Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all $x\in\mathbb{R}$ such that $\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x$. Find all $x\in\mathbb{R}$ such that:
$$
\left( \sqrt{2-\sqrt{2} }\right)^x+\left( \sqrt{2+\sqrt{2} }\right)^x=2^x\,.
$$
Immediately we notice that $x=2$ satisfies the equation.
Then we see that $LHS=a^x+b^x... | It is not difficult (formula for the double angle) to show that $$\sin \left(\frac{ \pi }{8} \right)= \sqrt{ \frac{2- \sqrt{2} }{4} }$$ which in combination with the trigonometric one gives $$\cos^2\left(\frac{ \pi }{8} \right)=1-\sin^2\left(\frac{ \pi }{8} \right)=\frac{2+ \sqrt{2} }{4}\Rightarrow \cos \left( \frac{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\fra... | We only need to prove this inequality for $n=3.$ Write the inequality as homogeneous form$:$
$${\dfrac {{a}^{2}}{b}}+{\dfrac {{b}^{2}}{c}}+{\dfrac {{c}^{2}}{a}}+{
\dfrac { 9\left( ab+bc+ac \right) ^{3}}{ \left( {a}^{2}+{b}^{2}+{c}^{2}
\right) \left( a+b+c \right) ^{3}}}\geqslant {\dfrac {6(ab+bc+ac)}{a+b+c}}\quad (\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all positive integers $a,b,c,x,y,z$ satisfying $a+b+c=xyz$ and $x+y+z=abc$.
Find all positive integers $a,b,c,x,y,z$ satisfying $$a+b+c=xyz,\tag{1}$$and$$x+y+z=abc,\tag{2}$$ where $a\ge b\ge c\ge 1$ and $x\ge y\ge z\ge 1$.
My try: I think this problem is unique in a way that there are 6 variables and 2 equation... | If all the integers are greater than or equal to $2$, then
$$
a+b+c < abc = x+y+z < xyz = a+b+c.
$$
Contradiction. Therefore, assume WLOG that $z=1$. You get
$$
a+b+c = xy, \qquad x+y+1 = abc.
$$
Now assume again, $a, b, c, x, y\geq2$, you get
$$
a+b+c < abc = x+y+1 \leq xy + 1 = a+b+c+1.
$$
Therefore, $abc=a+b+c+1$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k. For $a,b,c\geqslant 0;ab+bc+ca>0.$ Find $k_\max$ and proving in that case$:$
$$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{kabc(a^2+b^... | Find $k_{\max}.$ Let $a=b=1$ then
$$(2c+1)\left[\frac 14+\frac{2}{(c+1)^2}\right] \geqslant \frac 94+\frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
equivalent to
$$\frac{c(c-1)^2}{2(c+1)^2}\geqslant \frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
or
$$c(c-1)^2\left[k - \frac{(2c+1)(c+2)^3}{2(c+1)^2}\right] \leqslant 0.$$
Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3814954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $ has exactly three solutions. Find the sum of all values of the parameter $ b $ for which the equation $$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $$ has exactly three solutions.
I tried to represent this equation in two axes and draw this graph in order to analyze the cases wh... | Just reminding some basics:
*
*To plot $f(|x|)$, draw $f(x)$, ignore the left hand side part of the y-axis, reflect the graph to the right of the y-axis in the y-axis.
*To plot $|f(x)|$, draw $f(x)$, reflect any part of graph that was below the x-axis in the x-axis.
Let's draw $4|4|x| - b^2| = |16|x| - 4b^2|$:
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In an acute angled triangle with angles $A,B$ and $C$, prove that $\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac6\pi$ In an Acute Angled Triangle Prove That : -
$$\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac{6}{\pi}$$
A , B , C represent the angles of the tri... | Let $$f(x) = \frac{\sin x}{x}, \quad x \in \left[0, \frac{\pi}{2}\right].$$
We have
$$f'(x) = \frac{x \cos x - \sin x}{x^2} = \frac{g(x)}{x^2},$$
and
$$g'(x)= - x\sin x < 0.$$
So $g(x) < g(0) = 0,$ deduced $f'(x) < 0,$ or
$$f(x) > f\left(\frac{\pi}{2}\right) = \frac{2}{\pi}.$$
Therefore
$$\frac{\sin A}{A} + \frac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.
I have tried the question like this.
\begin{align*}
\frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x... | From the point you get to, $y = \frac{48\cdot x}{7x-48}$
As HCF of $x, y$ is $4$, try $x = 8, 12, ... \, (\ge 7)$. $8$ is not possible as that gives $y = 48$ and hence HCF of $8$. So next try with $x=12$ and you get $y=16$ and those are your numbers.
Your second method gives $b = 3, a = 4$ which is the same. In your se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $1 + \sum_{k=1}^{n-1} \frac{k}{n-k} { 2n-k-1 \choose n} = C_n$ where $C_n$ is the $n$'th Catalan number? Prove that $$1 + \sum_{k=1}^{n-1} \frac{k}{n-k} { 2n-k-1 \choose n} = C_n$$ where $C_n$ is the $n$'th Catalan number?
Wrote a program to validate it, appears to be correct.
This looks awfully like the for... | We seek to show that with $C_n$ the Catalan number we have
$$C_n = 1 + \sum_{k=1}^{n-1} \frac{k}{n-k} {2n-k-1\choose n}.$$
This holds by inspection when $n=0$ and $n=1$ when the sum is zero so we
may assume that $n\ge 2.$ Observe that
$$\frac{k}{n-k} {2n-k-1\choose n}
= k \times \frac{(2n-k-1)!}{n! \times (n-k)!}
= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$ I am trying to compute the Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$. When I compute the Cesàro means, I get the following sequence
$$\left(1, 1, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \frac{6}{9}, \frac{7}{10}, \frac{8}{11}, \frac{8... | Let $a_n = (-1)^n$ for $n\ge 0$. The partial sum $$s_k = \sum_{n = 0}^k a_n$$ for $k\ge 0$ is $1 , 0 , 1 , 0,\dots$ which clearly shows $s_k$ is divergent. Let $$t_n = \frac{1}{n}\sum_{k = 0}^{n-1}s_n$$for $n\ge 1$. So we have $t_1 = \frac{1}{1} , t_2 = \frac{1}{2} , t_3 = \frac{2}{3} , t_4 = \frac{2}{4} , t_5 = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding rational $p$ and $q$ that satisfy $(p+3\sqrt 7)(5+q\sqrt 7)=9\sqrt 7-53$
How do I find rational $p$ and $q$ that satisfy this equation?
$$(p+3\sqrt 7)(5+q\sqrt 7)=9\sqrt 7-53$$
The equation is equal to:
$$5p+15\sqrt 7+pq\sqrt 7+3q\sqrt 7=9\sqrt 7-53$$
What can I do next to find what $p$ and $q$ are equal to?
... | Whoa! IMPORTANT $(3\sqrt 7)\cdot (q\sqrt 7) = 3q\sqrt 7^2 = 3q\cdot 7=21q\ne 3q\sqrt 7$.
So we have $5p + 15 \sqrt 7 +pq \sqrt 7 + 21q = 9\sqrt 7 -53$
Now rewrite that as $(15+pq)\sqrt 7 + (5p+21a) = 9\sqrt 7 - 53$
The trick is that no matter how you add up the rational multiples of $\sqrt 7$ you will always get a a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$ The problem :
*
*find the minimal polynomial of $\sqrt{5} + \sqrt{7}$
*what is the degree of $ [ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ] $
*conclude $\mathbb{Q}( \sqrt{5},\sqrt{7} ) = \mathbb{Q}( \sqrt{5} + \sqrt{7} )$
My question and... | Once you know that $[ \mathbb{Q}( \sqrt{5},\sqrt{7} ) : \mathbb{Q} ]=4$, with basis $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$, you can proceed as follows, without finding the minimal polynomial of $\sqrt{5}+\sqrt{7}$.
Let $\alpha=\sqrt{5}+\sqrt{7}$. Then
$$
\begin{pmatrix} 1 \\ \alpha \\ \alpha^2 \\ \alpha^3 \end{pmatrix}
=
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \righ... | As an alternative
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )=$$
$$=n\left(\frac12 \frac1n+\frac12 \frac1n- \frac{1}{n+1} + \frac12\frac{1}{n+2}+\frac12\frac{1}{n+2}-\frac{1}{n+3}+\frac12\frac{1}{n+4}+\ldots\right)=$$
$$=\frac12+n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Find limit at 0 of cosine function with embedded sine I'm getting stuck on the following exercise where I have to find the limit as x approaches zero, for this cosine function:
$$\lim_{x \to 0}\cos\left(\frac{\pi\sin^2(x)}{x^2}\right)$$
The graph shows that there should be a limit of $-1$ at $0$, but I can't find a nic... | \begin{align*}
\lim_{x \rightarrow 0} &{}\cos \left( \frac{\pi \sin^2 x}{x^2} \right) \\
&= \cos \left( \lim_{x \rightarrow 0}\frac{\pi \sin^2 x}{x^2} \right) & &\text{cosine is continuous} \\
&= \cos \left( \pi \lim_{x \rightarrow 0}\frac{\sin^2 x}{x^2} \right) & &\text{constant multiple} \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving $x+\sqrt{a+\sqrt{x}}=a$ for $x$. I have to solve the following equation for $x$, for all values of $a$:
$$x+\sqrt{a+\sqrt{x}}=a$$
Clearly $x\gt0$
For $a\lt0,x\in\phi$ as RHS$\lt0$ but LHS$\gt0$.
$a=0\Rightarrow x=0$
Now for $a\gt0$
$$a-x=\sqrt{a+\sqrt{x}} ,x\in(0,a)$$
$$(a-x)^2=a+\sqrt{x}.....(1)$$
Now plottin... | Let $x=(y^2-a)^2$ with $y^2>a$ then
$$x+\sqrt{a+\sqrt{x}}=a \iff (y^2-a)^2+y=a \iff y^4-2ay^2+y-a+a^2=0$$
and since the $y^3$ term is equal to zero we can guess
$$y^4-2ay^2+y-a+a^2=(y^2+y+A)(y^2-y+B)=$$
$$=y^4+(A+B-1)y^2+(B-A)y+AB$$
which leads to $A+B-1=-2a$, $B-A=1$, $AB=a^2-a$ that is
$$y^4-2ay^2+y-a+a^2=( y^2 + y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Please give your explanation.
I've tried this and got stuck
\begin{align}|f(x) - f(u)|
&= \Big|\frac{2x}{1+2x^2} - \frac{2u}{1+2u^2}\Big| \\
&=\Big|\frac{(-4xu+2)(x-u)}{(1+2x^2)(1+2u^2... | Actually any continuous function $f$ on $\mathbb R$ such that $\lim_{x\to \pm\infty}f(x)=0$ is uniformly continuous on $\mathbb R.$ We have that hypothesis satisfied with $f(x)=2x/(1+2x^2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inverse z-transform by convolution method I've been asked to find the poles, impulse response of a system (in digital signal processing).
The transfer function was,
$$H(z) = \frac{z^2+z}{z^2-z+0.5}$$
I solved it by the partial fractions method as follows.
$$\color{blue}{H(z) = \frac{z(z+1)}{z^2-z+0.5}}$$
$$\begin{align... | Thanks to my professor for providing me a hint (To split $\dfrac{z+1}{z-b}$ into $\dfrac{z}{z-b}$ and $\dfrac{1}{z-b}$ and then convolute)
$$ H(z) = \dfrac{z(z+1)}{z^2-z+0.5} = \dfrac{z(z+1)}{\left(z-(0.5+0.5i)\right)\left(z-(0.5-0.5i)\right)}$$
Let $a= 0.5+0.5i$ and $b= 0.5-0.5i$.
$$H(z) = \dfrac{z}{z-a}\cdot\dfrac{z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +... | Let $a$ be a positive integer that is not a cube, and let $x$, $y$ and $z$ be positive integers such that
$$(xz+1)(yz+1)=az^3+1.$$
Expanding the left hand side and rearranging a bit then shows that
$$az^2-xyz-(x+y)=0,\tag{1}$$
so $z$ is an integral root of a quadratic with discriminant $x^2y^2+4a(x+y)$. In particular t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find all functions $f:\mathbb N_0\to \mathbb N_0$ such that $f(a^2+b^2)=f(a)^2+f(b)^2$ I think the answer is that there are only 2 such functions: the zero function and the identity function, but I'm not able to prove it.
A few findings:
*
*$f(0)=0$ and thus $f(a^2)=f(a)^2$.
*If $f(1)=0$, then $f(2^n)=0$ for all $n... | I found a much simpler proof, heavily based on and inspired by the ideas of Hagen von Eitzen's solution. For the sake of completeness, we first note that $f(0) = 2f(0)^2$ and because $1 \neq 2f(0)$, we must have $f(0) = 0$. Next we find $f(1)^2 = f(1)$ and so $f(1) \in \{ 0,1 \}$. For convenience, denote $u = f(1)$. Mo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt$ I am preparing for a master's degree entrance exam. One of the questions from the past exam asks to show the following formula. Could someone provide some hint?
$$\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt = \frac{\pi}{6} (e^{-x} - e^{-2x})$$ for $x \geq... | First $$I(x):=\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt$$ then applying a Laplace transform we get $$\mathscr{L}_{x\to s}\{I(x)\}=\int_0^{\infty} e^{-sx}\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dtdx$$ here by the $x\to s$ I just denote which vairables I change to which, to avoid confusion. Now since all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How do I evaluate $\int_{-1}^1\frac{dx}{(1+x^2)(e^x+1)}$? How do I evaluate:
$$\int_{-1}^1\frac{dx}{(1+x^2)(e^x+1)}$$
The answer given in the book is written below.
$$\int_{-1}^1\frac{dx}{(1+x^2)(e^{-x}+1)} \tag{1}$$
$$=\int_{-1}^1\frac{e^x}{(1+x^2)(e^x+1)}dx \tag{2}$$
On adding $(1)$ and $(2)$, we get:
$$\Rightarrow... | Answer:
$I= \int\limits_{-1}^{1} \frac{dx}{(1+x^2)(e^x+1) } $=
We put $x=-y$
$I=\int\limits_{1}^{-1} - \frac{dy}{(1+y^2)(e^{-y}+1)} =\int\limits_{-1}^{1} \frac{e^{y}}{(1+y^2)(e^{y} +1)} =
\int\limits_{-1}^{1} \frac{e^{y}+1-1}{(1+y^2)(e^{y} +1)} $
$2I=\int\limits_{-1}^{1} \frac{1}{1+y^2 } $
$I=\frac{[arct(y) ]_{-1}^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equati... | Use the fact that\begin{align}x^3-4x^2-4x+16=0&\iff x(x^2-4)-4(x^2-4)=0\\&\iff(x-4)(x^2-4)=0.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
reduce a differential equation $y^{'}=\dfrac{1+y}{1-x}$ I want to reduce a differential equation.
$$y^{'}=\dfrac{1+y}{1-x}$$
I reduce this, but my answer don't much "Wolfram alpha".
Please tell me what is wrong.
divide both sides of the equality by $(1+y)$
$$\dfrac{y'}{1+y}=\dfrac{1}{1-x}$$
integrate both sides by $x$... | From your calculations
$$1+y = \dfrac C {1-x} \implies y = \dfrac C{1-x} - 1$$
which is already the solution.
If you want to obtain the form given by WolframAlpha, notice that:
$$\frac C{1-x} - 1 = \frac C{1-x}-\frac 1{1-x} + \frac x{1-x} = \frac{C-1}{1-x}+\frac x{1-x}$$
and $C-1$ is absorbed into $C$ since $C$ can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $ I have been trying to derive
$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$
with $G$ being the the Catalan constant.
I noticed that a similarly-looking integral is posted and solved here. Although the solution is applic... | Write the numerator of the integrand as
$$\ln(1-x)\ln(x+x^2)=\ln(1-x)\ln x+\ln(1+x)\ln\frac{1-x}{1+x}+\ln^2(1+x)
$$
and, correspondingly
$$I=\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx
= I_1+I_2+I_3
$$
where
\begin{align}
I_1 &= \int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx
= \frac12\int_0^1\frac{\ln^2(1-x)+\ln^2x-\ln^2\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$... | Recall $\ln(a/b) = \ln(a) - \ln(b)$, so
$$\ln\left|\dfrac{x^2+2x+3}{2} \right| = \ln\dfrac{|x^2+2x+3|}{|2|} = \ln|x^2+2x+3|-\ln2$$
so, distributing the $-\dfrac{1}{2}$, we obtain
$$\ln|x^2+2x+3|-\dfrac{1}{2}\ln|x^2+2x+3|-\dfrac{1}{2}\ln2 - \dfrac{\sqrt{2}\arctan\frac{x+1}{\sqrt 2}}{2}+C$$
which is just
$$\dfrac{1}{2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate the infinite sum of $\frac{1}{ (n^2+a)(n^2+b)(n^2+c)}$ Playing around with Mathematica, I found a rather nice result for this summation
\begin{align}
\sum_{n=0}^\infty \frac{1}{(n^2 + a)(n^2 +b) (n^2 + c)}
&= \frac{1}{2 a b c}
- \frac{ \pi \left[
\sqrt{ab}(a-b) \coth(\pi\sqrt{... | Partial fraction decomposition gives
$$\frac1{(n^2+a)(n^2+b)(n^2+c)}=\sum_{\mathrm{cyc}}\frac1{(a-c)(b-c)}\cdot\frac1{n^2+c}$$
Now use the well-known sum
$$\sum_{n=0}^\infty\frac1{n^2+k}=\frac{1+\sqrt k\pi\coth\sqrt k\pi}{2k}$$
which should lead you to the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $x^{3}-3 x=\sqrt{x+2}$
Solve the equation $x^{3}-3 x=\sqrt{x+2}$
In the solution of this author wrote -
Let $-2 \leq x \leq 2 .$ Now Setting $x=2 \cos a, 0 \leq a \leq \pi$....
But here i did not understand the reason behind taking $ 0 \leq a \leq \pi$,i mean if we can take $a$ to be any real numbe... | In the answer by Dan Fulea, there is a cubic factor as "Just as a digression," namely $x^3 + x^2 - 2x - 1.$ This is well known, the roots are real numbers
$$ 2 \cos \left( \frac{2 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{4 \pi}{7} \right) \; \; , \; \;
2 \cos \left( \frac{8 \pi}{7} \right) \; \; . \; \; $$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I ... | Hint: If $\gcd(n+1,n^2+1)=2$ then $n$ is odd, say $n=2k+1$, and so you get
$$a^2=\frac{n^2+1}{2}=\frac{4k^2+4k+2}{2}=k^2+(k+1)^2=k^2+b^4,$$
which is a Pythagorean triple. Can you continue from here? [There is still quite some work to be done!]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Is there a way to rewrite this expression as a sum of the previous two terms?
The context of this question:
Consider for ${n \in \mathbb{N}, n\geq 0}$ a rectangular board divided into a ${2 \times n}$ square grid.
A ${2 \times 1}$ domino can cover any two horizontally or vertically adjacent squares. Prove by inductio... | One thing to do in a situation like this (especially with such a very complicated chain of reasoning) is to check the results for some small input values.
For example, if $n = 6,$ then
$M = \frac14(2(6) + (-1)^6 - 1) = 3,$ and
\begin{align}
\sum_{k=0}^{k=M} \binom{n-k}{k} &= \sum_{k=0}^{k=3} \binom{6-k}{k}\\
&= \bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| $ 0 \leq (b-c)^{2} =b^{2}+c^{2}-2bc=3-3bc$ so $bc \leq 1$. Now $(b+c)^{2}=b^{2}+c^{2}+2bc=3+bc \leq 4 $ Now take square root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$ Evaluate the limit:
$$\lim_{n\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$$
Using the fact that ${(1 + x)^{1/2} \approx 1 + x/2}$ for "small" x,
I have that $\sqrt{n+1}\approx\sqrt{n}(\fr... | Answer :
$\sqrt{n^3} (\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3}) =\sqrt{n^3}\frac{(2\sqrt{n^2 +3n+2}-2\sqrt{n^2 +3n})}{\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3}}=\frac{4 \sqrt{n^3} }{(\sqrt{n^2 +3n+2}+\sqrt{n^2 +3n})(\sqrt{n+1}+\sqrt{n+2}+\sqrt{n}+\sqrt{n+3})}=\frac{4 \sqrt{n^3} }{n \sqrt{n + 3} + \sqrt{n^3 + 4 n^2 + 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Are there restrictions to applying $\lim\limits_{x\to0}\frac {\tan x} {x}$? Question
Evaluate $$\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2}\ .$$
My working
So far, I have gotten to the step where
\begin{align}
\lim\limits_{x\to0}\ \left(\frac {\tan x} {x}\right)^\frac {1} {x^2} & =
\exp\left(\... | Indeed in the first step we have that
$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 \frac {\tan x} {x}} $$
and since $\frac {\tan x} {x} \to 1$ we reduce to study the following
$$\lim\limits_{x\to0} \frac{x\sec^2 x - \tan x }{ 2x^3 } $$
this one is a perfectly fine step but in the subsequent steps this is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove this geometric inequality Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$
My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have t... | Let $K$ and $N$ be placed on the sides $AB$ and $AC$ respectively.
Also, let $\frac{AK}{AB}>\frac{AN}{AC}$ and $G\in NC$ such that $KG||BC$, $KG\cap AD=\{P\}$ and $KG\cap AE=\{Q\}$.
Thus, $KP=PQ=QG$, which says that it's enough to solve our problem, when $K\equiv B$.
Now, let $AN=kAC$, where $0<k<1$ and $F\in EC$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A question on probability theory with 3 urns with return Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color ... | I am unsure of where your math came from, and I am unfamiliar with the "Law of total probability". The following is how I would compute the probability.
Let $p(k)$ represent the probability that the 1st ball came from urn $k : k \in \{1,2,3\}.$
Then, the chance that the new drawing will also be a white ball, since the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I ... | Since the expression is cyclic, we can, WLOG, reduce it into two cases:
Suppose $a\ge b\ge c > 0$. We have
$$\frac{a^2-c^2}{b+c} = \frac{a^2-b^2}{c+b} + \frac{b^2-c^2}{c+b} \ge \frac{a^2-b^2}{c+a} + \frac{b^2-c^2}{a+b}$$
Now suppose $a \ge c \ge b > 0$. We then have
$$\frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb N... | The solution given by "@ player3236" has $k=325$.
The integer $325$ can be written as sum of two squares
in three way's & is shown below:
$k= x_1^2+x_2^2=x_3^2+x_4^2=x_5^2+x_6^2$
($x_1,x_2,x_3,x_4,x_5,x_6$)=$(15,10,17,6,18,1)$
$325= 15^2+10^2=17^2+6^2=18^2+1^2$
Above can help in arriving at a parametric solution.
Also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Let $A = \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix} $. Find $A^n$ I am new here so please let me know if I must resentence the exercice. I considered it too short not to include it in the title too.
Let $ A= \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatri... | Let $X=aI,Y=bJ$ and $Z=cJ^2$, where $J$ denotes the $3\times3$ nilpotent Jordan block. Then $X,Y,Z$ commute. Since $J^3=0$, among all degree-$n$ monomials of them, only $X^n,\,X^{n-1}Y,\,X^{n-1}Z$ and $X^{n-2}Y^2$ are nonzero. It follows that
\begin{aligned}
A^n
&=(X+Y+Z)^n\\
&=X^n+nX^{n-1}Y+nX^{n-1}Z+\binom{n}{2}X^{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum binomial coefficients Show that
$$\sum_{k=0}^{n}\binom{n}{k}\left(\binom{m+k+1}{m-n}+(-1)^{k+1}\binom{m+2(n-k)+1}{m}\right)=0$$
for all integers $m$ and $n$ with $m\ge n\ge 0$.
I tried induction on $n$, but there's not a very nice way to change the LHS from the $n$ case to $n+1$.
| We get for the first of the two pieces
$$\sum_{k=0}^n {n\choose k} {m+k+1\choose m-n}
= [z^{m-n}] (1+z)^{m+1} \sum_{k=0}^n {n\choose k} (1+z)^k
\\ = [z^{m-n}] (1+z)^{m+1} (2+z)^n.$$
The second piece is
$$\sum_{k=0}^n {n\choose k} (-1)^{k+1} {m+2(n-k)+1\choose m}
\\ = - [z^m] (1+z)^{m+2n+1}
\sum_{k=0}^n {n\choose k} (-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing Joint Moment Generating Function Good morning, Stack Exchange. My problem is that I have a joint distrubtion of discrete random variables with the following PMF and support:
$f(x,y) = \frac{2^{x-y} e^{-3}}{x!(y-x)!}$ when $x = 0, 1, 2, ... y$ and $y = 0, 1, 2, ... \infty $ and $f(x,y) = 0$ otherwise
So, my p... | You have PMF:
$$
f(x,y) =
\begin{cases}
\frac{2^{x-y} e^{-\frac{3}{2}}}{x!(y-x)!},& x = 0, 1, 2, \ldots y\text{ and } y = 0, 1, 2,\ldots\\
0& \text{otherwise}
\end{cases}
$$
And the MGF is:
\begin{align}
M_{X,Y}(t_1,t_2)
&=
E(e^{t_1x+t_2y})\\
&=
\sum\limits_{y=0}^\infty\sum\limits_{x=0}^y
e^{t_1x+t_2y}\frac{2^{x-y} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think ... | $(x+y) ^2 =16$$\Rightarrow$$ x^2 +y^2 +2xy=16$$\Rightarrow$$ x^2 +y^2 =12$
Because $xy=2$
$x^2 +y^2 =12$ $\Rightarrow $ $ (x^2 +y^2)^3 =12^3$$\Rightarrow $$x^6+y^6+3(xy)^2(x^2 +y^2) =1728$$\Rightarrow $$x^6+y^6=1728 - 3(4)(12)=1584$
Finally :
$x^6+y^6 =1584$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
... | From
$$x\ne-\sqrt{x+1}$$
you are deducing
$$x^2\ne|x+1|$$
But this is invalid! You can't deduce $a^2\ne b^2$ from $a\ne -b$, because $a$ may be equal to $b$. (And in fact $x$ is equal to $\sqrt{x+1}$ when $x=\frac{1+\sqrt 5}{2}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Stuck solving this PDE problem I need to solve $$uu_x+yu_y=x, u(x,1)=2x$$ i tried with characteristics method but got stuck solving the ODEs $$\frac{dx}{dt}=u ~,~\frac{dy}{dt}=y~,~\frac{du}{dt}=x$$ with initial values $x(0)=s,y(0)=1,u(0)=2s$. From the second one i get $y=e^t$, but how to solve the first two ODEs? I tri... | According to your results we have
$$
\cases{
u^2-x^2-3s^2=0\\
y-\frac{x+\sqrt{x^2+3s^2}}{3s}=0
}
$$
or
$$
\cases{
u^2-x^2-3s^2=0\\
(3sy-x)^2-x^2-3s^2=0
}
$$
So solving the first for $s$ and substituting after choosing the feasible one into the second, we have
$$
u = x\frac{3y^2+1}{3y^2-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove Cauchy Sequence and limit Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.
Let $\{x_n\}$ be recursively defined as $ \left\{
\begin{array}{ll}
x_{n+1} =\frac{1-x_n}{4} \\
x_1 = \frac{1}... | For (a) and (c) you proofs are correct But your argument for (b) is wrong. $x_{n+1}-x_n \to 0$ does not imply that $(x_n)$ is a Cauchy sequence. Use the fact that $|x_{n+m}-x_n| \leq |x_{n+m}-x_{n+m-1}|+|x_{n+m-1}-x_{n+m-2}|+\cdots +|x_{n+1}-x_{n}|$ to show that $x_{n+m}-x_n \to 0$ as $n,m \to \infty$. [You will have t... | {
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"url": "https://math.stackexchange.com/questions/3901321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$ I am working on this problem:
$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
After I used Vietta's Formula,expanded $S^3$ and used many algebraic t... | Note $X^3+Y^3+Z^3 - 3XYZ = (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)$.
Using $X^3=ab, Y^3=bc, Z^3 = ca$, as $X, Y, Z$ are distinct (why?), $X^2+Y^2+Z^2> XY+YZ+ZX$ and hence the sign of $X+Y+Z = \sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}$ is the same as the sign of $X^3+Y^3+Z^3-3XYZ = -24-3=-27$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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maximum of the function $S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi\sqrt{2+2\cos\phi}$ In a problem I have the function:
$$S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi \sqrt{2+2\cos\phi}$$
in which $\phi \in [0,\pi]$.
I have to find the point of maximum and calculate the maximum value of the function.
$\frac{dS}{d \phi}= 2 \pi*sin(\... | Note
\begin{align}
S(\phi) &= \pi \sin^2\phi+\pi\sin\phi \sqrt{2+2\cos \phi}\\
&= \pi \sin^2\phi+2\pi\sin\phi \cos \frac\phi2\\
&= 4\pi\left(\sin\frac\phi2 + \sin^2\frac\phi2-\sin^3\frac\phi2-\sin^4\frac\phi2\right)\tag1
\end{align}
Set $S’(\phi)= 0$ to get
\begin{align}
&2\pi \cos\frac\phi2\left(
4\sin^3\frac\phi2 +3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \si... | For $x\in\left(0,\frac\pi2\right]$, the concavity of $\sin(x)$ says
$$
\frac2\pi\le\frac{\sin(x)}x\le1
$$
Therefore,
$$
\underbrace{\left(\frac4\pi\sqrt{x}-\sqrt{x}\right)^x}_{\left(\frac4\pi-1\right)^x\sqrt{x^x}}\le\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac1x\right)\right)^x\le\underbrace{\left(2\sqrt{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.
In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ .
What I Tried: Here is a picture:-
Let... | Let $E=(0,0)$, $B=(-\alpha,0)$, $C=(\alpha,0)$, $D=(0,\beta)$. Then we have
$$\alpha^2+\beta^2=7^2=49,\quad\tan \angle ACB=\frac{\beta}{\alpha}\in(0,\sqrt3).$$
Let $t=\tan \angle ACB$, then
$$y_{AB}=\frac{2t}{1-t^2}(x+\alpha),\quad y_{AC}=-tx+\beta.$$
And we have
$$A=\left(-\alpha\frac{1+t^2}{3-t^2},\alpha\frac{4t}{3-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3906150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Legendre symbols So for odd prime $p$, I have Legendre symbol $(\frac{-1}{p}) = (-1)^{(p-1)/2}$, and let $n\geq 3$ be an odd positive integer.
I have to show that for odd positive integers $n_1$ and $n_2$, we have
$\frac{n_1n_2-1}{2}\equiv \frac{n_1-1}{2} + \frac{n_2-1}{2} \pmod{2}$
And from this I have to deduce $(\fr... | You already know the result when $n$ is prime, so assume that $n$ is composite and positive. Then there exist positive integers $n_1$ and $n_2$ such that $2\le n_1\le n_2<n$ and $n_1 n_2 = n.$ What determines whether a power of $-1$ is $1$ or $-1$ is the exponent modulo $2.$ By what you have already proven, $$(-1)^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3912980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find the eigenvalues of $x x^T - y y^T$?
If $x$ and $y$ are two linearly independent column $n$-vectors, where $n \geq 2$, find all the eigenvalues of $x x^{T} - y y^{T}$
Could you please tell me whether what I have written is correct or not?
Using the identity
\begin{align*}
\lambda^n\det(\lambda I_{(m)} - A... | Every vector simultaneously orthogonal to $x$ and $y$ is an eigenvector of the matrix $xx^T - yy^T$ of eigenvalue 0 (note that the matrix has rank 2, so his kernel must have dimension $n-2$ and this is always an eigenspace). So, the remaining eigenvectors must be linear combination of $x$ and $y$.
Solve
$$ (xx^T - yy^T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving linear in $\langle a_a, a_a \rangle =
\begin{cases}
\frac{1}{9}, & \text{if $a$ $\ne$ $a$}\
\end{cases}$
| To begin with, let us consider the linear combination
\begin{align*}
\alpha_{1}a_{1} + \alpha_{2}a_{2} + \ldots + \alpha_{k}a_{k} = 0
\end{align*}
If we take the inner product of such linear combination with $a_{1}$ we get the equation
\begin{align*}
2\alpha_{1}b + \alpha_{2} + \alpha_{3} + \ldots + \alpha_{k} = 0\\\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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JEE Circles Doubt Can you derive a general formula for length of intercept made by a circle $x^2+y^2+2gx+2fy+c=0$ on the line $ax+by+c=0$? If no, then is there any other method to solve these type of questions?
| \begin{align}
x^2+y^2+2gx+2fy+d &= 0
\tag{1}\label{1}
,\\
ax+by+c &= 0
\tag{2}\label{2}
.
\end{align}
Expression \eqref{1} represent a circle
with radius $r=\sqrt{g^2+f^2-d}$
centered at $O=(-g,-f)$:
\begin{align}
(x+g)^2+(y+f)^2
&=g^2+f^2-d
\tag{3}\label{3}
.
\end{align}
Intersection points $X_+,\ X_-$
of the circle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to prove that $a^3 + b^3 \geq a^2b + ab^2$? Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--
For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$
I have no i... | Note
$$\frac{a^3+b^3}{a^2b+ab^2}=\frac{a^2-ab+b^2}{ab}=\frac{(a-b)^2}{ab}+1\ge 1
$$
Thus
$$a^3+b^3\ge a^2b+ab^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Domain of a function defined with another function Suppose $f(x)=\frac{1}{1+\frac{1}{x}}$, and $g(x)=\frac{1}{1+\frac{2}{f(x)}}$. To find the points not in the domain of $g(x)$, we obviously have to check the $x-$values where $f(x)$ becomes $0$ and $-2$, and count them first.
But where is the intuition behind consideri... | We want to find when $g$ is undefined and this is when the denominator, $1+ \frac {2}{f(x)} = 0 $ which is only true when $f(x) = -2$. So then we check $-2 = \frac {1}{1+\frac{1}{x}} \implies 1+\frac {1}{x} = -\frac{1}{2}\implies x=-\frac {2}{3}.$ So $g(-\frac{2}{3})$ is not defined and $x = -\frac {2}{3}$ is not in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3915964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating limits using a series I'm trying to use a Taylor series centered at $0$ to evaluate this limit:
$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$
I rewrote the function as its Maclaurin series:
$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$
In expanded form:
$$\sum_{k=1}^\infty... | As $x\to +\infty$, it should be
$$4x^3\left(e^{\color{blue}{\frac{-2}{x^3}}}-1\right)=4x^3\sum_{k=1}^{\infty}\frac{(\color{blue}{\frac{-2}{x^3}})^k}{k!}=
4x^3\left(-\frac{2}{x^3}+o(1/x^3)\right)=-8+o(1).$$
Then what is the limit?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find eigen values of such kind of matrices ( Matrices which are bigger than $ 3\times 3$) The following question is part of a masters exam for which I am preparing and I don't have any methodology on how this type of matrices are solved.
This was question asked:
I don't know how to find eigenvalues of matrices (... | In addition to the existing answers I'd like to say that it's not hard to calculate the characteristic polynomial using Laplace expansion:
\begin{align}p(x)&=\det\begin{bmatrix} x & 0 & 0 & -1 & 0 & 0\\
0 & x & 0 & 0 & -1 & 0\\
0 & 0& x& 0 & 0& -1\\
-1 & 0 &0&x& 0 & 0\\
0 & -1 & 0 & 0&x &0\\
0 & 0& -1 & 0 & 0 &x
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is $s_n$ a multiple of $2^n$ for all $n \geq 0$? Let's define:
$$a_{0,0} = 1$$
$$\forall k \ne 0, a_{0,k} = 0$$
$$\forall n \geq 0, \forall k, a_{n+1,k} = -(4n-2k+3)a_{n,k-1} + (2k+1)a_{n,k}$$
| 0 1 2 3 4 ...
----+-----------------------------
0 | 1
1 | 1 -1
2 | 1 -8 3
3 | 1 ... | Starting with $$s_n = 2^n R_n(1)$$
All we need to prove then is that $R_n(1)$ is an integer for all $n$
Clearly $s_n$ is a sequence of integers, implying that if $R_n(1)$ is not an integer, then it must be a fraction with denominator $2^m,m\leq n$.
Remember
$ R_0(x)=1, R_{n+1}(x) = \left( R_n(x) \times \frac{2x}{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$
Let $x \in \Bbb R$ and $n \in \Bbb Z$ with $n \gt 0$. Show that $\left\lfloor\dfrac{\lfloor x \rfloor}{n} \right\rfloor = \left\lfloor\dfrac{x}{n}\right\rfloor$;
in particular, $\left\lfloor\dfrac{\left\lfloor\df... | Let $m=\lfloor x\rfloor$. Then $m\leq x < m+1$. So $\dfrac{m}{n}\leq \dfrac{x}{n}< \dfrac{m+1}{n}$. By the division algorithm, $m =qn+r$ for some integers $q,r$ with $0\leq r\leq n-1$. Thus, $$q\leq\dfrac{m}{n}\leq\dfrac{x}{n}<\dfrac{m+1}{n}=\dfrac{qn+r+1}{n}\leq q+1.$$
So $\lfloor \tfrac{m}{n} \rfloor=\lfloor\tfrac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.
When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac... | Disclaimer: Here is a solution which is not mine. I just slightly adapted it from the very same question, asked as Problem 5642, by R. Redheffer, American Mathematical Monthly 76, (1969), p.422, entitled "A delightful inequality"...
Let us fix $x$ with $0 < x < 1$:
Starting from the classical infinite product for the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| Let $t = \tan (x/2)$. Then $\sin x = 2t/(1+t^2)$ and $\cos x = (1-t^2)/(1+t^2)$. Your equation looks quartic after multiplying both sides by $(1+t^2)^2$.
$$12t(1-t^2)+4((1-t^2)^2-4t^2)-(2-2t^2-12t)(1+t^2)-6(1+t^2)^2$$
but it simplifies to $-4(3t-1)^2=0$, so $t = 1/3$, giving $x = 2(\tan^{-1} t + n\pi) = 2(\tan^{-1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Expected projected length of radial vectors of n-sphere Situation
In $n$-dimensional Euclidean space rests a unit $(n-1)$-dimensional sphere that is orthographically projected onto a $(n-1)$-dimensional plane. The topological definition of a sphere is used, i.e. only the surface points belong to a sphere.
Question
What... | Disclaimer: I used Mathematica for part of the problem.
As an expected value, this would be $$\mathbb{E}\left[ \sqrt{x_1^2+x_2^2 + ... + x_{n-1}^2} \right]$$
where $\left(x_1, x_2, ..., x_n\right)$ is a random point on the surface of the sphere. Since it is known that $x_1^2 + x_2^2 + ... + x_n^2 = 1$, this is equival... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Evaluating Contour Integral of square Evaluating $\frac{1}{2i\pi} \oint_C \frac{z^2}{z^2+4} dz$ where $C$ is the square with vertices at $\pm 2$ and $\pm 2 + 4i$
So, I rewrote the contour integral:$\frac{1}{2i\pi} \oint_C \frac{z^2}{(z+2i)(z-2i)} dz$
then used the Cauchy Integral Theorem:
$\frac{1}{2i\pi} \oint_C \frac... | You have done the integration incorrectly; the answer is $i$ (backed up by numerical evaluation using mpmath).
The contour encircles a simple pole of the integrand at $2i$. By Cauchy's integral formula applied to $f(z)=\frac {z^2}{z+2i}$
$$f(2i)=i=\frac1{2\pi i}\oint_C\frac{f(z)}{z-2i}\,dz=\frac1{2\pi i}\oint_C\frac{z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Primes of the form $x^2+31y^2$ and $x^2+xy+8y^2$ and quadratic forms While working on an exercise in algebraic number theory, I encountered the following problem:
A prime number $p$ can be represented as $x^2+31y^2$ if and only if it can be represented as $x^2+xy+8y^2$.
In this answer by Will Jagy, he claims that th... | It turns out I was thinking too difficult. "Completing the square" gives $$x^2+xy+8y^2=(x+\tfrac12 y)^2+31(\tfrac12 y)^2.$$
Let $n$ be an odd integer such that $n=x^2+xy+8y^2$ for some integers $x,y$. Reducing modulo 2 gives $1\equiv x(x+y)\bmod{2}$, hence $x$ and $x+y$ are odd, so in particular $y=(x+y)-x$ is even.
Co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Stuck at solving differential equation by using integrating factor This is the text:
Solve the following differential eqution: $$(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$$
This particular differential equation has to be solved by finding an integrating factor. I first tried making this equation into a Darboux ... | $(2x^3+3x^2y+y^2-y^3)~dx+(2y^3+3x^2y+x^2-x^3)~dy=0$
$\dfrac{dy}{dx}=-\dfrac{2x^3+3x^2y+y^2-y^3}{2y^3+3x^2y+x^2-x^3}$
Let $y=xu$ ,
Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=-\dfrac{2x^3+3x^3u+x^2u^2-x^3u^3}{2x^3u^3+3x^3u+x^2-x^3}$
$x\dfrac{du}{dx}=\dfrac{(u^3-3u-2)x-u^2}{(2u^3+3u-1)x+1}-u$
$x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that $\frac{1}{1-z}= \frac{1}{2}\left(1 + i\cot(\theta/2)\right)$
Question.(source) If $z=\cos\theta+i\sin\theta$, where $\theta$ is real, show that
$$\frac{1}{1-z}= \frac{1}{2}\left( 1+ i\cot\frac{\theta}{2}\right)$$
I have been trying to rearrange the righ... | Note that $e^{i\theta}=\cos\theta+i\sin\theta$.
$$\frac{1}{1-e^{i\theta}}$$
$$=\frac{1}{2}\times \frac{-2}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \frac{e^{i\theta}-1-(e^{i\theta}+1)}{e^{i\theta}-1}$$
$$=\frac{1}{2}\times \left(1-\frac{e^{i\theta}+1}{e^{i\theta}-1}\right)$$
Note that $e^{i\theta}+1=e^{\frac{i\theta}{2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Need help solving Laplace Transform Question For context: I am just starting to learn LaPlace Transform and am stumped with this question.
Find the PS of $x''+x'-12x=sin(3t) + e^{-4t}+e^{3t}$
Where $x(0)=0, x'(0)=0$ using LaPlace Transform
My Approach:
Let X(s)=$\mathscr{L}[x(t)]$
Apply the LaPlace Transform to both si... | It's more simple to decompose fractions from this step:
$$X(s)[s^2+s-12] = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)(s+4)(s-3) = \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}$$
$$X(s)=\dfrac 1{(s+4)(s-3)}\left ( \cfrac{3}{s^2+9}+\cfrac{1}{s+4}+\cfrac{1}{s-3}\right)$$
And you have:
$$\dfrac 1{(s+4)(s-3)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$ I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$
I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x... | Just use the definition: you need to verify that
$$\Bigg|\frac{x}{x^2-2}-1\Bigg| < \epsilon$$
gives you a neighbourhood of $2$.
This means to solve
$$\begin{cases}
\dfrac{x}{x^2-2}-1 < \epsilon \\\\
\dfrac{x}{x^2-2}-1 > -\epsilon
\end{cases}
$$
Solve this and you will get
$$\sqrt{\dfrac{2\epsilon-4}{\epsilon-1}} < x < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3930532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Calculus Integral Question
Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.
I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabo... | The reqyired parabola should be of the type $y=-kx^2+hx, k>0$ if it passes through $(a,b)$ then $h=(b+ka^2)/a$. So the eq. of the parabola is $$y=-kx^2+\frac{(b+ka^2)}{a}x$$
This cots $x$ axis at $x=0$, $x=x_0=a+b/(ak)$, so the area made by the parabola on x-axis is
$$A(k)=\int_{0}^{x_0}\left(-kx^2+\frac{(b+ka^2)x}{a}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding a function $f:(0, \infty) \to(0, \infty)$ satisfying the differential equation ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$
Let $f:(0, \infty) \to(0, \infty)$ be a differentiable function such that ${f}'\left( \frac{a}{x} \right)=\frac{x}{f\left( x \right)}$ where $a$ is positive constant and $... | Let $\varphi(x) := f(x) f\left(\frac{a}{x}\right)$. Differentiating we get, for every $x > 0$,
$$
\varphi'(x) = f'(x) f\left(\frac{a}{x}\right)+f(x) \left[f\left(\frac{a}{x}\right)\right]'
= \frac{a}{x f\left(\frac{a}{x}\right)} f\left(\frac{a}{x}\right)- \frac{a}{x^2} f(x) \frac{x}{f(x)} = 0.
$$
Hence there exists $c ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$ I have to solve the following statements, finding the tangent planes to $f$ and its points of tangency:
that contains the line in $\mathbb{R}^3$ that passes through the points $(3,0,3)$ and $(0,-3,3)$
and this other statement:
that contains the point $(0,0,2)$... | Given
$$
\cases{
f(x,y,z)=x^2+y^2+z-1=0\\
\pi(x,y,z) = a x+b y + z c + d=0\\
p=(x,y,z)\\
p_1=(3,0,3)\\
p_2=(0,-3, 3)
}
$$
Construct $\pi(x,y,z)$ such that it contains the line $L \to p = p_1+\mu(p_2-p_1)$ and is tangent to $f(x,y,z)$.
At the tangency point $p_0$ we have
$$
\cases{
\nabla f(p_0) = \lambda \nabla \pi(x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t... | Using @Bernard solution
$$y=\frac{2(1-u^2)}{(1+u^2)\bigl(u^2+2u+3\bigr)}$$ using partial fractions
$$y=-\frac{1+i}{2(u-i)}-\frac{1-i}{2 (u+ i)}+\frac{\sqrt{2}+2 i}{2 \left(\sqrt{2} u+\sqrt{2}-2 i\right)}+\frac{\sqrt{2}-2 i}{2 \left(\sqrt{2} u+\sqrt{2}+2 i\right)}$$
$$\int y\,du=\left(-\frac{1}{2}-\frac{i}{2}\right) \lo... | {
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"url": "https://math.stackexchange.com/questions/3946129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How to find asymptotic expansion of $\frac{1}{\cos(z)}$
Find the asymptotic expantion of coefficients of the exponential generating function $f(z)=\frac{1}{\cos(z)}$ using all of its poles.
My work:
I know that:
$$
\frac{1}{\cos \left( z \right)}=\sum_{n\ge 0}{E_{2n}}\frac{z^{2n}}{\left( 2n \right) !}
$$
To find the ... | You found that
$$
\frac{1}{{\cos z}} = \sum\limits_{k = 0}^\infty {\left( {\frac{{( - 1)^k }}{{z + \left( {k + \frac{1}{2}} \right)\pi }} - \frac{{( - 1)^k }}{{z - \left( {k + \frac{1}{2}} \right)\pi }}} \right)} = \sum\limits_{k = 0}^\infty {( - 1)^k \frac{{(2k + 1)\pi }}{{\left( {k + \frac{1}{2}} \right)^2 \pi ^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$ Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$
How would I go about rearranging this to deduce a value for x?
Thanks for the help.
| Here I propose an alternative way to solve it for the sake of curiosity.
More precisely, what about polar coordinates? If we set $x = r\cos(\theta)$ and $y = r\sin(\theta)$, one gets that
\begin{align*}
\sqrt{x^{2} + ay^{2}} - \sqrt{x^{2}+y^{2}} = z & \Longleftrightarrow r\sqrt{\cos^{2}(\theta) + a\sin^{2}(\theta)} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
alternative way of proving $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ converges to $\ln 2$ without using integrals I found a lot of answered to this problem using Riemann - sums, I myself solved it rewriting the logarithm sum, but there is another way where you should use the following hints.
Show that... | Using the inequality$$\frac{1}{(n+1)}<\ln(\frac{n+1}{n})<\frac{1}{n}$$( Since $\frac{1}{(n+1)}*1<\ \int_{n}^{n+1}\frac{1}{x} \,dx <\frac{1}{n}*1$ (think area graphically) ),we can write $\ln(\frac{n+1}{n})<\frac{1}{n}$$<$$\frac{1}{n-1}<ln(\frac{n-1}{n-2})$. Then sum from $n$ to $2n$ and take $\lim_{n\to\infty}$.
$ \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$ Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1... | A trick: use $\int_0^1f(x)dx=\int_0^1f(1-x)dx$ to average two versions of your integral, reducing it to$$\tfrac12\left(\int_0^1\tfrac{xdx}{x^2+(1-x)^2}+\int_0^1\tfrac{(1-x)dx}{x^2+(1-x)^2}\right)=\tfrac12\int_0^1\tfrac{dx}{x^2+(1-x)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\a... | If $\arg(f(z))=\pi /2$, that means that $\Re(f(z))=0$ and as well $0 < \Im (f(z))$: that's a more "sure" way to go.
The steps in detail are
$$
\eqalign{
& \arg \left( {{{z - 8} \over {z - 2}}} \right) = {\pi \over 2}\quad \Rightarrow \quad
\left\{ \matrix{
{\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can't find the inverse of Laplace Transform. Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
... | $$F(s)=\dfrac{2}{s(s^2+4)}$$
$$F(s)=\dfrac{1}s\dfrac 2{(s^2+4)}= \mathcal {L}(1) \mathcal{L} ( \sin (2t)$$
You can also use the theorem of convolution :
$$f(t)=1* \sin (2t)=\int_0^t 1 \times \sin(2\tau) d\tau$$
$$f(t)= \dfrac {-\cos (2 \tau)}{2} \bigg |_0^t$$
$$f(t)= \dfrac 12 -\dfrac {\cos (2 t)}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If A is an $m\times n$ matrix, show that $||A||_2 \le \sqrt{||A||_1*||A||_\infty}$
If $A$ is an $m\times n$ matrix, show that $$\| A \|_2 \le \sqrt{\|A\|_1 \, \| A \|_\infty}$$
I reduced this to:
$$\rho(A^TA) \le||A||1*||A||_\infty$$
I created a matrix for experimenting:
$$A = \begin{bmatrix}a&b\\c&d\end{bmatrix} \\... | The conventional proof is
$$
\|A\|_2^2=\rho(A^\ast A)\le\|A^\ast A\|_\infty\le\|A^\ast\|_\infty\|A\|_\infty=\|A\|_1\|A\|_\infty,
$$
where the first inequality is due to the fact that the spectral radius is the infimum of all submulticative norms, and the second inequality is due to the submultiplicativity of the induce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Write $x$ in terms of $y$ in inequality ${x}^{2}+x\le y$ How can I write the inequality ${x}^{2}+x\le y$ such that $?\le x\le ?$ (in terms of $y$) where $-2\le y\le 1$ ?
I tried $x(x+1)\le y\Rightarrow x\le \frac{y}{x+1}$ then became stuck.
Any help will be appreciated.
| Rewrite the inequality $$x^2 + x \le y $$ as follows: $$ x^2 + x - y \le 0 $$ Then, by discriminant formula we get $$x_{1,2} = \frac{-1 \pm \sqrt{1+4y}}{2} $$ and therefore $$\left(x - \frac{-1 + \sqrt{1+4y}}{2}\right)\left(x - \frac{-1 - \sqrt{1+4y}}{2}\right) \le 0$$ Thus, $$\frac{-1 - \sqrt{1+4y}}{2} \le x \le \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to read and execute $\sum_{1 \leq \ell
How to read and execute this sum?
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$$
I am having trouble to understand where is my error.
The question does not say, but I am assuming that $\ell$ starts at $1$, $m$ at $2$, and so $n$ at $3$.
This is essential a product o... | The index region of the sum
\begin{align*}
\sum_{\color{blue}{1\leq \ell <m<n}}\frac{1}{5^{\ell}3^m2^n}\tag{1}
\end{align*}
is specified by the inequality chain
\begin{align*}
1\leq \ell <m<n
\end{align*}
which has $1$ as lower limit and $n$ as upper limit. We have two indices $\ell$ and $m$, which means we can write i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Proof by induction, $2^{n} > n^{2} - 2$ So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.
I start of with the "basecase" for n = 3:
RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$
Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where ... | I think this is a little clearer.
$2^{k+1}\ge2k^2-4$
$=k^2+2k+1-2+k^2-2k-3$
$=(k+1)^2-2+(k-3)(k+1)$
$\ge(k+1)^2-2$ for $k\ge3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is Below are the steps I have done so far.
Find $a$ in terms of $c$
$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$
Find $b$ in terms of $c$
$$
ab = 432 ... | $bc=96$ and $c<9$ constrains $c$ to be one of $1,2,3,4,6,8$. Since $\gcd(432,96)=48$, $c$ cannot be $1$ or $3$, leaving four cases: $(a,b,c)=(9,48,2),(18,24,4),(27,16,6),(36,12,8)$. Of these, $(18,24,4)$ achieves the smallest $a+b+c$ of $46$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{... | If $m+n$ is even, then $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor = \frac{m+n}{2}$ and as $m-n=m+n -2m$ is also even, $\Bigl\lfloor \frac{m- n+1}{2}\Bigr\rfloor = \frac{m-n-1}{2}$.
So $\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor = m$
If $m+n$ is odd, then $\Bigl\lfloor \frac{m+ n}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $
I found the minimum using Cau... | Since $\frac{x}{10}, \frac{y}{10}, \frac{z}{10}\in[0; 1]$, we have that $\left(\frac{x}{10}\right)^2\leqslant\frac{x}{10}, \left(\frac{y}{10}\right)^2\leqslant\frac{y}{10}, \left(\frac{z}{10}\right)^2\leqslant\frac{z}{10}$. Thus \begin{align*}2x^2+y^2+z^2&=100\cdot \left(2\cdot\left(\frac{x}{10}\right)^2+\left(\frac{y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x... | Since the divisor polynomial is of degree $3$, the remainder will be of atmost degree $2$. Hence, instead of $Ax+B$ that you have taken, take $Ax^2+Bx+C$ as the remainder and you also have $3$ equations to plug the values.
Note that $f(1)=A+B+C=3$, $f(-3)=9A-3B+C=-1$ and $f(-2)=4A-2B+C=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've fou... | After you substitute $(x,y)=(\cos{\theta},\sin{\theta})$, you just need to prove the followings:
$$
\begin{aligned}
\sin{\theta}+\cos{\theta}&\geq\sqrt{2}\\
\\
1-\sin{\theta}\cos{\theta}&\geq\sin{\theta}\cos{\theta}
\end{aligned}
$$
Which are easy. First one is well known, second one is just double angle sine identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 9
} |
Is there a reason why this technique is invalid? What is $ \lim_{x \rightarrow 0} \frac{1 - \cos x}{x}$? A simple way to evaluate this limit is to substitute $0$ for $x$ in the numerator to obtain
$ \displaystyle \lim_{x \rightarrow 0} \frac{1 - 1}{x} = \lim_{x \rightarrow 0} ( \frac{1}{x} - \frac{1}{x} )
= \lim_{x \r... | A counterexample:
$$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac12,\quad\enspace\text{not }0.$$
Indeed $\;1-\cos x=2\sin^2\tfrac x2$, so
$$\frac{1-\cos x} {x^2}= \frac{2\sin^2\frac x2}{4\bigl(\frac x2\bigr)^2}=\frac12\biggl(\underbrace{\frac{\sin\frac x2}{\frac x2}}_{\underset{\textstyle 1}{\downarrow}}\biggr)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Convergence/Divergence proof for Infinite products Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.
A close relative of the infinite series is the infinite product.
\begin{align*}
\prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdo... | For (a) it says "explicit formula". Yours is not. Here it is:
$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)
\cdots\left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)
\\
=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{n}{n-1}\cdot\frac{n+1}{n}
= \frac{n+1}{1} = n+1
$$
Eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\fra... | Hint: Rewrite as follows:
$$\begin{align*}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n+1-1}{n+1}-1-\frac{1}{n} \\
&=1-\frac{1}{n+1}-1-\frac{1}{n} \\
&=-\left(\frac{1}{n+1}+\frac{1}{n}\right).
\end{align*}$$
As $n\to\infty$, what do $1/n$ and $1/(n+1)$ approach?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Die and coin variance of random variable question. Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.
My work:
$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} + 3 \cdot \cfrac{1}{6} + 4 ... | $Y\perp\!\!\!\!\!\!\perp Z$ thus $V(X)=V(Y)+V(Z)$
Your error is this
$$E(X^2)=E[(Y+Z)^2] \ne E(Y^2)+E(Z^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Please explain how to derive the PGF of a 6 sided dice I understand the PGF is
$$ P(s) = \frac{s}{6} \frac{1-s^6}{1-s} $$
Can someone explain all the steps to get there
| We want
$$f(x)=x+x^2+x^3+\cdots +x^6.$$
By long division we can prove that
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$
If we multiply the above expression by $1-x^7$ we are going to see a pattern where all the $x^n$ exponents will cancel out after $x^6$:
$$\begin{align}&(1-\color{red}{x^7})+(x-\color{blue}{x^8}) +(x^2-x^9)+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a closed form for the recurrence relation: $a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1$ Find a closed form for the following recurrence relation:
$$a_{n+2}-a_n=\sin\frac{n \pi }{2} \;\;\;a_0=a_1=1\tag{$n \ge 0$}$$
This is what I've done so far:
Define a function $f(x):=\sum_{n\ge0}^{ }a_{n}x^{n}$,then:
$$\s... | Since a generating function answer has already been provided, I'll instead provide the direct proof. Note that
$$a_{n+2}=a_n+\sin(n\pi /2)=a_{n-2}+\sin((n-2)\pi/2)+\sin(n\pi/2).$$
But $\sin(n\pi/2-\pi)=-\sin (n\pi/2)$, so $a_{n+2}=a_{n-2}$. Hence the sequence is 4-periodic, and it suffices to know the first four entrie... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\fra... | Integration by parts
$$\int x \sqrt{1-x^2} \arcsin x\, dx$$
take
$$u'(x)=x \sqrt{1-x^2};\;v(x)=\arcsin x$$
so
$$u(x)=\int x \sqrt{1-x^2} \, dx=-\frac{1}{2}\int (-2x)(1-x^2)^{-1/2}\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}+C$$
and
$$v'(x)=\frac{1}{\sqrt{1-x^2}}$$
apply IBP formula
$$\int u'(x)v(x)\,dx=u(x)v(x)-\int u(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Another way to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$ The question is to prove that $81^{3^{n}}+4^{2n+1}$ is not prime for any positive integer $n$.
The easy way to solve this is by taking the last digit of each number:
*
*$81^{3^{n}}$ ends in $1$ because $81$ also ends in $1$.
*$... | \begin{align}
81^{3n} + 4^{2n+1} &= (3^4)^{3n} + (2^2)^{2n+1} \\
&= (3^{6n})^2 + (2^{2n+1})^2 \\
&= (3^{6n})^2 + (2^{2n+1})^2 + 2 \times 3^{6n} 2^{2n+1} - 2 \times 3^{6n} 2^{2n+1} \\
&= (3^{6n} + 2^{2n+1})^2 - 3^{6n} 2^{2n+2} \\
&= (3^{6n} + 2^{2n+1})^2 - (3^{3n} 2^{n+1})^2 \\
&= (3^{6n} + 2^{2n+1} - 3^{3n} 2^{n+1}) (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding Pythagorean triple where the squares of sides are weighted Suppose $a, b, c \in \mathbb{Z}_{>0}$. The following is a variation of a Pythagorean triple, but with weighted squares:
$$a^2 + 3 b^2 = 4 c^2$$
Can I find something like Euclid's formula that can generate all solutions for this formula? I'm not sure how... | After discovering Find all integers satisfying $m^2=n_1^2+n_1n_2+n_2^2$, the paper Abdelalim, S., & Dyani, H. (2014). The Solution of the Diophantine Equation x2+ 3y2= z2. International Journal of Algebra, 8(15), 729–732. gives an easy answer:
Theorem 2.1 Let $E : x^2 + 3 y^2 = z^2$ diophantine equation and $(x, y, z) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$ Find all $n$ such that ${n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}$ is a factor of $2^{2018}$.
I got this interesting question from a friend. I made an assumption and was able to get $n$ as $3$ and $7$ bu... | Let
$$u(n)={n \choose 0}+{n \choose 1}+{n \choose 2}+{n \choose 3}=\frac{(n+1)(n^2-n+6)}6$$
First, we must have $n\ge0$, otherwise the left hand side is zero.
If $u(n)$ is a factor of $2^{2018}$, that means $u(n)$ must be a power of two. Since $3$ must divide either $n+1$ or $n^2-n+6$, this means that either:
$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$n-$section formulas in goniometry without the calculus We know that the formulas of bisection of an angle $\alpha$ it is:
$$\sin^2\left(\frac{\alpha}2\right)=\frac{1-\cos(\alpha)}{2}, \qquad \cos^2\left(\frac{\alpha}2\right)=\frac{1+\cos(\alpha)}{2}$$
In particular
$$\sin\left(\frac{\alpha}2\right)=\pm\sqrt{\frac{1-\c... | I doubt that there are such elementary general formulas. I will derive a possible (restricted) formula as an infinite series using De Moivre's formula, though this may not satisfy your requirement of calculus free derivation. One value of the multivalued expression $\cos\left(\dfrac{\theta}{n}\right)+i\sin\left(\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3987946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Question about generating functions about an exercise I have a question about generating functions,
I've understood that generating functions are also representing some series of numbers, but is this series could be formed by 2 different functions?
For example, an exercise that I've encountered goes like this:
Find the... | Your approach is fine. We can show the generating functions are equal: $f=g$.
We have
\begin{align*}
\color{blue}{g(x)}&=\left(\left(x^2 + x^4\right) + \left(x^8 + x^{10}\right) + \left(x^{14} + x^{16}\right) + ... \right)^k \\
&=\left(x^2\left(1+x^2\right)+x^8\left(1+x^2\right)+x^{14}\left(1+x^2\right)+\cdots\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $ \lim_{n \to \infty}n \log\left(1+ \left(\frac{f(x)}{n}\right)^p\right)$ Find $$\lim_{n \to \infty}n \log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)$$ where $0<p<1$
my attempt :By using L-Hospital Rule i got
$$\lim_{n\to \infty} \dfrac {\dfrac{1}{1+\left( \dfrac{f(x)}{n}\right)^p}
\cdot\dfrac {f(x)^p}{-n^{2... | Let $y = f(x)$. Consider the essential case $y \neq 0$.
\begin{align*}
n\log\left(1 + \left(\frac{y}{n}\right)^p\right) &
= y^pn^{1-p}\log \left(\left( 1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right).
\end{align*}
Since $\left\{\left(\frac{n}{y}\right)^p \right\}_{n \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| Denote $y = \tan x \cot 2x = \frac{1-\tan^2x}2$ and express $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$ as a system of equations in $x,y$
$$\tan x+\frac{2\tan x}{1-\tan^2x}=\left(1+\frac1y \right)\tan x=\frac{2}{\sqrt{3}}
$$
Then, eliminate $\tan x$ to get
$$\frac3{y^3} -\frac{13}{y}-6=0$$
which is a depressed cubic equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{... | Let $$x = \sum_{m=2}^{n}2^{m-2},$$ and note that $$1 + x = 1 + 1 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{1}.$$ Then, multiplying $(1)$ by $2$, we have $$2+2x = 2 + 2 + 4 + \dotsb + 2^{n-2} + 2^{n-1}\tag{2}.$$ We can rewrite $(1)$ as $$1 + x = 2 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{3}.$$ Subtracting $(3)$ from $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Simplifying a trig relation My physics teacher wanted to prove Young's equation (my problem is pure mathematics), and in the derivation, he got the following expression:
$$\tag{1}
\begin{align}
\left(\gamma_{s l}-\gamma_{s o}\right) \cdot\left(\sin ^{2} \theta+\cos \theta\left(-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+... | The coefficient of $\gamma$ can be simplified as,
$\begin{align}(1-\cos\theta)\frac{2+2\cos\theta - 2- \cos\theta}{1+\cos\theta} = \boxed{\cos\theta\ \left[\frac{1-\cos\theta}{1+\cos\theta}\right]} \end{align}$
Now, the coefficient of $\gamma_{sl} - \gamma_{so}$ can be simplified as,
$\begin{align}\sin^2\theta -\cos\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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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.