Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Bhaskara approximation of cos(x) In my knowledge the best aproximation of $\sin$ and $\cos$ is Bhaskara approximation which is :
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
and
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$
what is t... | In terms of approximation, you can do a lot of things depending on how many terms you want to include in the expression. Most of the time, this is done using Padé approximants.
For example, for your question, using the simple $[3,2]$ Padé approximant built around $x=-\frac \pi 2$, we get
$$\cos(x)=\left(x+\frac{\pi }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit multivarible How do I solve this limit?
$$\lim_{x,y,z) \to (0,0,0)} \frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2+xyz} $$
This is equal to $$\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\times\frac1{1+\frac{xyz}{x^2+y^2+z^2}}$$
The first one is a standard limit with value one, but I'n not sure about the other term.
| $0 \leq | \frac {xyz} {x^{2}+y^{2}+z^{2}}|\leq \frac 1 2 |z| \to 0$ since $|xy| \leq \frac 1 2{(x^{2}+y^{2})}$. Hence the given limit is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why substitution in irrational equation doesn't give equivalent equation? I have two examples of irrational equations:
The first example: $\sqrt[3]{3-x} + \sqrt[3]{6+x}=3$
In solution, they take cube of both sides and do following:
\begin{eqnarray*}
&\sqrt[3]{3-x} &+ \sqrt[3]{6+x}=3\\
&\iff&
3-x+ 3\sqrt[3]{(3-x)(6+x)}(... | Very good question! The first example is misleading in writing equivalences everywhere; the second equivalence
$$3-x+ 9\sqrt[3]{(3-x)(6+x)}(\sqrt[3]{3-x} + \sqrt[3]{6+x})+6+x$$
$$\iff$$
$$3-x+ 9\sqrt[3]{(3-x)(6+x)} +6+x=27$$
should (at least conceptually) be an implication $(\implies)$. It is true that it is in fact an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$. I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!
When $m$ is even, put $m=2k$.
$2^m-1=(2^k-1)(2^k+1)$
By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime.
So, these ar... | From $p^n + 1 \ge p+1 \ge 4$ we have $m \ge 2$.
Hence $p^n = 2^m - 1 \equiv -1\pmod 4$, and thus $p^n$ cannot be a square.
This forces $n$ to be odd.
From $p^n + 1 = (p+1)(p^{n-1} - p^{n-2} + \dots + 1) = 2^m$, we see that
$$(p^{n-1} - p^{n-2} + \dots + 1) \mid 2^m$$
Since $p$ is odd, this factor, which is a sum of $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the coefficient of $x^n$ in the generating functions: $g(x) = \frac{x^3}{(1+x)^5 (1−x)^6}$ One of the problems in my Discrete Math course states that we need to find the coefficient of $x^n$ in generating function $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$
I separated $g(x) = \frac{x^3} {(1+x)^5 (1−x)^6}$ into $x^3$ * ... | The key is the generalized binomial theorem
in the handy special case
$(1-x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}x^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}x^k
$.
This implies that
$(1+x)^{-s}
=\sum_{k=0}^{\infty} \binom{s+k-1}{k}(-1)^kx^k
=\sum_{k=0}^{\infty} \binom{s+k-1}{s-1}(-1)^kx^k
$
and,
for any $m$,
$(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^... | $$\sin3A-\cos3A$$
$$=\sqrt2\sin\left(3A-\dfrac\pi4\right)$$
$$=\sqrt2\sin\left(3A+\dfrac{3\pi}4-\pi\right)$$
$$=-\sqrt2\sin3\left(A+\dfrac\pi4\right)$$
Similarly, $\sin A+\cos A=\sqrt2\sin(?)$
Finally for $\sin B\ne0,$ $$\dfrac{\sin3B}{\sin B}=\dfrac{\sin B(3-4\sin^2B)}{\sin B}=3-2(1-\cos2B)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Unable to represent the function as a power series Assuming I have this function below :
$$f(x) = \frac{(x-1)}{x^2+1}$$
How do I represent it as a power series ?
The only solution that I know is when using the geometric series :
$\sum_{k=1}^\infty q^k = \frac{1}{1-q}$ but this is not possible in our case .
| $$\frac{x-1}{x^2+1}=\frac{x}{x^2+1}-\frac{1}{x^2+1}$$
Integrate
$$\int\left(\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)\,dx=\frac{1}{2}\log(1+x^2)+\arctan x+C$$
Series for these functions are known, we get
$$\frac{1}{2}\log(1+x^2)-\arctan x+C=\frac{x^{10}}{10}-\frac{x^9}{9}-\frac{x^8}{8}+\frac{x^7}{7}+\frac{x^6}{6}-\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Expression of sum of squares as a sum of a specific form This question is a follow-up of this one.
Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$, and that $x,y$ are not both odd.
Do there exist $a,b,c,d \in \mathbb{Z}$ such that
$
(a+d)^2+(b-c)^2=x^2+y^2
$
and $ad-bc=1$?
This question is motivated by an a... | Let $x=0$ and $y=3$ so that $x^2+y^2=9$. Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that
$$(a+d)^2+(b-c)^2=x^2+y^2=9,\tag{1}$$
then $\{a+d,b-c\}=\{0,\pm 3\}$ for some for some choices of the signs. Then of course the matrices
$$\begin{pmatrix}-a&\hphantom{-}b\\\hphantom{-}c&-d\end{pmatrix},
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$. I am about to solve for the Laurent series of the function $f(z) = \frac{z-12}{z^2+z-6}$ that is valid for $1<|z-1|<4$.
What I did is I use the partial fraction decomposition to rewrite the $f(z)$ into $f(z) = \fra... | $$
\begin{align}
\frac{z-12}{z^2+z-6}
&=\frac3{z+3}-\frac2{z-2}\\
&=\frac3{(z-1)+4}-\frac2{(z-1)-1}\\
&=\frac34\underbrace{\sum_{k=0}^\infty\left(\frac{1-z}4\right)^k}_\text{converges for $|z-1|\lt4$}-2\underbrace{\sum_{k=1}^\infty\left(\frac1{z-1}\right)^k}_\text{converges for $|z-1|\gt1$}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
GCD of $n^2-3n-1$ and $n-4$ $n$ is a natural number and after trying the division algorithm
$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$
For the last part I'm not sure whether it does equal to $\gcd(n-1,3)$.
If yes, then should I take the cases where $n$ is $3k+i, 0\le i\le2$?
| Using this implementation of the Extended Euclidean Algorithm, rotated $90^{\large\circ}$ and applied to polynomials, gives
$$
\begin{array}{c|c|c}
\color{#00F}{n^2-3n-1}&1&0\\
\color{#00F}{n-4}&0&1\\
\color{#090}{n-1}&\color{#C00}{1}&\color{#C00}{-n}&n\\
\color{#090}{-3}&\color{#C00}{-1}&\color{#C00}{n+1}&1
\end{array... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$ find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$
I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be... | By plugging in $x=16$, we find $P(16)=0$. So $P$ has some positive roots. If $P$ is not $\equiv 0$, it has a minimal positive root $x_0$.
Plug in $\frac 12x_0$ to find $0=16(\frac12x_0-1)P(\frac12x_0)$ and so $x_0=2$ as $P(\frac12x_0)\ne 0$. By plugging in $x=2$, we then find $P(4)=0$, next $P(8)=0$ and $P(16)=0$. Simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simultaneous algebra question Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$
Find $$x^3+3y^3$$
Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$
I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$
| Being lazy, I changed a little the notations
$$ax+by=v \tag 1$$ $$ax^2+by^2=w \tag 2$$
From $(1)$, you have $y=\frac{v-a x}{b}$. Plug in $(2)$ to face
$$\left(v^2-b w\right)-2 a vx+a\left(a+ b\right)x^2=0$$
$$x_\pm=\frac{av\pm\sqrt{a b \left(w (a+b)-v^2\right)}}{a (a+b)}\qquad \qquad y_\pm=\frac{v-a x_\pm}{b}$$
Now, ju... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried ... | $$\overline{xyz138}=1000\overline{xyz}+138\equiv 6\overline{xyz}-2\pmod 7\\ \overline{138xyz}=138000+\overline{xyz}\equiv 5+\overline{xyz}\pmod{13}\\ \overline{x1y3z8}\equiv 10x+1+10y+3+10z+8\equiv 1-x-y-z\pmod{11}$$
Denote $a:=\overline{xyz}$, then you have the linear system
$$\begin{cases}6a-2\equiv 0\pmod 7\\5+a\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$ For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$
I couldn't solve this question and hence looked at the solution. The solution su... | Here I will show the thoughts that could help you to do all the steps above.
*
*The first substitution is unnecessary, but it simplifies the algebra, since you have $(\sqrt{a})^3+(\sqrt{b})^3+(\sqrt{c})^3=3$
*Now, for the inequality we want to prove we have $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3$ on the left side. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality with mean inequality
If $a,b,c > 0$ and $a+b+c = 18 abc$, prove that
$$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{c^{2}}}\geq 9$$
I started writing the left member as $\frac{\sqrt{3a^{2}+ 1}}{a}+ \frac{\sqrt{3b^{2}+ 1}}{b} + \frac{\sqrt{3c^{2}+ 1}}{c}$ and I applied AM-QM inequal... | $$\sqrt{3 +\frac{1}{a^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}+\sqrt{3 +\frac{1}{b^{2}}}$$
$$
= \sqrt{
9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\frac{1}{c^{2}} +
2\sum_{cyc}
{\sqrt{
\left(3 +
\frac{1}{a^{2}}\right)
\left(3 +\frac{1}{b^{2}}\right)}}
}
$$
$$\tag{By C-S}\geqslant \sqrt{9 +
\frac{1}{a^{2}} +
\frac{1}{b^{2}} +
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer
I said that since $a+b+c=0$ then $c=-a-b... | you have $$ 2017 + \left( a^2 + ab + b^2 \right)^2 = w^2 $$ or
$$ w^2 - \left( a^2 + ab + b^2 \right)^2 = 2017 $$
Difference of squares on the left, factors, meanwhile 2017 is prime. Thus
$$ w + a^2 + ab + b^2 = 2017 , $$
$$ w - a^2 - ab - b^2 = 1 , $$
or
$$ w = 1009$$
$$ a^2 + ab + b^2 = 1008 = 2^4 3^2 \cdot 7 = 7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$ The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?
| $$\sin(x)\cos(x)=\sin(x)+\cos(x)$$
$$\frac12 (\sin(2x))=\sin(x)+\cos(x)$$ $$\frac{\sin(2x)}{2}=\sin(x)+\cos(x)$$ $$\frac12 (\sin(2x))=\frac{2}{\sqrt2}(\sin(x+\frac{π}{4}))$$ $$\sin(2x)=\frac{4}{\sqrt2}\left(\sin(x+\frac{\pi}{4}) \right)$$ $$\sin(2x)=2\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ $$2\sin^2\left(x+\frac{π}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions? When factoring the quadratic $x^2 + bx + c $, where $b$ and $c$ are integers, why do we not consider the case where $u$ and $v$ in the factored form $(x+u)(x+v)$ are fractions?
We wish to write this as $(x + u) (x + v)$. ... | Notice that if $u = \frac{p}{q}$ and $v = \frac{m}{n}$ for some integers $p,q,m,n$ then you obtain the equation $$qn\left(x + \frac{p}{q}\right)\left(x + \frac{m}{n}\right) = qnx^2 + (mq + pn)x + pm.$$ In the expression $qnx^2 + (mq + pn)x + pm$ we can conclude that $qn, mq+pn$, and $pm$ are all integers. If we use t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
| Well, in the worst case when we are stuck we can do cases:
Case 1: $x+y\ge 0$ and $x-y \ge 0$ then
$x\ge -y$ and $x \ge y$ so $x \ge \max(y,-y) = |y|$.
And $|x+y| + |x-y| = (x+y)+(x-y) = 2x = 2$ so $x = 1$ and $|y| \le x=1$ so $-1 \le y \le 1$.
$x^2 - 6x +y^2 = -5+y^2$ and the max that $y^2$ can be is $1$ so the max of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability -- Proving Cantelli’s inequality by setting up a $E[φ(X)]$ and using Markov's inequality I'm working on proving cantellis inequality. We're told to use this setup:
$P(X >= a) ≤ E[φ(X)]/ φ(α)$ where $var[X] = σ^2$ and $E[X] = 0$ and $φ(x) = (x+c)^2$
An intermediate step was proving that $P(X ≥ α ) ≤ (σ^2+c^2... | Given that $$\mathbb P(X\geqslant\alpha)\leqslant\frac{\sigma^2+c^2}{(\alpha+c)^2},$$
we consider the map $f:[0,\infty)\to\mathbb R$ defined by $f(c) = \frac{\sigma^2+c^2}{(\alpha+c)^2}$. Now, $f(0) = \left(\frac\sigma\alpha\right)^2$, and $\lim_{c\to\infty}f(c)=+\infty$. It is clear that $f$ is differentiable on $(0,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $a+b+c+d$ where $(ab+c+d =13),\ (bc+d+a = 27),\ (cd+a+b = 30),\ (da+b+c = 17)$ Compute $(a+b+c+d)\ $ if $\ a,b,c,d\ $ satisfy the system of equations below:
$$\begin{cases}
ab+c+d = 13\\\\
bc+d+a = 27\\\\
cd+a+b = 30\\\\
da+b+c = 17
\end{cases}$$
.................................................................... | Let $u = a+b+c+d$, the set of equations can be rewritten as
$$
\begin{cases}
(a-1)(b-1) + a + b + c + d &= 14\\
(b-1)(c-1) + a + b + c + d &= 28\\
(c-1)(d-1) + a + b + c + d &= 31\\
(d-1)(a-1) + a + b + c + d &= 18
\end{cases}
\iff
\begin{cases}
(a-1)(b-1) &= 14- u\\
(b-1)(c-1) &= 28 -u\\
(c-1)(d-1) &= 31 -u\\
(d-1)(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Easy way to see that $(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$? As the title suggests, is there an easy way to see that$$(x^2 + 5x + 4)(x^2 + 5x + 6) - 48 = (x^2 + 5x + 12)(x^2 + 5x - 2)$$that doesn't require expanding in full? Is there a trick?
| For $a=x^2+5x+4$ we get to factor $a(a+2)-48=a^2+2a-48=(a+8)(a-6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
the limit of $\frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt2}+\frac{3}{1+\sqrt2+\sqrt3}+\dots+\frac{n}{1+\sqrt2+\sqrt3+\dots+\sqrt n})$ as $n\to\infty$ I need to find:
$$\lim_{n \to +\infty} \frac{1}{\sqrt{n}}(1+\frac{2}{1+\sqrt{2}} + \frac{3}{1+\sqrt{2}+\sqrt{3}} + \ldots + \frac{n}{1+\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}}), n \i... | The Stolz–Cesàro theorem indeed suffices, with a derivative-like limit (the last one below).
Let $a_n=\sum\limits_{k=1}^n\sqrt{k}$ and $b_n=\sum\limits_{k=1}^n k/a_k$; we're computing
\begin{align*}
\lim_{n\to\infty}\frac{b_n}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{b_n-b_{n-1}}{\sqrt{n}-\sqrt{n-1}}
\\&=\lim_{n\to\infty}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that the unique solution of $P_n(X) = 0$ is $x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)}$ I need to how that when $ k \neq 0$, the unique solution of $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i} = 0$ is $ x=i\frac{e^{2i\theta}+1}{e^{2i\theta}-1}=\frac{1}{\tan(\theta)} $ with $\theta=\frac{k... | We want the solutions to $$(x+i)^{2n+1} - (x-i)^{2n+1} =0.$$
Assume $x$ is real and $z=x+i = r e^{i\theta}$ so that $z^*=x-i=re^{-i\theta}$ is the conjugate of $z$.
Substituting and rearranging:
$$z^{2n+1} = {z^\ast}^{2n+1}$$
$$e^{i(2n+1)\theta}=e^{-i(2n+1)\theta}$$
$$e^{i2(2n+1)\theta}=1=e^{2\pi k i}$$
so $$\theta = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits... | If you define $g(x) = f(x) + e^{-x}$, then the equation becomes $g(x + y) = g(x) + g(y)$.
It is well-known that all continuous solutions to that equation are linear, i.e. there exists $c$ such that $f(x) = cx$.
The rest should be easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3920512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
We have $AB = BC$, $AC = CD$, $\angle ACD = 90^\circ$. If the radius of the circle is '$r$' , find $BC$ in terms of $r$ .
What I Tried: Here is a picture :-
Let $AC = CD = x$. As $\angle ACD = 90... | The given answer is larger than $r$, which clearly isn't true, as mirroring $B$ across $AC$ shows $|AB|<|AO|$, where $O$ is the center of the circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that $\lim\limits_{n \to +\infty}(\sin(\frac{1}{n^2})+\sin(\frac{2}{n^2})+\cdots+\sin(\frac{n}{n^2})) = \frac{1}{2}$ Show that the sequence defined as
$$x_n = \sin\left(\frac{1}{n^2}\right)+\sin\left(\frac{2}{n^2}\right)+\cdots+\sin\left(\frac{n}{n^2}\right)$$
converges to $\frac{1}{2}$.
My attempt was to evaluate... | Because $\sin x$ is convex on $x \in (0, \pi)$ then the sum of $\sin (kx)$ inside that interval will be not less than $n$ times the linear average on the secant $(\sin x+sin (nx))/2$ and not greater than $n$ times the linear average on the tangent $n \sin((n+1)/2 x)$, so in this case
$$
\eqalign{
& n\left( {{{\sin \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
} |
Brick wall with maximum height 3 Given n same-sized rectangular bricks. We want to build a wall with these constraints:
*
*All bricks should be horizontal.
*We can put a brick on two other bricks, such that the middle of the top brick is on the border of two other bricks.
*The maximum height of the wall should be 3... | This is only a partial answer, deriving a sixth-order linear recurrence.
A wall with $k$ bricks in the bottom row can be decomposed into $k$ diagonals with slope $-1$. The first diagonal contains only the leftmost brick in the bottom row. After that each diagonal can be at most one brick longer than the previous one, u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Small angle approximation on cosine The problem is
Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant
I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{... | By Taylor, as the function is even the development to second order is $f(0)+\dfrac{f''(0)}2x^2$ or
$$(3-2\cos(0)+4\cos^2(0))+(2\cos(0)+8\sin^2(0)-8\cos^2(0))\frac{x^2}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$ Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is ... | Using the binomial theorem,
\begin{align*}
\left( 1 - \frac{x}{3} \right)^5 &= \sum_{i=0}^{5} \binom{5}{i} 1^{5-i} \left(\frac{-x}{3} \right)^i \\
&= \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i \text{ and } \\
\left( 1 + \frac{2}{x} \right)^3 &= \sum_{j=0}^{3} \binom{3}{j}1^{3-j} \left(\frac{2}{x} \right)^j \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3932563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove the inequality $4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$ $$4(a+b+c)^3\ge 27(a^2b+b^2c+c^2a)$$, $a,b,c\ge0$
I tried to work out this inequality the way I did
So we have to show $$ 4(a + b + c) ^ 3 \ge 27 (a ^ 2b + b ^ 2c + c ^ 2a + abc) $$
One way is to use cyclic symmetry, and WLOG assumes $ a $ is $ a, b, c $ minute.... | The last inequality should be
$$9(x^2-xy+y^2)a+4x^3-15x^2y+12xy^2+4y^3 \geqslant 0,$$
or
$$9(x^2-xy+y^2)a+(4x+y)(x-2y)^2 \geqslant 0.$$
It's called BW method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$ If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$
Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that t... | Let $x = a^2+b^2, y = c^2+d^2 \implies x+y \le 1+xy \implies (x-1)(1-y) \le 0$, which is true since $ 0 \le x,y \le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $x^2 + py^2 = 4z^3$ with a fixed $z$ Given an odd prime $p$ and a positive integer $z$, I'd like to find all positive integers $x$, $y$ such that
$$
x^2 + p y^2 = 4 z^3.
$$
Because the left is positive definite there is only a finite number of solutions.
I could solve by incrementally guessing values of $y$ and... | I recommend Lehman as an introduction, simultaneous to binary quadratic forms and quadratic number fields.
I'm going to ignore the $4.$ It is there for a reason, but things begin more simply without it.
$$\left( x^2 + p y^2 \right)^3 = \left( x^3 -3 px y^2 \right)^2 + p \left( 3x^2 y - p y^3 \right)^2 $$
When $p \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$... | Suppose that we have $$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
and we want $Q$ such that $AQ =B$. Let
$$ Q = \begin{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing a double integral over $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$
Compute the integral $\int_{D} f \ dx \ dy$, when $f : D \to \mathbf{R}, f(x,y)= y^2x+x^3$ and $D= \{(x,y) \in \mathbf{R}^2 : \sqrt{x^2+y^2} < 5, y>0 \}$
It seems that $D$ defines a semicircle with radius $5$ in the $xy$-plane... | Note that
$$
\begin{split}
I
&= \int_0^\pi \int_0^5
\left(r^2\sin^2\theta \cdot r\cos\theta +r^3\cos^3\theta\right) r\ dr \ d\theta \\
&= \int_0^5 r^4 dr
\int_0^\pi \left(\sin^2\theta +\cos^2 \theta\right) \cos\theta d\theta \\
&= \frac{5^5}{5} \int_0^\pi \cos\theta d\theta
\end{split}
$$
which does eva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Laurent Series Expansion of $f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$ at $z=0$ For my function $$f(z) = \frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz}$$ I have determined that it contains $3$ poles at $z=0,z=-a,z=-\frac{1}{b}$
I believe these to be the only singularities.
I found them by factoring the denominator
$$f(z) = \f... | Before factoring take notice of the similarity of the coefficients in the numerator and denominator. This suggests that there is a short cut that can be taken before attempting to partial fraction that would make our lives easier.
$$\frac{bz^2+2z+b}{bz^3+(1+b^2)z^2+bz} = \frac{bz^2+(1+b^2)z+b+(1-b^2)z}{bz^3+(1+b^2)z^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to simplify the fraction? How to simplify the fraction $ \displaystyle \frac{\sqrt{3}+1-\sqrt{6}}{2\sqrt{2}-\sqrt{6}+\sqrt{3}+1} $ to $ (\sqrt{2}-1)(2-\sqrt{3}) $?
I've checked it in the calculator and both give the same value.
I was able to simplify it until $\displaystyle \frac{1+\sqrt2-\sqrt3}{3+\sqrt2+\sqrt3}... | \begin{align}
\frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} &= \frac {1 + \sqrt 2 - \sqrt 3}{3 + \sqrt 2 + \sqrt 3} \times \frac {3 - \sqrt 2 - \sqrt 3}{3 - \sqrt 2 - \sqrt 3}\\
&=\frac {3 - \sqrt 2 -\sqrt 3+3\sqrt 2-2-\sqrt 6-3\sqrt 3+\sqrt 6+3}{9 - (2 + 2\sqrt 6 + 3)}\\
&=\frac {4+2\sqrt 2-4\sqrt 3}{4 -2\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{... | First of all let investigate terms of inside the parentheses. When $x$ approaches to infinity $\displaystyle \frac{3x+2}{4x+3}=\frac{3}{4}$ then the limit transformed into $\displaystyle \lim_{x \to \infty} \left(\frac{3}{4}\right)^{x}$. Rewritre $\frac{3^x}{4^x}$. Obviously denomiter is greater than numerator. So $\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Calculating series expansions within a matrix: matrix exponential I have a $(3 \times 3)$ matrix $$ Y = \begin{pmatrix} 0 & - e^{-i \theta} & 0 \\ e^{i \theta} & 0 & - e^{-i \theta} \\ 0 & e^{i \theta} & 0 \end{pmatrix} $$ for which I would like to calculate the matrix exponential $\exp(t Y) = I + t Y + \frac{t^2 Y^2}{... | Setting $z = e^{i \theta}$ is a good idea. It becomes a bit clearer if $(- e^{-i \theta})$ is replaced by $-1/z$ instead of $-\overline z$ (and it makes the result correct even for complex $\theta$).
So we have
$$
Y = \begin{pmatrix}
0 & -1/z & 0 \\
z & 0 & -1/z \\
0 & z & 0
\end{pmatrix}
$$
and the first pow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Frenet frame field of a curve $\vec{C}(t)$ = $(\frac{t^3}{3}, 2t-1,t^2+2)$ Working out the unit tangent $T$
$\vec{r} = \frac{t^3}{3}i + (2t-1)j + (t^2+2)k$
$\frac{dr}{dt} = t^2 +2j+2tk$
$\frac{ds}{dt} = |\frac{dr}{dt}| = \sqrt{t^4+4+2t^2} = t^2+2$
$T = \frac{dr}{ds} = \frac{dr}{dt}/\frac{ds}{dt} = \frac{t^... | You've missed an $i$ in your calculations for $\frac{d\vec{r}}{dt}$.
$$\frac{d\vec{r}}{dt} = t^2i +2j+2tk$$
Even though the final result for $\frac{ds}{dt}$ is good, the rightmost term in the radical sign is incorrect.
$$\frac{ds}{dt} = \left| \frac{dr}{dt} \right| = \sqrt{t^4+4+(2t)^2} = t^2+2$$
Luckily, you put back... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim\limits_{x \to 0}(\frac{\sin x}{x})^{\cot ^2x}$
Calculate
$$\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\cot ^2x}$$
I tried to use the natural logarithm and L'hopital rule but it didn't help me.
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\sin x}{x})=\ln L$
$\lim\limits_{x \to 0}\cot ^2x\ln(\frac{\... | Composing Taylor series one piece at the time
$$y=\Bigg[\frac{\sin (x)}{x}\Bigg]^{\cot ^2(x)} \implies \log(y)={\cot ^2(x)}\log\Bigg[\frac{\sin (x)}{x}\Bigg]$$
$$\frac{\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\Bigg[\frac{\sin (x)}{x}\Bigg]=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Bound on subsequent terms of inductively defined sequence Let $(a_n)$ be the sequence defined by $a_0=1$ and $a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}$. Show that $|a_{n+1}-a_n|\leq(\frac{1}{2})^{n+1}$.
Using the induction hypothesis I can show that
$$|a_{n+1}-a_n|\leq \frac{|a_n-a_{n-1}|}{2} + |\frac{1}{a_{n-1}}- \frac{1... | Assume you want $|a_{n+1}-a_n| \le \frac{1}{2^{n+1}}$
First $a_n\ge 0$. If $n\ge 1$ then $a_n \ge 2\sqrt{\frac{a_{n-1}}{2} \cdot \frac{1}{a_{n-1}}} = 2\sqrt{\frac 12}=\sqrt 2$ via AM-GM.
$a_{n+1} - a_n = \frac{1}{a_n} - \frac{a_n}{2}=\frac{2-a_n^2}{2a_n} \le 0$
Lastly,
$$a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}, a_n=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\sqrt{x^2-1}=(x+5)\sqrt{\frac{x+1}{x-1}}$ Solve the equation $$\sqrt{x^2-1}=(x+5)\sqrt{\dfrac{x+1}{x-1}}.$$
I think that radical equations can be solved by determining the domain (range) of the variable and at the end the substitution won't be necessary which is suitable for roots which aren't very ... | The equation's defined for $\;x>1\;\;\text{or}\;\;x\le-1\;$ , so now square the whole thing:
$$x^2-1=(x+5)^2\,\frac{x+1}{x-1}\iff x^3-x^2-x+1=x^3+11x^2+35x+25\iff$$
$$12x^2+36x+24=0\iff12(x+1)(x+2)=0\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank y... | Hint:
Multiply for $12\sqrt{x^2-1}$ your initial equation,
you have $$12(\sqrt{x^2-1})x+12x=35\sqrt{x^2-1} \iff 12x(\sqrt{x^2-1}+1)=35\sqrt{x^2-1}$$
and square (two times to delete the sqrt) to the left and the right.
$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$
$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Integrate $x^3\sqrt{x^2-9}\,dx$ with trig substitution $\newcommand{\arcsec}{\operatorname{arcsec}}$This expression can be integrated without trig substitution, but I wanted to try using it anyways and got a different answer.
Starting with finding the indefinite integral $$\int x^3\sqrt{x^2-9} \, dx$$
I then substitute... | sometimes its easier to break it up into other expressions first:
$$I=\int x^3\sqrt{x^2-9}$$
$x=3u\Rightarrow dx=3du$
$$I=\int(3u)^3\sqrt{3^2(u^2-1)}(3du)$$
$$I=3^5\int u^3\sqrt{u^2-1}du$$
now we want to look for an identity that matches this, we know:
*
*$\cosh^2\theta-\sinh^2\theta=1$
*$\cos^2\theta+\sin^2\theta=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Dirichlet's Integral to find Volume with gamma function values Evaluate volume $\iiint_D dx dy dz$ over a domain $D$ where $D$ is the region bounded by $x\ge 0$, $y\ge 0$ and $z\ge 0$ and ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$.
My approach:
I used the RMS$\ge$AM inequality (since $x,y,z\ge 0$ a... | Did you just replace the ellipsoid $(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2\le 1$ with the octahedron $|\frac{x}{a}|+|\frac{y}{b}|+|\frac{z}{c}|\le\sqrt{3}$ ?
No wonder you are getting a different result for its volume. The RMS$\ge$AM
inequality guarantees that the ellipsoid is inscribed in the octahedron, i.e.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation using $f(x)=\frac{\sin\pi x}{x^2} $ If $f(x)= \frac{\sin \pi x}{x^2}$, $x>0$
Let $x_1<x_2<x_3<\cdots<x_n<\cdots$ be all the points of local maximum of $f$.
Let $y_1<y_2<y_3<\cdots<y_n<\cdots$ be all the points of local minimum of $f$.
Then which of the options is(are) correct.
(A) $x_1<y_1$
(B) $x_{n+1}-x_n>2$... | $$f'(x)=\frac{\pi x \cos (\pi x)-2 \sin (\pi x)}{x^3}$$
$$f'(x)=0\to \sin(\pi x)=\pi x\cos(\pi x)\tag{1}$$
Second derivative is
$$f''(x)=-\frac{\left(\pi ^2 x^2-6\right) \sin (\pi x)+4 \pi x \cos (\pi x)}{x^4}$$
To understand which are max/min we use second derivative test using relation $(1)$ in the second deriv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{... | You can convert the recurrence equation into a differential equation by multiplying it with $x^n$, summing over $n=1,2,3,...$ and defining $f(x)=\sum_{n=0}^\infty a_{n+1} x^n$, which gives the first order DE $$\sum_{n=1}^\infty a_{n+1}x^n = 2\sum_{n=1}^\infty a_n x^n + \sum_{n=1}^\infty \frac{a_n}{n} x^n - \sum_{n=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| For $|x| \le 1$ the Taylor series
$$
\cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
is an alternating series with terms that decrease in absolute value. It follows that for these $x$
$$
\cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{x^2}{4!} = 1 - \frac{11}{24... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
How many rectangles can be found in this shape?
Question: How many rectangles can be found in the following shape?
My first solution:
Let $a_n$ be the number of rectangles in such a shape with side length $n$. Then, the number of the newly added $1 \times k$ rectangles while expanding the shape with side length $n-1... | Label the lines starting from $1$ on the left to $n+1$ on right/above. Extend the lines by $1$ unit to create tetherings. There are $n+3$ of them, label these from $1$ to $n+3$. The diagram is for $n=5$.
There is a bijection between the coordinates of four vertices of any rectangle and the coordinates at end of tether... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=... | To get things going you can calculate the inverse of $[3]_{100}$ and then try relating it to powers of $[3]_{100}$, keeping in mind that $10^2 \equiv 0 \pmod{100}$.
In $\text{modulo-}100$ you know that the inverse has the form $[10n +7]_{100}$ and solving you can write out
$\tag 1 3 \cdot 67 \equiv 1 \pmod{100}$
We ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3971549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2 \\
f(n+1) &= 3 \cdot f(n) + 8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}... | $$d(n+1)=3d(n)+8n-12,\\d(0)=4.$$
The homogeneous solution is
$$d_h(n)=3^nd_h(0).$$
Plugging the initial condition,
$$3^0d_h(0)=3\cdot4+8\cdot0-12=0.$$
Then with the ansatz $d_a(n)=an+b,$
$$an+a+b=3an+3b+8n-12$$ or by identification,$$d_a(n)=4-4n=d(n).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find the volume of the set in $\mathbb{R}^3$ defined by the equation $x^2 + y^2 \leq z \leq 1 + x + y$. Let $D=\left\{\left(x,y,z\right) \in \mathbb{R}^3: x^2 + y^2 \leq z \leq 1 + x + y \right\}$. I tried to integrate with cartesian coordinates, so that
$$\iiint_Ddxdydz = \int_{x=\frac{1-\sqrt{5}}{2}}^{{\frac{1+\sqrt{... | $\textbf{Hint}$: Use this modified cylindrical coordinate system:
$$\begin{cases}x = r\cos\left(\theta-\frac{\pi}{4}\right)\\ y = r\sin\left(\theta-\frac{\pi}{4}\right)\\ z = z \\\end{cases}$$
What does the change do? Does it alter the Jacobian (which measures how much the volume is distorted between coordinate systems... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Since dividing $x=x^6$ by $x$ gives $1=x^5$, how can I get to $x=0$ as a root? this might sound like a stupid question,
bear with me it probably is.
I know the solutions for $x=x^6$ are 1 and 0.
Now, since $1 \cdot x = 1 \cdot x^6 $ and it follows $ 1 \cdot x = 1 \cdot x \cdot x^5$ and I divide both sides by $ 1 \cdot... | You find $x=0$ by using the property that if $A\cdot B=0$, and $A,B$ are real numbers (or elements of a field), then $A=0$ or $B=0$.
Write $x=x^6\iff x^6-x=0\iff x(x^5-1)=0$
Either $x^5-1=0$ or $x=0$. The last thing shows that $x=0$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find $\frac{a+b}{ab}$ such that $\int_{-1/2}^{1/2} \cos x\ln\frac{1+ax}{1+bx}dx=0$ Let $f(x) = \cos(x) \ln\left(\frac{1+ax}{1+bx}\right)$ be integrable on $\left[-\frac{1}{2} , \frac{1}{2}\right]$. Let
$$\displaystyle \int_{-1/2}^{1/2}f(x)\operatorname{dx}=0$$
where $a$ and $b$ are real numbers and not equal. Find the ... | To be zero the integral, $h(x)=\log \left(\frac{1+a x}{1+b x}\right)$ must be odd, thus
$$\log \left(\frac{1+a x}{1+b x}\right)=-\log \left(\frac{1-a x}{1-b x}\right)$$
$$\log \left(\frac{1+a x}{1+b x}\right)+\log \left(\frac{1-a x}{1-b x}\right)=0$$
$$\log\frac{1-a^2x^2}{1-b^2x^2}=0$$
$$1-a^2x^2=1-b^2x^2\to b=-a$$
sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3977522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$
Evaluate the series $$S=\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$$
I have tried many values of $(a,b,c)$ and plugged into Wolframalpha, it always converges. I know that for particular values of $a,b,c$, we solve it by forming a telescoping series by using the fact... | Given quadratic $ar^2 +br+c$, suppose that we can find some function $f(r)$ such that
$$
(\dagger)\quad\quad\frac{1+f(r+1)f(r)}{f(r+1)-f(r)} = ar^2 + br + c.
$$
Then as $$\cot^{-1}(ar^2+br+c)=\arctan\frac{1}{ar^2+br+c}\\
=\arctan \frac{f(r+1)-f(r)}{1+f(r+1)f(r)}=\arctan(f(r+1))-\arctan(f(r)),
$$
we can proceed to find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Prove the polynomial $x^4+4 x^3+4 x^2-4 x+3$ is positive Given the following polynomial
$$
x^4+4 x^3+4 x^2-4 x+3
$$
I know it is positive, because I looked at the graphics
and I found with the help of Mathematica that the following form
$$
(x + a)^2 (x + b)^2 + c^2(x + d)^2 + e^2
$$
can represent the polynomial with t... | if $x\ge 0$ by AM-GM $$x^4+3=x^4+1+1+1\ge 4x$$ and the inequality $$\color{blue}{x^4+3-4x}+4x^3+4x^2>0$$ is obvious as equality of am-gm and $x=0$ is simultaneously is not achieved.
If $x\le 0$ then replace $x$ by $-t $ and we have to prove $$t^4-4t^3+4t^2+4t+3>0$$which is true as $$t^4+4t^2\ge 4t^3$$ by AM-GM and as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to get from $x^2 + y^2 = e^2(d + x)^2$ to $(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$ I'm working through a problem in my pre-calc book and I don't know how to get the equation into the same form as the book. We start with $x^2 + y^2 = e^2(d + x)^2$ and the book gets to the f... | To be completed
*
*given equation
*you missed a term $dx$ inside the bracket on the RHS
*$x^2+y^2 = e^2d^2 + 2e^2dx + e^2x^2$
*$x^2 - e^2x^2 - 2e^2dx = e^2d^2 -y^2$ (same logic with coefficient corrected)
*$(1-e^2)x^2 - 2e^2dx= e^2d^2 - y^2$
*$$x^2 - \frac{2e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$$
*$$x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the Maximum Trigonometric polynomial coefficient $A_{k}$
Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$
Find the maximum value of $A_{k}$.
I don't know if this ... | My second answer:
Edit 2021/03/11
According to NTstrucker@AoPS's result (https://artofproblemsolving.com/community/c6h2477208), I give the following conjecture.
Conjecture 2: $\max A_1 = 2\cos \frac{\pi}{n+2}$.
When $n = 2$, $\max A_1 = \sqrt{2} = 2 \cos \frac{\pi}{4}$.
When $n = 3$, $\max A_1 = \frac{\sqrt{5} + 1}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Prove by induction that $9^{n+1}+2^{6n+1}$ is divisible by 11 I need to prove that $9^{n+1}+2^{6n+1} $ is divisible by $11$ $\forall n\in N$
Steps I did:
*
*basis step
$$\\ P(0)\ 11\mid(9+2 ) \ (True)$$
*inductive step
$$\\ P(n)\implies P(n+1) \ where: \\ P(n): 9^{n+1}+2^{6n+1} = 11k\\P(n+1): 9^{n+2}+2^{6n+7} = 11... | Hypothesis gives that if $9^{n+1} \equiv r \pmod{11}$ then $2^{6n+1} \equiv -r \pmod {11}$
The induction step follows:
$$\begin{align}P(n+1) &=9^{n+2}+2^{6n+7} \\
&= 9\cdot 9^{n+1}+2^6\cdot2^{6n+1} \\
&\equiv 9 r + 2^6(-r) \\
&= - 55r \\
&\equiv 0 \pmod{11}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$
Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$
The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.
The first graph looks pretty wei... |
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$,$x+y=23$
The obvious way to solve it is by substituting for one variable.
No, cart before the horse. The first step is to move one of the radicals to the RHS and square both sides. This gives:
$$(x^2 + 12y) = 1089 + (y^2 + 12x) - 66\sqrt{y^2 + 12x}.\tag1$$
At this point, there a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$ Find a $g\in K$ such that $g^2=x^3+x+1$ Let $K=\mathbb{Z}_2[x]/\langle x^4+x^2+x \rangle$. Find a $g\in K$ such that $g^2=x^3+x+1.$
I tried $x^3+x+1$ itself but unfortunately the degree is only $2$. I don't know how I can multiply something and get a polynomial of degre... | A nice trick to know.
$\newcommand{\d}{\frac{\mathbb Z_2[x]}{\langle x^4+x^2+x\rangle}}$In $\mathbb Z_2[x]$, we have the nice identity that $(a+b)^2 = a^2+2ab+b^2 = a^2+b^2$, because $2ab \equiv 0$ in $\mathbb Z_2[x]$! This naturally becomes $(a+b+c)^2 = a^2+b^2+c^2$ in $\mathbb Z_2[x]$ as well.
We can use this to grea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfra... | In fact you are very close to the right answer. I would suggest you check Euler - Maclaurin formula first - plus, not minus in the second term, though it does not change the result that you got $(+\frac{1}{2})$: $$\sum\limits_{k=1}^nf(\frac{k}{n})=n\int_1^nf(t)dt+\frac{1}{2}\left(f(\frac{1}{n})+f(1)\right)+\sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3 - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3 - x^2 + x + 1)$.
S... | Note that
$$
(a^2-a^3)^6-1=(a^6 - 2a^5 + a^4 - a^3 + a^2 + 1)(a^4 - a^3 - a^2 + a + 1)(a^3 - a^2 + 1)(a^3 - a^2 - 1)(a^2 - a + 1),
$$
so that the claim of the title follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the I? This is question is inspired from this
find the value of $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}\sqrt{ \frac{u^4}{a^2} +u^2} du dv$$
My attempt : $$I=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} du dv$$
after that im not able to solved this
| Before I worked out a complete solution, @TitoEliatron has left a key step in his comment.
\begin{align}
I&=\int_{0}^{2\pi} \int_{0}^{\sqrt 2 a}u\sqrt{ \frac{u^2}{a^2} +1} \,du \, dv\\
&=\int_0^{2\pi}dv \; \frac12 \int_0^{\sqrt2 a} \sqrt{ \frac{u^2}{a^2} +1} \, d(u^2) \\
&= \pi \int_0^{2a^2} \sqrt{x/a^2 + 1} \, dx \tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Logarithm to trigonometry conversion I have an expression like
$$\frac{1}{2} i \log \left(\frac{a-b+i c}{a+b-i c}\right)$$
I was wondering if it is possible to write it in terms of trig functions. I guess it is possible to write all logs in some form of trig due to Euler's relation.
As an attempt to solve the problem, ... | $\frac {i}{2}\ln \frac {a+b+ic}{a-b-ic} = x\\
\ln \frac {a+b+ic}{a-b-ic} = -2ix\\
\frac {a+b+ic}{a-b-ic} = e^{-2ix}\\
a+b+ic = e^{-2ix}(a-b-ic)\\
a+b+ic = e^{-2ix}(a-b-ic)\\
(ic-b)(1+e^{-2ix}) = a(e^{-2ix} - 1)\\
\frac {ic - b}{a} = \frac {(e^{-2ix} - 1)}{e^{-2ix} + 1}$
At this point the right hand side equals very nea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Estimate a value with square roots squared without calculating If $n$ is a natural number and $n<(6+\sqrt{29})^2<n+1$, find the value of $n$.
This problem is, of course, very easy to solve with a calculator. But I am looking for a solution that does NOT need to find the value of $(6+\sqrt{29})^2$. Is there a way to do... | Use the difference of two squares to get:
$$6+\sqrt{29} = \frac{(6+\sqrt{29})(6-\sqrt{29})}{6-\sqrt{29}} = \frac{7}{6 - \sqrt{29}}$$
and so:
$$n<\frac{49}{(6-\sqrt{29})^2}<n+1 \Rightarrow \frac{1}{n+1} < \frac{(6 - \sqrt{29})^2}{49} < \frac{1}{n}.$$
Adding $49$ times this inequality with the original gives:
$$n + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola I searched but couldn't find the proof.
Isosceles hyperbola equation:
$${H:x^{2}-y^{2} = a^{2}}$$
And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this po... | For any point on the hyperbola, $x^2 - y^2 = a^2$
Foci of the hyperbola are $(\pm a\sqrt2,0)$ and the center is $(0, 0)$.
So product of distance of point $P(x,y)$ on the hyperbola to foci is
$\sqrt{(x-a\sqrt2)^2 + y^2} \times \sqrt{(x+a\sqrt2)^2 + y^2}$
$\sqrt{x^2 + 2a^2 + y^2 - 2 \sqrt2 a x} \times \sqrt{x^2 + 2a^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inverse of fourth root function I have a function in the form
$$y=a\sqrt[^4]{b^2-(x-c)^2}, \qquad a, b, c\in \mathbb{R}$$
but I'm having trouble finding its inverse. That is, solving for $x$. The solution should seem pretty trivial with putting both sides to the $4^{\text{th}}$ power, rearranging, and applying the quad... | \begin{align}
y=a\sqrt[^4]{b^2-(x-c)^2} &\implies y^{4} = a^{4}(b^{2} - (x-c)^2)\\
&\implies \frac{y^{4}}{a^{4}} = b^{2} - x^{2}+2cx-c^{2}\\
&\implies x^{2} - 2cx + \left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right) = 0\\
&\implies x = \frac{2c \pm \sqrt{4c^2-4\left(\frac{y^{4}}{a^{4}}-b^{2}+c^{2}\right)}}{2}\\
&\implies\box... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplify conditional probability tree There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".
This question is similar to my last question except the desired outcome is rolling one of each number, the ... | There is no easy way to do it with a probability tree. For any $k\in \{0,\dots,6\}$, the number of rolls with $k$ gold faces is equal to the number of permutations of $\{1,\dots,6\}$ with $k$ fixed points, which is in turn equal to $\binom{6}k$ times the number of derangements of a set of size $6-k$. The number of dera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
John Wallis and calculating ratios I was asked the following question:
Using Wallis's method, calculate the values $n=0,\frac12,1,\frac32,2,\frac52$ in row $p=\frac32$ of his ratio table.
The Wallis rereferred to here is John Wallis of England (1616-1703). The ratio table looks like this: $$\begin{matrix}p/n&0&1&2&3&... | The ratio table represents the ratio of the unit square over the area under the stated curve, $y = (1-x^{1/p})^n$. We shall assume $p, n \geqslant 0$ and adopt the convention that the area is $1$ when $p=0$.The area of the unit square is always $1$, while the area under the curve is
\begin{align}
A_{p,n} = \int_0^1 (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions over $\mathbb{R}$ The problem goes as follows:
Using elementary methods prove that $x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0$ has no solutions for $x \in \mathbb{R}$.
I first came across the problem in a school book targeted towards students just expo... | We may observe that our polynomial has the form
$$ (x^2+x+1)\left(x^3 - \frac{5}{2}\right)^2 + \frac{1}{4} (3x^2+3x-1) $$
The first term is always non-negative (and positive for $x \neq \sqrt[3]{\frac{5}{2}}$, that case we can check separately), and the second one is non-negative outside interval $(-2,1)$, so we just n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a ... | Unfortunately, I wasn't able to pick up from where you left off.
Starting from the following inequality:
$$a(b+c-a) \leq b(a+c-b) \leq c(a+b-c) $$
which is true given your initial WLOG $a\geq b\geq c$.
And note that $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$.
You can use Rearrangement Inequality so that:
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $ Prove $ abc \geq 8 $ If $ a,b,c \gt 0 $ and $ \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 1 $
Prove $ abc \geq 8 $
I have tried AM-GM and substituting the 1, without results.
| This simplifies to $$a+b+c+2=abc$$
$$⇒\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}=1$$
Applying AM-GM inequality we get
$$\frac{1}{4}=\frac{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{2}{abc}}{4}≥(\frac{2}{a^3b^3c^3})^\frac{1}{4}$$
Riasing both sides to the power of 4 we get
$$\frac{1}{256}≥(\frac{2}{a^3b^3c^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $ax^2+3x+b$ that goes through the given points using the least squares The points are $ (2,1.5),(−1,1.7),(1,1.9)$.
I did:
$A = \left [ \begin{matrix}
1 & 2 & 4 \\
1 & -1 & 1 \\
1 & 1 & 1
\end{matrix} \right ] \\
x = \left [ \begin{matrix}
b \\
3 \\
a
\end{matrix} \right ]\\
b ... | $3$ is not an unknown variable, so your solution is incorrect.
We have
$$
\left\{\begin{array}{rrrrr}
2^2\cdot a &+3\cdot 2&+b&= &1.5 \\
(-1)^2\cdot a&+3\cdot(-1)&+b&= &1.7 \\
1^2\cdot a&+3\cdot 1&+b&= &1.9 \\
\end{array}\right.
$$
or
$$
\left\{\begin{array}{rrrrr}
4a&+&b&= &-4.5 \\
a&+&b&= &4.7 \\
a&+&b&= &-1.1 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | Divisibility by $7$ is congruence to zero modulo $7.$
So we might get some insights by looking at the numbers' congruences mod $7$.
Note that $x^{2^{k+1}} = x^{2^k\cdot 2} = \left(x^{2^k}\right)^2.$
That is, every time we add $1$ to the exponent $k$ in
$2^{2^k}$ or $4^{2^k}$, we square the number.
So let's try applying... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 0
} |
Prove that $\sqrt{13}$ is irrational in 3 ways. I am asked how to prove that the $\sqrt{13}$ is irrational in 3 ways. I only know of one, where we assume $\sqrt{13} = \frac{a}{b}$ where $a, b$ are coprime, and we prove with contradiction that $13 \mid a$ and $13 \mid b$.
However, I am not sure where to begin with the o... | \begin{align}
\sqrt{13} & = 3 + \text{a fractional part, since } 3^2<13<4^2 \\[8pt]
& = 3 + (\sqrt{13}-3) \\[8pt]
& = 3 + \frac{4}{\sqrt{13}+3} \text{ by rationalizing the numerator} \\[8pt]
& = 3 + \frac 1 {(\sqrt{13}+3)/4}
\end{align}
Since $3<\sqrt{13}<4,$ we have $1 < (\sqrt{13}+3)/4 < 2,$ so $(\sqrt{13}+3)/4$ is b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem in proving Cauchy sequence - how to tackle inequalities and changing indices [EDITED] We are given this sequence:
$a_1 =1, a_{n+1} = 1 + \frac{1}{a_{n}} \forall \; n\geq 1$
Now before marking this question as "duplicate", consider that I have already referred these sources:
How to find the convergence of this... | Step 1. By induction show that $$\alpha:=\frac32 \le a_n \le \frac{15}8.$$
(The upper-bound is the main trick!)
Step 2. We have
$$
|a_{n+1}-a_n| =
|\frac{1}{a_n}-\frac{1}{a_{n-1}}| =
\frac{|a_n-a_{n-1}|}{a_na_{n-1}} \le
\frac{|a_n-a_{n-1}|}{\alpha^2}.
$$
So by induction we have
$$|a_{n+1}-a_n|\le \alpha^{-2(n-1)}|a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $? How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$
I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to :
$$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}... | Substitute $t=\tan\frac x2$
\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\
=&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt
=\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5
\end{align}
where, per the results
\begin{align}
I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt
= -G \ln 2-K+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alp... | $$\sin^22\theta=4\sin^2\theta\ (1-\sin^2\theta)$$
$$=\sin\theta\ (\sin\theta+(3\sin\theta-4\sin^3\theta))$$
$$=\sin\theta\ (\sin\theta+\sin 3\theta)=\sin^2\theta+\sin 3\theta\sin \theta.$$
Therefore, $\cos^22\theta=\cos^2\theta-\sin 3\theta\sin\theta.$ Finally,
replace $\theta$ by $2\alpha.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Optimizing a Conical Frustum Using Partial Differentiation I am working on minimizing the surface area of a frustum with a known volume, however, the equations for the volume and surface area (listed below) have three variables. I am trying to use partial differentiation to optimize surface area with volume as a constr... | Define the Lagrangian function
$$\Lambda(r,R,h;\lambda):=\pi(r+R) \sqrt{(r-R)^2+h^2}+\pi r^2+\pi R^2 -\lambda(V_0-(\pi h) (r^2+rR+R^2)/3), $$
with $V_0$ denoting the allocated volume.
The partial derivatives are
$$\begin{align}\Lambda_r&=\pi \sqrt{h^2+(r-R)^2}+\frac{\pi (r-R) (r+R)}{\sqrt{h^2+(r-R)^2}}+\frac{1}{3} \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
compute $ \iiint_Kxyz\ dxdydz$ The question is:
$$ \iiint_Kxyz\ dxdydz\quad k:=\{(x,y,z):x^2+y^2+z^2\leq1, \ \ x^2+y^2\leq z^2\leq 3(x^2+y^2), \ x,y,z\geq 0\} $$
Here how i have tried to solve this:
$$\iint_{x^2+y^2\leq1}\int_{\sqrt{x^2+y^2}}^{\sqrt{3(x^2+y^2)}}xyz \ dz\ dxdy=\frac{1}{2}\iint xy\left(3(x^2+y^2)-(x^2+y... | Basically we are trying to find the volume of the sphere $x^2+y^2+z^2 \leq 1$ between two cones $z^2 = x^2 + y^2$ and $z^2 = 3(x^2+y^2)$.
So the bounds are much easier to set up in spherical coordinates.
$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$
$0 \leq \rho \leq 1, 0 \leq \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
By Using the Binomial expansion as follow that the expression for the first three term is mathematically correct or not? Let the expression is defined as $(1-\frac{1}{i})^{2i}$.
For example: Using the Binomial expansion as follow $$(1-\frac{1}{i})^{2i}=\sum_{n=0}^{n}\frac{(2i)!}{n!(2i-n)!}(-1)^n(\frac{1}{i})^n$$
$$\ap... | That's fairly straight forward. When n= 0, n!= 0!= 1 and 2i- n= 2i so (2i- n)!= 2i!. The coefficient is $\frac{(2i)!}{n!(2i- n)!}= \frac{(2i)!}{(2i)!}= 1$. Of course, $(-1)^0= 1$ and $\left(\frac{1}{i}\right)^01= 1$ so the first term is 1
When n= 1, n!= 1!= 1 and $\frac{(2i)!}{(2i- n)!}$$= \frac{2i(2i-1)!}{(2i-1)!}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the Taylor and Laurent series with center at $z_{o}= 1$ of the function $\frac{\sinh(z)}{(z-1)^4}$ Let $$f(z)=\frac{\sinh(z)}{(z-1)^4}.$$
First I do the following:
$$\sinh(z)=\sinh((z-1)+1)=\sinh(z-1)\cosh(1)+\cosh(z-1)\sinh(1)$$
The expansions of $\sinh(z)$ and $\cosh(z)$ are
$$\begin{align*}
\sinh(z) &= z+\frac... | Your work is good but you can make life a bit easier starting with $z=t+1$ (this is what you implicitly did)
$$\frac{\sinh (z)}{(z-1)^4}=\frac{\sinh (t+1)}{t^4}$$ Expanding as you did
$$\sinh (t+1)=\sinh (1) \cosh (t)+\cosh (1) \sinh (t)$$ Now, using the expansions of $\cosh (t)$ and $\sinh (t)$ around $t=0$ you then h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conditional expectation of sum of numbers on 2 throws of a dice given their difference Let$ X$ and $Y$ be the numbers obtained on first and second throws of a fair die .
Calculate $E(X+Y|(X-Y)^2=1)$
I am doing basically $$E(X|(X-Y)^2=1)=E(Y|(X-Y)^2=1)$$
So $E(X+Y|(X-Y)^2=1)=2E(X|(X-Y)^2=1)$
$E(X+Y|(X-Y)^2=1)=2E(X|(X-Y... | The result is correct.
To "see" the solution observe that throwing 2 fair dice the events that correspond to
$$(X-Y)^2=1$$
are 10
$$\{(1,2);(2,1);(2,3);(3,2);(3,4);(4,3);(4,5);(5,4);(5,6);(6,5)\}$$
given this sample space, the sum $X+Y$ is a rv taking values in
$$ Z=
\begin{cases}
\frac{1}{5}, & \text{if $z=3$} \\
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration with trig substitution Trying to evaluate this using trig substitution:
$$\int \frac {1}{49x^2 + 25}\mathrm{d}x $$
Here's how I'm going about it, using $x = 5/7(\tan\theta)$
$$\int \frac {1}{49\left(x^2 + \frac{25}{49}\right)}\mathrm{d}x $$
$$=\int \frac {1}{49\left(\frac{25}{49}\tan^2\theta + \frac{25}{49}... | Welcome to MSE. The issue is that you left off $dx$:
$$\int \frac1{49x^2 + 25} dx = \int \frac1{49(5/7\tan(\theta))^2 + 25} d(5/7 \tan(\theta) = \frac57\int \frac1{25\tan^2(\theta) + 25} \sec^2(\theta)d\theta = \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
System of differential equations: why does it solve it the example? consider the system with conditions: $\quad y_1(0) = 1,\quad y_2(0) = -1,\quad y_3(0) = 0,\quad y_4(0) = 2$
$\begin{align}
&{y_1}'= y_1 + 6\,y_2\\
&{y_2}'= y_2 + 6\,y_3\\
&{y_3}'= y_3 + 6\,y_4\\
&{y_4}'= y_4
\end{align}$
what can be written as matrix: ... | We are given the system
$$\begin{align}
&{y_1}'= y_1 + 6\,y_2 = 1\, y_1 + 6\, y_2 + 0 \,y_3 + 0\, y_4\\
&{y_2}'= y_2 + 6\,y_3 = 0 \,y_1 +1\, y_2 + 6\, y_3 + 0\, y_4\\
&{y_3}'= y_3 + 6\,y_4 = 0\, y_1 + 0\, y_2 + 1\, y_3 + 6\, y_4\\
&{y_4}'= y_4 = 0 \,y_1 + 0 \,y_2 + 0\, y_3 + 1\, y_4
\end{align}$$
This can be compactly ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac1{\sin A}+\frac1{\sin B}+\frac1{\sin C}-\frac12(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}) \ge \sqrt{3}$ Let $ABC$ be a triangle. Prove that: $$\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}-\frac{1}{2}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right) \ge \sqrt{3} $$
My attempt:
$$P=\fr... | You are not wrong because
$$
\frac32\sqrt{3}\ge \sqrt3.
$$
You just have proved a stronger inequality than required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Cross product and determinants The determinant of a $3 \times 3$ matrix gives the volume of the parallepiped formed by 3 vectors. With the cross product, say to find the torque, we find the $3 \times 3$ determinant value of displacement vector $\mathbf{r}$ and force $\mathbf{F}$,
\begin{align}
\boldsymbol{\tau} &= \ma... | When we write a cross-product as a determinant, we are really abusing notation. If you look at the expression,
\begin{align}
\mathbf{a\times b}
&= \begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
a_1&a_2&a_3\\
b_1&b_2&b_3\\
\end{vmatrix}
\\
&=
(a_2b_3 - a_3b_2)\mathbf{i} -(a_1b_3 - a_3b_1)\mathbf{j} +(a_1b_2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{ Th... | Assuming the nested radical is infinite we have $x=\sqrt{2+\sqrt{2+...+\sqrt{2}}}\Rightarrow x^2-2=x \iff x^2-x-2=0\Rightarrow x=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the probability of placing $5$ dots on an $8 \times 8$ grid. $5$ dots are placed at random on an $8 \times 8$ grid such that no cell has more than $1$ dot.
What is the probability that no row or column has more than $1$ dot?
I thought about this in the following way, the number of ways to place $5$ dots on $64$ ce... | Pick $5$ of $8$ rows to occupy. Sort those in ascending order. Hence the order they were picked in will be forgotten. This gives:
$$
8C5=\binom 85
$$
Pick $5$ of $8$ columns to occupy. Pair those with the ordered set of chosen rows in the order you pick them. Hence order counts this time, and we have:
$$
8P5=8!/(8-5)!=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
| Since
$$f(x,y)= \sqrt{(2y-3)^2+1}+\dfrac{\sqrt2}2\sqrt{(6x-3)^2+1}+\dfrac12\sqrt{(6x-4y+2)^2+(6x)^2},$$
then the substitutions
$$2y-3 = \sinh s,\quad 6x-3=\sinh t,\quad s,t\in\mathbb R,\tag1$$
transform $\;f(x,y)\;$ to
$$g(s,t)= \cosh s + \dfrac{\sqrt2}{2} \cosh t+\dfrac12\,r(s,t),\tag2$$
where
$$r(s,t)=\sqrt{\left(\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
Interesting logarithmic inequality Recently the following question was published on the site and soon after this deleted:
Prove the inequality
$$
\left(\log\frac{a^2+b^2}{2ab}\right)^n\le \left(\log\frac{a^2+c^2}{2ac}\right)^n+\left(\log\frac{b^2+c^2}{2bc}\right)^n\tag1
$$
for all positive $a,b,c$ and $n=\frac12$.
Th... | My attempt :
$\sqrt{ln (\frac{1+(\frac{b}{a}) ^2}{2\frac{b} {a} })}<\sqrt{ln (\frac{1+(\frac{c}{a}) ^2}{2\frac{c} {a} })}+
\sqrt{ln (\frac{1+(\frac{c}{b}) ^2}{2\frac{c} {b} })} $
Suppose $x=\frac{b}{a} $ and $y=\frac{c} {a} $
$\sqrt{ln (\frac{1+x^2}{2x})}<\sqrt{ln (\frac{1+y^2}{2y})}+\sqrt{ln (\frac{x^2+y^2}{2xy})}$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different? Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?
Probably a really dumb question but I just don't see it :(
| The difference is:
We can have $4 \ge x(x-4)$ if $x(x-4) \le 0$.
But we can not have $\frac 4{x(x-4)} \ge 1$ if $x(x-4) \le 0$.
(if $x(x-4) = 0$ then $\frac 4{x(x-4)}$ doesn't exist. And if $x(x-4) < 0$ then $\frac 4{x(x-4)} < 0 < 1$.)
(Otherwise, when $x(x-4) > 0$ then $\frac {4}{x(x-4)} \ge 1$ and $4 \ge x(x+4)$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4045039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Two maximum values for $|iz+3-4i|$ what went wrong? I need to find the maximum value of $|iz+3-4i|$ given that $|z|\leq4$
$|iz+3-4i|\leq |iz|+|3-4i|$
$|iz+3-4i|\leq 4+5$
But I could write an inequality for $|3-4i|$ whose maximum value would be 7, not 5 so I get the maximum value to be 11 when doing so rather than the o... | Put $z = a + bi\implies |iz+3-4i|= |i(a+bi)+3-4i|=|3-b+(a-4)i|=\sqrt{(a-4)^2+(b-3)^2}$. The problem translates to: Maximize $f(a,b) = \sqrt{(a-4)^2+(b-3)^2}$ given that $a^2+b^2 \le 16$. First observe that if the max occurs at an interior point inside the disk $D: a^2+b^2 \le 16$ at $M$, then consider the point of inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Antiderivative of $\frac{1}{x^2-1}$ for $x>1$ or $x<-1$ We know that the antiderivative of $\frac{1}{x^{2} - 1}$ on $(-1,1)$ is $\operatorname{artanh}(x)$. Is there a nice way of writing the antiderivative on the intervals $(-\infty,-1)$ and $(1,\infty)$?
I realize we could write $$\int_{a}^{x}\frac{1}{y^{2} - 1}\,dy$... | For the sake of completeness, here are the two suggested ways of writing the antiderivative:
First Way:
$$\boxed{\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.}$$
This holds because:
\begin{align}
\frac{d}{dx}\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| &=\frac{1}{2}\frac{d}{dx}\bigl(\ln|x-1| - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$ if $ab+bc+ca=3$ Let $a,b,c$ be real positive number, $ab+bc+ca=3$.
Prove that $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 12$$
My attempt: $$P=(a+b)(a+bc) + (b+c)(b+ca) + (c+a)(c+ab) \ge 3\sqrt[3]{(a+b)(a+bc)(b+c)(b+ca)(c+a)(c+ab)}$$
$$P \ge 3\sqrt[3]{\frac{8}{9}\cdo... | Hints :
We have :
$$P=(a+b)(b+bc) + (b+c)(c+ca) + (c+a)(a+ab) \ge 12$$
Using :$$(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0$$
Now we use Cauchy-Schwarz inequality to get :
$$P\geq \sum_{cyc}^{}(a+\sqrt{abc})^{2}\geq 12$$
Now expand and use the substitution $3u=a+b+c$,$3v^2=ab+bc+ca$ and $w^3=abc$
Conclude using uvw's method .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Functional Expectation Problem: Let $X$ be distributed over the set $\Bbb N$ of non-negative integers, with pmf: $$P(X=i)=\frac{\alpha}{2^i}$$
*
*$\alpha$
*$E[X]$
For $Y=X$ mod $3$, find:
*
*$P(Y=1)$
*$E[Y]$
*
*I will assume $\alpha=1$ for now to set the stage althought is there a way to find this value ra... |
*
In order to guarantee that $p_i=\frac{\alpha}{2^i}$ is a distribution, we have that:
$$\sum_{i=0}^{+\infty} p_i = 1 \Rightarrow \\\alpha \sum_{i=0}^{+\infty}\frac{1}{2^i} = \alpha \frac{1}{1 - \frac{1}{2}} = 2 \alpha = 1 \Rightarrow \alpha = \frac{1}{2}.$$
1.
$$P(Y=1) = P(X ~\text{mod}~ 3 = 1) = \sum_{i : i ~\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare ... | First Solution
By AM-GM inequality we have
$$
\frac{x^2+y^2}{2} \geq |xy| \geq xy \\
\frac{x^2+1}{2} \geq |x| \geq x \\
\frac{1+y^2}{2} \geq |y| \geq y \\
$$
Add them together.
Second Solution: By Cauchy-Schwarz we have
$$
\left( xy+1\cdot x+y\cdot 1 \right)^2 \leq ( x^2+1^2+y^2)(y^2+x^2+1^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
} |
Exact differential equation $\frac{y}{(y+x)^2}dx+ ( \frac{1}{y} - \frac{x}{(x+y)^2})dy=0$ I have to solve the differential equation $\dfrac{y}{(y+x)^2}dx+ ( \dfrac{1}{y} - \dfrac{x}{(x+y)^2})dy=0$
Let $ M= \dfrac{y}{(y+x)^2}$ and $ N=\dfrac{1}{y} - \dfrac{x}{(x+y)^2}$
Since it is an exact differential equation, the so... | Both solutions are correct. They differ by a constant, which you can see by pulling out $-1$ from the general constant:
$$\frac x{x+y}+C=\frac x{x+y}-1+C=\frac{x-(x+y)}{x+y}+C=-\frac y{x+y}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.