Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Expected value of dice game (with options) Setup
The question (modified from an old Jane Street interview):
(i) Two people throw a 6-sided dice. The person with a larger number wins $1. If the dice values are equal, neither person wins a reward. How much are you willing to pay to play this game?
(ii) A followup: say yo... | If you have rolled a $6$ and do not pay to add $2$, your expectation is $\frac 56 \approx 0.83$ because you win $1$ unless your opponent has also rolled a $6$. By paying $0.25$ you increase this to $1$, so your expectation of rolling a $6$ and buying the $2$ is $1-0.25=0.75$. As this is less than $0.83$ you should no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left... | Here $$ \frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} $$
you canceled $x^4$ in the numerator and denominator, so it should be $$\frac{-\frac{1}{3}+o\left(x^{-1}\right)}{-\frac{1}{2}+o\left(x\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, w... | One way is to start from $$a^2+ab+b^2=b^2\left(\left(\frac{a}{b}\right)^2+\frac{a}{b}+1\right)=0$$ so letting $x=\frac{a}{b}$ we get the quadratic equation $x^2+x+1=0$ which has no real roots. This is similar to the idea hinted at in the comment by RiverX15.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the volume between the sphere $x ^2 + y^2 + z^2 = 4$ and the plane $z = 1$ Suppose $y \geq 3$. I want to compute the volume between the sphere $x ^2 + (y − 2)^2 + z^2 = 4$
and the plane $y = 3$.
So I move left the sphere and and the plan, and rotate it counterclockwise. I got the new sphere and the new plan:
Suppo... | There are two mistakes in your work. At the intersection of the sphere and the plane,
$z = 2 \cos\phi = 1 \implies \phi = \pi/3$
Also the lower bound of $\rho$ is defined by the plane
$z = \rho\cos\phi = 1 \implies \rho = \sec\phi$
So the integral should be,
$ \displaystyle \int_{0}^{2 \pi} \int_0^{\pi/3} \int_{\sec\ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $z^3 =-1$ I need to find the solution of $z^{3} =-1$.
Let$$z=[ r,\theta ]$$
then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$
write the number $-1$ in modulus argument from,
$$[ 1,( 2n-1) \pi ]$$
the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$
Therefore,$$ r^{3} =1 \sp... | $z=re^{i\theta}$
$z^3 = -1$
$r^3 (e^{3 i \theta})= 1\cdot e^{i\pi}$
Hence, $ r=1$ and $\theta= \frac{i(\pi + 2k\pi) }{3}$ for $k=0, 1,2$
$z= e^{\frac{i\pi}{3}} , e^{i\pi}, e^{\frac{5i\pi}{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Rankine-Hugoniot for Burgers equation with Lorentzian initial data We have the following equation with initial condition
\begin{align*}
u_y + uu_x &= 0\\
u(x,0)&=\frac{1}{1+x^2}\quad x\in\mathbb{R}
\end{align*}
let's say $g(x)=\dfrac{1}{1+x^2}$ and corresponding to the manifold $\Gamma$ in the $xyz$ space given b... | The characteristic curves satisfy
$$
x= x_0 + g(x_0) y, \qquad u = g(x_0)
$$
where $g(x) = (1+x^2)^{-1}$ is the initial data, i.e.
$
u = g(x-uy)
$
in implicit form. The shock time is given by
$$
y_s = \frac{-1}{\inf_{x_0} g'(x_0) } = \frac{8}{3\sqrt{3}} \simeq 1.54
$$
for $x_0 = 1/\sqrt{3}$. Therefore, the results in O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4333069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of region ABC. For reference: In figure $T$ and $K$ are points of tangency,
$MT = a$ and $KN = b$; calculate area of region $ABC$.
(Answer:$2\sqrt{ab}(\sqrt a+\sqrt b)^2$)
My progress:
$$S_{ABC} = p \cdot r = \frac{r \cdot (AB+BC+AC)}{2}\\
AC +2R = AB+BC\\
S_{ABC} = AG \cdot GC \qquad \text{(property)... | I continue your process and we already have $\text{TK}=\sqrt{2} R$.
According to intersecting chords theorem, we have $$\text{AT} \cdot\text{TB}=\text{MT}\cdot \text{TN}$$$$\text{BK} \cdot\text{KC}=\text{KN} \cdot \text{MK}$$
So $$\text{AT}=\sqrt{2} a+\frac{\text{ab}}{R}$$$$\text{KC}=\sqrt{2} b+\frac{\text{ab}}{R}$$
B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the Variance of $(1-X)/X$ and the Expectation of $(1-X)/X^{2}$? Knowing that
$X = \begin{cases}
0.05 \ \ \text{with probability} \ \ 0.7\\
0.2\ \ \text{with probability} \ \ 0.3\end{cases}$
What would be the Variance of $\frac{1-X}{X}$ and the Expectation of $\frac{1-X}{X^2}$ ?
I have computed it and would lik... | Since
$$\operatorname{E}\left[\frac{1-X}{X}\right]=\sum_{i=1}^2 \frac{1-x_i}{x_i}p_i$$
$$=\frac{0.95}{0.05}\cdot 0.70+\frac{0.80}{0.20}\cdot 0.30=14.5$$
And
$$\operatorname{E}\left[\left(\frac{1-X}{X}\right)^2\right]=\sum_{i=1}^2 \left(\frac{1-x_i}{x_i}\right)^2 p_i$$
$$=\left(\frac{0.95}{0.05}\right)^2\cdot 0.70+\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of $a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$ Let $a,b,c$ be real numbers such that $a+b+c=3$, then find minimum value of
$$a^2 \cot (10^ \circ)+b^2\cot (70^ \circ)+c^2\cot (130^ \circ)$$
I have solved many such question using A.M.-G.M. inequality but since $\cot (130^ \circ)$ is nega... | There's neither a maximum nor a minimum value. Take $a=1,b=x\in\mathbb{R},c=2-x$. Then
$$a^2\cot(10^\circ)+b^2\cot(70^\circ)+c^2\cot(130^\circ)\\
=(\cot(130^\circ)+\cot(70^\circ))x^2-4\cot(130^\circ)x+4\cot(130^\circ)+\cot(10^\circ)\xrightarrow{x\to\infty}-\infty $$
since the above is a quadratic function with negative... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Geometry problem to prove Euler line of a triangle $I$ is the incenter of $\triangle{ABC}$ and $D$ is the tangent point of the incircle with side $BC$. Points $E$ and $F$ are on side $BC$ so that $\angle{BAE}=\angle{ACB}, \angle{CAF}=\angle{ABC}$. $G$ and $H$ are incenters of $\triangle{ABE}$ and $\triangle{ACF}$ respe... | Oh, this is a nice exercise, i will remember it for didactic purposes. Below, all details will be given, and the computations are done explicitly.
As references i am using
bary-full and bary-short. Notations like $X(2)$ for the centroid are as in ETC.
Fix a (non-degenerated) triangle $\Delta ABC$ with sides $a,b,c$.
Be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1... | Hints:
*
*For $x>0$ and $x\neq 1$, one of the two values, $x$ or $\frac1x$, is between $0$ and $1$, so the floor of that value will be $0$.
*If $x$ is a solution, then so is $\frac1x$. This means you can first only focus on $x>1$, to get half of the solutions, then know that this is just half of the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
(Dis)prove the inequality $2^{\frac{1}{2}}\cdot\left(f\left(xa\right)+f\left(\frac{1}{xa}\right)\right)^{\frac{1}{2}}\leq g(x)$ for $a,x>0$ Problem :
Let $a,x>0$ then define :
$$f\left(x\right)=x^{\frac{x}{x^{2}+1}}$$
And :
$$g\left(x\right)=\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}$$
Then prove or disprove that :
... | Sketch of proof :
Define :
$$n\left(x\right)=\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{-\frac{x}{2}}-\sqrt{x^{1.1}}-3x\left(x-1\right)\left(1-\frac{x}{x+1}\right)^{3}+3\left(0.5-\frac{x}{x+1}\right)$$
Then we have on $(-\infty,\infty)$ :
$$\frac{d}{dx}\left(\frac{d}{dx}n\left(e^{x}\right)\right)<0$$
So we have by Jen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4349487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Faster way to find the first four non-zero terms of the Maclaurin series for $\frac{1-x}{1+x}\cosh\sqrt{x}$ I want to find the first 4 non-zero terms for :
$$\frac{1-x}{1+x}\cosh\sqrt{x}$$
Before expanding, I rewrite this as $$(1-x)\left(\frac{1}{1+x}\right)\cosh\sqrt{x}$$
Then I expand to get $$(1-x)\left(1-x+x^2-x^3+... | It's unnecessary to break up $1-x$ and $\frac{1}{1+x}$ into two separate series. You can just differentiate $\frac{1-x}{1+x}$ directly. As J.G. pointed out in his comment, it may be easier to rewrite the expression as $\frac{2}{1+x} - 1$ or $2(x + 1)^{-1} - 1$. This gives you $\frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the sum (Includes binomial coefficients) I came across this question under summation:
Find the sum: $$\sum_{r=0}^n \left[\frac{r}{n} - \alpha \right]^2 {n \choose r}x^r(1-x)^{n-r} $$
To start with this, I wrote it as $\sum_{r=0}^n \left[\frac{r-n\alpha}{n} \right]^2 {n \choose r}\left(\frac{x}{1-x}\right)^r(1-x)^... | Correct, and here's an alternative approach that uses the binomial distribution. Let $X$ be a $\text{Bin}(n,x)$ random variable. Then $\mathbb{E}[X] = n x$ and $$\mathbb{E}[X^2] = \mathbb{V}[X] + \mathbb{E}[X]^2 = n x(1-x) + (nx)^2.$$
So
\begin{align}
\sum_{r=0}^n \left[\frac{r}{n} - \alpha \right]^2 {n \choose r}x^r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_{-1}^{1} u^2(1-u^2)^{3/2}du$ The integral I'm having difficulty with solving is:
$I = \int_{-1}^{1} u^2(1-u^2)^{3/2}du$
I arrived at this integral in trying to solve
$\int_{0}^{\pi} sin^4(x)cos^{2}(x)dx$
by making the substitution $u=cos(x)$
I attempted integration by parts as follows:
$F=u, \: F'=1 \\... | The integral expression looks like differnetial binom
$$
I = \int{u^m(a + bu^n)^pdu}.
$$
Comparing that to the initial integral,
$$
m = 2, n = 2, p = \frac{3}{2}, a = 1, b = -1.
$$
$$
p + \frac{m+1}{n} = \frac{3}{2} + \frac{2+1}{2} = 3
$$
is an integer. Then the following substitution is used
$$
au^{-n} + b = x^s, \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove $\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}$ by induction I'm not sure if my induction proof is correct. If anyone could kindly review it, I'd appreciate it.
Prove that $n \in \mathbb{N}$:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$
For $n=1$ the statemen... | There aren't a mistake in your third equalty. So you proof is right, clean and Nice, good work Enzo.
For $n=1$ the statement is true because $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
Here you are Right
Let's assume that the statement is true for some natural number $k$, then:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
| Just take the difference
of the expression and its
proposed limit
and conjugate that.
Works out very nicely.
$\begin{array}\\
d(n)
&=\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2})\\
&=(\sqrt{(n+a)(n+b)}-(n+\frac{a+b}{2}))\dfrac{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}{\sqrt{(n+a)(n+b)}+(n+\frac{a+b}{2})}\\
&=\dfrac{(n+a)(n+b)-(n+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proving $\sum_{i=1}^ni\sin{ix}=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{(n+1)\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}$ by induction Prove by induction that:
$$\sum_{i=1}^ni\sin{ix}=\frac{\sin{((n+1)x)}}{4\sin^2{\frac{x}{2}}}-\frac{(n+1)\cos{((2n+1)\frac{x}{2})}}{2\sin{\frac{x}{2}}}$$
I tried to prove it ... | Okay, as you say prove by induction...
Base case $n=1$, we need to prove:
$$\sin x=\frac{\sin{(2x)}}{4\sin^2{\frac{x}{2}}}-\frac{2\cos{(\frac{3x}{2})}}{2\sin{\frac{x}{2}}}.$$
We have
$$\frac{2\cos{(\frac{3x}{2})}}{2\sin{\frac{x}{2}}}=\frac{2\cos{(\frac{3x}{2})}\sin \frac x2}{2\sin^2{\frac{x}{2}}}=\frac{\sin 2x-\sin x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and... | You could make it faster using
$$I=\int \frac{d x}{\left(x^{n}+a\right)^{m}}=x\, a^{-m} \, _2F_1\left(m,\frac{1}{n};1+\frac{1}{n};-\frac{x^n}{a}\right)$$ which gives
$$J=\int_0^\infty \frac{d x}{\left(x^{n}+a\right)^{m}}=a^{\frac{1}{n}-m}\,\frac{\Gamma \left(1+\frac{1}{n}\right) \Gamma
\left(m-\frac{1}{n}\right)}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Geometrical meaning of $x^2+y^2+z^2-xy-xz-yz$ I am looking for a geometrical interpretation of the symmetrical expression
$$f=x^2+y^2+z^2-xy-xz-yz\tag{1}$$
with $x,y,z \in \mathbb{R}$. I could for example $f$ interprete as dot products of a vector with its permuted vector
$$f=\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{p... | Render $(x-y)^2=x^2-2xy+y^2$ and similarly for the $x,z$ and $y,z$ pairs. When you add up the squares you get
$2(x^2+y^2+z^2-xy-xz-yz).$
So $x^2+y^2+z^2-xy-xz-yz$ is half the sum of squared distances between three points on a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $x^2+2^{17}=y^3$ over the integers I am asked to solve the equation $x^2+2^{17}=y^3$ over the integers. The technique I was taught for such equations uses algebraic numbers and a little algebraic number theory/ring theory depending on how you look at it. However, I got stuck at some point because I have never had... | Note, your approach is about powers of $2$ that divide $x.$ How many factors of $\sqrt{-2}$ divide $x+512\sqrt{-2}$ is entirely about powers of two that divide $x.$
Review
Let $\nu_2(m)$ be the largest integer $n$ such that $2^n\mid m.$
A few key results:
$$\nu_2(m^k)=\nu_2(m)^k$$
and,
If $\nu_2(m_1)\ne \nu_2(m_2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Math Limits Questions with Square Roots I seek your help on this question.
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)$$
This is what I have done so far.
I tried to divide both numerator and denominator with the largest power but I am not too sure if I am doing it right. Please help... | We know that, by asymptotic relations:
$$(1+f(x))^{\alpha} \;\; \sim \;\; \alpha \cdot f(x) \;\; \text{when}\;\; f(x) \to 0$$
One can write:
$$\lim_{x\to\infty}\left(\sqrt{9x^6+7x^3+5x}-\sqrt{9x^6+6x^2+3x}\right)=\lim_{x\to\infty}\sqrt{9x^6+6x^2+3x}\cdot\left(\sqrt{\frac{9x^6+7x^3+5x}{9x^6+6x^2+3x}}-1\right)=\lim_{x\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$
I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}... | Hint: $x^{4}+y^{4} \geq \frac 1 2 ({x^{2}+y^{2}})^{2}$. Now use polar coordinates to see that the limit is $0$.
This allows you to get rid of $\theta$. You will also need the fact that $t^{4}e^{-t} \to 0$ as $ t \to \infty$ (Take $t=\frac 1r$). This fact is proved using L'Hopital's Rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$.
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. Then $z+\frac{1}{y}=\frac{m}{n}$, where $m$ and $n$ are coprime. Find... | Just substitute the values of $y$ and $z$ in $xyz=1$.
$$\frac{x(29x-1)}{x(5-x)}=1$$
$$29x-1=5-x$$
$$x=\frac{1}{5}$$
$$y=\frac{29/5-1}{1/5}=24$$
$$z=\frac{1}{5-1/5}=\frac{5}{24}$$
$$z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$$
Thus, $$m+n+1=1+4+1=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with somethi... | As Spivak points out in his Calculus, the proofs of limit theorems are strategies for implementing the definition of a limit.
Here, guided by the trick
\begin{align*}
\frac{f(x + h)g(x + h) - f(x)g(x)}{h}
&= \frac{f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)}{h} \\
&= f(x + h)\frac{g(x + h) - g(x)}{h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Rectangle in a circle of radius a that maximizes x^n+ y^n
Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $$x^2 + y^2 = a^2,$$ and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $$z = x^n + y^n.$$ If $n = 2$, it is clear that $z$ has the constant... | Alternative way, without calculus.
*
*If $\,n \le 2\,$ then by the generalized means inequality $\,\left(\frac{x^n+y^n}{2}\right)^{1/n} \le \left(\frac{x^2+y^2}{2}\right)^{1/2}=\frac{a}{\sqrt{2}}\,$. The maximum is attained when the equality is achieved, for $\,x=y\,$ i.e. the square.
*If $\,n \gt 2\,$ then $\,x^n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the point... | Note that
$$ 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)= \frac{3-\sqrt2}3
$$
not ${3+5\sqrt{2}\over 3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Alternative way to solve a limit problem $$
\lim _{n \rightarrow \infty} \frac{1}{1+n^{2}}+\frac{2}{2+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
I want to find the limit of this infinite series which I found in a book. The answer is $1/2$.
The solution to this limit was given by Sandwich/Squeeze Theorem, which was basically th... | This answer uses the same idea as the answers of TurlocTheRed and Koro, but shows that one doesn't need infinite series or big-$O$ notation. Each denominator is extremely close to $n^2$, so let's compare explicitly:
\begin{align*}
\sum_{k=1}^n \frac{k}{k+n^2} &= \sum_{k=1}^n \frac{k}{n^2} - \sum_{k=1}^n \bigg( \frac{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Solving recursive formula dependent on a periodic function Say we have some initial values $a_1, a_2, \dots, a_k$ and a recursive formula $a_n = f(n) + c_{1}a_{n-1} + c_{2}a_{n-2} + \dots + c_{k}a_{n-k}$ Where $f(n)$ is periodic and $a_n,f(n),c_i \in \mathbb{N}$.
For example let $a_1 = 1$ and $a_n = f(n) + 2a_{n-1}$ wh... | Suppose $\,f(n)\,$ has period $\,K\,$, then eliminating $\,f(n+K)=f(n)\,$ between the two recurrence relations for $\,a_n+K\,$ and $\,a_n\,$ gives a linear homogeneous recurrence that no longer includes $\,f(n)\,$, which can be solved with the usual methods.
$$
\begin{cases}
\begin{align}
a_{n+K} &= f(n+K) + c_{1}a_{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$? I want to show that $\left(\sqrt{\binom{n}{1}}+\sqrt{\binom{n}{2}}+...+\sqrt{\binom{n}{n}}\right)^2≤n(2^n-1)$.
My attempt:
We know that $\sqrt{ab}≤\frac{a+b}{2}$ because $(\sqrt{a}-\sqrt{b})^2≥0$ and that $\binom... | Hint: Use the CS inequality: $(1\cdot x_1+1\cdot x_2+\cdots+1\cdot x_n)^2 \le (1+1+\cdots+1)(x_1^2+x_2^2+\cdots+x_n^2)$, with $x_k = \sqrt{\binom{n}{k}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How many ways you arrange $12$ marbles where two marbles of the same color are considered the same? Problem:
Suppose that we have $6$ red marbles, $4$ blue marbles, $2$ green marbles and
$2$ black marbles. How many different ways are there to arrange them?
Note: Two marbles of the same color are considered identical.
A... | Your solution is correct.
We can choose six of the twelve positions for the red marbles in $\binom{12}{6}$ ways, four of the remaining positions for the blue marbles in $\binom{6}{4}$ ways, and two of the remaining four positions for the green marbles in $\binom{4}{2}$ ways. Both of the remaining positions must be occ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}$ If $\sum^{\infty}_{r=1} \frac{8}{(2r-1)\sqrt{(2r+3)(2r+5)}+\sqrt{(2r+1)(2r-1)}(2r+3)}=\sqrt{a}+\sqrt{\frac{5}{3}}-\sqrt{c}$, where $a$ and $c$ are coprime numbers, then the value of $a+\frac{c}{4}$ is
(1) 3
(2) 5
(3) 8
(4) 4
I... | Hint: Use telescoping series. Multiply numerator and denominator by $\sqrt{\frac{2r+5}{2r+3}}-\sqrt{\frac{2r+1}{2r-1}}$, to get final simplified expression for $t_{r}$ as under.
$$\begin{aligned}t_{r}&=\frac{8}{(2r-1)(2r+3)\left[\sqrt{\frac{2r+5}{2r+3}}+\sqrt{\frac{2r+1}{2r-1}}\right]} \\ &= \sqrt{\frac{2r+1}{2r-1}}-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Differentiability of $f(x,y)= \frac{xy^3}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$.
Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with\begin{align*} f:=\begin{cases} \frac{xy^3}{x^2+y^2},&(x,y)\neq (0,0)\\0, & (x,y)=(0,0). \end{cases} \end{align*}
Show that $f$ is differentiable at $(0,0)$.
My approa... | Noting
$$ \frac{|x|}{\sqrt{x^2+y^2}}\le 1, \frac{y^2}{x^2+y^2}\le1, $$
one has
$$ \bigg|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\bigg|=|y|\frac{|x|}{\sqrt{x^2+y^2}}\cdot\frac{y^2}{x^2+y^2}\le|y|$$
which gives
$$ \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}=0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4393649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Limits Question $$\lim _{x\to 0}\frac{\left(\cos \left(\frac{\pi }{2\cos \left(x\right)}\right)\right)}{\sin \left(\sin \left(x\right)^2\right)}=l$$
then find the value of {l}, where {l} denotes fractional part.
I tried multiplying and dividing the denominator with $\sin \left(x\right)^2$ and $x^2$ to sim... |
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{\cos \left(x\right)}\right)}{x^2}=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(-\frac{1}{2}\right)}{x^2}$
This step is incorrect .It should have been:
$=\lim _{x\to 0}\frac{\frac{\pi }{2}\left(\frac{\left(\cos \:\left(x\right)-1\right)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving PDE using characteristic method without polar coordinate. I'm trying to solve this PDE using method of characteristic without coordinate transformation :
$xu_x+yu_y = \sqrt{x^2+y^2}$ for $x^2+y^2>1$, $u(x,y)=x$ on $x^2+y^2 = 1$.
I only know how to use the method of characteristic for this kind of PDE such tha... | $$xu_x+yu_y = \sqrt{x^2+y^2}$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{\sqrt{x^2+y^2}}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ :
$$y=c_1x$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{\sqrt{x^2+(c_1x)^2}}$
$$u-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What rationale am I missing with this simple boundary value problem? Consider $$u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0$$
Antidifferentiating the PDE with respect to $y$ yields $u_x = \frac{xy^2}{2} + a(x)$, and antidifferentiating this equation with respect to $x$ yields $u = \frac{x^2 y^2}{4} + A(x) + b(y)$, wher... | $$u(x,y) = \frac{x^2 y^2}{4} + A(x) + b(y) \quad\text{is OK.}$$
Condition $u(0,y) = 0$ :
$$u(0,y) = A(0) + b(y) \quad\implies\quad b(y)=-A(0)=\text{constant}$$
$$u(x,y) = \frac{x^2 y^2}{4} + A(x) -A(0)$$
Condition $u_x(x, 0) = 0$ :
$u_x(x,y)=\frac{x y^2}{2} + A'(x)$
$$u_x(x,0)= A'(x)=0$$
$$A(x)=\text{constant}=C\quad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can we prove that the equation $2^n+1=5^m$ has no integral solution other than $(2, 1)$? $\textrm{Assume, on the contrary, that }n\geq 4,\text{then }$
\begin{aligned} 2^{n} &=5^{m}-1 \\ &=(5-1)\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ &=4\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ 2^{n-2} &=5^{m-1}+5^{m-2}+\cdots+1 \\ 2^{n... | Another way to do this. Assume that $n \ge 3$. Then for the equation $5^m \pmod {2^n} = 1$ to hold for positive integral $m$, the equation $5^m \pmod 8 = 1$ must hold, which gives $m$ even, and $n \ge 3$ gives $m$ positive as well.
So it now remains to show that there are no positive integral [in $n$ and $m$] solutions... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I find the centralizer for this matrix?
Let $B=\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \in GL_2(\mathbb{R})$. Determine the centralizer for $B$.
Here's what I've done so far:
Let $D=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \in GL_2(\mathbb{R})$.
$\begin{align}
BD&=DB\\
\begin{pmatrix}
2 & 2 \\
... | $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\R}{\mathbb{R}}$Allow me to sketch a slightly more conceptual approach.
$B$ has clearly the eigenvalues $2$ and $1$, hence it can be diagonalised, and one sees immediately that
$$
T^{-1} B T = \begin{bmatrix}2 & 0\\0 & 1\end{bmatrix},
\quad
\text{for }
T
=
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving by Induction that $x_{2n+1} = 1+1/2+1/2^3+\cdots+1/(2^{2n-1})$ Let $x_1 = 1$, $x_2 = 2$ and $x_n = \frac{x_{n-1}+x_{n-2}}{2}$. The sequence $(x_n)$ is Cauchy, which I can easily prove. To find its limit I must first show (by Induction) the assertation
$$
P(n): x_{2n+1} = 1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac... | After re-checking the algebra I finally found the solution I was looking for. The rest of the proof is thus as follows.
By definition $x_{2k+2} = \frac{x_{2k}+x_{2k+1}}{2}$ and $x_{2k+1} = \frac{x_{2k}+x_{2k-1}}{2}$. Thus $x_{2k} = 2x_{2k+1}-x_{2k-1}$; noting that $x_{2k-1} = x_{2(k-1)+1}$ and replacing $x_{2k+2}$ in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4406976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The palindrome counting function Let $b \geq 2$ be a positive integer and consider the function $f_b : \mathbb{N}^+ \to \mathbb{N}^+$ given by $$f_b (n) = |\{ k \in \mathbb{N}^+ : k \leq n \mbox{ and } k \mbox{ is palindromic in base } b\}|.$$
A simple counting argument shows that:
$$f_b (b^k) = \begin{cases} (1+b) b^{... | The proof is by me, not from any source, sorry.
Let $n$ be $k$-digit number in base $b$. Let $n=a_{k-1}a_{k-2}\dots a_0$. We split the interval $[0,n]$ to $[0,b^{k-1}), [b^{k-1},a_{k-1}a_{k-2}\dots a_{\lfloor k/2\rfloor}0\dots 0),[a_{k-1}a_{k-2}\dots a_{\lfloor k/2\rfloor}0\dots 0,n]$.
We count the number of palindrome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding roots of the equation $8x^3+4x^2-4x-1=0$ which are in form of cosine of angles. So, I start this by reducing it to a cubic with 0 coefficient of the 2nd degree term which gives me :
$$\left(x + \frac{1}{6}\right)^3 - \frac{7}{12}\left(x + \frac{1}{6}\right)-\frac{7}{216}=0$$
Replacing $\left(x + \frac{1}{6}\rig... | if we take $t=2x$ we have $t^3 + t^2 - 2t - 1$
next, take complex number $w$ with $w \neq 1$ but $w^7 = 1.$ Factoring out a $w-1$ tells us $1 + w + w^2 + w^3 + w^4 + w^5 + w^6 = 0.$ Now take $t = w + w^6$ which is the same as $w + \frac{1}{w}.$ We find $ t^2 = w^2 + 2 + w^5,$ while $t^3 = w^3 + 3 w + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
... | I don't know if this may interest you or answer your question but if you play around with field different from $\mathbb{R}$, special things happens. For example working with field $\mathbb{F}$ with characteristic equal to 2 it's always true that
$$(a+b)^2 = a^2 + b^2, \space \forall a,b \in \mathbb{F}$$
And if we take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 6,
"answer_id": 4
} |
Sum of "alternating-like" series I tried to prove that the following series converges.
$$1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} +\frac{1}{9} - ...$$
I found a expression for this sum in sigma form by summing pair of numbers with the same sign:
$$1 + \sum_{n=1}... | $$S_p=1 + \sum_{n=1}^{p} \frac{8n +1}{16n^{2}+4n} - \sum_{n=1}^{p}\frac{8n -3}{16n^{2} - 12n +2} $$
Notice that
$$ \frac{8n +1}{16n^{2}+4n}=\frac{1}{4 n+1}+\frac{1}{4 n}$$
$$\frac{8n -3}{16n^{2} - 12n +2}=\frac{1}{4 n-1}+\frac{1}{4n-2}$$ Computing the four summations
$$\sum_{n=1}^{p}\frac{1}{4 n+1}=\frac{1}{4} \left(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Investigate whether the sequence $\frac{nx^n + 3\sin{2n\pi x}}{n}$ is uniformly convergent Let $(f_n)$ be a sequence of functions defined by
\begin{equation*}
f_n(x) = \frac{nx^n + 3\sin{(2n\pi x)}}{n}
\end{equation*}
for all $x \in [0,1]$ and $n \in \Bbb N$.
Show that $f_n$ is converges pointwise to a function $f:[0,1... | There is simple way of showing that the convergence is not uniform.
Note that $f_n(x)=x^{n}+g_n(x)$ where $g_n(x)=\frac {3\sin (2\pi nx)} n$. Now $g_n(x) \to 0$ uniformly so $f_n$ converges uniformly if and only if $x^{n}$ does. Can you show that $x^{n}$ does not converge uniformly?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that:... | Futhermore, I tried C-S inequality and saw something "tricky":
It is obvious to get equivalent inequality: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{ca+abc}}{b+\sqrt{ac+abc}}\ge4$$
By C-S: $$2(a+b+c)=\sqrt{\left(a(b+c)+bc+abc\right)\left((b+c-a)^2+4a(b+c)\right)}\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find the value of $x^2+y^2$, given that $x^3-3xy^2=3$ and $y^3-3x^2y=3\sqrt{2}$ Let $x,y$ be reals that satisfy the equations:
$$\begin{align*}
x^3-3xy^2&=3 \\
y^3-3x^2y&=3\sqrt{2}
\end{align*}$$
Find the value of $x^2+y^2.$
The terms in the above equations are similar to the expansion $(x\pm y)^3=x^3\pm y^3\pm3xy(x\... | $\begin{array}\\
(x+iy)^3
&=x^3+3ix^2y+3i^2xy^2+i^3y^3\\
&=x^3+3ix^2y-3xy^2-iy^3\\
&=x^3-3xy^2+i(3x^2y-y^3)\\
&=3-3i\sqrt{2}\\
&=3(1-i\sqrt{2})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(\cos(a)-i\sin(a))
\qquad \cos(a)=1/\sqrt{3}\\
&=3\sqrt{3}e^{ia}\\
x+iy
&=\sqrt{3}e^{ia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Length of the shadow of a tower at 4:00 pm compared to the length of the shadow at 10:00 am Consider a location on the Earth where the Sun is overhead at noon. Compared to its shadow at 10:00 am, the shadow of a tower at 4:00 pm would be
(1) twice longer
(2) three times longer
(3) four times longer
(4) eight times long... | You would compare the two after you calculate it.
Here is another solution:
$\frac{y}{x} = tan(60) = \sqrt 3$.
Therefore, $y = \sqrt 3 \times x$.
$\frac{z}{x}=tan(30)=\frac{1}{\sqrt3}$
Therefore, $z = \frac{x}{\sqrt 3}$.
$\frac{y}{z}=\frac{x \times \sqrt 3}{\frac{x}{\sqrt 3}} = 3$.
Therefore, the shadow will be 3 tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve this equation $12^x-5^y=19$ positive integers Find all $x,y$ be positive integers,such
$$12^x-5^y=19$$
I found $(x,y)=(2,3)$ is solution,maybe have other,so I consider case $x,y>3$ and $\pmod 9$,since
$$12^x\equiv 0\pmod 9,x\ge 2$$
then $5^y\equiv -1\pmod 9$,
$5^1\equiv 5 \pmod 9$
$5^2\equiv 7 \pmod 9$
$5^3\equi... | To show that there are no solutions with $x \ge 3$, let's consider the equation modulo $27$. When $x\ge 3$, the equation becomes $-5^y \equiv 19 \pmod{27}$, or $5^y \equiv 8 \pmod{27}$. This calculation is intimidating, but rewarding: $5$ has multiplicative order $18$ modulo $27$, and so we are able to conclude $y \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering:
$$\sqrt{6}-\sqrt{2} =
\sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}... | In the general case let solve: $$\sqrt{x\pm y}=\sqrt{a}\pm\sqrt{b}$$
Since LHS exsits and is positive we will also assume $x\ge y$ and $a\ge b$.
By squaring expressions we get the system $\begin{cases}x+y=a+b+2\sqrt{ab}\\x-y=a+b-2\sqrt{ab}\end{cases}\iff \begin{cases}x=a+b\\y=2\sqrt{ab}\end{cases}$
Solving for $a,b$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find minimum of the function using AM-GM Problem: Find the minimum of the function $f(x,y)=x + \frac{8}{y(x-y)}$, where $x>y>0$ using AM-GM.
My attempt:
$$f(x,y)=2\cdot \frac{x+\frac{8}{y(x-y)}}{2} \ge 2 \sqrt{\frac{8x}{y(x-y)}}$$
But here nothing cancels and I couldn't get a constant. I also tried partial fraction, bu... | First off $x= \frac{y+ (x-y)}{2} >= \sqrt{y(x-y)}$
AM-GM was used for the first time.
So now we have $x+ \frac{8}{\sqrt{y(x-y)}} >= x+ \frac{8}{x}$
Then one uses AM-GM inequality for the second time to conclude that
$x+ \frac{8}{x} >= 2\sqrt{x*\frac{8}{x}} = 4\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $x,y,z \in \mathbb{R}$ are such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$, prove that $(x-y)(y-z)(z-x)=1.$ If $x,y,z$ are real numbers that not equal to 0 such that $x^2+x=y^2$, $y^2+y=z^2$ and $z^2+z=x^2$ prove that $(x-y)(y-z)(z-x)=1$.
I solve this problem by adding each two equations, but I'm interested can w... | Adding the three equations we get
$$x+y+z=0.$$
So then, if we rewrite the equations as $y^2-x^2=x$, $z^2-y^2=y$, etc., we find $(y+x)(y-x)=x$, $(z+y)(z-y)=y$, $(x+z)(x-z)=z$, which in view of the above relation we can rewrite as $(x-y)z=x$, $(y-z)x=y$, $(z-x)y=z$. Multiplying these three equations, and using that $xyz ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_{1/\sqrt{2}}^1\frac{1}{(3y^2-1)\sqrt{2y^2-1}}\,\mathrm{d}y$; how do you avoid using a complex substitution? $\newcommand{\d}{\mathrm{d}}$The given exam question - I provide the beginning for context:
Let: $$I=\int\frac{1}{(b^2-y^2)\sqrt{c^2-y^2}}\d y$$Where $b,c\gt0$, and employ the substitution $y=\f... | Yes, the same kind of substitution can be used here. The substitution $t=\frac{y}{\sqrt{2y^2-1}}$ transforms the integral in question to
$$\int_1^\infty\frac{\mathrm{d}t}{t^2+1}=\tan^{-1}t\Big|_1^\infty=\frac{\pi}{4} $$
More generally, this substitution can be used to evaluate integrals of the form
$$\int\frac{\mathrm{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000} =\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}$ I am trying to show that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{1}{1999}-\frac{1}{2000}
=\frac{1}{1001}+\frac{1}{1002}+\ldots+\frac{1}{1999}+\frac{1}{2000}... | Let $$S(n) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1} - \frac{1}{2n}.$$
Then notice if we add $$T(n) = 2 \left(\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n}\right)$$ to $S(n)$, all the negative terms become positive and we get a nice sum:
$$S(n) + T(n) = 1 + \frac{1}{2} + \frac{1}{3} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $x^4+y^4+z^4=1$ Find the minimum value of $\sum_{cyc}\frac{x^3}{1-x^8}$. Let $x,y,z$ be positive reals such that $x^4+y^4+z^4=1$. Find the minimum value of
$$\sum_{cyc}\frac{x^3}{1-x^8}$$
First I tried Jensen. Let $$f(x)=\frac{x^3}{1-x^8}$$
Then $f$ is convex on $\mathbb R^+$, using Jensen $$\frac{1}{3}\sum_{cyc... | Consider the function $$f(x)=\frac{1}{1-x^8}$$
And note that $f$ is convex on $[0,1[$ hence it lies above its tangent line, $$f(x)\ge f'(a)(x-a)+f(a)$$
set $a=1/\sqrt[4]{3}$
$$\frac{1}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x$$
Meaning $$\frac{x^3}{1-x^8}\ge \frac{9}{8}\sqrt[4]{3}x^4$$
Now cycle sum to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find a point of the plane which distance to a line minimal is Let be $P \subseteq R^3$ (P is a plane)and $X_1, X_2 \subseteq R^3$ two lines given by:
$P = \Bigg\langle \begin{pmatrix} 2 \\ 0 \\ 1\end{pmatrix},\begin{pmatrix} 0 \\ 2 \\ 1\end{pmatrix} \Bigg\rangle$,
$X_1 = \Bigg\langle \begin{pmatrix} 2 \\ 0 \\ 1\e... | A bit late answer but I think worth mentioning it because it doesn't need any Gram-Schmidt:
Note that the direction vector $\begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}$ of $X_2$ is also a spanning vector of $P$. Hence, the line $X_2$ is parallel to $P$.
The line $X_2$ passes through point $(0 , -2 , 5)$. So, you need to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Writing the recurrence $O_t=-\frac1{T_w}\sum_{i=t-T_p}^{t-1}O_i-\frac1{T_i}\sum_{i=1}^{t-T_p-1}O_i+B_t$ in terms of its initial value I want to write the following solely in terms of its initial value $O_1$
$$
O_t = - \frac{1}{T_w} \sum_{i=t-T_p}^{t-1} O_{i}
- \frac{1}{T_i} \sum_{i=1}^{t-T_p-1} O_i
+ B_t
... | Hint.
Considering first $T_p = 3$ from
$$
\left\{
\begin{array}{rcl}
\frac{o_0+o_1+o_2}{T_w}+o_3 &=& b_3 \\
\frac{o_1+o_2+o_3}{T_w}+o_4&=&b_4 \\
\frac{o_1}{T_i}+\frac{o_2+o_3+o_4}{T_w}+o_5 &=& b_5\\
\frac{o_1+o_2}{T_i}+\frac{o_3+o_4+o_5}{T_w}+o_6 &=& b_6\\
\end{array}
\right.
$$
we can represent a matrix formula as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\sqrt[3]{9\sqrt3-11\sqrt2}$ Simplify $$\sqrt[3]{9\sqrt3-11\sqrt2}$$
How can we actually simplify this radical?
| Let $x = \sqrt[3]{9\sqrt 3 -11 \sqrt 2}$. Cubing both sides gives $x^3 = 9\sqrt 3 -11 \sqrt 2$.
Per the suggestions in the comments, assume $x = a\sqrt 3 + b \sqrt 2$. Then:
$$x^3 = (a\sqrt 3 + b \sqrt 2)^3$$
$$x^3 = (a\sqrt 3)^3 + 3(a\sqrt 3)^2(b \sqrt 2) + 3(a\sqrt 3)(b \sqrt 2)^2 + (b \sqrt 2)^3$$
$$x^3 = a^3(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Find number of positive integer solutions to the equation $a+b+c=15$ where a,b are even and c is odd This is equivalent to:
Coefficient of $x^{15}$ in $(1+x^2+x^4+\cdots+x^{14})^2(x+x^3+\cdots+x^{15})$
Or
Coefficient of $x^{14}$ in $(1+x^2+x^4+\cdots+x^{14})^3$
Is this approach correct?
| It's already been pointed out that the OP has an error in that $a,b,c$ are supposed to be positive, not non-negative.
But I would like to point out that the algebra is a little simpler if we use the constraints $a,b,c \ge 1$ without taking into account our knowledge that $a,b,c \le 15$. This gives the generating func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $x$ such that $\frac{1}{2^{p-1}} -\frac{1}{a^p}>x$? Let $a, p\in\mathbb{R}$ be such that $a\ge 2$ and $p>1$. Consider the quantity
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}.$$
Since $a\ge 2$, it seems clear to me that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>0.$$
My question is: it is possible to find a positive real nu... | $$\frac{1}{2^{p-1}} -\frac{1}{a^p}\geq \frac{1}{2^{p-1}} -\frac{1}{2^p}$$
So if you want strict inequality, we simply have:
$$\frac{1}{2^{p-1}} -\frac{1}{a^p} > \frac{1}{2}\left(\frac{1}{2^{p-1}} -\frac{1}{2^p}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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provided $f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}$, what is f(x)? $$
f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}
$$
Here is one of the solution from my workbook:
$$
\begin{array}{l}
f( x+1) \ =\ \lim _{n\rightarrow \infty }\left[\left( 1+\frac{2+x}{n-2}\right... | $$y=\left(\frac{n+x}{n-2}\right)^{n}=\left(\frac{(n-2)+(x+2)}{n-2}\right)^{n}=\left(1+\frac{x+2}{n-2}\right)^{n}$$ Taking logarithms
$$\log(y)=n \log\left(1+\frac{x+2}{n-2}\right)\sim n \frac{x+2}{n-2}=\frac{n}{n-2}(x+2)\sim x+2$$ So
$$y=f(x+1)\sim e^{x+2} \implies f(x)=e^{x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$
Find $x$ in the equation $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$.
My attempt:
$$\log\left(x\right)\ -\ \frac{1}{2}\ \log\left(x-\frac{1}{2}\right)\ =\ \log\... |
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
Note that $$\log\left(x^2\right)=t\kern.6em\not\kern-.6em\implies2\log(x)=t,$$ but that $$2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximal and minimal eigenvalues of a symmetric tridiagonal Toeplitz matrix
Given $m \times m$ symmetric tridiagonal Toeplitz matrices $$M=\begin{pmatrix}
4 & 1 & & \\ 1 & 4 & \ddots & \\ & \ddots & \ddots & 1\\ & & 1 & 4\end{pmatrix}, \qquad A=\begin{pmatrix} 2 & -1 & & \\ -1 & 2 & \ddots & \\ & \ddots & \ddots ... | According to the notation $$M=4I+J,\quad A=2I-J$$ Therefore
$$ M+h^{-\alpha}A =(4+2h^{-\alpha})\,I +(1-h^{-\alpha})\,J$$ Every eigenvalue of $M+h^{-\alpha}A$ is thus of the form $$ [4+2h^{-\alpha}]+(1-h^{-\alpha})\, \lambda\qquad (*)$$ where $\lambda$ is an eigenvalue of $J.$ The eigenvalues of $J$ are known, and are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$\forall a\ge 2, \exists n\ge 1$ such that $2n+1\mid a+n!$ Let $a\ge 2$ be an integer. Prove that $\exists n\ge 1$ such that $$2n+1\mid a+n! \text{ or }a-n!$$
I think we'll construct a solution for $n$ as a function of $a$. But I have no idea what that function looks like. I just took care of some trivial cases.
First ... | As you've already noted, $n = 1$ works with either $a \equiv 1 \pmod{3}$ (since $3 \mid a - 1$) or $a \equiv 2 \pmod{3}$ (since $3 \mid a + 1$). Also, as aschepler's question comment indicates, $n = 4$ works where $a$ has just one factor of $3$. This is because with $a = 3i$, then if $i \equiv 1 \pmod{3}$, we get $a + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Show that $\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ=\cos7^\circ$
Show that $$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ=\cos7^\circ$$
NOTE: I have seen the other questions and solutions for this problem. I have a particular question if my idea has a potential.
I decided to rearrange the LHS (with... | $$4\sin18^\circ\cos36^\circ\cos7^\circ=\frac{4\sin18^\circ\cos18^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\frac{2\sin36^\circ\cos36^\circ\cos7^\circ}{\cos18^\circ}=\\
\frac{\sin72^\circ\cos7^\circ}{\cos18^\circ}=\cos7^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all integer values of $m$ such that the equations $x+y+\sqrt{4-x^2}+\sqrt{4-y^2}=4$ and $x\sqrt{4-x^2}+y\sqrt{4-y^2}=m+2$ have a solution.
Consider the system of equation $\left\{ \begin{aligned} x + y + \sqrt{4 - x^2} + \sqrt{4 - y^2} &= 4\\ x\sqrt{4 - x^2} + y\sqrt{4 - y^2} &= m + 2 \end{aligned} \right. (x; y ... | $$\begin{align}
x + y + \sqrt{4-x^2} + \sqrt{4-y^2} &= 4 \tag 1\\
x\sqrt{4-x^2} + y\sqrt{4-y^2} &= m + 2 \tag 2
\end{align}$$
Obviously, $x,y\in[-2,2]$, that is we can represent $x$ and $y$ as
$$\begin{align}
x &= 2\sin a \\
y &= 2\sin b
\end{align}$$
which transforms (1) and (2) into
$$\begin{align}
\sin a + \sin b +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Criteria for $3 \times 3$ matrix to positive definite Here it is said that a $2\times 2$ matrix $A$ is positive definite if and only if $tr(A) >0$ and $det(A)>0$. This will not work if $A$ is $3\times 3$. But is there any way to enforce the positive definiteness of the matrix $A$ via the trace and determinant of $A$, i... | No, it is not possible. Take any two real numbers $a$ and $b$ such that $a^2+b^2=1$ and the matrix
$$
A=
\begin{pmatrix}
1 & a & b \\
a & a^2+1 & a b \\
b & a b & -a^2+2
\end{pmatrix}.
$$
Then $\operatorname{tr}(A)=4$, $\det(A)=1$, and $A$ is definite positive.
But if you take two real numbers $a$ and $b$ such that $a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find $ \int \frac{dx}{(x^2 + a^2)^3} $ I would like to find the anti-derivative
$$\displaystyle \int \dfrac{dx}{(x^2 + a^2)^3 }$$
My attempt:
By substitution: $ x = a \tan(\theta) \Rightarrow dx = a \sec^2(\theta) d\theta$
Then the integral becomes
$$\displaystyle \int \dfrac{ a \sec^2(\theta) d\theta }{a^6 \sec^6(\t... | Alternatively, integrate by parts to avoid the trig substitution
\begin{align}
I_3=&\int \frac{1}{(x^2 + a^2)^3 }dx
=\int \frac{1}{4a^2x^3}d\left(\frac{x^4}{ (x^2 + a^2)^2}\right)
= \frac{x}{4a^2 (x^2 + a^2)^2}+\frac3{4a^2}I_2\\
I_2=&\int \frac{1}{(x^2 + a^2)^2}dx
=\int \frac{1}{2a^2x}d\left(\frac{x^2}{ x^2 + a^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Eigenvalue and spectral condition
Let $A=\begin{pmatrix} 1& 1 \\ a^2 &1 \end{pmatrix} \text{ with } a\in (0,\frac{1}{2}]$. Show $$cond_2(A)=||A||_2 \cdot ||A^{-1}||_2\leq 4(1-a^2)^{-1}$$ by first showing $||A||^2_2\leq||A||_1||A||_{\infty}$.
$||A||^2_2$ is the maximal eigenvalue of $A^TA$ and $||A||_1||A||_{\infty}=2... | $0 \lt 1-a^2 = \det\big(A)=\lambda_1\cdot \lambda_2 = \sigma_1\cdot \sigma_2$
i.e. the product of singular values gives the modulus of the determinant (and in particular when the determinant positive it is equal to the product of singular values)
proof: using polar decomposition
$\det\big(A\big) =\det\big(UP\big)=\det... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Evaluate/asymptotic of the sum $\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$ I am trying to evaluate the sum and its asymptotic limit as $N \rightarrow \infty$ of
$$\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\rig... | I get
$ \frac{N}{2}(1-2\ln(2))+O(L(N))
$.
So we want
$m \le N/(2a+1) \lt m+1
$
or
$m/N \le 1/(2a+1) \lt (m+1)/N
$
or
$N/m \ge 2a+1 > N/(m+1)
$
or
$(\frac{N}{m}-1)/2 \ge a \gt (\frac{N}{m+1}-1)/2
$
or
$\frac{N-m}{2m} \ge a \gt \frac{N-m-1}{2(m+1)}
$.
The number of these is about
$\begin{array}\\
\frac{N-m}{2m} -\frac{N-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x$ Latest Edit
As suggested by Mr Claude Leibovici, I go further to investigate the integral in general,
$$I_{m}:=\int_{0}^{\frac{\pi}{2}} x^{m} \ln (\cos x) d x$$
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac... | Using $(*)$, I think that we could even make the problem more general using the incomplete gamma function or the exponential integral function
$$\int x^p \cos(2nx)\,dx=-\frac{1}{2} x^{p+1} (E_{-p}(-2 i n x)+E_{-p}(2 i n x))$$
$$K_p=\int_0^{\frac \pi 2} x^p \cos(2nx)\,dx=$$ $$-i^{p+1} 2^{-p-2} n^{-p-1} \left(\Gamma (p+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solving the integral of $\cos^2x\sin^2 x$ Solving the integral of $\cos^2x\sin^2 x$:
My steps are: $(\cos x\sin x)^2=\left(\frac{\sin(2x)}{2}\right)^2$. Now we know that
$$\sin^2(\alpha)=\frac{1-\cos (2x)}{2}\iff\left(\frac{\sin(2x)}{2}\right)^2=\frac 14\sin^2(2x)=\frac 14\cdot \frac{1-\cos (4x)}{2}=\frac 18(1-\cos (4x... | Integrate by parts
\begin{align}
\int \cos^2x\sin^2 x \ dx =& -\frac1{16}\int\tan 2x\ d(\cos^2 2x)= -\frac1{16}\tan2x\cos^22x+\frac18 x+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{k=1}^{n} a_k \left(1-\frac{k^2}{n^2}\right) \to 1$
Problem : Given $a_n$ which satisfies $\displaystyle\sum a_n = 1$, prove that
$$\lim_{n\to\infty}\sum_{k=1}^na_k\left(1-\frac{k^2}{n^2}\right) = 1.$$
My Attempt
From $\displaystyle\sum a_n = 1$, I changed this problem to show $\displaystyle\lim\sum... | The standard approach to tacking abelian-type results like this is to consider $s_n = \sum_{k=1}^{n} a_k$ and write the sum in terms of $(s_n)$ using summation by parts:
\begin{align*}
\sum_{k=1}^{n} a_k \left( 1 - \frac{k^2}{n^2} \right)
&= \sum_{k=1}^{n} (s_k - s_{k-1}) \left( 1 - \frac{k^2}{n^2} \right) \\
&= \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$ So, my friend has challenged me to solve the following integral
$\displaystyle \tag*{} I=\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$
I started by doing $x^2=t$, so
$\displaystyle \tag{1} I=\frac 12 \int_0^\i... | Note that $a_\pm =\sqrt{\frac{1\pm i\sqrt3}{2}}=\frac{\sqrt3\pm i}2$
and
\begin{align}
I=&\ \frac 12 \int_0^\infty \frac{\arctan x\ln(1+x^2)}{x(x^4+x^2+1)} dx\\
=& \ \frac{1}{2\sqrt 3 i} \int_0^\infty \frac{\arctan x\ln(1+x^2)}{x(x^2+a_-^2)}-\frac{\arctan x\ln(1+x^2)}{(x^2+a_+^2)}\ dx\\
= & \ \frac\pi{4\sqrt3 i}\bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Proof $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition I'm trying to prove that $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition. Here is my proof so far:
My effort: Let $ \epsilon \gt 0 $, Choose $ \delta = min\{1, \}.$ Suppose $ 0 \lt \le... | $|\frac{x-3}{x^2-5x+6}-1|<\epsilon$
$ \Leftrightarrow$ $\frac{|x-3||x-3|}{|x-3||x-2|}<\epsilon$
$ \Leftrightarrow$$\frac{|x-3|}{|x-2|}<\epsilon$
let's put $|x-3|<\frac{1}{2}$ so $\frac{1}{|x-2|}<2$,so finally;
$|x-3|<\frac{\epsilon}{2}$ and $|x-3|<\frac{1}{2}$(our assumption),so we can put $\theta = min(\frac{\epsilon}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3... |
Squaring both sides was too complicated so I tried to substitute
$a = \sqrt{5−x}, ~b = \sqrt{x + 3}.$
This gave the new multivariable equation: $~a+b−2ab=−4$.
$a + b - 2ab = -4.$
$a^2 + b^2 = 8.$
This is nice work, so far.
Subtracting the 1st equation above, from the 2nd equation above gives
$(a^2 + b^2 + 2ab) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2}... | Thank you all for those who answered. Different amazing solutions!
But I'd still share my final solution, this is to be done as fast as possible so there's no proving needed and algebraic/arithmetic manipulation is probably the desired/suggested solution (not that mine is the optimal one, I just find this easier).
$x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
complex integral $\int_0^{2\pi} (\cos x)^n \;dx$ I tried to solve following integral $$\int_0^{2\pi} (\cos x)^n \;dx$$
by saying $z = e^{ix}$ we get $dx=\frac{-i \;dz}{z}$ and $\cos x = \frac{1}{2}\left( z + \frac{1}{z}\right)$
$$\int_0^{2\pi} (\cos x)^n \;dx = \frac{-i}{2^n}\oint_{|z|=1} \frac{1}{z}\left( z + \frac{1}... | The problem is that the $z^2+1$ in the numerator should be raised to the same power as the $2z$ in the denominator.
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_0^{2\pi}\cos^n(x)\,\mathrm{d}x
&=\int_{|z|=1}\left(\frac{z+\frac1z}2\right)^n\,\frac{\mathrm{d}z}{iz}\tag1\\
&=\frac{2\pi}{2^n}\Res_{z=0}\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the power series in powers of $x$ for the function Find the power series in powers of $x$ for the function
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt$$
Solution Verification Requesteed
$$f(x) = \int_0^x \frac{t^2}{1-t^4} \,dt \implies f'(x) = \frac{x^2}{1-x^4}$$
Now, we proceed with partial fraction decomposition. W... | Way faster:
$$ \frac{t^2}{1-t^4}=\sum_{n\geq 0} t^{4n+2} $$
for any $t\in(-1,1)$, so by termwise integration
$$ \int_{0}^{x}\frac{t^2}{1-t^4}\,dt = \sum_{n\geq 0}\frac{x^{4n+3}}{4n+3} $$
for any $x\in(-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sub-blocks of matrix quadratic equations Suppose the following matrix quadratic equation has at least one real solution for $X$:
$$
\begin{bmatrix}
A & a \\
0 & 1\\
\end{bmatrix} X^2 +
\begin{bmatrix}
B & b \\
0 & 0\\
\end{bmatrix} X +
\begin{bmatrix}
C & c \\
0 & 0\\
\end{bmatrix} = 0,
$$
where $A$, $B$ and $... | When $A=B=0$ but $C\neq 0$,
$A Y^2 + B Y + C = 0$ fails to have a solution.
However, letting $A,B,C$ be $1$-by-$1$ square matrices,
you can set $A=B=0\;$; $C=1\;$; $a=1\;$; $b=-1\;$; $c=0\;$; $X=\begin{bmatrix}
0 & 0 \\
1 & 0\\
\end{bmatrix}$
satisfying $\begin{bmatrix}
A & a \\
0 & 1\\
\end{bmatrix} X^2 +
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and ... | The Lagrange multiplier method is effective for this problem. First of all observe that the case $x=y=z$ is inadmissible. By symmetry we may assume that $z\neq x$ and $z\neq y.$ We have to solve for
$$\nabla(xyz)=\lambda \nabla(x+y+z)+\mu\,\nabla(x^2+y^2+z^2) $$ i.e.
\begin{eqnarray*}
yz &=& \lambda +2\mu x\\
xz &=& \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$ I want to show that:
$$\lim_{(x,y)\to(0,0)} (1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} = 1$$
Direct substitution doesn't work, because of the denominator of $\frac{1}{x^2 + y^2}$.
Usually these problems are solved by bounding the following expression... | Hint
Take the natural logarithm and use that $z-\frac{z^2}2\le \ln(1+z) \le z$.
Additionally we can switch to radial coordinates $x=r\cos\theta,\, y=r\sin\theta$ and let $r\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to solve such fraction differential equation? Here's my first-order differential equation:
$$
(x^3 - 2xy^2)dx + 3yx^2dy = xdy - ydx
$$
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.
| $$(x^3 - 2xy^2)dx + \color {red}{ 3yx^2dy} = xdy - ydx$$
$$(x^3 - 2xy^2)dx +\color {red}{ 2yx^2dy+yx^2dy} = xdy - ydx$$
Rearrange terms:
$$(x^3dx+x^2ydy) - 2xy(ydx - xdy )= xdy - ydx$$
$$x^2(xdx+ydy) = (xdy - ydx)(1-2xy)$$
$$xdx+yd y = (1-2xy)d \left (\dfrac {y}{x}\right)$$
$$d(x^2+y^2) = 2(1-2xy)d \left (\dfrac {y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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With $x_1^2+3x_2^2+2x_1x_2=32$, find max value of $|x_1-x_2|$
With $x_1^2+3x_2^2+2x_1x_2=32$, find the maximum of $|x_1-x_2|$.
I have tried with AM-GM, but can't solve it. With $d = |x_1-x_2|$ and $d^2 = 2x_1^2 + 4x_2^2 - 32$, then I wonder how to do next.
Thank a lot for helping!
| $(x_1+x_2)^2+2(x_2)^2=32.$
We are finding the value of $\max(|x_1-x_2|)=\max(|x_1+x_2-2x_2|)$.
Let $x_1+x_2=X, x_2=Y.$
Then, we can rewrite the problem:
If $X^2+2Y^2=32$, Find the maximum of $|X-2Y|$.
Let $\sin(A)=\dfrac{X}{4\sqrt{2}}.$
Then, $\cos(A)=\dfrac{Y}{4}.$
So, the problem changes:
Find the maximum of $|4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Evaluate the definite integral $I = \int_0^\pi \frac{\sin^2(\theta) d\theta}{10-6\cos(\theta)}$ I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omission... |
(2) if there are any better/faster ways of evaluating this integral.
Note that $$\int_a^b f(x)dx=\int_a^bf(a+b-x)dx$$ Also,
$\sin(\pi-x)=\sin x, \ \cos(\pi-x)=-\cos x.$ Thus,
$$I=\int_0^{\pi}\frac{\sin^2\theta}{10-6\cos\theta}d\theta=
\int_0^{\pi}\frac{\sin^2\theta}{10+6\cos\theta}d\theta$$ Thus,
$$2I=20\int_0^{\pi}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$ A\sin\alpha + B\cos\alpha=C$ solution confusion The solution of equation:
$$ A\sin\alpha + B\cos\alpha=C$$
according to python is:
$$\alpha=2\arctan\left[\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C}\right]$$
My question is why two answers or in otherwords in which situations we choose the postive and which the negative ? Can... | Starting with:
$$Asin(\alpha) + Bcos(\alpha) = C,$$
divide by $cos(\alpha)$ on both sides:
$$Atan(\alpha) + B = Csec(\alpha).$$
Then, square both sides:
$$A^2tan^2(\alpha) + 2ABtan(\alpha) + B^2 = C^2sec^2(\alpha).$$
Apply a variant of the Pythagorean identity to the RHS to get:
$$A^2tan^2(\alpha)+2ABtan(\alpha) + B^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Existence of Simsons Line using complex numbers Let $ABC$ be a triangle and $P$ is a point on its circumcircle. $X,Y$ and $Z$ are the feets of the perpendicular from $P$ on $BC, CA$ and $AB$, respectively. Prove that points $X, Y, Z$ are collinear.
Now proving this using angle chasing isn't hard. But I want to prove i... | Noting $\;\overline{p}=1/p \;$ we have
$$\frac{(b-a)(1-c\bar p)}{(c-a)(1-b\bar p)}=\frac{(b-a)(p-c)}{(c-a)(p-b)} =\frac{a-b}{a-c}\div \frac{p-b}{p-c}$$
which is real since $A,B,C,P \;$ are concylic. $\; \blacksquare$
Alternatively, it also suffices to verify
$$\frac{\left(\frac{1}{b}-\frac{1}{a}\right) \left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that for any $k$ between $2$ and $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$ I am trying to show that for any $k$ from $2$ to $n-2$ we have $\binom{n}{k} \ \ge \binom{n}2$.
What I tried
I started from the fact that $\binom{n}{k} \le \binom{n}{k+1}$ if $k \le \frac{n-1}{2}$ and by using the symmetry of bi... | Proof by induction:
(1) Base case: $n=6$, easy to verify, $2<k=3<6-2$,
$$~~~~~~~~~~~~~\binom{6}{3}>\binom{6}{2},~~\text{True.}$$
(2) Assume it is true for $n=m$, where $m>6$ and $m\in \mathbb{N^+}$, so we have:
$$\binom{m}{k}>\binom{m}{2}$$
(3) Induction step: when $n=m+1$,
$$\begin{align}
\binom{m+1}{k}&=\frac{(m+1)!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
If $x=\frac12(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}})$, what is the value of $(x+\sqrt{1+x^2})^3$?
I solved this problem as follow,
Assuming $\sqrt[3]{2009}=\alpha$ , we have $x=\frac12(\alpha-\frac1{\alpha})$ and $$... | Here's a rather elegant approach that involves pattern matching. Let $x=\sinh t$. Then
$$(x+\sqrt{1+x^2})^3 = (\sinh t + \cosh t)^3 = e^{3t}$$
And from comparing definitions
$$\sinh t = \frac{1}{2}(e^t-e^{-t})$$
$$x = \frac{1}{2}\left(\sqrt[3]{2009}-\frac{1}{\sqrt[3]{2009}}\right)$$
clearly, $e^t=\sqrt[3]{2009}$, thus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Analytical Solution to a nonlinear system of equations Is it possible to analytically solve for x,y,z in the following system, and if so how would one go about it?
$a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x = 0$
$a_{1} (z - 2y + x) - y + 2y^{3} - a_{2}y = 0$
$a_{1} (x - 2z + y) - z + 2z^{3} - a_{2}z = 0$
I've tried nume... | Let $\,s = x+y+z\,$ then the first equation can be written as:
$$
\begin{align}
0 &= a_{1} (y - 2x + z) - x + 2x^{3} - a_{2}x
\\ &= a_{1} (s - 3x) - x + 2x^{3} - a_{2}x
\\ &= 2x^{3} - (3 a_1 + a_2 + 1) x + a_{1} s
\end{align}
$$
The same equation is satisfied by $\,y\,$ and $\,z\,$. Assuming $\,x,y,z\,$ are distinct, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Interpreting the ratio test for $\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $ We want to find the limit of this.
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k} $$
I would have done it as follows:
$$\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n}\bigg| = \lim_{n\to\infty}\frac{\frac{n+1}{(n+1)^2+n}}{\frac{n}{n^2+n}... | Will you allow me a proof without the ratio test which I think you
didn't use it in the correct way. We are interested in the limit of the
sum $s_{n}=n[\dfrac{1}{n^{2}+1}+\dfrac{1}{n^{2}+2}+....+\dfrac{1}{n^{2}+n}]$.
Notice that $\dfrac{1}{n^{2}+1}<\dfrac{1}{n^{2}}$ and likewise .... till
$\dfrac{1}{n^{2}+n}<\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\... | *
*Continue with
\begin{align}
I=&\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx\\
=&\ 8\int_0^\infty \frac{\sqrt{t}(t+1)}{\left(t^2+2t+4\right)^2}dt\overset{x=\sqrt{\frac t2}}
=2\sqrt2\int_0^\infty \frac{2+\frac1{x^2}}{\left(x^2+\frac1{x^2}+1\right)^2}\overset{x\to\frac1x}{dx}\\
=& \ 3\sqrt2 \int_0^\infty \frac{1+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How many six-letter "words" can be formed from the alphabet $\{a-z\}$, if a "word" contains at least one vowel and at least one consonant?
How many six-letter “words” can be formed from the alphabet $\{a – z \}$?
*
*A “word” for this question must have at least one vowel $\{a, e, i, o, u\}$ and have at least one con... | I think that exponential generating functions can be used for this question. If there must be at least one consonant and at least one vowel , then there can be at most $5$ consonant or vowel.
*
*E.G.F for consonant : $$\bigg(21x+21^2\frac{x^2}{2!}+21^3\frac{x^3}{3!}+21^4\frac{x^4}{4!}+21^5\frac{x^5}{5!}\bigg)$$
*E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\binom{n}{0}\binom{n+1}{0}+\binom{n}{1}\binom{n+1}{1}+\ldots+\binom{n}{n-1}\binom{n+1}{n-1}+\binom{n}{n}\binom{n+1}{n}$ Here's a question that arose when I was trying to solve a probability problem in my textbook. I'm trying to calculate the sum:
$$\binom{n}{0}\binom{n + 1}{0} + \binom{n}{1}\binom{n + 1}{1}... | Let's use the binomial theorem to find:
$$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n$$
and
$$(x + 1)^{n + 1} = \binom{n + 1}{0}x^{n + 1} + \binom{n + 1}{1}x^n + \binom{n + 1}{2}x^{n - 1} + \cdots + \binom{n + 1}{n}x + \binom{n + 1}{n + 1}$$
Let's look at the coefficent of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Showing $\sum_{cyc} \frac{\cos(\frac{\alpha+\beta}{2})}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}=2$ when $\alpha+\beta+\gamma=\pi$ I saw this problem in a math magazine:
Let $\alpha,\beta$ and $\gamma$ be the angles of a triangle, so that
$$
\alpha+\beta+\gamma=\pi
$$
Show that
$$
\frac{\cos\left(\frac{\alpha+\beta}... | First off the bat, your formula in the first line is wrong, it should be, as per my comment, $$\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right) = \cos \frac{\pi}{2}\cos \frac{\gamma}{2}\color{purple}+\sin\frac{\pi}{2}\sin \frac{\gamma}{2}=\sin\frac{\gamma}{2}$$
I’ll take an alternative route. Expand $\cos\left(\dfrac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the ... |
$2\cos(\pi/8)=\sqrt{2+\sqrt{2}}$
$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}$
Start from your work,
$$2\cos(\pi/8)=2\sin(\frac{\pi}2-\frac{\pi}8)=2\sin(\frac{3\pi}8)$$
So
$$S=2\cos(\pi/8)+2\sin(\pi/8)=2\sin(\frac{3\pi}8)+2\sin(\frac{\pi}8)$$
Use formula:
$$\sin(x)+\sin(y)=2\sin({\frac{x+y}{2}})\cos(\frac{x-y}2)$$
We have
$$S=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
} |
Solve $(x^2+y^2+2z^2)^{1/2}=\cos z$ for $z$ near $(0,1,0)$
Can the equation $(x^2+y^2+2z^2)^{1/2}=\cos z$ be solved uniquely for $z$ in terms of $x$ and $y$ near $(0,1,0)$?
Put $F(x,y,z)=(x^2+y^2+2z^2)^{1/2}-\cos z$. Then $F(0,1,0)=0$,
$$
\nabla F(x,y,z)=(x(x^2+y^2+2z^2)^{-1/2},y(x^2+y^2+2z^2)^{-1/2},2z(x^2+y^2+2z^2)... | For this kind of problems, you need to use the multidimensional Taylor series taking directional derivatives.
Truncated to the second order, this gives
$$\frac{1}{2}x^2+y+\frac{3 }{2}z^2-1=0 \tag 1$$ Checking at $x=\frac 1{10}$, $y=\frac 9{10}$, we have
$$\sqrt{2 z^2+\frac{41}{50}}-\cos (z)=0$$ which, expanded, gives
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A question from Integration Bee 2022: Evaluate $\int^2_{1/2}\ln(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})})dx$ I am trying to evaluate the integral $$\int^2_{\frac{1}2}\ln\left(\frac{\ln(x+\frac{1}{x})}{\ln(x^2-x+\frac{17}{4})}\right)dx$$ by separating it as $$\int^2_{\frac{1}2}\ln\ln\left(x+\frac{1}{x}\right)d... | Separately
\begin{align}
\hspace{-10mm}\int^2_{1/2}\ln\ln(x^2-x+\frac{17}{4})dx\overset{x-\frac12\to x}
= \int^{3/2}_{0}\ln\ln(x^2+4)\ dx \\
\end{align}
\begin{align}
\int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)\overset{x\to \frac1x}{dx}
= &\ \frac12 \int^2_{1/2}\ln\ln\left(x+\frac{1}{x}\right)(1+\frac1{x^2})\overset{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Upper bound of the sum $\sum_{n=1}^\infty \tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)$ as $x\to \infty$ I am stuck at finding a Big-O upper bound of the following sum as $x\to \infty$ $$\sum_{n=1}^\infty \tan^{-1}\left(\frac{x}{1-(x-n)^2}\right)$$
I tried: Since $$ \sum_{n=1}^\infty\tan^{-1}\left(\frac{x}{1-(x-n)^2}\righ... | To get the leading asymptotics term, we can switch from summation to integration. As @Jack D'Aurizio mentioned, we are getting $O(1)$ term while evaluating $S_2$, therefore we can only try to catch the leading term. As was shown above, the leading term is $O(\sqrt x)$.
Indeed, implementing the program, denoting for a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equ... | Method. $(1)$, Use Mittag-Leffler's series for $\cot(x)$
$$\begin{align}
I&=\int_0^\frac{\pi}{2}x\ln(\sin(x))~dx=\frac{1}{2}\int_0^\frac{\pi}{2}\ln(\sin(x))~dx^2\\
\\
&=-\frac{1}2 \int_0^\frac{\pi}{2}\frac{x^2}{\tan(x)}~dx=-\frac{1}2\int_0^\frac{\pi}{2}x^2\left(\frac{1}x+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Finding all functions satisfying $f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}$ Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x f(x+y))+f(f(y) f(x+y))=(x+y)^{2}, \forall x,y \in \mathbb{R} \tag1)$$
My approach:
Let $x=0$, we get
$$f(0)+f\left((f(y))^2\right)=y^2$$
$\Rightarrow$
$$f\left((f(y))^2\right)=y^... |
$f: \Bbb{R \to R}, f(xf(x+y))+f(f(y)f(x+y))=(x+y)^2.$
My attempt was to show that $f(0)=0$.. And I did it.
\begin{align}
P(0, y): \; & f(0)+f(f(y)^2)=y^2. \\
P(0, 0): \; & f(0)+f(f(0)^2)=0. \\
& \text{let } f(0)=k. \\
\ \\
\Rightarrow \; & k+f(f(y)^2)=y^2, k+f(k^2)=0. \\
\ \\
& \text{let } f(a)^2=f(b)^2. \\
P(0, a): ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.