frac{\sin\theta}{\theta}$$
Figured I have to use L'Hospital's Rule, but I think I don't get how to calculate the deri... | Two ways to go about doing this:
Taylor Series
We can use the Taylor expansion of $\sin(x)$
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$
$$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^3}{5!}-\frac{x^6}{7!}+\ldots$$
So $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$
Squeeze Theorem
The area of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles.
In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality:
$$(a-c)(a+c)^2+bc(a+c)=ab^2$$
Find the value in degrees the other two angles of the triangle.
... | $$(a-c)(a+c)^2+bc(a+c)=ab^2$$
$$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$
$$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$
$$\implies (a-b+c)(a(a+b)-c^2)=0$$
Dividing the both sides by $a-b+c\gt 0$ gives
$$c^2=a^2+ab$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| Using the approximation $\sqrt3\approx1.7$, one can do a bit of mental arithmetic and see that
$${3\sqrt3-4\over7-2\sqrt3}\approx{1.1\over3.6}\lt1\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}\approx{-2.9\over-2.4}={2.9\over2.4}\gt1$$
This only works, of course, because the two numbers are not at all close. It's al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \... | Integration by parts yields
\begin{equation}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx=x\arctan\sqrt{x+2}\ \bigg|_{-1}^{1}-\frac12\int_{-1}^{1}\frac{x}{(x-1)\sqrt{x+2}}\ dx
\end{equation}
Making substitution $u^2=x+2$, then
\begin{align}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx&=\frac{\pi}{3}+\frac{\pi}{4}-\int_{1}^{\sqrt3}\frac{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
A System of Infinite Linear Equations Suppose that $\{a_{i}\}_{i=-\infty}^{\infty}$ with $\sum_{i=-\infty}^\infty a_{i} \lt \infty$ is known and that $\{b_i\}_{i=-\infty}^{\infty}$ is such that
$$\sum_{i=-\infty}^\infty a_{i}b_{-i} =1,$$
and that, for all $k \in \mathbb Z/\{0\}$,
$$\sum_{i=-\infty}^\infty a_{i}b_{-i+k}... | Note: This did not help, but was fun:
Let us try this for a finite sum:
$$
1 =
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
b_1 \\
b_0 \\
b_{-1}
\end{pmatrix}
=
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
b_{-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| $$\begin{align}
& {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\ge 1\,\,\,\,\quad,\,\,\,\,\,\,\,\,\,\frac{x}{y}>0\,\,\,\,\,\,\Rightarrow \,\,\,\,\theta =k\pi +\frac{\pi }{2}\, \\
& {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\le -1\,\,\,\,\,,\,\,\,\frac{x}{y}<0\,\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $2^x+4^x=2$ This is the equation, but the result is different from wolframalpha:
$$2^x+4^x=2$$
$$2^x+2^{2x}=2^1$$
$$x+2x=1$$
$$x=\frac{1}{3}$$
WolframAlpha: $x=0$
Where is the error?
| $2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$ $\Rightarrow $ $1 + 2^x = 2 ^{1 - x}$ $\Rightarrow$ $1 + 2^x = 2 ^{- x} \times 2$
Now Set $ y = 2^x$; then we have
$1 + y = y^{-1} \times 2 $ $\Rightarrow$ $y^2 + y -2 = 0$. Which has solutions $y = 1$ and $y = -2$.
$y = -2 $ is unacceptable, because $y = 2^x$ is a po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}$ The question arise in connection with this problem
Prove that
$$\lim_{n\rightarrow \infty}\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+... | We have
\begin{align}
&\sqrt{\frac{1}{n}} - \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} - \cdots + \sqrt{\frac{4n-3}{n}} - \sqrt{\frac{4n-2}{n}} + \sqrt{\frac{4n-1}{n}} \\
=& \sqrt{\frac{1}{n}} +\sum_{k=1}^{2n-1}(\sqrt{\frac{2k+1}{n}}-\sqrt{\frac{2k}{n}}) \\
=& \sqrt{\frac{1}{n}} + \frac{1}{\sqrt{n}}\sum_{k=1}^{2n-1}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)
i need to show, in a simple way, that:
$\displaystyle \int_0^... | I finally get a solution (i swear i didn't know it when i have posted the question)
Define for $x\in [0,1]$ the function $F$:
$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$
Notice that $F(1)=-\dfrac{\pi^2}{12}$
(use Taylor's development)
and, after performing the change of variable $y=\dfrac{t}{x}$,
$\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt... | The suggestion about $u\ge\sqrt{u}$ helps in removing one level of square roots. Now you have
$$
x+\sqrt{x^2-10x+9}\ge1
$$
or
$$
\sqrt{x^2-10x+9}\ge1-x
$$
If you notice that the expression under the square root is $(x-1)(x-9)$, you can possibly use the clever trick shown in another answer. On the other hand, such an in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How prove this inequality: $\frac1{1-a}+\frac1{1-b}+\frac1{1-c}\ge \frac1{ab+bc+ac}+\frac1{2(a^2+b^2+c^2)}$ for $a+b+c=1$? Let $a,b,c>0$ and such $a+b+c=1$ show that
$$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}\ge \dfrac{1}{ab+bc+ac}+\dfrac{1}{2(a^2+b^2+c^2)}$$
Let $p=a+b+c=1,ab+bc+ac=q,abc=r$
$$\Longleftrightarrow ... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $\frac{9u^2+3v^2}{9uv^2-w^3}\geq3u\left(\frac{1}{3v^2}+\frac{1}{18u^2-12v^2}\right)$,
which is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for an extremal value of $w^3$.
$a$, $b$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What's the mistake on my answer for this inequality $ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $ Good evening to everyone! I have the following inequality: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $$. I don't know what's wrong with my answer: $$ \... | The first part is correct. The inequality holds iff $$(x+1)\sqrt{x^2+4}>(x+2)\sqrt{x^2+1}\ \ (*)$$ Hence (squaring), provided $x+1>0$ (and hence $x+2>0$) we have $(x+1)^2(x^2+4)>(x+2)^2(x^2+1)$ which is equivalent to (expanding etc) $$2x(2-x^2)>0$$ which implies $0<x<\sqrt2$.
On the other hand if $-2<x<-1$ we have $x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this nice rational equation I am trying solve this equation
$$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried
We rewrite the given equation in the form
$$\dfrac{... | The equation of tangent line of the grap of the funtion $ y = \dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}$ at the point $x=-1$ is $y=\dfrac{x}{2}+\dfrac{5}{2}$ and the equation of tangent line of the grap of the funtion $ y = \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}$ at the point $x=-1$ is $y=\dfrac{5}{2}-\dfrac{x}{2}$. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Questions regarding dot product (possible textbook mistake) I am given the following exercise:
Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and
only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and
point to t... | Both of your questions can be answered using the fact that
$$
\lVert a\rVert^2 = a\cdot a
$$
So,
$$
\begin{align}
\lVert a + b\rVert^2 &= (a + b)\cdot (a + b) = \lVert a\rVert^2 + \lvert b\rVert^2 + 2a\cdot b.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the following Harmonic Number Identity true?
Is the following identity true?
$$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right] + \operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(... | The sum in question has a closed form in terms of polylogarithms. The proof is complicated, and I don't intend to reproduce it as I derived it some 15 years ago, and polylogs are not a primary interest now. You can always differentiate both sides and use polylog IDs in Lewin.
$$\sum_{k=1}^\infty \frac{y^k}{k^3}H_k=\z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| $$(x^2 + a y^2) (x^2 + b y^2) = x^4-11x^2y^2+y^4$$
Hence,
$$a + b = -11 \qquad \qquad \qquad a b = 1$$
In SymPy,
>>> a, b = symbols('a b')
>>> solve_poly_system([a + b + 11, a*b - 1], a, b)
____ ____ ____ ____
11 3*\/ 13 11 3*\/ 13 11 3*\/ 13 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{... | We have $$\int_{1/e}^{\tan\left(x\right)}\frac{t}{1+t^{2}}dt=\frac{1}{2}\int_{1/e}^{\tan\left(x\right)}\frac{2t}{1+t^{2}}dt
$$ $$=\frac{1}{2}\log\left(1+\tan^{2}\left(x\right)\right)-\frac{1}{2}\log\left(1+\frac{1}{e^{2}}\right)
$$ and $$\int_{1/e}^{\cot\left(x\right)}\frac{1}{t\left(1+t^{2}\right)}dt=\int_{1/e}^{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.
| ${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$
*
*${n^5-5n^3+4n = (n-2)(n-1)(n)(n+1)(n+2)= 5!\binom{n+2}{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $... | As a (possibly) different approach: start by remarking that $$3^3=4\times 7-1$$
It follows that $$3^{33}=(4\times 7-1)^{11}=7N-1$$
Where $N$ is a simple expression in powers of $4$ and $7$ (with some binomial coefficients thrown in for good measure). Specifically $$N=\sum_{i=1}^{11}\binom {11}i (-1)^{11-i}4^i7^{i-1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$? How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$ ?
Can someone please tell me a few steps as to how to approach these category of problems? I know $2x+3>0$ and $3x+7>0$ is a must.
| HINT:
Using Laws of Logarithms,
$$\log_{2x+3}(6x^2+23x+21)=\log_{2x+3}(2x+3)+\log_{2x+3}(3x+7)=1+\log_{2x+3}(3x+7)$$
Let $\log_{2x+3}(3x+7)=a$
$$\log_{3x+7}(4x^2+12x+9)=\log_{3x+7}(2x+3)^2=2\log_{3x+7}(2x+3)=\dfrac2a$$ as $\log_{3x+7}(2x+3)\cdot\log_{2x+3}(3x+7)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the determinant using colum or row operations I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.
| Maybe it's unlikely that you would have learned of this method, but via Dodgson condensation, you can reduce this to a problem of computing the determinants of several $2\times2$ matrices, which you might find easier to compute strictly via row operations.
$$\begin{vmatrix}1&1&1\\a&b&c\\ab&bc&ca\end{vmatrix}=\begin{vma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x)... | First note that, for every $y$, $2y/(1+y^2)\in[-1,1]$.
Consider $g(y)=\frac{1}{2}\arcsin(2y/(1+y^2))$, so
$$
g'(y)=\frac{1}{\sqrt{1-\dfrac{4y^2}{(1+y^2)^2}}}
\cdot\frac{1+y^2-2y^2}{(1+y^2)^2}=\frac{2}{1+y^2}\frac{1-y^2}{|1-y^2|}
$$
Therefore $g'(y)$ does not exist for $y=\pm1$ and
$$
g'(y)=\begin{cases}
\dfrac{1}{1+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
| $$2x^3+1=2x\cos(a)\implies x^3-x\cos(a)+{1\over 2}=0$$Let $x=y+{b\over y}\implies y={1\over 2}(x+\sqrt{x^2-4b})$ Using back substitution$$\left(y+{b\over y}\right)^3-\left(y+{b\over y}\right)\cos(a)+{1\over 2}=0$$$$\implies {1\over 2}y^3+y^6+b^3+y^4(3b-\cos(a))+y^2(3b^2-b\cos(a))=0$$
Now substitute $b={\cos(a)\over 3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{k=2}^\infty\frac{1}{k^3-1}$ I was considering a specialization of the Cauchy product
$$ \left(\sum_{n=1} ^\infty x^n \right) \left(\sum_{n=1}^\infty (-1)^n x^n \right)=\frac{-x^2}{1-x^2},$$
that converges for $0<x<1$.
The cited specialization is $x=\frac{1}{j^{3/2}}$ for integers $j\geq 2$. Then since... | The key is to use the identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}.\tag{1}$$
that leads to:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{1}{b-c}\left(\frac{\psi(a)-\psi(c)}{a-c}-\frac{\psi(a)-\psi(b)}{a-b}\right).\tag{2} $$
We have $k^3-1 = (k-1)(k-\omega)(k-\omega^2)$, hence by ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| Just to follow the answer above, "..for dealing with a numerator that is a power of positive definite quadratic polynomial".
Use integration by parts,
\begin{align*}
I_{n}:=\int\frac{dx}{\left(x^{2}+a^{2}\right)^{n}} & =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2n\int\frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}}dx\\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Circular permutation probability
A circular table has $9$ chairs that $4$ people can sit down randomly. What is the probability for no two people sitting next to each other?
My current idea is to calculate the other probability, which is there are two people sitting next to each other, but I don't know enough conditi... | This problem can be approached using the Polya Enumeration Theorem
(PET). Note that the four people seated on the table create four gaps
between them that consist of empty chairs and the empty chairs in all
four gaps must add up to five, where we have rotational symmetry
acting on the gaps.
Now the cycle in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Quadratic Functional equations. Suppose $f$ is a quadratic ploynomial, with leading cofficient $1$, such that $$f(f(x) +x) = f(x)(x^2+786x+439)$$ For all real number $x$. What is the value of $f(3)$?
| Stupid method:
Let $f(x) = x^2 + ax + b$. Then the left hand side of the equation is
$$\begin{split}
f(f(x) + x) & =f( x^2 + ax + b + x) \\
&= (x^2+ (a+1)x +b)^2 + a(x^2+(a+1) x+ b) + b
\end{split}$$
If you compare the $x^3$ coefficient, this gives $2(a+1) = 786 +a$. Comparing the constant coefficient tells you $b$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Indefinite Integral - How to do questions with square roots?
$$\int \frac{dx}{x^4 \sqrt{a^2 + x^2}}$$
In the above question, my first step would be to try and get out of the square root, so I would take $ t^2 = a^2 + x^2 $. But that gets me nowhere. How would you solve this, and if you are going to take a substituti... | substitute $x=a\tan { \theta } ,dx=\frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } }$
so
$$\\ \\ \\ \int \frac { dx }{ x^{ 4 }\sqrt { a^{ 2 }+x^{ 2 } } } =\int \frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } { \left( a\tan { \theta } \right) }^{ 4 }\sqrt { a^{ 2 }+{ a }^{ 2 }\tan ^{ 2 }{ \theta } } } =\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3... | In general
$$x^p + y^p \geq \dfrac{(x+y)^p}{2^{p-1}}$$
which follows from Holder's inequality, which states that
$$\Vert a \Vert_p \Vert b \Vert_q \geq \vert a\cdot b \vert$$
where $\dfrac1p + \dfrac1q = 1$. To obtain your result, take $a = (x,y)$ and $b=(1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\... | \begin{align}
&\int \frac{\sqrt{64x^2-256}}{x}\,dx\\
=& \int \bigg( \frac{8x}{\sqrt{{x^2}-4}}-\frac{32}{x^2\sqrt{1-\frac4{x^2}}}\bigg)\,dx
=\ 8 \sqrt{{x^2}-4}+16\sin^{-1}\frac2x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving equations system: $xy+yz=a^2,xz+xy=b^2,yz+zx=c^2$ Solve the following system of equations for $x,y,z$ as $a,b,c\in\Bbb{R}$
\begin{align*}xy+yz&=a^2\tag{1}\\xz+xy&=b^2\tag{2}\\yz+zx&=c^2\tag{3}\end{align*}
My try: Assume that $x,y,z\ne 0$ (it is easy to check the case where some are zero).
Subtract first equa... | Adding we get $$2(xy+yz+zx)=a^2+b^2+c^2$$
$$2xy=a^2+b^2+c^2-2c^2=a^2+b^2-c^2$$ etc.
Muliplying we get $$8x^2y^2z^2=\prod_{\text{cyc}}(a^2+b^2-c^2)$$
$$xyz=\pm\sqrt{\dfrac{\prod_{\text{cyc}}(a^2+b^2-c^2)}8}$$
and $xy=\dfrac{a^2+b^2-c^2}2$
$$z=\dfrac{xyz}{xy}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my... | $\dfrac{1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}-\dfrac{1}{n^{\frac{2}{3}}}+\dfrac{1}{n} = \dfrac{-n(-1)^n+n + n^{2/3}(-1)^n}{n^{5/3}(n^{2/3}+n^{1/3}+(-1)^n}$
The right hand side has $n$ in the numerator and $n^{7/3}$ in the denominator and hence is $O\left(\frac{1}{n^{4/3}}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax... | In general, if $F$ is a field, whose algebraic closure is $\bar{F}$, and $q(X)\in F[X]$ is a polynomial of degree $2$ with leading coefficient $k\neq 0$, then all solutions to $$f\big(\alpha\,f(X)+X\big)=f(X)\,q(X)\,,\tag{*}$$ where $\alpha\neq 0$ is a fixed element of $F$ and $f(X)\in \bar{F}[X]$, are given by $$f(X)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the eigenvalues of the following matrix where $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ as well as the associated eigenvector
Find the eigenvalues of the following matrix $\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix}$
The eigenvalues are $\lambda_1 < \lambda_2 < \lambda_3 < \lambda... | Note that your matrix is the block diagonal matrix
$$
M =
\left[\begin{array}{rr}
A & 0\\ 0 & B
\end{array}\right]
$$
where
\begin{align*}
A &=
\left[\begin{array}{rr}
-12 & -6 \\
8 & 2
\end{array}\right]
&
B &=
\left[\begin{array}{rr}
-14 & -9 \\
42 & 25
\end{array}\right]
\end{align*}
Next, note that the eigenvalues... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| Another way which makes life easier when you face products, quotients, powers : logarithmic differentiation $$m=\frac{x}{\sqrt{4x-3}}\implies \log(m)=\log(x)-\frac 12 \log(4x-3)$$ Differentiate $$\frac{m'} m=\frac 1x-\frac 12 \times \frac 4{4x-3}=\frac{2x-3}{x(4x-3)}$$ $$m'=m\times \frac{2x-3}{x(4x-3)}=\frac{x}{\sqrt{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to solve this inequality problem? Given that $a^2 + b^2 = 1$, $c^2 + d^2 = 1$, $p^2 + q^2 = 1$, where $a$, $b$, $c$, $d$, $p$, $q$ are all real numbers, prove that $ab + cd + pq\le \frac{3}{2}$.
| HINT:
For real $a-b,c-d,p-q;$
$$(a-b)^2+(c-d)^2+(p-q)^2\ge0$$
More generally for real $a,b;$
$$a^2+b^2=(a-b)^2+2ab\ge2ab\iff2ab\le a^2+b^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$
$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors.
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx... | $x^2 = k^{k-1}, k$ must be odd, or k is a perfect square.
k is odd.
$k = 2n+1\\
x,y = (2n+1)^n, (2n+1)^{n+1}$
$k$ is a pefect square. $k = n^2$
$x^2 = n^{2(2n-1)}\\
x,y = n^{2n-1}, n^{2n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $$... | According to your generating function approach let's assume we want to find the number of compositions of $5$ having at least two parts.
Using generating functions we have to look for the coefficient of $x^5$ in
\begin{align*}
&(x+x^2+x^3+\cdots)^2+(x+x^2+x^3+\cdots)^3+(x+x^2+x^3+\cdots)^4\\
&\qquad+(x+x^2+x^3+\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-... | Consider
$$
\frac{t^2-t}{2-t}=(-1-t)+\frac{2}{2-t}
$$
so your integral is
$$
\int\left(-\frac{2}{u^2-2}-u^2-1\right)\,du
$$
and your reduction is (almost) good, but a sign went wrong. Now your computation of partial fractions is really messed up:
$$
\frac{2}{u^2-2}=\frac{A}{u-\sqrt{2}}+\frac{B}{u+\sqrt{2}}
$$
so you ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$.
Expanding, the inequality becomes
$$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$
which is
$$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$
We can try using AM-GM:
$$x^2+xy+xy\geq 3\sqrt[3]{x^4y^2}$$
This is close to ... | $x^2+xy+y^2 = (x+\sqrt{xy}+y)(x-\sqrt{xy}+y)$, so divide through and get
$3(x-\sqrt{xy}+y) \ge x+\sqrt{xy}+y$
$2(x+y) \ge 4\sqrt{xy}$
$\frac{x+y}{2} \ge \sqrt{xy}$
which is AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions to $5m^2-6mn+7n^2 = 1985$
Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$
Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?
| The discriminant of the quadratic form is $6^2-4\cdot 5\cdot 7=-2^3\cdot \color{blue}{13}$ and the equation is equivalent to:
$$ (5m-3n)^2+26 n^2 = 9925 \tag{1}$$
that gives:
$$ 6\equiv 9925 \equiv (5m-3n)^2 \pmod{\color{blue}{13}}\tag{2} $$
but $6$ is not a quadratic residue $\!\!\pmod{13}$, since $13\equiv 5\pmod{8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int \frac{x}{x^4+5}dx$
$$\int \frac{x}{x^4+5}dx$$
$u=x^2$
$du=2xdx\Rightarrow \frac{du}{2}=xdx$
$$\int \frac{x}{x^4+5}=\frac{1}{2}\int \frac{du}{u^2+5}$$
I want to get to the expression in the form of $\frac{da}{a^2+1}$ so I factor out $5$ to get to:
$$\frac{1}{2}\int \frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}... | Your last line isn't correct, you can procced as follows:
$$\int \frac{1}{2}*\frac{1}{5}\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}=\frac 1 {10}\int\frac{du}{u^2/5+1}$$
Set $t=\frac{u}{\sqrt5}$ and $dt=\frac{1}{\sqrt 5}du$
$$=\frac{1}{2\sqrt 5}\int\frac{dt}{t^2+1}$$
$$=\frac{\arctan t}{2\sqrt 5}=\frac{\arctan \left(u/\sqrt 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
... | Define the sequence given by $b_1=1,b_2=1$ and $b_{n+2}=3b_{n+1}-b_{n}$ for $n>0$.
To prove that $a_n=b_n^2$ we just have to show:
$b_{n+2}^2=7b_{n+1}^2-b_{n}^2-2$.
Of course $b_{n+2}^2=9b_{n+1}^2-6b_{n+1}b_n+b_{n-1}^2$
So to finish we just need to prove $3b_{n+1}b_n=b_{n+1}^2+b_n^2+1$.
Notice $b_{n+1}^2+b_n^2+1=b_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Triple Integration in vector calc NIn the following example I have been asked to find the volume V of the solid bounded by the sphere $ x^2 + y^2 +z^2 = 2 $ and the paraboloid $ x^2 + y^2 = z $ by using triple integration.
I am not quite sure how to set up this triple integration and what each of the integrands should ... | It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis).
In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x +... | $$\begin{aligned}\int \frac{1}{\frac{\sin(x)}{\cos(x)}\:+\:\frac{\cos(x)}{\sin(x)}\:+\:\frac{1}{\sin(x)}\:+\:\frac{1}{\cos(x)}}dx
& = \int \:\frac{\sin \:\left(2x\right)}{2\left(\cos \:\left(x\right)+\sin \:\left(x\right)+1\right)}dx
\\& =\frac{1}{2}\cdot \int \:\frac{\sin \left(2x\right)}{\cos \left(x\right)+\sin \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help req... | $$0>4x^2+4ax+4a^2+24a=(2x+a)^2+3a^2+24a$$
Now $1<x<2\iff2+a<2x+a<4+a$
$\implies(2x+a)^2<$min$\{(2+a)^2,(4+a)^2\}$
Case$\#1:$ If $(4+a)^2\ge(2+a)^2\iff a\ge-3,(2x+a)^2<(2+a)^2$
$$0>(2+a)^2+3a^2+24a=4(a^2+7a+1)$$
We need $a^2+7a+1<0$
As the roots of $a^2+7a+1=0$ are $\dfrac{-7\pm\sqrt{49-4}}2=\dfrac{-7\pm3\sqrt5}2$
$\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Eliminating $x=-3,5$ when we solve the system $y=|3-x|, 4y-(x^2-9)=-24$ $$y=|3-x|$$
$$4y-(x^2-9)=-24$$
I use two methods to solve it,
Method 1: Squaring both sides and eventually get a equation like this$$(x-5)(x-7)(x+3)(x+9)=0$$, $x=-9,-3,5,7$.
Method 2:Using $$y=3-x$$ and $$y=-(3-x)$$, and finally also get $x=-9,-3,5... |
Is there any method to eliminate $x=-3,x=5$ from the real answers?
For method 1 :
You have to have
$$x^2-33=4y=4|3-y|\ge 0\implies x^2\ge 33$$
For method 2 :
As f'' comments, note that $|3-x|=3-x$ holds only when $3-x\ge 0,$ i.e. $x\le 3$ and that $|3-x|=-(3-x)$ holds only when $3-x\lt 0$, i.e. $x\gt 3$.
So, if you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | By using identities
$$\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}$$ and
$$\tan^{-1}{\alpha}-\tan^{-1}{\beta}=\tan^{-1}(\frac{\alpha-\beta}{1+\alpha\beta}),$$
after a little calculation, we get $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)=\tan^{-1}(\frac{1}{\sqrt{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Compute Triple integral Compute $$\iiint\limits_{x^2+y^2+(z-2)^2\leq1}\frac{dx\,dy\,dz}{x^2+y^2+z^2}$$
I tried to convert it to spherical cordinates and I got $$\iiint \sin\theta \,dr\,d\theta \,d\varphi$$
However I am lost on how to find the boundaries of the new domain, can anyone help?
| Change coordinates by letting $z'=z-2$. Then, we have
$$\begin{align}
\iiint_{x^2+y^2+(z-2)^2\le 1} \frac{1}{x^2+y^2+z^2}\,dx\,dy\,dz&=\iiint_{x^2+y^2+z^2\le 1} \frac{1}{x^2+y^2+(z+2)^2}\,dx\,dy\,dz\\\\
&=\int_0^{2\pi}\int_0^\pi\int_0^1 \frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,dr\,d\theta\,d\phi \\\\
&=2\pi\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\fra... | You have $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\sqrt[3]{\tan x}}dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sqrt[3]{\tan x}}d\tan x$$
Replacing $t = \tan x$, you have the integration equals
$$\int_{1}^{\sqrt{3}} \frac{1}{\sqrt[3]{t}}dt = \frac{3}{2}t^{\frac{2}{3}} \mid_{1}^{\sqrt{3}} = \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\displaystyle \sum_{1\leq kProve that
$$\sum_{1\leq k < j\leq n}\tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4}$$
| It is enough to recall Cauchy's proof of the Basel problem, relying on the identity:
$$ \binom{2n+1}{1}t^n-\binom{2n+1}{3}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{2n+1}=\prod_{k=1}^{n}\left(t-\cot^2\frac{k\pi}{2n+1}\right)\tag{1} $$
If we consider the "reciprocal polynomial"
$$ \binom{2n+1}{2n+1}t^n-\binom{2n+1}{2n-1}t^{n-1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $x+y+z=1$ and $x,y,z>0\,$ find $\min xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$
If $x+y+z=1$ and $x,y,z>0\,$ find the minimum of $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$
$\bf{My\; Try::}$ We can write expression as $xy(1-z)^2+yz(1-x)^2+zx(1-y)^2$
$ = xy(1+z^2-2z)+yz(1+x^2-2x)+zx(1+y^2-2y)$
$ = xyz(x+y+z)+(xy+yz+zx)-6xyz= xy+yz+zx-5xy... | The minimum does not exist, but arbitrarily small values can be reached by letting $x,y\to0$, $z\to1$; i.e., the infimum is $0$.
Explicitly, $x=y=\varepsilon$ and $z=1-2\varepsilon$ gives
$$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\leq3\times\varepsilon\cdot1\cdot(1+1)^2=12\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The integral $\int_0^1 x^n \log(2-x) \, dx$ In this question I had stuck on the integral:
$$\int_0^1 x^n \log(2-x) \, {\rm d}x$$
Claude Leibovici says that $\displaystyle \int_0^1 x^n \log(2-x) \, {\rm d}x=\frac{\, _2F_1\left(1,n+2;n+3;\frac{1}{2}\right)}{2 (n+1) (n+2)}=a_n+b_n \log 2$ where $a_n, b_n$ are sequences of... | By integration by parts
$$ I_n=\int_{0}^{1}x^n\log(2-x)\,dx = \frac{1}{n+1}\int_{0}^{1}\frac{x^{n+1}}{2-x}\,dx \tag{1}$$
so it is enough to write $x^{n+1}$ as $(x^{n+1}-2^{n+1})+2^{n+1}$ to get:
$$\begin{eqnarray*} I_n &=& \frac{1}{n+1}\left(2^{n+1}\log 2-\int_{0}^{1}\sum_{k=0}^{n}2^k x^{n-k}\,dx\right)\\&=&\color{blue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
generating function: how many possibilities are there to throw 10 different dice so that their sum is 25 My discrete math textbook has the following solution to the problem using generating functions (the solution has to use generating functions):
$$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
Once we start developing the forma... | The coefficient of $x^{25}$ in $(x+x^2+\ldots+x^6)^{10}$ equals the coefficient of $x^{15}$ in
$$ (1+x+\ldots+x^5)^{10} = (1-x^6)^{10}\cdot\frac{1}{(1-x)^{10}} $$
so the answer is given by
$$ [x^{15}] \left(\sum_{k=0}^{10}\binom{10}{k}(-1)^k x^{6k}\right)\cdot\left(\sum_{j\geq 0}\binom{9+j}{j}x^j\right) $$
and it is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to prove that no prime factor of $x^2-x+1$ is of the form $6k-1$ Consider sequence $x^2-x+1$ ($1,3,7,13,21,31,43,57,73,91,\dots$). Let's consider prime factorization of each term.
$$3=3$$
$$7=7$$
$$13=13$$
$$21=3\times7$$
It seems that the only prime factors we ever get are 3 and those of the form $6k+1$. In fact,... | First, some facts about numbers in your sequence. Let $n=x^2-x+1$. If $x=3k-1$, then
\begin{align*}
n=x^2-x+1 &= (3k-1)^2-(3k-1)+1\\
&= 9k^2-6k+1-3k+1+1\\
&= 9k^2-9k+3\\
\implies n/3 &= 3k^2-3k+1
\end{align*}
$k^2$ and $k$ have the same parity. Therefore
$n/3=1\mod 6$. If $x$ is any other integer then $n=1\mod 6$ as ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalen... | Using this:
$$
\frac{\sum_{k=1}^N k^2}{N}= \frac{1}{6} (N+1) (2 N+1)=K^2
$$
the condition can be written as (as you pointed out):
$$(N+1) (2 N+1)=6K^2$$
through numerical search, the last part of the problem is solved
$$N=1,337,65521$$
Not really a satisfying answer i know.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Question on the inequality Question.
prove that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } } $$
Proof
$$\\ \left( { a }_{ 1 }+{ a }_{ 2 }+...+{... | You have $a_i\cdot \frac1{a_i}$ for each $i$; these sum up to $n$.
Apart from that, you have for each of the $n\choose 2$ choices of $i<j$ the summands $\frac{a_i}{a_j}+\frac{a_j}{a_i}$, which is $\ge 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Am I correct when I do this in questions about induction? I have a problem which is:
If $n$ is a fixed positive integer and greater than 1, show by induction that for each positive integer $r$, where $2\le r\le n$,
$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{r-1}{n})$$
Denote this inequality... | Maybe you could prove the inequality,
$$1-\frac{r(r+1)}{n} \le 1-\frac{r^2}{n}+\frac{r^2(r-1)}{n^2}$$
directly (without induction). Infact by expanding you obtain
$$1-\frac{r^2}{n}-\frac{r}{n}\le 1-\frac{r^2}{n}+\frac{r^3}{n^2}-\frac{r^2}{n^2},$$
that is, $-1\le \frac{r^2}{n}-\frac{r}{n},$ or $-n\leq r(r-1)$ which triv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 >... | @ZackNi As David indicated, you are assuming "infinity minus infinity = 0,which works out well because both radicals are squareroots and the headcoefficients of the polynomials are both 1.You can also get"infinity minus infinity = 0 from a squareroot and cuberoot, but using your approach will then yield to a wrong answ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
| By integration by parts you have that
$$I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx=\frac{1}{n+1}\left[x^{n+1}\arctan(x)\right]_0^1-\frac{1}{n+1}\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx\\=\frac{1}{n+1}\left(\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx\right)$$
Hence
$$(n+1)I_{n}-\frac{\pi}{4}=-\int_{0}^{1} \frac{x^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Functional equation of type linear expression
If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(xy+1) = f(x)f(y)-f(y)-x+2\;\forall x, y\in \mathbb{R}$ and $f(0) = 1\;,$ Then $f(x) $ is
$\bf{My\; Try::}$ Put $x=y=0\;,$ We get $f(1) = (f(0))^2-f(0)-0+2 = 1-1+2=2$
Similarly put $x=y=1\;,$ We get $f(2) = (f(1))^2-f(1)-1+2 ... |
$$f(xy+1) = f(x)f(y)-f(y)-x+2$$
Let $y=0$. Then $$f(1)=f(x)f(0)-x+2$$
Let $f(1)=a$. Then
$$f(x)=x+a-2$$
Check:
$xy+1+a-2=(x+a-2)(y+a-2)-(y+a-2)-x+2$
$xy+a-1=xy+ax-2x+ay+a^2-2a-2y-2a+4-y-a+2-x+2$
$(a-2-1)x+(a-2-1)y+(a^2-6a+9)=0$
$(a-3)x+(a-3)y+(a-3)^2=0$ $(\forall x,y \in \mathbb R)$
Then $a=3$
Answer: $$f(x)=x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin... | So, $\cos b=+\sqrt{1-\left(\dfrac1{\sqrt{10}}\right)^2}=\dfrac3{\sqrt{10}}$
$\implies\tan b=\dfrac{\dfrac1{\sqrt{10}}}{\dfrac3{\sqrt{10}}}\implies b=\arctan \dfrac13$
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then how are $x$ and $y$ related?
If $$\cos x + \cos y - \cos(x+y) = \frac{3}{2}$$ then
*
*(a) $x + y = 0$
*(b) $x = 2 y$
*(c) $x = y$
*(d) $2 x = y$
It is problem of trigonometry, and I have the solution of the problem. However, after seeing the so... | Using Prosthaphaeresis Formula,
$$\cos x+\cos y=2\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
Double angle formula says:
$$\cos(x+y)=2\cos^2\dfrac{x+y}2-1$$
So we have $$2\cos^2\dfrac{x+y}2-2\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\dfrac12=0\ \ \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{x+y}2$ which is real,
so the discrim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Problem solving recurrence equation for $a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0=a_1=1$ I want to try the generating function $A(x)$ and a formula for $a_n$ for the following sequence:
$$a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0 = a_1 = 1$$
So I multiplied the above equation on both sides by $x^n$ and summed them for each $n \... | As the recurrence relation is linear with constant coefficients, let us look for a solution of the form $a_n = z^n$. We obtain
$$z^n (z^2 - z - 2) = 0$$
Given the nonzero initial conditions, $z^n \neq 0$. Hence, $z^2 - z - 2 = 0$. The solutions are $-1$ and $2$.
$$a_n = \beta_1 (-1)^n + \beta_2 2^n$$
From the initial c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given a matrix $B$, what is $det(B^4)$? My task is this:
Compute det$(B^4)$, where
$$B =\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix}$$
My work so far:
It can be shown that
$B =
\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\... | I don't know how you have obtained
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}$$
But making elementary operations ($R2\leftarrow R2-R1$, $R3\leftarrow R3-R1$) I obtain
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 2 & 0
\end{pmatrix}$$
and then ($R3\leftarrow R3-2\cdot R2$):
$$\begin{pmatrix}
1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow... | Induction will work here. Assume $1+2^{2^n} + 2^{2^{n+1}}$ is divisible by 7.
Note that $2^{2^{n+2}} = 2^{2^n4} = 16^{2^n} = (14+2)^{2^n} = 14M+2^{2^n}$ for some integer $M$. Then $1+2^{2^{n+1}} + 2^{2^{n+2}} = 1+2^{2^{n+1}}+(14+2)^{2^n} = 1+2^{2^{n+1}} +2^{2^n}+14M$, which must be a multiple of 7, since the first th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
| HINT:
$$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$
Now using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=\cos(A+B)\cos(A-B)+\cos^2C+1$$
Now as $\cos(A+B)=\cos(\pi-C)=-\cos C,$
$$F=-\cos C\cos(A-B)+\cos C\{-\cos(A+B)\}=1-\cos C(2\cos A\cos B)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$
Find
$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$$
My attempt :-
$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$
$$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$
$$\implies 2x^2 -7x - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2 + 10x + 1)... | Hint: $2 x^2 - 7 x - 4 = (2x+1)(x-4)$, so you may guess that the desired result is of the form $a \sqrt{2x+1} + b \sqrt{x-4}$. Now square that, and see what $a$ and $b$ would give you $\dfrac{3}{2}(x-1) + \sqrt{2x^2-7x-4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter exam... | The binomial coefficient is $2^n (2^n -1)/2 = 2^{n-1} (2\cdot 2^{n-1} -1)$.
For the divisibility to hold you need this to be $-1 \pmod{7}$.
Now the sequence nonnegative powers of $2$ modulo $7$ have a period of at most $6$, by Fermat's little theorem, but actually it is just has period $3$, and it starts
$2^0 = 1, 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to prove each element of the following sequence is a perfect square? Sequence $\{a_n\}$ satisfies the following formula:
$a_{n+2}=14a_{n+1}-a_n+12$,
and $a_1=1, a_2=1$.
It is easy to check that $a_3=25$ and $a_4=361$.
The question is how to prove each element of the sequence $\{a_n\}$ is a perfect square?
| The above proofs are elegant. The following is what I was thinking:
Like what Steven did, we can use generation functions and write out $$a_n = c(2+\sqrt{3})^{2n} + d(2-\sqrt{3})^{2n} - 1$$ for some constants $c$ and $d$. Note that $c>0$ since $a_n$ increases as $n$ increases. If $d>0$, then $a_n = (\sqrt{c}((2+\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| You can express $x^4+y^4$ in terms of $s=x+y$ and $p=xy$:
$$
x^4+y^4=as^4+bs^2p+cp^2
$$
(the degree of $s$ is $1$ and the degree of $p$ is $2$). We can determine $a,b,c$ by using particular values for $x$ and $y$.
*
*If $x=1$ and $y=0$, then $x^4+y^4=1$, $s=1$ and $p=0$
*If $x=1$ and $y=1$, then $x^4+y^4=2$, $s=2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Positive number whose square exceeds its cube by the greatest amount
Find the positive number whose square exceeds its cube by the greatest amount.
Could you please help me with this question? It seems pretty easy at first glance, but I struggled to produce any answer. Thank you in advance!
UPD:
Sorry, I didn't m... | Here is a method that does not require calculus:
Lemma: If $x+y+z = k$ (where $k$ is a constant), the product $xyz$ is maximized when $x=y=z= \dfrac k3 $
Proof:
By $AM-GM$, we have
$$\dfrac {x+y+z}{3} \ge \sqrt[3] {xyz}$$
$$\dfrac {k}{3} \ge \sqrt[3] {xyz}$$
$$\left(\dfrac {k}{3}\right)^3 \ge xyz$$
We can see that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which ... | Observe first that if must be $\;0\le x\le 1\;$ , so the given interval is within this constraint. Second, square:
$$\sqrt{1-x}-\sqrt x\ge\frac1{\sqrt3}\implies1-x-2\sqrt{x(1-x)}+x\ge\frac13\implies$$
$$2\sqrt{x(1-x)}\le\frac23\implies x(1-x)\le\frac19\implies 9x^2-9x+1\ge0\implies$$
The last quadratic's roots are give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Constructing a base $b$ repunit that is $a$ $\pmod p$ Suppose I want to construct a number of the form ($b^n-1$)/($b-1$) $=$ $a$ $\pmod p$ with $n$ and $p$ prime. What are the limitations on the base $b$ $>$ $1$ and prime $n$?
For example, if ($b^n-1$)/($b-1$) $=$ $4$ $\pmod 5$, what are possible choices for $b$ and (... | $(b^n-1) = (b-1)(b^{n-1} + b^{n-2} + ... + b^2 + b + 1)$
So (assuming $b \neq 1$!) it might be easier to directly consider
$$(b^{n-1} + b^{n-2} + ... + b^2 + b + 1) = a \mod p$$
If $p | b$, the above is true only if $a = 1$; and in that case it is true for all $n$, prime or not. So let's assume that $p$ is not a facto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$4$ digit integers such that none of the digits appear more than twice
How many $4$ digit integers can be chosen such that none of the digits
appear more than twice?
$4$-digit integers with distinct digits= $9\times9\times8\times7 = 4536$
$4$-digit integers with two alike digits $(XXYY) = 9\times9\times2\times1\tim... | Direct approach. There are three possibilities. The first is ABCD, where all four digits are distinct. There are $9 \times 9 \times 8 \times 7 = 4536$ of these.
The second is AABC/ABAC/ABCA/BAAC/BACA/BCAA. Each of these has $9$ choices for the first position, and then $9 \times 8$ choices for the other two distinct ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.