Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve for $\sqrt{x}+\sqrt[4]{y}=\sqrt{p}$ with $x$, $y$ are positive integers and $p$ is a prime number For that given problem, I have constructed a solution to it:
Squaring both sides, we get $x+\sqrt{y}+2\sqrt[4]{x^2y}=p$. Suppose $\sqrt{y}$ and $\sqrt[4]{x^2y}$ are integers, we can let $y=y'^2$, $y'$ being a positi... | There are not solution. In fact we have
$$y=(\sqrt p-\sqrt x)^4=p^2+6px+x^2-4(p+x)\sqrt{px}$$ One has three possible ways:
►If $x=p$ then $y=8p^2-8p^2=0$
►If $p+x=0$ then $y=2p^2-6p^2\lt0$
►If $x=pz^2$ then $y=p^2(z-1)^4$ which implies the given equation becomes
$(z+z-1)\sqrt p=\sqrt p\Rightarrow z=1\Rightarrow x=p$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Trigonometric Substitution Absolute value issue Evaluate $ \, \displaystyle \int _{0}^{4} \frac{1}{(2x+8)\, \sqrt{x(x+8)}}\, dx. $
$My\ work:-$
by completing the square and substitution i.e. $\displaystyle \left(\begin{array}{rl}x+4 & = 4\sec (\theta )\\ dx & = 4\sec \theta \tan \theta \, d\theta \end{array}\right) \q... | If $\theta$ lies in 4th quadrant, you need to consider the values of $\text{arcsec}$ in fourth quadrant(as the inverse trig functions are multivalued).
$\text{arcsec }(2) = \theta_1 \Rightarrow \theta_1 = \dfrac{5\pi}3$
$\text{arcsec} (1) = \theta_2 \Rightarrow \theta_2 = 2\pi$
So, $-\dfrac{1}{8}(\theta_1-\theta_2) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Q: Epsilon-delta proof for $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$ In proving $\lim_{x\to1} \frac{x}{x^2+1}=\frac{1}{2}$, I have done this working:
Let $ϵ>0$.
$0<|x-1|< \Rightarrow|\frac{x}{(x^2+1)}-\frac{1}{2}|<$
$ |\frac{x}{(x^2+1)}-\frac{1}{2}|< = |\frac{2x-x^2+1}{2(x^2+1)}| = |\frac{-2(x-1)^2}{x(x-1)^2+4x}| = |\... | As $ x $ goes to $ 1$, we can always assume that
$$|x-1|<2$$
and we look for $ \delta$ such that
$$|x-1|<2 \text{ and } |x-1|<\delta \implies$$
$$ |\frac{-(x-1)^2}{2x^2+2}|<\epsilon$$
but
$$2x^2+2\ge 2$$
and
$$|x-1|<2\implies $$
$$|\frac{(x-1)(x-1)}{2x^2+2}|\le |x-1|$$
So, we just need a $ \delta<2$ such that
$$|x-1|<\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\displaystyle\lim_{x\rightarrow0}\dfrac{a^{2x}-2}{x^x}=-1$ I tried like this:
Let $y=a^{2x}-2\Rightarrow a^{2x}=y+2\Rightarrow 2x\ln a=\ln\left(y+2\right)\Rightarrow x=\dfrac{\ln\left(y+2\right)}{2\ln a}$
Also if $x\longrightarrow0,$ then $y\longrightarrow a^{2(0)}-2=-1.$
But we I put each and every t... | Given
\begin{equation}
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x}=-1
\implies
\lim_{x\rightarrow 0}\dfrac{a^{2x}-2}{x^x} +1 =0
\end{equation}
You say,"then I get hanged due to $^$.." but $0^0=1$ and we can even eliminate the denominator if that isn't enough.
\begin{align*}
(\large{x}^x)
\small{\bigg(\dfrac{a^{2x}-2}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$ I found some difficulty with this exercise:
Calculate
$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$
I use change of Variables in Polar Coordin... | Take $$I=\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
Consider the change of variables $(x,y)\to(r\cos\theta,r\sin\theta)$.
Let $\partial D=\{(x,y)\in\mathbb{R}^2:\frac{x^2}2+y^2=1\}$. Then for all $(x,y)=(r\cos\theta,r\sin\theta)\in\partial D$, $r^2(1-\frac12\cos^2\theta)=1$. Hence,
$$I=\int_0^{2\pi}\left(\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I get to this formula for this recursive sum and is my reasoning in proving this correct? I have the following sequence:
$n=1$ $$1$$
$n=2$ $$2+3+4$$
$n=3$ $$5+6+7+8+9$$
$n=4$ $$10+11+12+13+14+15+16$$
$$...$$
So abstracting these sums, we can write:
$n_1+(n_1+1)+...+(n_2-1)+n_2$
, where $n_1$ is the first term an... | The sum is equal to the sum of integers up to $n^2$ which is trivially given by $\frac{1}{2}n^2(n^2+1)$, then we could prove at first the general result by induction and then use it for this particular case.
Refer also to the related
*
*Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$
Edit
Following your way, for the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How would you evaluate the following expression? It seems very difficult to simplify the trig.
I have tried to many different ways and I also end up with a tan inside an arctan which I do not know how to simplify. Please suggest how I can solve this. Am I missing something simple or is it quite tricky and need some so... | By symmetry we can reduce the interval to $[0,\frac{\pi}4]$, the coefficient before the integral sign becomes $8$.
$$I=\frac 8{\pi}\int_0^{\frac{\pi}4}\dfrac{\cos(4x)^2+1}{a\cos(2x)^2+1}dx$$
Then by substitution $u=2x$:
$$I=\frac 4{\pi}\int_0^{\frac{\pi}2}\dfrac{\cos(2u)^2+1}{a\cos(u)^2+1}du$$
The reason behind reducin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex Matrix and its Conjugate terms If $z = \left| {\begin{array}{*{20}{c}}
{3 + 2i}&1&i\\
2&{3 - 2i}&{1 + i}\\
{1 - i}&{ - i}&3
\end{array}} \right|\& \left| {z + \overline z } \right| = k\left| z \right|$, find the value of k
My approach is as follow
$ \Rightarrow z = - \left| {\begin{array}{*{20}{c}}
1&{3 + 2i}&... | You already exchanged the first two columns, which is a good start. The corresponding matrix
$$
A = \begin{pmatrix}
1&{3 + 2i}&i\\
{3 - 2i}&2&{1 + i}\\
{ - i}&{1 - i}&3\end{pmatrix}
$$
has the property that $A^T$ is the elementwise conjugate of $A$, so that
$$
\det(A) = \det(A^T) = \overline{\det(A)} \, .
$$
It follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compund Inequality with Reciprocal $0<(2x-4)^{-1}<\frac{1}{2}$ I need help with this exercise: $0<(2x-4)^{-1}<\frac{1}{2}$.
$$0<(2x-4)^{-1}<\frac{1}{2}$$
$$0<\frac{1}{2x-4}<\frac{1}{2}$$
I separate this into two inequalities:
*
*$0<\frac{1}{2x-4}$
*$\frac{1}{2x-4}<\frac{1}{2}$
The first one I had no problem solving... | \begin{align*}
0<(2x-4)^{-1}<\frac{1}{2}\\ \\
\implies
0<\frac{1}{2x-4}<\frac{1}{2}\\ \\
\implies 0< 2< 2x-4\\ \\
\implies 4< (2)+4< (2x-4) + 4 \\ \\
\implies 4< 6< 2x \\
\implies 2< 3< x\\
\implies 3< x\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$ Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$
My attempt:
Method:- 1
Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4}$ and $\theta \in [0,\frac{π}{2}]$
Therefore $$\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt 7}{4}$$
Now $\tan(\frac{1}{2}\sin^{... | So you have $\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \sqrt{1-\cos^2\frac{\theta}{2}}\cos\frac{\theta}{2}=\frac{3}{8}$ Solving $(1-x^2)x^2=\frac{9}{64} \implies x^4-x^2+\frac{9}{64}=0 \implies x^2=\frac{1+\sqrt{1-\frac{9}{16}}}{2}=\frac{4+\sqrt{7}}{8} $$\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}\cos\frac{\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is rolling a 60 face die (D60) the same as rolling 10 6 faced ones (D6)? I understand that if you roll $10$ $6$-faced dice (D6) the different outcomes should be $60,466,176$ right? (meaning $1/60,466,176$ probability)
but if you are playing a board-game, you don't care about that right? what matters is the sum of the ... | No, the probability distribution of the sum of the faces of rolling $10$ $6$-sided dice is different than that of rolling a single $60$-sided die.
For obvious reasons, this is because you can roll any number between $1$ and $9$ with a $60$-sided die, but cannot attain such a sum with $10$ $6$-sided dice.
The probabilit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Radical expression for $\tan 14.4^\circ$ Given that
$\displaystyle\cos 20^\circ=\frac{\sqrt[3]{\frac{1-i\sqrt{3}}{2}}+\sqrt[3]{\frac{1+i\sqrt{3}}{2}}}{2}$
and
$\displaystyle\sin 20^\circ=\frac{i\left(\sqrt[3]{\frac{1-i\sqrt{3}}{2}}-\sqrt[3]{\frac{1+i\sqrt{3}}{2}}\right)}{2}$
It turns out that there is a nice radical ex... | Try using:$$\tan(x)=\pm\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$ because you already know $\cos(x),x=14.4^\circ$.
$$\cos 14.4^\circ=\frac{\sqrt[5]{\frac{-1+\sqrt{5}-i\sqrt{10+2\sqrt{5}}}{4}}+\sqrt[5]{\frac{-1+\sqrt{5}+i\sqrt{10+2\sqrt{5}}}{4}}}{2}$$
Therefore
$$\tan(14.4)=\tan\frac{2\pi}{25}= \frac{\sqrt{1-\bigg(\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find value of $x^{5} + y^{5} + z^{5}$ given the values of $x + y + z$, $x^{2} + y^{2} + z^{2}$ and $x^{3} + y^{3} + z^{3}$ If$$x+y+z=1$$
$$x^2+y^2+z^2=2$$
$$x^3+y^3+z^3=3$$
Then find the value of $$x^5+y^5+z^5$$
Is there any simple way to solve this problem ? I have tried all my tricks tried to multiply two equations ,... | HINT
Multiply the third relation by the second so that you obtain
\begin{align*}
(x^{2} + y^{2} + z^{2})(x^{3} + y^{3} + z^{3}) & = x^{5} + y^{5} + z^{5} + x^{2}(y^{3} + z^{3}) + y^{2}(x^{3} + z^{3}) + z^{2}(x^{3} + y^{3})
\end{align*}
Now we can rearrange the last expression as
\begin{align*}
x^{2}y^{2}(x + y) + x^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256912",
"timestamp": "2023-03-29T00:00:00",
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How to find recurrence relation for $n$-digits number where $1$ and $3$ occur an odd number of times
Let $h(n)$ be the number of $n$-digits number with each digit odd, where the digits $1$ and $3$ occur an odd number of times. Find a recurrence relation for $h(n)$.
I have found the recurrence relation:
\begin{equatio... | The following does not derive $(1)$ from your recurrence but instead finds an explicit formula for $h_n$, which is perhaps the motivation for $(1)$. First replace $b$ with $a$:
\begin{align}
h_n &= 2a_{n-1} + 3h_{n-1}\\
a_n &= g_{n-1} + h_{n-1} + 3a_{n-1}\\
g_n &= 2a_{n-1} + 3g_{n-1}\\
\end{align}
With generating func... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove a Limit Using the $\varepsilon - N$ definition I'm having trouble proving the following limit using the $\varepsilon-N$ definition:
$$\lim_{n \to \infty} \frac{3n}{\sqrt{9n^2 + 11n + 4}} = 1.$$
Here is my work so far.
Take arbitrary $\varepsilon$. We need to prove the existence of a natural number, $N$, such that... | Let us assume that $n\geq n_{\varepsilon}$. Then it results that
\begin{align*}
\left|\frac{3n}{\sqrt{9n^{2} + 11n + 4}} - 1\right| & = \left|\frac{3n - \sqrt{9n^{2} + 11n + 4}}{\sqrt{9n^{2} + 11n + 4}}\right|\\\\
& \leq\left|\frac{3n - \sqrt{9n^{2} + 11n + 4}}{3n}\right|\\\\
& = \left|\frac{11n + 4}{3n[3n + \sqrt{9n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What did I do wrong in my proof of $4\nmid (n^2+3)\implies 2\nmid (n^4-3)$? A homework problem is asking me to prove the following statement:
For any $n\in\mathbb{Z}$, $2\mid (n^4-3)$ if and only if $4\mid (n^2+3)$
Quick note: we're not allowed to use the theory of congruences in our solution.
I figured the best way ... | $n^2+3=(2k)^2+3=4k^2+3$ for even $n$, so remainder is $3$.
$n^2+3=(2k+1)^2+3=4k^2+4k+4$ for odd $n$ and is divisible by $4$, so it doesn't qualify.
In other words assumption $4\nmid n^2+3$ implies $n$ is even and $n^4-3$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim _{x\to \infty }\tan\left(x^2\sin\frac{\pi }{4x^2}\right)$ without L'hopital's rule How do I solve this without using L'hopital's rule?
$$\lim _{x\to \infty }\left(\tan\left(x^2\sin\left(\frac{\pi }{4x^2}\right)\right)\right)$$
Problem
EDIT:
My progress so far:
$$\lim _{x\to \infty }\left(\tan\left(x^2\... | You've already taken care of one important bit, namely rewriting
$$\tan\left(x^2\sin\frac{\pi}{4x^2}\right)$$
as
$$\tan\left(\frac{\pi}{4}\cdot\frac{\sin\frac{\pi}{4x^2}}{\frac{\pi}{4x^2}}\right)$$
As you noted, $\lim_{x\to 0}\frac{\sin(x)}{x}=1$, so we should make use of this somehow. We can achieve this goal by makin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Nice problem: Prove that: $ab+bc+ca \ge \sum{\sqrt{a^2+b^2+3}}$ Problem: Let $a,b,c>0:a+b+c=abc.$ Prove that: $$ab+bc+ca\ge \sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}$$
Please help me give a hint to get a nice proof!
My attempts after squaring both side, it is desired to show: $a^2b^2+b^2c^2+c^2a^2+4(ab+bc+ca)\... | We need to prove that:
$$ab+ac+bc\geq\sum_{cyc}\sqrt{\frac{abc}{a+b+c}\cdot\left(a^2+b^2+\frac{3abc}{a+b+c}\right)}$$ or
$$(ab+ac+bc)(a+b+c)\geq\sum_{cyc}\sqrt{abc((a^2+b^2)(a+b+c)+3abc)}.$$
Now, by AM-GM and C-S
$$\sum_{cyc}\sqrt{(a^2+b^2)(a+b+c)+3abc}=\sum_{cyc}\sqrt{(a^2+b^2)(a+b)+(a^2+b^2+3ab)c}\leq$$
$$\leq\sum_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Given two pythagorean triples, generate another I don't know if this has been asked before, but I could not find any existing answer.
I noticed that for any pair of primitive pythagorean triples (not necessarily distinct), let's say:
a² + b² = c²
d² + e² = f²
Then there is at least another primitive triple where:
g² +... | You can always find new triplets this way, although they are not necessarily primitive. This is just the Brahmagupta–Fibonacci identity in action.
For example, for $(5,12,13)$ and $(8,15,17)$, the identity states that:
$$(5\cdot8 - 12\cdot15)^2 + (5 \cdot 15 + 12 \cdot 8)^2 = (-140)^2 + (171)^2$$
$$= (5\cdot8 + 12\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ Show that $1998 < \sum_1^{10^6}\frac{1}{\sqrt{n}}<1999$ by first recalling the inequality:
$$2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})$$ when $n=1, 2, 3...$
What I have tried to show using the inequality:
By the first step of induction to se... | More directly, multiply and divide by the conjugate:
$2 ( \sqrt{n+1} - \sqrt{n} ) = 2 \frac{ ( \sqrt{n+1} - \sqrt{n} ) ( \sqrt{n+1} + \sqrt{n} ) }{ ( \sqrt{n+1} + \sqrt{n} )} = \frac{2}{ ( \sqrt{n+1} + \sqrt{n} )} < \frac{2}{2\sqrt{n}} = \frac{1}{ \sqrt{n}}. $
Same for the RHS, can you complete it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Dependent equations, with Trigonometry. My questions is: do the first two equations imply the third ?
$$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+\frac{z}{y}+\cos^2\alpha$$
$$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+\cos^2\alpha$$
$$\left(z+\frac{1}{z}\right)\sin\alpha=
\frac{x}{y}+\frac{y}{x... |
do the first two equations imply the third ?
No, take $x=\frac 12,y=4,z=2$ and $\alpha=\frac{\pi}{2}$ for which the first two equations hold and the third doesn't.
Subtracting
$$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+1-\sin^2\alpha\tag1$$
from
$$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $2\tan^{-1}\left(\sqrt{\frac{a}{b}}\tan\frac{x}{2}\right)=\sin^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}$ $$
2\tan ^{-1}\left(\sqrt{\frac{a}{b}}\tan \frac{x}{2}\right)=\sin ^{-1}\frac{2\sqrt{ab}\sin x}{\left(b+a\right)+\left(b-a\right)\cos x}
$$
I know within inverse of trigonometric ... | Let $$\tan A=\sqrt{\frac{a}{b}}\tan\frac{x}{2}$$ and
$$\sin B=\frac{2\sqrt{ab}\sin x}{(b+a)+(b-a)\cos x}$$
You wish to show $2A=B$. Calculate
$$\sin 2A=2\sin A\cos A=2\frac{\tan A}{\sec^2 A}=2\frac{\tan A}{1+\tan^2 A}$$
so substtitute your value for $\tan A$ into this expression and simplify using $$\tan^2 \frac{x}{2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $f(x)$ irreducible over $\mathbb{Z}_3$? I am looking for irreducible polynomials to construct finite fields. In this case I need a degree $4$ polynomial over $\mathbb{Z}_3$. Is $f(x)=x^4+x+2$ irreducible over $\mathbb{Z}_3$? I don't have much practice proving the irreducibility of a polynomial. So I am not sure what... | Assume that we have a factorization
$$x^4 + x + e= (x^2 + a x + b)(x^2 + c x + d)$$
Looking at the coefficient of $x^3$ we get $c=-a$. Now the coefficient of $x^2$ on RHS is $-a^2 + b + d$ so $d = a^2- b$. Now we get
$$(x^2 + a x + b)(x^2 - a x + (a^2 - b))= x^4 + (a^3 - 2 a b)x + a^2 b - b^2 $$
Now we get the system
$... | {
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For positive real numbers $a, b, c$, maximise $(a-3)(b-2)(c-1)$ given $a+b+c=18$. For positive real numbers $a, b, c$, if $a + b + c = 18$, then max value of $(a-3)(b-2)(c-1) =$?
My approach:-
$a-3=x$
$b-2=y$
$c-1=z$
I get $x+y+z=12$ , and I need to find $\max(xyz)$.
So I applied AM-GM inequality here,
$\dfrac{x+y+z}{3... | The maximum value of $102$ can be proven if we relax the constraints slightly, such that $a,b,c$ are all non-negative reals, instead of positive reals.
Case $1$: All $3$ terms are greater than or equal to zero.
By AM-GM,
\begin{align}
\sqrt[3]{(a-3)(b-2)(c-1)} & \leq \dfrac{a-3+b-2+c-1}{3} \\
& =4.
\end{align}
This giv... | {
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Is there a natural number $n$ for which $\sqrt[n]{22-10\sqrt7}=1-\sqrt7$ Is there a natural number $n$ for which $$\sqrt[n]{22-10\sqrt7}=1-\sqrt7$$
My idea was to try to express $22-10\sqrt7$ as something to the power of $2$, but it didn't work $$22-10\sqrt7=22-2\times5\times\sqrt7$$ Since $5^2=25, \sqrt7^2=7$ and $25+... | Alternate method : Binomial theorem
We have,
$$\begin{align}&\sqrt[n]{22-10\sqrt7}=1-\sqrt7\\
\implies &\left(1-\sqrt 7\right)^n=22-10\sqrt 7\end{align}$$
$22-10\sqrt 7<0$ tells us that, $n$ must be an odd integer. This implies $n≥3$ and $n=2k+1,\thinspace k\in\mathbb Z^{+}$.
Let, $N(n)$ be the integer part of $\left(... | {
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What is the value of $\sqrt[3]{9+2\sqrt5+3\sqrt{15}}$?
$$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=?$$
Here is my work:
I considered the value is equal to $a+b$. Hence
$$9+2\sqrt5+3\sqrt{15}=(a+b)^3=a^3+b^3+3a^2b+3ab^2$$
$$2\sqrt5+3(3+\sqrt{15})=a^3+b^3+3(a^2b+ab^2)$$
We have
$$ \begin{cases} {a^2b+ab^2=3+\sqrt{15}} \\ {a^3+b^3... | Remark: Write $$9+2\sqrt{5}+3\sqrt{15}=(a+b\sqrt{3}+c\sqrt{5})^3.$$
Then the equations in rational $a,b,c$ yield immediately a contradiction.
Of course, there might be more complicated expressions. But for such questions, usually the answer is simple.
Edit: The equations are
\begin{align*}
0 & =6abc - 3\\
0 & = 3a^2c +... | {
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How many negative real roots does the equation $x^3-x^2-3x-9=0$ have? How many negative real roots does the equation $x^3-x^2-3x-9=0$ have ?
My approach :-
f(x)= $x^3-x^2-3x-9$
Using rules of signs, there is 1 sign change , so there can be at most 1 positive real root
f(-x)= $-x^3-x^2+3x-9$
2 sign changes here, indicat... | $x^3 - x^2 - 3x - 9=0$
$3x = x^3 - x^2 - 9$
When $x<0$ every term on the right hand side is less than 0. If we are gong to find equality, $3x$ must be very negative.
When $-3< x < 0, |3x| > |-9|$
When $x< -3, |3x| < |x^2|$ and $|3x| < |x^3|$
There is no way that $3x$ can be sufficiently negative.
| {
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"answer_id": 4
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Let $a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}$. Prove that $a(2n) \neq 1$ for any value $n$. Let $$a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}.$$
I experienced this function while doing a problem, I could do the problem, but I got stuck at a point where I had to prove that $a(2n)<1$ for all $n$. I proved that $a(2n) \le 1... | Here's a solution that leverages the identity
$$\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}=\sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}$$
I'd suggest you try and prove this yourself. If that's not an option, I've provided a proof below:
Proof:\begin{align} \sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r} &= \lef... | {
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Taylor series of $f(x) = \cos x$ centered at $\frac{\pi}{4}$ I am asked to find the Taylor Series that represent the function $f(x) = \cos x$ centered at $\frac{\pi}{4}$.
My process
Finding the first few derivatives and establishing a pattern. Given that sine and cosine functions go back and forth, there needs to be tw... | You basically want the periodic sequence
$$
a_0=1,a_1=1,a_2=-1,a_3=-1,a_4=1,a_5=1,a_6=-1,\dotsc
$$
which obeys the recursion $a_{k+4}=a_k$, with the given initial terms. The characteristic equation is $t^4-1=0$, so the roots are $1,i,-1,-i$ and the general solution is therefore
$$
a_k=\alpha+\beta(-1)^k+\gamma i^k+\del... | {
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"answer_id": 1
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Looking for a way to compute the discrete version of a continuous random distribution I just learn the way to compute the cumulative probability distribution (CDF) of a distribution in terms of some known results. For example, a random variable $Y$ is written in terms of a uniform random variable $U(0,1)$ as
$$
Y \si... | Since you calculated the original expected value correctly, you already know this, but just to be thorough for other readers, the CDF in the question is not properly specified, as $P(U < \frac{5}{y} - 1)$ is only $\frac{5}{y} - 1$ if this quantity is between 0 and 1. Otherwise, the probability is either 0 or 1. The app... | {
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Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square. Find $x ,y$ satisfying : $x,y$ are $2$ positive integers and $(xy+x+2)(xy+1)$ is a perfect square.
I have solved this problem and will post the solution as soon as possible. Hope everyone can check my solution! Thanks ver... | $(xy+x+2)(xy+1) = a^2$
$\Rightarrow (2xy+x+3)^2 = (2a)^2 +(x+1)^2 $
Because $2 | 2a$ , there exist $3$ positive integers $m , n,d$ satisfying :
$(m,n) = 1$ and : $2a = 2dmn ; x+1 = d(m^2-n^2) ; 2xy+x+3 = d(m^2+n^2)$
$\Rightarrow xy+1 = dn^2 $
$\Rightarrow y = \frac{dn^2-1}{d(m^2-n^2)-1} $
$\Rightarrow \frac {y-1}{d}... | {
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Extracting coefficients I stuck at the following problem:
Let
\begin{equation}
f(z) = \frac{1 - \sqrt{1 - 4z}}{2}.
\end{equation}
Find $[z^n]$ of $f(z)^r$ for $r \in \mathbb{N}$. Where $[z^n]f(z)$ is the $n$-th coefficient of the power series $f(z) = \sum_{n \geq 0}{a_n z^n}$ therefore $[z^n]f(z) = a_n$.
So far I got
\... | At page 200 of the the renowned book
"Concrete Mathematics: a foundation for computer science" R. L. Graham - D.E. Knuth - O. Patshnik
the authors analyze the properties of what they call Generalized Binomial Series
(but it seems not to be a widely accepted denomination) defined as
$$
\eqalign{
& {\cal B}_{\,t} (z) =... | {
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Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Find the maximum of $\sin(A)+\sin(B)+\sin(C)$ for $\triangle ABC$. (Without Jensen Inequality)
Proof of Jensen Inequality:
\begin{align}
&\text{let } f(x)=\sin x. \\
\ \\
\Rightarrow & f(A)+f(B)+f(C) \\
& =3 \bigg( \frac 1... | Note that
$$\sin A+ \sin B=2\sin \frac{A+B}2\cos\frac{{A-B}}2$$
Thus for any fixed $C$ the first 2 terms obtains a maximum for $A=B$. Similarly if we fix $A$ we get that there is a maximum when $B=C$. It follows that the maximum is obtained when:
$$
A=B=C=\frac\pi3
$$
| {
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Polynomial inequality: show $p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n\ge 0$ Consider the following polynomial in $x$:
$$p(x) = \sum_{n=0}^{2k} x^n - \frac{1}{2k+1} \sum_{n=0}^{2k} (-x)^n.$$
I want to show that $p(x)\geq 0$ for real $x$. It is trivial to show that $p(-1) = 0$, and by graphing $... | $$p(x) = \frac{2}{ 2k+ 1} ( k + (k+1) x + k x^2 + (k+1)x^3 + \ldots + (k+1) x^{2k-1} + k x^{2k} ) $$
By considering signs (IE Replace $x$ with $-x$), we WTS that for $ x \geq 0$,
$$ k( 1 + x^2 + \ldots + x^{2k}) \geq (k+1)( x + x^3 + \ldots + x^{2k-1}) .$$
Since the number of terms on each side are equal to $k(k+1)$, a... | {
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"answer_id": 2
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proving $\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1} $ The identity is
$$\binom{n-1}{k} - \binom{n-1}{k-2} = \binom{n}{k} - \binom{n}{k-1}. $$
I wrote a matlab code and numerically checked it. It is right.
It is also not difficult to prove it by brute force. But can anyone come up with a simple pr... | Hint: You might also find appealing that algebraically this binomial identity is encoded by the relation
\begin{align*}
1-z^2=(1+z)(1-z)
\end{align*}
Denoting with $[z^k]$ the coefficient of $z^k$ of a series we can write
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n\tag{1}
\end{align*}
Using (1) and noting that $[z^p]z^qA... | {
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How do 3 fractions having respectively $(a-x)^2$ , $(a+x)^2$ and $(a-x)$ as denominators add up to a fraction having $(a^2 - x^2)^2$ as denominator? Source : Olry Terquem, Exercices de mathématiques élémentaires , $1842$
( At Archive.org https://archive.org/details/exercicesdemath00terqgoog/page/n85/mode/1up)
I'm worki... | $\require{cancel}$
We do not need the product of all denominators if one is a factor of another. For instance
$$\quad \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{8}\\
=\dfrac{8}{24}+\dfrac{6}{24}+\dfrac{3}{24}$$
where the denominator is
$\space LCM(3,4,8)=24$ and not $3\times4\times8=96$
$$LCM \big((a-x)^2 , (a+x)^2, (a-x)\big... | {
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Calculate the sum of the series $\sum_{n=1}^{\infty} \frac{n+12}{n^3+5n^2+6n}$ They tell me to find the sum of the series
$$\sum a_n :=\sum_{n=1}^{\infty}\frac{n+12}{n^3+5n^2+6n}$$
Since $\sum a_n$ is absolutely convergent, hence we can manipulate it the same way we would do with finite sums. I've tried splitting the g... | As you said, for every $N \in \mathbb{N}^*$, one has
\begin{align*}\sum_{n=1}^N a_n &=\sum_{n=1}^N \frac{2}{n}-\frac{5}{n+2}+\frac{3}{n+3}\\
&=2 \sum_{n=1}^N \frac{1}{n} - 5 \sum_{n=3}^{N+2} \frac{1}{n} + 3 \sum_{n=4}^{N+3} \frac{1}{n}\\
&=2 \left(1 + \frac{1}{2} + \frac{1}{3} + \sum_{n=4}^N \frac{1}{n} \right) - 5 \le... | {
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Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge 5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge 5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\... | I would preface by saying that we can prove a stronger inequality, namely $x^2+y^2+z^2 \geq \frac{25}3$.
You can use the Cauchy-Schwarz inequality. For three variables, it states that $(a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$. Using $a_1=a_2=a_3=1$ and $b_1=x,b_2=y,b_3=z$ gives $25\leq 3(x^... | {
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"source": "stackexchange",
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Find the Volume of a Solid Revolution around the y axis Having trouble with this question from my OpenStax Calculus Volume 1 Homework, It is question 89 of Chapter 6 about Solid Revolution. I put my math below:
y=4-x, y=x, x=0 Find the volume when the region is rotated around the y-axis.
$$y= 4-x \Rightarrow x = 4 - y ... |
It is symmetrical about $y=2$.
For $0 \le y \le 2$, we have $R(y)=y$.
\begin{align}
V &= 2\pi \int_0^2 y^2 \, dy\\
&= 2\pi \left.\frac{y^3}{3}\right|_0^2\\
&= \frac{16\pi}{3}
\end{align}
Note that $R(y)$ should be the radisu when $y$ takes a certain value.
| {
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"question_score": "2",
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"answer_id": 0
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Compute explicitly $\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}(\frac{1}{2})^{x+y}$ It is possible to compute explicitly the following series?
$$\sum_{x=1}^\infty\sum_{y=1}^\infty \frac{x^2+y^2}{x+y}\Big(\frac{1}{2}\Big)^{x+y}$$
@Edit I tried to sum and subtract $2xy$ in the numerator.
In this way I get the ... | Little easier way is to do a substitution (since $n+m$ arrives many times). What I mean is, we can rewrite (where $x \in (0,1)$ let's say to not bother about changing the order of summation (but any $|x| < 1$ will be good due to absolute convergence of double series) $$ \sum_{m =1}^\infty \sum_{n=1}^\infty \frac{n^2 + ... | {
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"source": "stackexchange",
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"answer_id": 0
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Solving a nonlinear autonomous system using polar coordinates I have the nonlinear system:
\begin{equation}
\begin{array}
fx'=-y-x\sqrt{x^2+y^2}\\
y'=x-y\sqrt{x^2+y^2}
\end{array}
\end{equation}
which I solve by polar coordinates transformation. $x=r\cos\phi \ y=r\sin\phi$ and $x'=dr\cos\phi-r\sin\phi$ and $y'=dr\sin\p... | $dr=-r^2$ cannot be integrated as you do. You need to write it under the form
$$- \dfrac{dr}{r^2}=1$$
giving :
$$\dfrac{1}{r}=\varphi+c$$
($c$ arbitrary constant), otherwise said:
$$r=\frac{1}{\varphi+c}$$
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve the equation $(2x^2-3x+1)(2x^2+5x+1)=9x^2$ Solve the equation $$(2x^2-3x+1)(2x^2+5x+1)=9x^2$$
The given equation is equivalent to $$4x^4+4x^3-11x^2+2x+1=9x^2\\4x^4+4x^3-20x^2+2x+1=0$$ which, unfortunately, has no rational roots. What else can I try?
| We have, $$((2x^2+x+1)-4x)((2x^2+x+1)+4x)=9x^2$$
$$(2x^2+x+1)^2-16x^2=9x^2\;\Rightarrow\; (2x^2+x+1)^2=(5x)^2$$
Hence, $\;2x^2+x+1=\pm5x$. Can you take it from here?
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{... | Let $x=\cos 2 \theta$, for $0\leq 2\theta < \pi$, then $d x=-2 \sin 2 \theta d \theta$ and
$$
\begin{aligned}
I &=\int \frac{-2 \sin 2 \theta d \theta}{\sqrt{1+\cos 2 \theta-\sqrt{1-\cos 2 \theta}}} \\
&=-2 \int \frac{\sin 2 \theta d \theta}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}} \\
&=-\sqrt{2} \int \frac... | {
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"source": "stackexchange",
"question_score": "4",
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Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$ Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$.
First off, it's easy to see that $\lim_{x\to \infty}\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}}$ = 1. Therefore,... | Let the limit exist and equal $L$. Then $\ln L$ equals:
$$\lim_{x \to \infty} x\left(\ln\left(2^{1/x}+3^{1/x}\right)-\ln\left(4^{1/x}+5^{1/x}\right)\right)$$
and making the substitution $u \mapsto 1/x$ gives:
$$\lim_{u \to 0^+} \frac{1}{u} \left(\ln\left(2^u+3^u\right)-\ln\left(4^u+5^u\right)\right)$$
Considering the f... | {
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Is $\lfloor{\frac{a+b+c+d}{4}}\rfloor=\lfloor\frac{\lfloor{\frac{a+b}{2}}\rfloor+\lfloor{\frac{c+d}{2}}\rfloor}{2}\rfloor$ for $a,b,c,d\in\mathbb R$?
Does the following hold $\forall a,b,c,d\in\mathbb R$?
$$\quad\left\lfloor{\frac{a+b+c+d}{4}}\right\rfloor=\left\lfloor\frac{\left\lfloor{\frac{a+b}{2}}\right\rfloor+\le... | The result is false. As pointed out by J.G., we can apply change of variable $x=\frac{a+b}{2}$ and $y=\frac{c+d}{2}$ to get:
\begin{equation*}
\lfloor \frac{x+y}{2}\rfloor = \lfloor \frac{ \lfloor x \rfloor+\lfloor y\rfloor}{2}\rfloor
\end{equation*}
But this cannot hold for every real $x$ and every real $y$: just put ... | {
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Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do ... | By using the substitution $\sqrt[3]t=a$, we have, $$\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}=\dfrac{a^3-1}{a-1}+\dfrac{a^3+1}{a+1}=(a^2+a+1)+(a^2-a+1)=2a^2+2$$
Which is equal to $2+2\sqrt[3]{t^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $4\lfloor x \rfloor=x+\{x\}$. What is wrong in my solution? Can anyone tell? $\lfloor x \rfloor$ = Floor Function, and $\{x\}$ denotes fractional part function
Solve for $x$
$$4\lfloor x \rfloor= x + \{x\}$$
$ x = \lfloor x \rfloor + \{x\}$
$\implies x - \{x\} = \lfloor x \rfloor$
$\implies 4x - 4\{x\} = x + ... | We have :
$$4 \lfloor x \rfloor = x + \{x\} = \lfloor x \rfloor + 2 \{x\}$$
then :
$$2 \{x\} = 3 \lfloor x \rfloor \in \mathbb{Z}$$
We deduce that :
$$\{x\} = 0 \text{ or } \{x\} = \dfrac{1}{2}$$
*
* If $\{x\} = 0$ then $3 \lfloor x \rfloor = 2 \{x\} = 0$. It means that $\lfloor x \rfloor = 0$ then $x = \lfloor x \rf... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Get derivative with square root in the denominator Given:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &=
\end{align}$$
The expected answer is:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &= -\frac{21\sqrt{2}}{4x\sqrt{x^7}}
\end{align}$$
But what steps would I need to take to ... | This would use the chain rule. Rewrite the function as:
$$f(x) = 3(2x^7)^{-\frac{1}{2}}$$
Using the chain rule:
$$\frac{df}{dx} = -\frac{1}{2} *3(2x^7)^{-\frac{3}{2}} * 14x^6$$
$$\frac{df}{dx} = -\frac{21x^6}{2^\frac{3}{2} x^\frac{21}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}x^\frac{9}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Easy question on probability I know this is a trivial question but I want to make sure I'm not missing anything:
We have a biased 6-sided die, which brings any of the 6 numbers with equal probability in the first roll, but in the second and all subsequent rolls, brings the previous result with probability $\frac {1}{2}... | There are actually two cases:
(1) Get a 4 in the 2nd roll and the 3rd roll
(2) Get a 4 in the 3rd roll but not in the 2nd roll
The probability of (1) is $\frac{1}{4}$ as you have correctly calculated.
In the case of (2), it is $\frac{1}{2} \times \frac{1}{10} = \frac{1}{20}$ because the probability of not getting a 4 i... | {
"language": "en",
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Proof that $a^n-b^n$ is non prime if n is non prime Hello guys I want to know if my proof is valid, and whether it is coherent and where I can improve it.
Prove that for all natural numbers, $a,b, n$, where $a>b$, and $n>1$ is a not a prime, $a^n-b^n$ is not a prime number.
My attempt:
Since $n$ is a non-prime greater ... | I don't see anything wrong with your proof. Perhaps I could mention some shortcuts or alternatives that you might find useful.
Let $d$ be a non trivial divisor of $n$, then we may write $n = dk$ for some integer $k > 1$. From $a^d \equiv b^d \pmod{a^d - b^d}$, we obtain $a^n \equiv b^n \pmod{a^d - b^d}$ by raising each... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Upper bound for a series $\sum_{n=0}^{\infty}2^n r^{2^n}$ for $r\in (0,1)$ It is clear that for $r\in (0,1)$ fixed, the series $$f(r)=\sum_{n=0}^{\infty}2^n r^{2^n}$$ is convergent, and also it is not to hard show that $f(r)$ is not uniformly convergent in $(0,1)$. Since $\sum_{n=0}^{\infty}r^n=\frac{1}{1-r}$, I think ... | This is a different approach with an arbitrary good bound for $x\in(0,1)$.
$$f(x)=\sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty 2^{n-1} x^{2^n} = \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty \underbrace{x^{2^n} + ... + x^{2^n}}_{2^{n-1} \text{ terms}} \\
\leq \sum_{n=0}^N 2^n x^{2^n} + 2\sum_{n=N+1}^\infty x^{2^{n... | {
"language": "en",
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How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}... | Thanks to Mr Nejimban, I am now going to evaluate the integral by converting it into a Wallis integral by the substitution $x=\tan \theta$, which yields
$$I_{n}(1)= \int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}} =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \theta d \theta}{\sec ^{2 n} \theta} =\int_{0}^{\frac{\pi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$S_n = \frac{12}{(4-3)(4^2-3^2)} + \frac{12^2}{(4^2-3^2)(4^3-3^3)} + ... + \frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$
So I know that in order to find the series we need to change the form because we can cancel the terms $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} = \frac{A}{4^n-3^n} + \frac{B}{4^{n+1}-3^{n+1}}$$
But my q... | $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$$
$4^n=a$, $3^n=b$, $12^n=ab$
$$\frac{ab}{(a-b)(4a-3b)}$$
Partitional fractioning in two variables is not unique. Make polynomial partitional fractioning considering $a$ as independent variable
$$\frac{ab}{(a-b)(4a-3b)}=\frac{A}{a-b}+\frac{B}{4a-3b}$$
$$ab=4aA-3bA+Ba-Bb=(4A+B)a... | {
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Why is this secant substitution allowed? On Paul's Math Notes covering Trig Substitutions for Integrals we start with an integral:
$$\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}}$$
Right away he says to substitute $x=\frac{2}{5}\sec(θ)$. Why is that allowed?
Looking further down onto how he approaches the problem, it s... | To add some information to Greg Martin's answer, this is because in some cases, integrals with $\sqrt{\pm(ax)^2 \pm b^2}$ can be solved using trigonometric substitutions. Recall the identities $$\sin^2\theta + \cos^2\theta = 1, \\ 1 + \cot^2\theta = \csc^2\theta, \\ \tan^2\theta + 1 =\sec^2\theta.$$
In the case $\sqrt{... | {
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First proof writing practice: writing and style feedback, and is it correct? I'm not even sure if this is correct because there was no answer in the book and I'm new to proofs by induction. I'm mostly interested in feedback about the style and any proof-writing faux-pas'sss I may have made. It felt right to put the lem... | Instead of editing I decided to make an answer, so you could compare better. In companion to the previous answers and comments, here is a shortened version of the results. Of course, a lot is personal taste how to write things down and you will learn a lot along the way.
Lemma 1. $n^2 + n$ is even for all $n \in \mathb... | {
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How do I evaluate $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$ How do I evaluate the following integral when where $C$ is the square with vertices at $\pm2, \pm2+4i$,
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$$
Using Cauchy integral:
$\frac{z^{2}}{z^{2}+4}=\frac{z^2}{(z+2i)(z-2i)}=\frac12\frac{... | Here you don't have to use partial fractions but use $f(z) = \frac{z^2}{z+2i}$ for $f(a)$ in the Cauchy Integral Formula, since $f$ is holomorphic in the smallest open disk that fits your $C$.
Then you get $f(2i) = \frac{(2i)^2}{4i} = i$
| {
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"question_score": "2",
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divisibility problem and gcd Let a and b be a positive integers. Proof that if number $ 100ab -1 $ divides number $ (100a^2 -1)^2 $ then also divides number $ (100b^2 -1)^2 $.
My attempt:
Let's notice that \begin{split} b^2(100a^2-1)^2 -a^2(100b^2-1)^2 & =(100a^2b-b)^2-(100ab^2-a)^2\\
& =(100a^2b-b-100ab^2+a)(100a^2b-... | For a modular arithmetic argument:
Let $N=100ab-1$ and work modulo $N$.
Starting with $(100a^2 -1)^2\equiv 0$, multiply by $10^4b^4$:
$$(10^4a^2b^2-100b^2)^2\equiv 0.$$
Since $10^2ab\equiv 1$, we have $(1-100b^2)^2\equiv 0.$
| {
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When we classify groups of order $n$, can I skip to check if the associativity law holds after I complete Cayley Table? For example, classify groups of order $4$.
Let $G=\{a,b,c,d\}$ be a group whose order is $4$.
$G$ must have an identity element $e$.
Without loss of generality, we can assume $d$ is the identity eleme... | In general one has to check associativity as pointed out in the comment by @ancientmathematician. However, as you have managed to find, there are often quick ways to do this by spotting that the group is simply some well known group.
Moreover, no one is going to ask you to classify groups of order $n$ by completing a C... | {
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Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead ... | To show:
$$\text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right) ~~<~~ \frac{\pi}{2}.$$
In fact this conclusion is immediate by the following analysis.
Let $~\displaystyle \theta ~~=~~ \text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right).$... | {
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Limit of a series with interchanging signs Let $S_x=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}(1-\frac{1}{x^n+1})$. The problem is to find the following limit: $\displaystyle\lim_{x\to1^{-}}S_x$.
I've tried to tackle the problem in several methods so far. Proving that this series is convergent is fairly easy. For eval... | Write
$$ S_x = \frac{x}{1+x} \sum_{n=1}^{\infty} \underbrace{\frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 + x^{2n})}}_{=T_{n,x}}. $$
Then we find that
$$ T_{n,x} \geq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-2})(1 + x^{2n})} = \frac{1}{1+x^{2n}} - \frac{1}{1+x^{2n-2}} $$
and
$$ T_{n,x} \leq \frac{x^{2n-2} - x^{2n}}{(1+x^{2n-1})(1 ... | {
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Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.
Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.
My Approach:
I could not find any elegant meth... | Let $\alpha, \beta$ be roots of this polynomial. If $\alpha=\beta$, then $\alpha^2-2=\alpha$, $\alpha=-1, 2$, and then $(p,q)=(2,1)$ or $(-4, 4)$, both choices work.
Suppose $\alpha\ne \beta$.
If $\alpha^2-2=\beta$, $\beta^2-2=\alpha$, then $(\alpha^2-2)^2-2=\alpha$, $\alpha^4-4\alpha^2-\alpha+2=0$, same for $\beta$. T... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$ I am trying to evaluate this integral:
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$$
I coverted the region by using polar coordinate system for xy-plane and it be... | Very similar steps from (Almost) Impossible Integrals, Sums, and Series, pages 261-261, can be applied where there we have that
$$\int_0^1\left( \int_0^1 \sqrt{x^2+y^2}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\textrm{d}x\right)\textrm{d}y$$ $$=\int_0^1 \left(\int_0^1 \left(\int_0^1 \frac{x^2+y^2}{x^2+y^2+z^2}\textrm... | {
"language": "en",
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$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$
Evaluate $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$$
Wolfram alpha gives $\dfrac{-7}{20}$.
Here is my work
For $x \to 0 $
$\tan x \sim x$
$\sin(x+\tan x) \sim \sin2x$
$\sin2x \sim 2x$ (I wasn't sure about this but I evaluated the limit ... | You have used the asymptotic relations like $\sin x\sim x$ in an incorrect fashion. The meaning of such relation is $$\lim_{x\to 0} \frac{\sin x} {x} =1$$ It does not mean that you can replace $\sin x$ with $x$ in any limit evaluation as $x\to 0$. I have explained this in detail in this answer of mine.
The simplest a... | {
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In how many ways we can place 9 different balls in 3 different boxes such that in every box at least 2 balls are placed?
In how many ways we can place $9$ different balls in $3$ different boxes such that in every box at least $2$ balls are placed?
Approach 1:
Let $x_1$, $x_2$, $x_3$ denote the number of balls in boxe... | $\mathbf{\text{For Alternative Approach:}}$
Distinct balls into distinct boxes with restriction , then it is good time to use exponential generating functions , if each boxes will have at least two objects , then the e.g.f. of each boxes is equal to $$\bigg(\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382752",
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show this equation $3x^4-y^2=3$ has no integer solution show this diophantine equation
$$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer solution,then $3|y$,let $y=3y'$,then we have
$$x^4-1=3y'^2\tag{1}$$
and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if... | $(1)$ WLOG we study the impossibility of $A^4-3B^2=1$ with $B\ne0$. Clearly $(A,B)=1$ and by modulo $5$ and modulo $16$ we have $A$ odd and $10|B$.
$(2)$ We need only consider non-negative solutions. The fundamental unit of the ring $\mathbb Z[\sqrt3]$ is $2+\sqrt3$ so all (positive) solution $(x_n,y_n)$ of the equatio... | {
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Proof for combinatorial Identity How to prove the following combinatorial identity?
$$\sum_{k=2}^n {k+1\choose3}{2n-2-k\choose n-2}2^k= \frac{1}{3}\cdot\frac{(2n)!}{(n-2)!n!}$$
My attempt:
The LHS is equal to the coefficient of $x^{n-2}$ in $\frac{1}{(1-x)^4\cdot(1-2x)^{n-1}}$. But this doesn't help. I don't see any ot... | We seek to prove that
$$\sum_{k=2}^n {k+1\choose 3} {2n-2-k\choose n-2} 2^k
= \frac{1}{3} \frac{(2n)!}{(n-2)! \times n!}.$$
The LHS is
$$4\sum_{k=0}^{n-2} {k+3\choose 3} {2n-4-k\choose n-2} 2^k
= 4\sum_{k=0}^{n-2}
{k+3\choose 3} {2n-4-k\choose n-2-k} 2^k.$$
Writing
$$4 \sum_{k=0}^{n-2} 2^k [w^k] \frac{1}{(1-w)^4} [w^{n... | {
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Which solution to $\int \frac{x^3}{(x^2+1)^2}dx$ is correct? I tried solving the following integral using integral by parts :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
but I got a different answer from Wolfram Calculator This is the answer that I got :
$$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$
... | Both are correct. $\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C=\frac 1{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C'$ where $C'=C-\frac 1 2$
[Just transfer $\frac 1{2(x^2+1)}$ to the left side and combine it with $\frac{-x^2}{2(x^2+1)}$].
| {
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Proving segments equal $(O)$ is internally tangent to a circle $(O')$ at $F$ i.e $(O')$ is inside $(O)$. Let $CE$ be a chord of $(O)$ which touches $(O')$ at $D$. Let $CO$ meet $(O)$ again at $A$ and $CO'$ meet $AE$ at $B$. It is given that $OO' \perp AC$. Prove that $AB=CD$.
From working backwards and using power of ... | Here's a "horrible" proof using algebraic manipulations.
Set things in coordinate plane as in the picture. That is, $O$ is the unit circle centered at origin and center of $O'$ lies on the $y$-axis. Let the circle $O'$ have radius $r$. Denote the x-coordinates of points $D, E, B$ as $x_1, x_2, x_3$, respectively and l... | {
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Union of sets having $\binom{n - 1}{2}$ elements with each two of them having $n - 2$ common elements has cardinality of at least $\binom{n}{3}$.
Let $S_1, S_2, \dots, S_n$ sets that have each of them $n - 1 \choose 2$ elements, with $n - 2$ common elements for each two of them. Prove that their union has at least $n ... | Here is an alternative approach based on @BillyJoe's formulation. Let $x_j$ be the number of elements that appear in exactly $j$ sets. Then the two conditions in the problem imply the following two linear equality constraints:
\begin{align}
\sum_{j=1}^n j x_j &= n\binom{n-1}{2} = 3\binom{n}{3} \tag1 \\
\sum_{j=1}^n \... | {
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} |
How do I finish proving this inequality?
Given $a, b,c >0$ satisfying $abc=1$. Prove that: $\dfrac{1}{{{a^2} + 2{b^2} + 3}} + \dfrac{1}{{{b^2} + 2{c^2} + 3}} + \dfrac{1}{{{c^2} + 2{a^2} + 3}} \le \dfrac{1}{2}$?
This is my try:
Using the AM–GM inequality, I get: $a^2+b^2 \ge 2ab$ and $b^2+1 \ge 2b$
Therefore, $a^2+2b^... | You've done all the hard work! Now you only need to prove $$\frac{1}2\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\leq\frac{1}2$$
HINT: utilize $abc=1$.
$$\frac{1}{ab+b+1}=\frac{c}{abc+bc+c}=\frac{c}{1+bc+c}$$
$$\frac{1}{ca+a+1}=\frac{bc}{abc^2+abc+bc}=\frac{bc}{c+1+bc}$$
Therefore, $$\frac{1}{ab+b+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4399988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why this Taylor expansion of $ \dfrac{1}{1+\cos x}$ is wrong? I want to expand the function $$ \dfrac{1}{1+\cos x} $$ at $x = 0$
The way I did this is first regard $\cos x $ as a whole and denote it by $y$
The expansion will become $$ \dfrac{1}{1+y}=1-y+y^2+o(y^2)$$
and then return $y=\cos x$ back and I want to expand ... | This is a good method, and it will work if applied correctly! As Jean-Claude is pointing out in the comments, what you need to do is expand $\frac{1}{1 + y}$ centered at $y = 1$, because $\cos 0 = 1$.
In other words, you need to use the Taylor series in terms $a_0 + a_1 (y - 1) + a_2 (y-1)^2 + \cdots$. Currently you ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac... | As an aside, reversing (or repeating) the change of variables further simplifies the work.
Starting off similar to OP's / River Li's work, we WTS
$$\sqrt{ ( x+y+z) (\frac{x}{x+y} + \frac{y}{y+z} + \frac{z}{z+x}) } \leq \frac{ xy+yz+zx} { \sqrt{2} }. $$
This is equivalent to
$$ \sum x + \sum \frac{xz}{x+y} \leq \frac{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx... | $$ I = \displaystyle \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{1}{2}\int_{0}^{\infty} \frac{\ln(x)}{\cosh x +\frac{1}{2}} dx $$
Consider the "discrete" Laplace transform
$$ \sum_{k=1}^{\infty}e^{-kt}\sin(kz) = \frac{1}{2}\frac{\sin z}{\cosh t -\cos z} \quad t>0 $$
If we put $z = \frac{2\pi}{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Does FOSD + log concavity of $f(x)$ and $g(x)$ imply MLRP? I am looking for a result on the ordering of distribution functions.
The probability density functions $f(x)$ and $g(x)$ bear the Monotone Likelihood Ratio Property (MLRP) if
$$ \frac{f(x)}{g(x)} $$
is increasing in $x$.
By First Order Stochastic Dominance (FOS... | A counterexample:
Let
\begin{align*}
f(x) &= \frac{2x^2 + 2x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}, \\
g(x) &= \frac{2x^2 + x + 4}{5\sqrt \pi}\mathrm{e}^{-x^2}.
\end{align*}
We have $f, g > 0$ for all $x\in (-\infty, \infty)$. We have, for all $x \in (-\infty, \infty)$,
\begin{align*}
(\ln f)'' &= - \frac{2x^4 + 4x^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \... | There is a much simpler solution. At the intersection of $y = mx + c$ and $y = x^2$,
$$x^2 - mx - c = 0$$
We are given $|AP - PB| = 1$ and I am taking the case where $P$ is interior to segment $AB$. Just for completeness sakes, if $P$ is exterior to segment $AB$, then we still have $|AP - PB| = 1$ if we consider $AP$ a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number...
We have three players A, B and C. They roll an honest die, first A, then B, and lastly C until someone is the first to get an even number. This player will be the winner of the g... | Your answer is correct to the number of decimal places to which you rounded. The probability that player $A$ wins is
\begin{align*}
\sum_{k = 1}^{\infty} \frac{1}{2}\left(1 - \frac{1}{2}\right)^{3(k - 1)} & = \frac{1}{2}\sum_{k = 1}^{\infty} \left(\frac{1}{2}\right)^{3(k - 1)}\\
& = \frac{1}{2}\sum_{k = 1}^{\infty} \le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating function for $a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$ Let $a_n$ be a sequence following the recurrence relation $$a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$$ with initial conditions $a_0 = a_1 = 1$.
We have to find the generating function for $a_n$ that does not contain an infinite series.
Let $f(x) = \sum_{k=0}^{n}... | Naturally, you would consider the square of the generating function:
$$f^2(x) = \sum_{i=0}^\infty a_ix^i \sum_{j=0}^\infty a_jx^j = \sum_{n=0}^\infty \sum_{k=0}^na_ka_{n-k}x^n = a_0^2 + 2a_0a_1x + \sum_{n=2}^\infty a_{n+2}x^n =$$
$$= 1+2x + x^{-2} \sum_{n=2}^\infty a_{n+2}x^{n+2} = 1+2x+x^{-2}(f(x) - a_0-a_1x-a_2x^2-a_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $Re{\{z}\}>0$ then $|z+\sqrt{z^{2}-1}|\geq 1$ In Complex Analysis - Bruce P . Palka - Ex: 4.12 I have not been able to prove that $|z+\sqrt{z^2-1}|^2\geq 1+ Re{\{\overline{z}\sqrt{z^2-1}}\},$ what I get is:
$|z+\sqrt{z^2-1}|^2=(z+\sqrt{z^2-1})\overline{(z+\sqrt{z^2-1})}= (z+\sqrt{z^2-1})(\overline{z}+\over... | Suppose $w = re^{i\theta}$ is a complex number. Then $|\sqrt{w}|^2 = |r^{1/2}e^{i\theta/2}|^2 = r^{1/2}e^{i\theta/2}r^{1/2}e^{-i\theta/2} = r = |w|$. Now \begin{align*}
|z + \sqrt{z^2 - 1}|^2 &= |z|^2 + 2\mathfrak{R}\left\{\overline{z}\sqrt{z^2 - 1}\right\} + \left|\sqrt{z^2 - 1}\right|^2\\ &= |z|^2 + 2\mathfrak{R}\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$ How can I find the intervals of increase and decrease of $\frac{x^4 - x^3 -8}{x^2 - x - 6}$?
I tried to find the derivative by the quotient rule to obtain the critical points but the formula was getting complicated, I know that $D_f = ... | One can simplify calculation of derivative by partitioning fraction:
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=Ax^2+Bx+C+\frac{D}{x-3}+\frac{E}{x+2}$$
$$\frac{x^4 - x^3 -8}{x^2 - x - 6}=x^2+6+\frac{46}{5(x-3)}-\frac{16}{5(x+2)}$$
$$\left(\frac{x^4 - x^3 -8}{x^2 - x - 6}\right)'=2x-\frac{46}{5(x-3)^2}+\frac{16}{5(x+2)^2}$$
Unf... | {
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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$... | Another approach:
By direct division we find:
$3n^2+4n+5=(\frac 32 n+\frac 54)(2n+1)+\frac{15}4$
multiplying both sides by $4$ and simplify we get:
$12n^2+16n+5=(6n+5)(2n+1)$
Or:
$\frac{4(3n^2+4n+5)}{2n+1}=6n+5+\frac {15}{2n+1}$
Therefore $2n+1$ must divide $15$ that is it must be equal to one of its divisors which are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2... | From your working step $\left(x+\frac{b}{2a}\right)^2=\left(\frac{b^2-4ac}{4a^2}\right),$
take square root (permissible because non-negative);
then apply your definition $\sqrt{x^2}=|x|,$ which implies that $\sqrt{x^2}=\pm x$ (what does $\pm x$ mean);
then simplify;
finally, verify (by substitution) that your two solut... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in... | If you accept that $$f(x)=\sum_{k=1}^\infty \frac{x^{k}}{k\,16^k} \binom{4k}{2k} = f_+(x) + f_-(x)$$
where
$$f_{\pm}(x) = 2\log(2) - 2 \log\left(1+\sqrt{1\pm\sqrt{x}}\right) \, ,$$
which shouldn't be too difficult to obtain (see RobPratt), then
$$I_- =\int_0^1 \frac{f_-(x)}{x} \, {\rm d}x \stackrel{u=\sqrt{1-\sqrt{x}}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integration of a piecewise defined discontinuous function In a proof i posted recently on this site (link) i made the mistake of thinking that a function $f$, which is bounded on a closed interval $[a,b]$, would assume its infimum at a certain $x$ in the domain of $f$. I was presented with a counter example which, for ... | Yes, the proof seems good and rock solid to me. If I could add something would be this: are you familiar with taking limits of successions?
One equivalent definition of (Riemann) Integrable function would be to check that $\lim_{n\to \infty}L(f,P_n)=\lim_{n\to \infty}U(f,P_n)$ , and using this definition of integrable ... | {
"language": "en",
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Showing that $(a^2 - b^2)^2$ $ \ge $ $4ab(a-$ $b)^2$ An inequality problem from Beckenbach and Bellman:
Show that $(a^2 - b^2)^2 \ge 4ab(a-b)^2$
The given answer is simply
Equivalent to $(a - b)^4 \ge 0$
I have tried two approaches, one which agrees with the given answer, and the other which does not.
Approach one.... | Let's examine your second attempt in more detail and check to see what's going wrong with it.
Observation 1
Whenever we are dividing using algebraic terms, it is vital that we make sure that we are not dividing by zero. Otherwise, the proof can end up going very wrong. There are many examples of this going wrong online... | {
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Finding rational solutions to $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but $abc\notin\mathbb{Z}$ I'm trying to find $a,b,c\in\mathbb{Q}$ such that $0\neq a^3c + b^3c^2\in\mathbb{Z}$ but that $abc\notin\mathbb{Z}$.
I tried to pick specific values for $a, b$
$$
b^3x^2 + a^3x - n= 0
$$
where I tried different integers $n\neq 0$... | I don't know offhand of any way to use your approach with $b^3x^2 + a^3x - n= 0$. Instead, since you're looking for just one set of values, consider that $a = \frac{3}{2}$, $b = \frac{5}{2}$ and $c = 1$ gives $a^3c + b^3c^2 = \frac{3^3}{2^3} + \frac{5^3}{2^3} = \frac{27 + 125}{8} = 19 \in \mathbb{Z}$, but $abc = \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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A problem involving a tetrahedron Let $ABCD$ a tetrahedron. We know the angle
$$\angle{ACB}=45^\circ$$the sum
$$\overline{AD}+\overline{BC}+\frac{\overline{AC}}{\sqrt2}=90$$and that the volume is $4500.$ We also know (but I don't know if this can be useful for the solution) that $\overline{CD}^2$ is integer: how much i... | This is an expanded version of the comment of Zerox. Let $h$ be the (length of the) height from $D$, the distance between $D$ and the plane of $\Delta ABC$. Then:
$$
\begin{aligned}
30^3 &= 27000
=6\operatorname{Volume}[ABCD]
=h\cdot 2\operatorname{Area}[ABC]
=h\cdot AC\cdot BC\cdot\sin 45^\circ
\\
&=h\cdot \frac{AC}{\... | {
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"timestamp": "2023-03-29T00:00:00",
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Graph of a function $(x^2+2)x(x-2)$ Consider $f(x)=(x^2+2)x(x-2)$. It has two real roots: $0$ and $2$.
If $x<0$, then $f(x)$ is positive, and if $x>2$, then also $f(x)$ is positive.
If $x$ is in $(0,2)$, then $f(x)$ is negative.
I am trying to understand why $f(x)$ has unique local minima in the interval $(0,2)$, so th... | For the general case.
Let $f(x) = (x^2+2)g(x)$ with $g(x) = x(x-2)(x-3)(x-4)...(x-2n)$.
Then $f'(x) = 0$ and $f(x)\neq 0 \iff (x^2+2)g(x) + 2xg'(x) = 0$ and $x\neq 0$ and $g(x)\neq 0 \iff \dfrac{g'(x)}{g(x)} + \dfrac{x^2+2}{2x} = 0$
But $\dfrac{g'(x)}{g(x)} = \dfrac 1 x + \dfrac 1 {x-2} + \ldots + \dfrac 1{x-2n}$.
So ... | {
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"timestamp": "2023-03-29T00:00:00",
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Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{... | Letting $x\mapsto \frac{\pi}{2}-x$ changes
$$
\begin{aligned}
I=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right)^{2}\left(\frac{\pi}{2}+x\right)^{2}}{\cos ^{2} x} d x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\pi^{2}}{4}-x^{2}\right)^{2} d(\tan x)
\end{aligned}
$$
Integration by parts... | {
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A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation...
A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation of the top of the tower are $\alpha$... | As the figure shows the distances between the base of the tower $P$ and each of the three vertices $A,B,C$ of equilateral $\triangle ABC$ is
$ PA = h \cot \alpha = h p $
$ PB = h \cot \beta = h q $
$ PC = h \cot \gamma = h r $
Now using Barycentric coordinates, and taking $A$ to be the origin, we can express point $P$ ... | {
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Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2... | $$S_k = \frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \dots + \frac{1}{(2k+1)^2}$$
$$T_k = \frac1{1^2} + \frac1{3^2} + \frac1{5^2} + \frac1{7^2} + \dots + \frac{1}{(2k+1)^2}$$
Look at $\lim_{k \to \infty} {S_k - T_k}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the value of expression $Q = \frac{x + 1}{y}$ when $xy > 1$ and expression $P = x + 2y + \frac{5x + 5y}{xy - 1}$ reaches its maximum value.
Consider two positives $x$ and $y$ where $xy > 1$. The maximum value of the expression $P = x + 2y + \dfrac{5x + 5y}{xy - 1}$ is achieved when $x = x_0$ and $y = y_0$. C... | With $P = x+2y+5\frac{x+y}{xy-1}$ and $Q=\frac{x+1}{y}$ and forming the lagrangian
$$
L(x,y,\lambda,s) = P+\lambda(x y-1-s^2)
$$
the stationary points are the solutions for
$$
0 = \nabla L = \cases{
\lambda y-\frac{y (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+1 \\
\lambda x-\frac{x (5 x+5 y)}{(x y-1)^2}+\frac{5}{x y-1}+... | {
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Fractional part and greatest integer function Here are a few questions on the fractional part and the greatest integer function.
*
*Find out $[\sqrt[3]{2022^2}-12\sqrt[3]{2022}]$
*If $\{x\}=x-[x],$ find out $[255\cdot x\{x\}]$ for $x=\sqrt[3]{15015}.$
For question 1, using a calculator, I know that the answer is $8.... | Solution for part 1. Let $a:=\sqrt[3]{2002}$. Then $0<a-12<1$ and
$$0<(a-12)^3 = a^3-36(a^2-12a)-12^3<1.$$
Rearranging the inequality we get
$$
8<\frac{2022-12^3-1}{36}<a^2-12a<\frac{2022-12^3}{36}<9
$$
| {
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"url": "https://math.stackexchange.com/questions/4446271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Computing limits using Taylor expansions and $o$ notation on both sides of a fraction Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved... | Observe that
$$
\frac{o(x^5)}{x^4} = \frac{o(x^5)}{x^5}x \to 0 \quad \text{as } x \to 0$$
since, by definition of $o$, we have $o(x^5)/x^5 \to 0$ as $x \to 0$.
Similarly,
$$
\frac{o(x^6)}{x^4} = \frac{o(x^6)}{x^6}x^2 \to 0 \quad \text{as } x \to 0.
$$
This explains why
$$
\lim\limits_{x \to 0} \frac{\frac{1}{2}+\frac{o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find $\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx$? I was trying to solve this indefinite integral;
$$\displaystyle\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx.$$
I tried using the online site; Integral Calculator. But it gave a very weird answer:
$$\int{\frac{\cos2x}{\sqrt{\sin^3 x}}}dx=2\operatorname{\Pi}\left(2\,;\dfrac... | First, note that $d(\sin x\cos x)/dx = d(\sin(2x)/2)/dx = \cos(2x)$. So we can integrate by parts using $u = \cos^{3/2}x$ and $v = (\sin x \cos x)^{-1/2}$. This gives
\begin{multline}
\int\frac{\cos(2x)}{\sin^{3/2}x}dx = \int\cos^{3/2}x\frac{\cos(2x)dx}{(\sin x\cos x)^{3/2}}dx = -\frac{2\cos x}{\sin^{1/2}x}+\int\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a mistake in an old question on this site: "Where does the sum of sin(n) formula come from?" https://math.stackexchange.com/a/1119137/29156
Shouldn't Abel's answer be:
we will use the fact that $$2 \sin 1 \sin k = \cos(k-1) -\cos(k + 1)$$let $S = \sin 1 + \sin 2 \cdots + \sin n,$ then
$\begin{align}
2S \sin 1 ... | hint
Yes, Abel's result seems to be false.
$$2S\sin(1)=\sum_{k=1}^n\Big((\cos(k-1)-\cos(k))+(\cos(k)-\cos(k+1))\Bigr)=$$
$$1-\cos(n)+\cos(1)-\cos(n+1)=$$
$$2\sin^2(\frac{n+1}{2})+\cos(1)-\cos(n)=$$
$$2\sin(\frac{n+1}{2})\Bigl(\sin(\frac{n+1}{2})+\sin(\frac{n-1}{2})\Bigr)=$$
$$4\sin(\frac{n+1}{2})\sin(\frac n2)\cos(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $z=x+iy$ is the answer of the equation $\cos(z)=-3$ What is the value of $x$?
If $z=x+iy$ is the answer of the equation $\cos(z)=-3$. What is the value
of $x$?
$1)x=2k\pi-\frac{\pi}2 , k\in \mathbb{Z}$
$2)x=2k\pi+\frac{\pi}2 , k\in \mathbb{Z}$
$3)x=k\pi+\frac{\pi}2 , k\in \mathbb{Z}$
$4)x=k\pi-\frac{\pi}2 , k\in \m... | \begin{align}
z &= 2\pi k \pm \cos^{-1}(-3), \; k \in \mathbb{Z}\\
&= 2\pi k \pm \frac{\pi}{2} \pm \sin^{-1}(3)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(\sqrt{1 - 3^2} - 3i\right)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln\left(i\left(2\sqrt{2} - 3\right)\right)\\
& = 2\pi k \pm \frac{\pi}{2} \pm i\ln(i) \pm i\ln\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Locus of middle points of the chords of conicoid which are parallel to $xy$ plane and touch the given sphere I have the following conicoid before me:
$ax^2+by^2+cz^2=1$
I have to find the locus of the middle points of the chords which are parallel to the plane $z=0$ and touch the sphere $x^2+y^2+z^2=a^2$.
This is what ... | I have figured out the answer on my own.
Here it goes:
Dividing final equation by $lm$,we get:
$\dfrac{l}{m}.2AB -(B^2+C^2-a^2)(\dfrac{l}{m})^2-(A^2+C^2-a^2)=0$
From equation $alA+bmB=0$, we get:
$\dfrac{l}{m}=-\dfrac{bB}{aA}$
Putting this value of $\dfrac{l}{m}$ in first equation, we get:
$\dfrac{2B^2b}{a}+(B^2+C^2-a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is My Solution Valid? Question from the 1999 Bulgarian Math Olympiad:
Find all pairs $(x,y)\in\mathbb{Z}$ satisfying $$x^3=y^3+2y^2+1$$
My first approach was to take the cube root of both sides: $$x=\sqrt[3]{y^3+2y^2+1}$$
For the sake of comfort, I will switch $x$ and $y$, as I can switch back in the end: $$y=\sqrt[3]{... | With some minor adjustments, your approach seems sensible (you do not have to take cube roots so explicitly and you seem to have swapped $x$ and $y$)
*
*on the reals, cubing is an increasing bijective function
*$2y^2+1 >0$ so $x^3=y^3+2y^2+1 \implies x^3> y^3\implies x>y$
*so any integer solution has $y+1 \le x$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
General formula for logarithmic cosine integral I am trying to find a general expression for
$$ \int_0^{\pi/2} (\ln(\cos(x)))^n dx $$ for integer $n$, but have not been able to find it online or derive it. I have a good idea that the general form will involve terms with $\zeta(n)$, but little else.
For reference, here... | Recall that $$\int_{0}^{\frac{\pi}{2}} \sin^p(x) \cos^q(x) \: dx = \frac{ \Gamma(\frac{p+1}{2}) \Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{p+q}{2}+1)} \\ \frac{d^n}{dq^n} \cos^q(x) = \cos^q(x)\ln^n(\cos(x)) $$
so that $$ \int_{0}^{\frac{\pi}{2}} \cos^q(x)\ln^n(\cos(x)) \: dx = \frac{d^n}{dq^n} \frac{ \Gamma(\frac{1}{2})\G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sharp Exponential Inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ I am trying to find a sharp inequality of $\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}}$ s.t. we can write: $$\frac{e^{-a\sqrt{x^2+n^2}}}{\sqrt{x^2+n^2}} \leq \frac{e^{-ax}}{x}f(an) .$$ Here, $a,x>0, n \in \mathbb{N}. $ Any particular techniqu... | The RHS approaches infinity as x approaches 0. Also, if $f(an)$ has the value 1, then the RHS is always greater than LHS. The derivative of $\frac{e^{−ax}}{x}$ is negative for all $a,x>0$, and that of the LHS is also negative.
Dividing the LHS by $\frac{e^{−ax}}{x}$ and taking the limit as $x\rightarrow \infty$ would g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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.