id string | output_text string | type string | source_dataset string | source_config string | source_split string | source_row_index int64 | source_field string | metadata_json string |
|---|---|---|---|---|---|---|---|---|
latex-00029668 | M(x,e_\alpha) = \int_{-h/2}^{h/2} x_3 e_3 \times t (x,x_3,e_\alpha) dx_3, \alpha=1,2, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,946 | latex_formula | {"original_latex": "\\begin{align*} M(x,e_\\alpha) = \\int_{-h/2}^{h/2} x_3 e_3 \\times t (x,x_3,e_\\alpha) dx_3, \\alpha=1,2,\\end{align*}"} |
mixed-00033142 | If $a,b \in \mathbb{Z}$, then
$$
2a^{2} + b^{2} + 2ab\sqrt{2} = 3c^{2},
$$
and then
\begin{align}
(*)\ \ \ \ ab\sqrt{2} = \frac{3c^{2}-2a^{2}-b^{2}}{2} .
\end{align}
The number $\sqrt{2}$ is irrational;
so $ab \neq 0$ leads to a contradiction,
and hence $ab = 0$.
We claim that $a=b=c = 0$; without loss of generality,
l... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,494 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1498091", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 1, "answer_id": 0}} |
latex-00049976 | \sharp F(a,b,\varepsilon,B,r)&=\sum_{e_1f_1|b,e_2f_2|a,e_3f_3|b-a}(\prod_{i=1}^3\mu(e_i))\sharp F(e_1,e_2,e_3,f_1,f_2,f_3,a,b,\varepsilon,B,r), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 50,673 | latex_formula | {"original_latex": "\\begin{align*}\\sharp F(a,b,\\varepsilon,B,r)&=\\sum_{e_1f_1|b,e_2f_2|a,e_3f_3|b-a}\\left(\\prod_{i=1}^3\\mu(e_i)\\right)\\sharp F(e_1,e_2,e_3,f_1,f_2,f_3,a,b,\\varepsilon,B,r),\\end{align*}"} |
latex-00029565 | f_{n} := \coprod_{D \in \mathcal{D}_n} G~\times_{\bar{D}}~ G \times~ X_{n,D} \to \coprod_{D \in \mathcal{D}_n} G~\times_{\bar{D}}~ E_{n,D} \times G ~\colon~ ~ (g,h,x)\to (g,f_{n,D}(h,x),gh) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,843 | latex_formula | {"original_latex": "\\begin{align*}f_{n} := \\coprod_{D \\in \\mathcal{D}_n} G~\\times_{\\bar{D}}~ G \\times~ X_{n,D} \\to \\coprod_{D \\in \\mathcal{D}_n} G~\\times_{\\bar{D}}~ E_{n,D} \\times G ~\\colon~ ~ (g,h,x)\\to (g,f_{n,D}(h,x),gh)\\end{align*}"} |
mixed-00024186 | max and min of $f(x,y)=y^8-y^4x^6+x^4$ The function is continuous in $\mathbb{R}^2$ and $f(x,y)=f(x,-y)=f(-x,y)=f(-x,-y)$.
If I consider $f(x,0)=x^4$ for $x\rightarrow +\infty$ $f$ is not limited up so $\sup f(x,y)=+\infty$.
But the origin is absolute min?
$f(x,x)=x^8-x^{10}+x^4\rightarrow -\infty$ if $x\rightarrow \i... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,043 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2490870", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0}} |
latex-00013128 | \Omega : \sigma \to 2\pi -\sigma , {\rm or} z \equiv e^{\tau +i\sigma} \to \bar z , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 13,156 | latex_formula | {"original_latex": "\\begin{align*}\\Omega :\\ \\sigma \\to 2\\pi -\\sigma\\ , \\ \\ {\\rm or}\\ \\ \\ z \\equiv e^{\\tau +i\\sigma} \\to \\bar z \\ ,\\end{align*}"} |
normal-00029737 | Dendroaspis polylepis is a large , round @-@ bodied , slender , but powerful snake . It tapers smoothly towards the tail , but is of markedly more robust build than its distinctly gracile congeners Dendroaspis angusticeps and Dendroaspis viridis . The head is often said to be " coffin @-@ shaped " with a somewhat prono... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 71,310 | text | {} |
latex-00029223 | \omega=\omega_{n}x^{\delta_{1,P_{2}}^{+}}+o(x^{\delta_{1,P_{2}}^{+}}), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,493 | latex_formula | {"original_latex": "\\begin{align*}\\omega=\\omega_{n}x^{\\delta_{1,P_{2}}^{+}}+o(x^{\\delta_{1,P_{2}}^{+}}),\\end{align*}"} |
mixed-00021964 | Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series
$\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 12,691 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/645560", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
mixed-00032390 | Let $p$ be the $(n+1)$st prime. Then $a_n\le (-8)+(1+3+5+7+9+\ldots +p-2)=(\frac{p-1}{2})^2-8$ provided $p\ge 11$ (the smaller cases can be dealt with by checking manually). So with $b_n:=\lfloor \sqrt {a_n}\rfloor $ we have $b_n< \frac{p-1}{2}$. Then $a_n<(b_n+1)^2=b_n^2+2b_n+1< a_n+p=a_{n+1}$ as was to be shown. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,039 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/939747", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0}} |
latex-00031949 | \alpha =(\tilde{z}'_2,f_1)+(\tilde{z}_1,f_2)+(\tilde{z}_3,f_3)+(\pi,f_4) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 32,257 | latex_formula | {"original_latex": "\\begin{align*} \\alpha =(\\tilde{z}'_2,f_1)+(\\tilde{z}_1,f_2)+(\\tilde{z}_3,f_3)+(\\pi,f_4)\\end{align*}"} |
mixed-00032616 | Lets call this polynomial $P(x)$ than by the conditions you have that the polynomial $P(x)$ can be written as $$P(x)=(x-1)Q_1(x)+2\\P(x)=(x+1)Q_2(x)+6\\P(x)=(x-2)Q_3(x)+3$$ From this you can see that $P(1)=2,P(-1)=6,P(2)=3$
Now you can also write your polynomial as
$$P(x)=(x-1)(x+1)(x-2)Q_4(x)+ax^2+bx+c$$
Where $ax^2+b... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,174 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1103023", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0}} |
latex-00030418 | & \big(P_{\xi}f\big)(x)=f(x) (0<x<\xi, f\in L^2_{m_2}(0,\ell)), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 30,706 | latex_formula | {"original_latex": "\\begin{align*} & \\big(P_{\\xi}f\\big)(x)=f(x) (0<x<\\xi, f\\in L^2_{m_2}(0,\\ell)),\\end{align*}"} |
normal-00003052 | The passage of the Act in the Irish Parliament was ultimately achieved with substantial majorities , having failed on the first attempt in 1799 . According to contemporary documents and historical analysis , this was achieved through a considerable degree of bribery , with funding provided by the British Secret Service... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 7,352 | text | {} |
latex-00011318 | \begin{array}{c}sc=c\partial c , \ sb=-(\partial b)c-2b\partial c .\end{array} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 11,342 | latex_formula | {"original_latex": "\\begin{align*}\\begin{array}{c}sc=c\\partial c\\;, \\\\ sb=-(\\partial b)c-2b\\partial c\\;.\\end{array}\\end{align*}"} |
mixed-00038364 | How many $4$ digit integer elements $X$ having no digit $0$ are in the set $C$ such that $X$ has exactly one $1$ or $X$ has exactly one $5$? Let $A = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~\text{has exactly one}~1\}$.
Let $B = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,710 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1680250", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
mixed-00037487 | Note that all even numbers in the numerator cancel out with the denominator. But in your calculation, $2k-2$ from denominator doesn't cancel out.
The correct way to arrive at the answer is:
$$\begin{align}1 \times 3 \times 5 \cdots \times (2k-3) &= 1 \times \dfrac{2}{2} \times 3 \times \dfrac{4}{4} \times \cdots \times... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,160 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/949313", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
normal-00013072 | The Clean Tech Revolution was published by Collins as a 320 @-@ page hardcover book on June 12 , 2007 . An e @-@ book version was published by HarperCollins on June 7 , 2007 . In 2008 , a revised paperback edition was published , with a new sub @-@ title : Discover the Top Trends , Technologies and Companies to Watch .... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 31,733 | text | {} |
latex-00021982 | \mathcal{H}^1( J \cap B_{2s\rho} ) \geq \mathcal{H}^1( J \cap B_{2 r_x}(x)) = \mathcal{H}^1( J \cap B_{\lambda_x}(x)) \geq \eta \lambda_x > \eta(1-s)\rho, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 22,138 | latex_formula | {"original_latex": "\\begin{align*} \\mathcal{H}^1\\left( J \\cap B_{2s\\rho} \\right) \\geq \\mathcal{H}^1\\left( J \\cap B_{2 r_x}(x)\\right) = \\mathcal{H}^1\\left( J \\cap B_{\\lambda_x}(x)\\right) \\geq \\eta \\lambda_x > \\eta(1-s)\\rho, \\end{align*}"} |
latex-00027915 | \sum_{s=1}^{n-k+1}\frac{x_s}{s!(n-s)!}B_{n-s, k-1}(x_1, x_2, \ldots) (n(n-1)-ks(n-1)) = 0 | latex | OleehyO/latex-formulas | cleaned_formulas | train | 28,167 | latex_formula | {"original_latex": "\\begin{align*}\\sum_{s=1}^{n-k+1}\\frac{x_s}{s!(n-s)!}B_{n-s, k-1}(x_1, x_2, \\ldots) (n(n-1)-ks(n-1)) = 0\\end{align*}"} |
latex-00019116 | t_{k,m} = \sum_{(k',m')<_\mathrm{time}(k,m)} 2^{-2k'} < 1, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 19,197 | latex_formula | {"original_latex": "\\begin{align*}t_{k,m} = \\sum_{(k',m')<_\\mathrm{time}(k,m)} 2^{-2k'} < 1,\\end{align*}"} |
mixed-00006689 | Here is another method to solve the problem by using residue. Suppose $\alpha\in(0,\pi/2)$. Using $x=\sin(\theta)$ and $u=\tan(\theta), z=e^{i\theta}$, one has
\begin{eqnarray}
&&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=&\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\math... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 3,464 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/550145", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 4, "answer_id": 2}} |
mixed-00047139 | Use the integration formula $\frac{1}{a}\arctan\frac{x}{a}$ to solve $\frac{1}{2} \int_{-1}^1 \mathrm{ \frac{dx}{1+\sqrt{2}x+x^2} }\, $ As question states, I am trying to figure out how to use the integration formula to solve the integral. My issue is that the integral isn't of the form $\frac{dx}{a^2+x^2}$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 28,103 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/741497", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0}} |
mixed-00026058 | A generalisation for non-negative integer $n$ is provided in this answer. The expansion there can be written as
\begin{align*}
&\,\,\color{blue}{(a+b)^n-\left(a^n+b^n\right)}\\
&\quad\,\,\color{blue}{=ab(a+b)\sum_{k=1}^{\lfloor n/2\rfloor}\left(\binom{n-k}{k}+\binom{n-k-1}{k-1}\right)(-ab)^{k-1}(a+b)^{n-2k-1}}\tag{1}\\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 15,189 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4563475", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3}} |
normal-00046058 | Love Kraft was released on CD , SACD , vinyl and as a digital download on 22 August 2005 in the United Kingdom and was the band 's last release for Sony 's Epic imprint before they moved to independent label Rough Trade . The album reached # 19 in the UK Albums Chart . In America the album was released on 13 September ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 111,885 | text | {} |
latex-00017024 | U_t^{ph}(h,h')=(\frac{l}{i\hbar t})^{r/2}\sum\limits_{\hat{s}\in W_A}(\kappa (h)\kappa (\hat{s}h'))^{-1}\exp(\frac{i\pi l(h-\hat{s}h')^2}{\hbar t} +itE_0) , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 17,066 | latex_formula | {"original_latex": "\\begin{align*}U_t^{ph}(h,h')=\\left(\\frac{l}{i\\hbar t}\\right)^{r/2}\\sum\\limits_{\\hat{s}\\in W_A}(\\kappa (h)\\kappa (\\hat{s}h'))^{-1}\\exp\\left(\\frac{i\\pi l(h-\\hat{s}h')^2}{\\hbar t} +itE_0\\right)\\ ,\\end{align*}"} |
mixed-00005797 | $$T_n = \frac{n^2}{4n^2-1} = \frac{1}{4}\left( \frac{4n^2-1+1}{4n^2-1} \right) = \frac{1}{4}\left( 1 + \frac{1}{4n^2-1} \right)=\frac{1}{4}\left( 1 + \frac{1}{(2n-1)(2n+1)} \right). $$
Decomposing with partial fractions, we have:
$$ \frac{1}{(2n-1)(2n+1)} = \frac{\frac{1}{2}}{2n-1} - \frac{\frac{1}{2}}{2n+1} = \frac{1}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 2,949 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4461891", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0}} |
normal-00002055 | Following acts are also considered as violation of the seventh commandment : price manipulation to get advantage on the harm of others , corruption , appropriation of the public goods for personal interests , work poorly carried out , tax avoidance , counterfeiting of checks or any means of payment , any forms of copyr... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 5,016 | text | {} |
normal-00022591 | List of ship classes of the Second World War | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 54,330 | text | {} |
normal-00036255 | C. pedatifolia , origin : Querétaro , Mexico | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 87,428 | text | {} |
normal-00041309 | Ursa Minor and Ursa Major were related by the Greeks to the myth of Callisto and her son Arcas , both placed in the sky by Zeus . In a variant of the story in which Boötes represents Arcas , Ursa Minor represents a dog . This is the older tradition , which explains both the length of the tail and the obsolete alternate... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 100,136 | text | {} |
latex-00022174 | \ell\sum_{i=1}^N\sum_{m\in\mathbb{Z}}\Delta(\frac{m}{\ell})=L. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 22,332 | latex_formula | {"original_latex": "\\begin{align*} \\ell\\sum_{i=1}^N\\sum_{m\\in\\mathbb{Z}}\\Delta\\left(\\frac{m}{\\ell}\\right)=L.\\end{align*}"} |
normal-00017747 | In 1899 Madge met her future husband Edgar Syers , a figure skater and coach who was 18 years her senior . Edgar was an exponent of the international skating style , which was freer and less rigid than the traditional English style , and encouraged Madge to adopt this style . Madge and Edgar completed together in pairs... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 42,549 | text | {} |
latex-00034702 | Z_f(s, \chi)=\dfrac{1}{1-q^{-(\omega+2)-(n+l)s}}Z_f(s, \chi, A^c). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 35,061 | latex_formula | {"original_latex": "\\begin{align*}Z_f(s, \\chi)=\\dfrac{1}{1-q^{-(\\omega+2)-(n+l)s}}Z_f(s, \\chi, A^c).\\end{align*}"} |
normal-00006693 | The episode received generally mixed @-@ to @-@ positive reviews . Ryan McGee of Zap2it wrote that " this episode wasn 't a stinker by any measure , but after the run of early episodes , this is the first that really didn 't hold its own when compared to the others " , adding that " The moth imagery / metaphor just bea... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 16,325 | text | {} |
normal-00021120 | Upon the death of Emperor Dawit , his older brother Tewodros ordered Zara Yaqob confined on Amba Geshen ( around 1414 ) . Despite this , Zara Yaqob 's supporters kept him a perennial candidate for Emperor , helped by the rapid succession of his older brothers to the throne over the next 20 years , and left him as the o... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 50,614 | text | {} |
latex-00031466 | k_{-}(u) = \{ \begin{array}{lcl}1, & \mbox{for} & u \leq \frac{X-Y -2}{4}, \ \frac{-4u}{Y} + \frac{X-2}{Y}, & \mbox{for} & \frac{X-Y-2}{4} \leq u \leq \frac{X-2}{4},\ 0, & \mbox{for} & \frac{X-2}{4} \leq u.\end{array} . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 31,762 | latex_formula | {"original_latex": "\\begin{align*}k_{-}(u) = \\left\\{ \\begin{array}{lcl}1, & \\mbox{for} & u \\leq \\frac{X-Y -2}{4}, \\\\ \\displaystyle \\frac{-4u}{Y} + \\frac{X-2}{Y}, & \\mbox{for} & \\frac{X-Y-2}{4} \\leq u \\leq \\frac{X-2}{4},\\\\ 0, & \\mbox{for} & \\frac{X-2}{4} \\leq u.\\end{array} \\right. \\end{align*}"} |
latex-00024716 | L'_{g,d,k}=|L'_{+}\setminus \psi(L'_{-})|. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 24,906 | latex_formula | {"original_latex": "\\begin{align*}L'_{g,d,k}=|L'_{+}\\setminus \\psi(L'_{-})|.\\end{align*}"} |
normal-00010556 | Philologist , archeologist , and Dead Sea Scrolls scholar John Marco Allegro postulated that early Christian theology was derived from a fertility cult revolving around the entheogenic consumption of A. muscaria in his 1970 book The Sacred Mushroom and the Cross , but his theory has found little support by scholars out... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 25,812 | text | {} |
normal-00002021 | Respect and care is required for non @-@ combatants , wounded soldiers and prisoners . Soldiers are required to disobey commands to commit genocide and ones that violate universal principles . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 4,946 | text | {} |
mixed-00038984 | Let us rewrite
\begin{align}
\frac{x-\sin^{[n]}x}{x^3}&=\frac{x-\sin x+\sin x-\sin\sin x+\ldots+\sin^{[n-1]} x-\sin^{[n]} x}{x^3}=\\
&=\frac{x-\sin x}{x^3}+\frac{\sin x-\sin\sin x}{x^3}+\ldots+\frac{\sin^{[n-1]}x-\sin^{[n]}x}{x^3}
\end{align}
and calculte the limit of each fraction separately.
*
*Since
$$
\sin t=t-\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 23,094 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2262015", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2}} |
mixed-00003306 | The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,676 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2247408", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2}} |
latex-00019745 | \frac{(n-r) \cdot (n-r-1) \cdots (n-k+1)}{n^{k-r}} = \frac{(n-r)_{(k-r)}^{}}{n^{k-r}} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 19,829 | latex_formula | {"original_latex": "\\begin{align*}\\frac{(n-r) \\cdot (n-r-1) \\cdots (n-k+1)}{n^{k-r}} = \\frac{(n-r)_{(k-r)}^{}}{n^{k-r}}\\end{align*}"} |
normal-00044294 | In reserve , Farnese deployed a large battalion made up by the German regiments of Hannibal d 'Altemps and Georg von Frundsberg , flanked on its right by troops of reiters under Duke Francis of Saxe @-@ Lauenburg , elder brother of Duke Maurice , John Casimir 's former lieutenant , and on its left by lancers under Pier... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 107,298 | text | {} |
mixed-00019987 | As $-1\le x\le1$
WLOG $x=\cos2t,0\le2t\le\pi,\sin2t=\sqrt{1-x^2}$
$$\implies\sqrt{1+\sin2t}[(2\cos^2t)^{3/2}+(2\sin^2t)^{3/2}]=2+\sin2t$$
As $\sin t,\cos t\ge0$ and $(\sin t+\cos t)^2=1+\sin2t$
$$2\sqrt2(\cos t+\sin t)(\cos^3t+\sin^3t)=2+\sin2t$$
$$\sqrt2(1+\sin2t)(2-\sin2t)=2+\sin2t$$ which is on rearrangement, a Quad... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 11,479 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3239071", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "11", "answer_count": 7, "answer_id": 3}} |
latex-00034057 | E_{k}(t_{u,k},t_{d,k}) = E_{u,k}(t_{u,k}) + \beta E_{BS,k}(t_{d,k}), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 34,406 | latex_formula | {"original_latex": "\\begin{align*} E_{k}(t_{u,k},t_{d,k}) = E_{u,k}(t_{u,k}) + \\beta E_{BS,k}(t_{d,k}),\\end{align*}"} |
normal-00048687 | About 19 isotopes and 8 nuclear isomers are known for americium . There are two long @-@ lived alpha @-@ emitters , 241Am and 243Am with half @-@ lives of 432 @.@ 2 and 7 @,@ 370 years , respectively , and the nuclear isomer 242m1Am has a long half @-@ life of 141 years . The half @-@ lives of other isotopes and isomer... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 118,433 | text | {} |
latex-00041125 | \lim_{A \to \infty}\liminf_{x \to \infty}\frac{nJ_1(x)}{\overline{\mu^{n*}}(x)}= \lim_{A \to \infty}\limsup_{x \to \infty}\frac{nJ_1(x)}{\overline{\mu^{n*}}(x)}= 1. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 41,721 | latex_formula | {"original_latex": "\\begin{align*}\\lim_{A \\to \\infty}\\liminf_{x \\to \\infty}\\frac{nJ_1(x)}{\\overline{\\mu^{n*}}(x)}= \\lim_{A \\to \\infty}\\limsup_{x \\to \\infty}\\frac{nJ_1(x)}{\\overline{\\mu^{n*}}(x)}= 1.\\end{align*}"} |
latex-00030854 | f(t):=F(\gamma(t))=\frac{1}{\sqrt{N}}\sum_{\mu\in\mathcal{E}}a_{\mu}e^{2\pi i\langle\mu,\gamma(t)\rangle}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 31,144 | latex_formula | {"original_latex": "\\begin{align*}f(t):=F(\\gamma(t))=\\frac{1}{\\sqrt{N}}\\sum_{\\mu\\in\\mathcal{E}}a_{\\mu}e^{2\\pi i\\langle\\mu,\\gamma(t)\\rangle},\\end{align*}"} |
latex-00027070 | \Lambda(R(F))v=\sum_j \lambda_j \Lambda(F)^j (v) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 27,315 | latex_formula | {"original_latex": "\\begin{align*} \\Lambda(R(F))v=\\sum_j \\lambda_j \\Lambda(F)^j (v) \\end{align*}"} |
latex-00004338 | c=r+12Q^2=r(1+h(h+1)(b+1/b)^2). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 4,341 | latex_formula | {"original_latex": "\\begin{align*}c=r+12Q^2=r(1+h(h+1)(b+1/b)^2). \\end{align*}"} |
latex-00023517 | a-b &= \int_0^1 \int_0^1 (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx \ &= \int_0^1 \int_0^x (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx \ &\quad+ \int_0^1 \int_x^1 (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,687 | latex_formula | {"original_latex": "\\begin{align*} a-b &= \\int_0^1 \\int_0^1 (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx \\\\ &= \\int_0^1 \\int_0^x (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx \\\\ &\\quad+ \\int_0^1 \\int_x^1 (1-py)^{-2}(1-px)^{-1}y^jx^{j-1}(y-x)dydx.\\end{align*}"} |
mixed-00036381 | Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 21,481 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4608272", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 6, "answer_id": 1}} |
latex-00017489 | \tilde{b}_2^+=e^{-W}b_2^+ e^W={1\over{4\omega C_{\cal R}}}\sum_{\mu\in{\cal R}}(\tilde{\ell}_{\mu}^+)^2+VT, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 17,535 | latex_formula | {"original_latex": "\\begin{align*}\\tilde{b}_2^+=e^{-W}b_2^+\\,e^W={1\\over{4\\omega C_{\\cal R}}}\\sum_{\\mu\\in{\\cal R}}(\\tilde{\\ell}_{\\mu}^+)^2+VT,\\end{align*}"} |
latex-00014823 | F_{\theta\bar{\psi}_{2}} = 2\sin\theta\cos\theta, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 14,853 | latex_formula | {"original_latex": "\\begin{align*}F_{\\theta\\bar{\\psi}_{2}} = 2\\sin\\theta\\cos\\theta,\\end{align*}"} |
normal-00040871 | X. Biedler , one of the Alder Gulch and Helena vigilante enforcers wrote about his vigilante activities in his personal journals . They weren 't available until well after his death when Helen F. Sanders , the daughter @-@ in @-@ law of Wilbur Sanders finally got them published in 1957 . Nathaniel Langford , also a mem... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 99,055 | text | {} |
latex-00023430 | \varphi_j(p) := \frac{\int_0^\infty e^{-\lambda s} p_j(s) ds}{\int_0^\infty e^{-\lambda s} \big(\sum_{k=j}^\infty p_k(s)\big) ds} = \begin{cases} \dfrac{1}{j+1}, & p=0, \ 1-\dfrac{ \int_0^{1} (1-p y)^{-1} y^{j} dy}{\int_0^1 (1-py)^{-1} y^{j-1} dy}, & 0<p<1, \end{cases} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,600 | latex_formula | {"original_latex": "\\begin{align*} \\varphi_j(p) := \\frac{\\int_0^\\infty e^{-\\lambda s} p_j(s) ds}{\\int_0^\\infty e^{-\\lambda s} \\big(\\sum_{k=j}^\\infty p_k(s)\\big) ds} = \\begin{cases} \\dfrac{1}{j+1}, & p=0, \\\\ 1-\\dfrac{ \\int_0^{1} (1-p y)^{-1} y^{j} dy}{\\int_0^1 (1-py)^{-1} y^{j-1} dy}, & 0<p<1, \\end{... |
latex-00044333 | I_{Np}x-C_1\sim \begin{bmatrix}W(x) & 0\\0 & I_{(N-1)p}\end{bmatrix} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 44,959 | latex_formula | {"original_latex": "\\begin{align*}I_{Np}x-C_1\\sim \\begin{bmatrix}W(x) & 0\\\\0 & I_{(N-1)p}\\end{bmatrix}\\end{align*}"} |
normal-00012291 | Ambassador Soval is summoned before Administrator V 'Las and the High Council to face punishment over his use of a mind meld . Since the act is widely considered to be criminal by the Vulcan authorities , Soval is summarily dismissed from the Ambassadorial service . Meanwhile , Captain Archer and Commander T 'Pol are q... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 29,860 | text | {} |
normal-00000653 | The synagogue occupied the greater part of the plot , facing west . It receded from the street regulation @-@ line in accordance with the rule then still enforced in Austria – Hungary , prohibiting non @-@ Catholic places of worship from having a public entrance from the street . The synagogue had a wider and slightly ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 1,597 | text | {} |
latex-00045668 | Q_a^{\varepsilon}(t)^*Q_a^{\varepsilon}(t)f(x)=(2\pi)^{-\nu n}\iint e^{i(x-y)\xi}q_a^{\varepsilon}(t,x,\xi)q_a^{\varepsilon}(t,y,\xi)f(y)dyd\xi. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 46,313 | latex_formula | {"original_latex": "\\begin{align*}Q_a^{\\varepsilon}(t)^*Q_a^{\\varepsilon}(t)f(x)=(2\\pi)^{-\\nu n}\\iint e^{i(x-y)\\xi}q_a^{\\varepsilon}(t,x,\\xi)q_a^{\\varepsilon}(t,y,\\xi)f(y)dyd\\xi.\\end{align*}"} |
mixed-00018232 | For $n=2$ and $n=3,4$ we can give infinite families using a Pell equation and elliptic curves, respectively,
$n=2$:
$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$
where $p,q$ solve $p^2-d^3q^2=1\tag1$.
$n=3$:
$$(a y)^2 + (ma)^3 + a^4 = a^5$$
and the elliptic curve solvable for an appropriate constant $m$,
$$a^3 -... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 10,407 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1619089", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 4, "answer_id": 0}} |
mixed-00049968 | I'm assuming you meant the three diagonal elements of $S+R$ are equal.
Let $R = u v^\top$. You have $5$ equations to solve for $6$ variables $u_1,u_2,u_3,v_1,v_2,v_3$ (of course there is redundancy here): the coefficients of $\lambda^0$ to $\lambda^2$ in $\det(S+R-\lambda I) - \det(S-\lambda I)$ are $0$, $S_{11} + R_{1... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 29,822 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3193599", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0}} |
mixed-00033484 | Not sure, if I missed out anything here. Take a look.
For non negative, $X,Y,Z$,
We can perhaps use Titu's inequality (a mix of Holder and CS), sometimes called Titu's screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality).
\begin{equation}
\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,704 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1775572", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "124", "answer_count": 8, "answer_id": 2}} |
latex-00048149 | E(Iu_\lambda(0)) = \frac{1}{2}\|Iu_\lambda(0)\|^2_{\dot{H}^{k/2}_x} +\frac{1}{4}\|Iu_\lambda(0)\|^4_{L^4_x}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 48,816 | latex_formula | {"original_latex": "\\begin{align*}E(Iu_\\lambda(0)) = \\frac{1}{2}\\|Iu_\\lambda(0)\\|^2_{\\dot{H}^{k/2}_x} +\\frac{1}{4}\\|Iu_\\lambda(0)\\|^4_{L^4_x}. \\end{align*}"} |
latex-00038640 | \frac{d\sigma(s,t)}{dt} = \frac{1}{16 \pi s^2} \{ A_1^2(s, t) +A_2^2(s, t) - 2A_1(s, t) A_3(s, t) \} . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 39,130 | latex_formula | {"original_latex": "\\begin{align*}\\frac{d\\sigma(s,t)}{dt} = \\frac{1}{16 \\pi s^2} \\left\\{ A_1^2(s, t) +A_2^2(s, t) - 2A_1(s, t) A_3(s, t) \\right\\} \\;.\\end{align*}"} |
normal-00035506 | main ( i.e. lower ) counterweight . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 85,521 | text | {} |
latex-00030586 | E_0=\{u\in E | \exists m\in M \mbox{~such that~} \sigma_1(u)=0_1(m), \sigma_2(u)=0_2(m)\}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 30,874 | latex_formula | {"original_latex": "\\begin{align*}E_0=\\{u\\in E\\,|\\,\\exists m\\in M \\mbox{~such that~} \\sigma_1(u)=0_1(m), \\sigma_2(u)=0_2(m)\\}.\\end{align*}"} |
normal-00008328 | The following single , a rendition of the Bee Gees 's " To Love Somebody " , also reached the UK Top 10 in 1969 . " The House of the Rising Sun " was featured on Nina Simone Sings the Blues in 1967 , but Simone had recorded the song in 1961 and it was featured on Nina at the Village Gate ( 1962 ) , predating the versio... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 20,382 | text | {} |
mixed-00038540 | Finding an Explicit Formula for $a_n$ from a Recurrence Relation I have the following recurrence relation
$$\beta ^n n(n+2)a_n = \sum_{k=0}^{n-1} \beta^k (\alpha+k+1) a_k , \quad a_0=1, \quad n \ge 1 \tag{1}$$
where $\beta$ and $\alpha$ are positive real numbers.
I know that one can easily find $a_n$ by back substituti... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,819 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1850930", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 1, "answer_id": 0}} |
latex-00012386 | \frac{1}{\sqrt{-i\pi (a-i\varepsilon )}} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 12,413 | latex_formula | {"original_latex": "\\begin{align*}\\frac{1}{\\sqrt{-i\\pi (a-i\\varepsilon )}}\\end{align*}"} |
mixed-00002572 | Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,307 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1767200", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 0}} |
normal-00020515 | As Emperor , Domitian strengthened the economy by revaluing the Roman coinage , expanded the border defenses of the Empire , and initiated a massive building program to restore the damaged city of Rome . Significant wars were fought in Britain , where his general Agricola attempted to conquer Caledonia ( Scotland ) , a... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 49,209 | text | {} |
normal-00023424 | The three known specimens of the 1873 @-@ CC quarter , without arrows by the date , and the only known dime of that description , may have been salvaged from assay pieces , as the remainder of those coins had been ordered melted as underweight . A similar mystery attends the 1894 Barber dime struck at San Francisco ( 1... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 56,172 | text | {} |
latex-00018001 | \|f\|_{\dot{\mathcal F}^{\alpha,q}_{p,{\rm D}}}&=\bigg\{\sum_{k\in \Bbb Z} \Big({\mathfrak R}^{k\alpha}\Big\|q_{k}\Big(\sum_{j\in \Bbb Z} \sum_{Q\in Q^j}\omega(Q){\widetilde D}_j(x,x_{Q})D_j(f)(x_{Q})\Big)\Big\|_p\Big)^q\bigg\}^{1/q}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 18,055 | latex_formula | {"original_latex": "\\begin{align*}\\|f\\|_{\\dot{\\mathcal F}^{\\alpha,q}_{p,{\\rm D}}}&=\\bigg\\{\\sum_{k\\in \\Bbb Z} \\Big({\\mathfrak R}^{k\\alpha}\\Big\\|q_{k}\\Big(\\sum_{j\\in \\Bbb Z} \\sum_{Q\\in Q^j}\\omega(Q){\\widetilde D}_j(x,x_{Q})D_j(f)(x_{Q})\\Big)\\Big\\|_p\\Big)^q\\bigg\\}^{1/q}.\\end{align*}"} |
mixed-00016186 | You wrote
$$\int^{2a^2}_{4a^2\sin^22θ}
\frac{\sqrt u}{2}\,du=\color{red}{\frac{2a^3}{3}(\sqrt 2−4\sin^3θ)}
$$ in your arguments.
However, it should be $$\frac{2a^3}3(\sqrt 2−4\left|\sin^32θ\right|)
.$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 9,165 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4426895", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
latex-00016511 | l_n(\xi_1,\ldots,\xi_n):=\{\begin{array}{lll}l_n'(\xi_1,\ldots,\xi_n),& \mbox{if all $\xi_i\in L'$,}\\l_n''(\xi_1,\ldots,\xi_n),& \mbox{if all $\xi_i\in L''$, and}\\0,& \mbox{otherwise}.\end{array}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 16,549 | latex_formula | {"original_latex": "\\begin{align*}l_n(\\xi_1,\\ldots,\\xi_n):=\\left\\{\\begin{array}{lll}l_n'(\\xi_1,\\ldots,\\xi_n),& \\mbox{if all $\\xi_i\\in L'$,}\\\\l_n''(\\xi_1,\\ldots,\\xi_n),& \\mbox{if all $\\xi_i\\in L''$, and}\\\\0,& \\mbox{otherwise}.\\end{array}\\right.\\end{align*}"} |
mixed-00030371 | Prove $\sum_{i=1}^{n-1} \left[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \right] =(n-1)^2$? Apparently the following expression
$$
\sum_{i=1}^{n-1} \Bigg[\frac{n}{i(i+1)} + \frac{n(n-1)}{i(i+1)} (n(H_{n-2} - H_{n-i-1}) - (i-1))) \Bigg] \\
$$
simplifies to $(n-1)^2$, where $H_i$ is the i... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 17,812 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3666784", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
normal-00038543 | The current 7 @,@ 500 @-@ square @-@ foot ( 700 m2 ) Woodstock Library building was completed in 2000 . It has a " lantern @-@ like " quality and has received multiple awards for its design . In addition to offering the Multnomah County Library catalog , which contains two million books , periodicals and other material... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 93,337 | text | {} |
normal-00033847 | After graduating from high school , Maulbetsch joined the Ann Arbor Independents , a football team made up of Michigan " varsity eligibles " and " townies . " Maulbetsch was once reportedly called upon to drive across the goal line for the Independents in a game in which a large crowd , including a farmer with his plow... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 81,590 | text | {} |
normal-00003815 | Through the sad heart of Ruth , when , sick for home , | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 9,050 | text | {} |
latex-00026816 | \bigg(\sum_{j=1}^N |p_{1,j}-q_{1,j}||w_j|\bigg)^2 \le \bigg(\sum_{j=1}^N (p_{1,j} + q_{1,j})|w_j|\bigg)^2 \le 2 \sum_{j=1}^N (p_{1,j} + q_{1,j}) |w_j|^2 , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 27,052 | latex_formula | {"original_latex": "\\begin{align*}\\bigg(\\sum_{j=1}^N |p_{1,j}-q_{1,j}||w_j|\\bigg)^2 \\le \\bigg(\\sum_{j=1}^N (p_{1,j} + q_{1,j})|w_j|\\bigg)^2 \\le 2 \\sum_{j=1}^N (p_{1,j} + q_{1,j}) |w_j|^2 \\,, \\end{align*}"} |
normal-00031528 | Life during the pipeline construction project was characterized by long hours , poor conditions , and limited entertainment compensated by excellent benefits and pay . Each worker was handed a small booklet of 23 camp rules , but the rules ( including no alcohol or smoking ) were frequently broken and became the target... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 75,643 | text | {} |
normal-00012343 | Citizens in the south were opposed to a centralised government , and to the decrees of its rule , which resulted in rebellion . Prior to the revolution France had been divided into provinces with local governments . In 1790 the government , the National Constituent Assembly , reorganised France into administrative depa... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 30,005 | text | {} |
normal-00043622 | 1541 – Garamond is advanced money to cut the Grecs du Roi type . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 105,722 | text | {} |
mixed-00041600 | Does an integer $9
Does an integer $9<n<100$ exist such that the last 2 digits of $n^2$ is $n$? If yes, how to find them? If no, prove it.
This problem puzzled me for a day, but I'm not making much progress. Please help. Thanks. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 24,713 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/292651", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0}} |
latex-00002652 | {\hat G}^{(i)}_{p_i+1}\equiv (d+s) {\hat B }^{(i)}_{p_{i+1}}+ { K}^{(i)}_{p_{i+1}}(\hat B ) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 2,653 | latex_formula | {"original_latex": "\\begin{align*}{\\hat G}^{(i)}_{p_i+1}\\equiv (d+s) {\\hat B }^{(i)}_{p_{i+1}}+ { K}^{(i)}_{p_{i+1}}(\\hat B )\\end{align*}"} |
mixed-00043362 | You can find a computation method for the eigenvalues and eigenvectors in the following recent document: "Eigendecomposition of Block Tridiagonal Matrices" by
A. Sandryhaila and J.M.F. Moura (arxiv.org/pdf/1306.0217) | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,786 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1758958", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 1, "answer_id": 0}} |
mixed-00047409 | For example, for the denominator $$2\times 4\times 6\times 8=2\cdot 1\times 2\cdot 2\times 2\cdot 3\times 2\cdot 4=2\times 2\times 2\times 2\cdot 1\times 2\times 3\times 4=2^4\cdot 4!$$
I let you adapt this hint to your exercise. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 28,262 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/949313", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2}} |
mixed-00038777 | *
*A behaviour as $n \to \infty$. Laplace's method ($3.$, p. $2$)
may be applied here, $$ \begin{align} \int_a^bf(x)e^{-\lambda
g(x)}dx\sim f(a)e^{-\lambda g(a)}\sqrt{\frac{\pi}{2\lambda
g''(a)}},\qquad \lambda \to \infty, \tag1 \end{align} $$ with
$g'(a)=0$, $g''(a)>0$, $f(a)\neq 0$.
One may write $$ \begin{align}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,963 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2062149", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
normal-00049329 | mayh @-@ ga this @-@ ALL ' to here' | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 120,055 | text | {} |
mixed-00008736 | Prove$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}$$
Proof: Begin with the identity$$(1-z)^{a+b-c}\space_2F_1(a,b;c;z)=_2F_1(c-a,b-a;c;z)\tag1$$This can be easily proven... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 4,675 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2155025", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 1}} |
latex-00034700 | Z_f(s, \chi, A)=q^{-i-j-k}\int_{\tilde A}{\chi(ac (f(\pi^ix, \pi^jy, \pi^kz)))|f(\pi^ix, \pi^jy, \pi^kz)|^s|dxdydz|}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 35,059 | latex_formula | {"original_latex": "\\begin{align*} Z_f(s, \\chi, A)=q^{-i-j-k}\\int_{\\tilde A}{\\chi(ac (f(\\pi^ix, \\pi^jy, \\pi^kz)))|f(\\pi^ix, \\pi^jy, \\pi^kz)|^s|dxdydz|},\\end{align*}"} |
normal-00013516 | During his re @-@ examination of Iguanodon , David Norman was able to show that this posture was unlikely , because the long tail was stiffened with ossified tendons . To get the tripodal pose , the tail would literally have to be broken . Putting the animal in a horizontal posture makes many aspects of the arms and pe... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 32,739 | text | {} |
latex-00017307 | \omega _{ b}^{a}(x)\equiv \omega _{ b\mu }^{a}(x)dx^{\mu }, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 17,350 | latex_formula | {"original_latex": "\\begin{align*}\\omega _{\\;b}^{a}(x)\\equiv \\omega _{\\;b\\mu }^{a}(x)dx^{\\mu },\\end{align*}"} |
normal-00033154 | There was major restoration work done to the church in 1886 , and a large amount of the Chancel woodwork dates from this period . During the restoration the east window was also replaced ; it now depicts the stylised form of Saint Oswald , flanked on either side by Saint Aidan , and Saint Cuthbert , both Christian sain... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 79,942 | text | {} |
mixed-00019686 | Find the maximum of the $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$ with $|w|=1$ Let $w \in \mathbb{C}$, and $\left | w \right | = 1$. Find the maximum of the function $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=\cos x+i \sin x$. Then we have an... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 11,293 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2882231", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1}} |
latex-00039340 | (\tilde b\sigma)^s=s\nu_{\tilde b}(p)\sigma^s, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 39,839 | latex_formula | {"original_latex": "\\begin{align*}(\\tilde b\\sigma)^s=s\\nu_{\\tilde b}(p)\\sigma^s,\\end{align*}"} |
normal-00000594 | To the south @-@ east of the Great Cave is the second excavation , which faces east @-@ northeast . It includes a chapel at the north end . The front of this cave is completely destroyed ; only fragments of some semi @-@ columns remain . The interior has suffered water damage . The portico is 26 m ( 85 ft ) long and 11... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 1,472 | text | {} |
mixed-00010686 | Some quick observations : if $x$ is a root of $P(x)$ then so is $-x$. Thus if $(x^2-x+a)$ is a factor of $P(x)$, so is $(-x)^2-(-x)+a=x^2+x+a.$
Then $(x^2-x+a)(x^2+x+a)=x^4+(2a-1)x^2+a^2$ is a factor of $P(x)$. $\, a^2|36 \,$ $\Rightarrow a \in \{1,2,3,6\}$. (Here @Anwesha1729 observe in their answer that as $P(x)$ tak... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,828 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3965950", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
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