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-00008778 | \chi_\lambda(gx)=(-1)^\frac{q^2-1}{8}&\sum_{\mu\in M_q(\lambda_l)}(-1)^{L(c^{\lambda_l}_\mu)}(\chi_{(\mu,\lambda\backslash\{l\})}^+(x)+\chi_{(\mu,\lambda\backslash\{l\})}^-(x)). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 8,791 | latex_formula | {"original_latex": "\\begin{align*}\\chi_\\lambda(gx)=(-1)^\\frac{q^2-1}{8}&\\sum_{\\mu\\in M_q(\\lambda_l)}(-1)^{L(c^{\\lambda_l}_\\mu)}(\\chi_{(\\mu,\\lambda\\backslash\\{l\\})}^+(x)+\\chi_{(\\mu,\\lambda\\backslash\\{l\\})}^-(x)).\\end{align*}"} |
mixed-00016053 | Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\sum_{k=1}^{n-1} k\cdot3^k$$
I have a calculus-based solution: As a more general result, I replace $3$ with $x$ and denote the sum by $f(x)$. Then
$$f(x) = \sum_{k=1... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 9,083 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4238437", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 3}} |
mixed-00035374 | But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$
This is indeed wrong, since differentiating the equation at $x=-5$ gives $\displaystyle\frac{\sqrt6}{60}=-\frac{\sqrt6}{60}.$
Case $1:x>0$
the integral is definitely $\sec^{-1}x.$
Case $2: x<0$
the integral is $-\sec^{-... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 20,864 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3485512", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 4, "answer_id": 1}} |
mixed-00032373 | $$M=1+3+5+7+...+(2n-3)+(2n-1)\\M=(2n-1)+(2n-3)+...+7+5+3+1\\M+M=(1+2n-1)+(3+2n-3)+(5+2n-5)+....(2n-3+3)+(2n-1+1)\\n-term\\M+M=n(2n)\\2M=2n^2\\M=n^2
$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,026 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/922714", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 5, "answer_id": 2}} |
normal-00049219 | In the midst of feuding with Umaga , who defeated Hardy at The Great American Bash to retain the Intercontinental Championship in late July , Hardy was abruptly taken off WWE programming . He posted on his own website and in the forums of TheHardyShow.com that it was time off to heal , stemming from a bad fall taken in... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 119,828 | text | {} |
latex-00035128 | E_{u^\alpha}(f)=0 | latex | OleehyO/latex-formulas | cleaned_formulas | train | 35,492 | latex_formula | {"original_latex": "\\begin{align*}E_{u^\\alpha}(f)=0\\end{align*}"} |
normal-00022467 | Slay Tracks ( 1933 – 1969 ) ( also referred to as Slay Tracks ) is the debut extended play by the American indie rock band Pavement . Pavement , then consisting of founding members Stephen Malkmus ( guitar , vocals ) and Scott Kannberg ( guitar ) , recorded Slay Tracks with producer and future member Gary Young ( drums... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 54,013 | text | {} |
mixed-00018716 | Use Bernoulli:
$$\left(1+\frac{2}{n^2}\right)^n= \frac{1}{\left(\frac{n^2}{n^2+2}\right)^n}= \frac{1}{\left(1-\frac{2}{n^2+2}\right)^n}$$
And by Bernoulli
$$\left(1-\frac{2}{n^2+2}\right)^n \geq 1-\frac{2n}{n^2+2}=\frac{n^2-2n+2}{n^2+2}$$
Therefore
$$1 \leq \left(1+\frac{2}{n^2}\right)^n \leq \frac{n^2+2}{n^2-2n+2}$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 10,704 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2050128", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 1}} |
latex-00022193 | & r(x,z) + s(x,z) := m(r(x,z),z,s(x,z)), \ &-r(x,z) := m(z,r(x,z),z) \ & r(x,z) \cdot s(x,z) := r(s(x,z),z) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 22,351 | latex_formula | {"original_latex": "\\begin{align*} & r(x,z) + s(x,z) := m(r(x,z),z,s(x,z)), \\\\ &-r(x,z) := m(z,r(x,z),z) \\\\ & r(x,z) \\cdot s(x,z) := r(s(x,z),z) \\end{align*}"} |
mixed-00046743 | $$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$
$$x=1+{\cos^2\theta}+{\cos^4\theta}+\cdots\infty \;terms$$
this is infinite geometric series :
$S_{\infty}=\dfrac {a}{1-r}\;\;\,|r|<1$ ,here $\c... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 27,864 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/435803", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
normal-00017763 | While in session , the Senate had the power to act on its own , and even against the will of the presiding magistrate if it wished . The presiding magistrate began each meeting with a speech ( the verba fecit ) , which was usually brief , but was sometimes a lengthy oration . The presiding magistrate would then begin a... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 42,593 | text | {} |
normal-00022013 | In Brown 's arrangement of the genus , B. sphaerocarpa was placed between B. pulchella and B. nutans in taxonomic sequence ; that is , an order that places related taxa next to each other . No subdivision of the genus was given , other than to separate a single distinctive species into a subgenus of its own . Swiss bot... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 52,890 | text | {} |
mixed-00028274 | the inequality equals:
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4\ge 0 $
WLOG, let $d=min(a,b,c,d)$
$3(a+b+c+d)^2(a^2+b^2+c^2+d^2)+64abcd-(a+b+c+d)^4=2d\left(\dfrac{1}{3}\sum_{cyc (a,b,c)}(a-d)^2(2a+d)+\sum_{cyc (a,b,c)} a(a-d)^2+(a+b+c-3d)((a-b)^2+(b-c)^2+(a-c)^2)+3(a^3+b^3+c^3+3abc-\sum ab(a+b))+\dfrac{a^3+b^... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 16,539 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1698269", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 4, "answer_id": 2}} |
mixed-00006839 | Show $1+\cosθ+\cos(2θ)+\cdots+\cos(nθ)=\frac{1}{2}+\frac{\sin[(n+1/2)θ]}{2\sin(θ/2)}$ Show
$$1+\cosθ+\cos(2θ)+\cdots+\cos(nθ)=\frac12+\frac{\sin\left(\left(n+\frac12\right)θ\right)}{2\sin\left(\frac\theta2\right)}$$
I want to use De Moivre's formula and $$1+z+z^2+\cdots+z^n=\frac{z^{n+1}-1}{z-1}.$$ I set $z=x+yi$, but... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 3,553 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/684603", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 1}} |
latex-00026995 | & V^k_0=\sum_{j_k}^{j_{k+1}-1}W_{(\nu_j)}+\check h_0^k, \\& V^k_1=\sum_{j_k}^{j_{k+1}-1}\check \alpha_j (\Lambda W_{(\nu_j)})_{[\nu_j]}+\check g_1^k, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 27,239 | latex_formula | {"original_latex": "\\begin{align*}& V^k_0=\\sum_{j_k}^{j_{k+1}-1}W_{(\\nu_j)}+\\check h_0^k, \\\\& V^k_1=\\sum_{j_k}^{j_{k+1}-1}\\check \\alpha_j (\\Lambda W_{(\\nu_j)})_{[\\nu_j]}+\\check g_1^k,\\end{align*}"} |
latex-00025747 | \mathbb{P}(|\mathcal{C}(V_n)|>k)=\mathbb{P}(|\mathcal{A}_t|>0\forall t\leq k)&=\mathbb{P}(1+\sum_{i=1}^{t}(\eta_i-1)>0\forall t\leq k)\\&\leq \mathbb{P}(1+\sum_{i=1}^{t}(X_i-1)>0\forall t\leq k). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 25,946 | latex_formula | {"original_latex": "\\begin{align*}\\mathbb{P}(|\\mathcal{C}(V_n)|>k)=\\mathbb{P}(|\\mathcal{A}_t|>0\\forall t\\leq k)&=\\mathbb{P}(1+\\sum_{i=1}^{t}(\\eta_i-1)>0\\forall t\\leq k)\\\\&\\leq \\mathbb{P}(1+\\sum_{i=1}^{t}(X_i-1)>0\\forall t\\leq k).\\end{align*}"} |
mixed-00013091 | Yes, the gradient of line AB is $-1/2$ so the gradient of any line perpendicular to $AB$ is $2$. The line with gradient $2$ through $B = (3, 1)$ is $y = 2(x- 3)+ 1$ or $y= 2x- 5$. All points on that line, $(x, 2x- 5)$, satisfy the conditions for C. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 7,278 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1493877", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 2}} |
latex-00018513 | T(\rho_n, \vec{a}) &= e^{-i[\vec{a} \cdot (\hat{H}(\rho_n)\hat{f}_0 + \hat{P}(\rho_n)\hat{f}_1)]} , T(\rho_n, \vec{a})^{-1} = e^{i[\vec{a} \cdot (\hat{H}(\rho_n)\hat{f}_0 + \hat{P}(\rho_n)\hat{f}_1)]} , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 18,575 | latex_formula | {"original_latex": "\\begin{align*}T(\\rho_n, \\vec{a}) &= e^{-i[\\vec{a} \\cdot (\\hat{H}(\\rho_n)\\hat{f}_0 + \\hat{P}(\\rho_n)\\hat{f}_1)]} , T(\\rho_n, \\vec{a})^{-1} = e^{i[\\vec{a} \\cdot (\\hat{H}(\\rho_n)\\hat{f}_0 + \\hat{P}(\\rho_n)\\hat{f}_1)]} ,\\end{align*}"} |
latex-00027986 | \sqrt{n}\geq (F_Q)_j = \sum_{I \in \mathcal I_j} \frac{|I|}{|Q|} (F_I)_j > \frac{A}{\sqrt{n}}\sum_{I \in \mathcal I_j}\frac{|I|}{|Q|} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 28,241 | latex_formula | {"original_latex": "\\begin{align*}\\sqrt{n}\\geq (F_Q)_j = \\sum_{I \\in \\mathcal I_j} \\frac{|I|}{|Q|} (F_I)_j > \\frac{A}{\\sqrt{n}}\\sum_{I \\in \\mathcal I_j}\\frac{|I|}{|Q|} \\end{align*}"} |
latex-00035111 | f_{\gamma_M}(\gamma\big| N_Q=m)=\frac{f_{\gamma_{M,k}}(\gamma)}{1-F_{\gamma_{M,k}}(\Gamma_M)},\gamma \geq \Gamma_M, K_Q \neq 0 | latex | OleehyO/latex-formulas | cleaned_formulas | train | 35,475 | latex_formula | {"original_latex": "\\begin{align*}f_{\\gamma_M}(\\gamma\\big| N_Q=m)=\\frac{f_{\\gamma_{M,k}}(\\gamma)}{1-F_{\\gamma_{M,k}}(\\Gamma_M)},\\gamma \\geq \\Gamma_M, K_Q \\neq 0\\end{align*}"} |
normal-00013661 | Mosley was elected unopposed to his fourth term as president of the FIA in 2005 . In recognition of his contribution to road safety and motorsport , Mosley was made a Chevalier dans l ’ Ordre de la Légion d ’ honneur in 2006 . The Légion d 'honneur ( Legion of Honour ) is France 's highest decoration for outstanding ac... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 33,070 | text | {} |
latex-00033845 | \lim_{n \to +\infty} \frac{ e^{2T K(Q_n) }} {M_n^{\beta}} =0. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 34,191 | latex_formula | {"original_latex": "\\begin{align*}\\lim_{n \\to +\\infty} \\frac{ e^{2T K(Q_n) }} {M_n^{\\beta}} =0. \\end{align*}"} |
latex-00041783 | \tilde{U}(\cdot,U_0,\mathbf{W}(\cdot))=U(\cdot,U_0,\mathbf{W}_0(\cdot)). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 42,384 | latex_formula | {"original_latex": "\\begin{align*}\\tilde{U}(\\cdot,U_0,\\mathbf{W}(\\cdot))=U(\\cdot,U_0,\\mathbf{W}_0(\\cdot)).\\end{align*}"} |
latex-00007833 | b_{-+}(\xi,\eta) & = - \frac{i}{4}\frac{|\xi| \widetilde{m}_2(\xi,\eta)}{|\xi-\eta|^{1/2} |\eta|} - \frac{i}{4}\frac{|\xi|^{1/2} \widetilde{q}_2(\xi,\eta)}{ |\xi-\eta|^{1/2}|\eta|^{1/2}}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 7,842 | latex_formula | {"original_latex": "\\begin{align*}b_{-+}(\\xi,\\eta) & = - \\frac{i}{4}\\frac{|\\xi| \\widetilde{m}_2(\\xi,\\eta)}{|\\xi-\\eta|^{1/2} |\\eta|} - \\frac{i}{4}\\frac{|\\xi|^{1/2} \\widetilde{q}_2(\\xi,\\eta)}{ |\\xi-\\eta|^{1/2}|\\eta|^{1/2}},\\end{align*}"} |
normal-00024376 | While driving through a rural area in Georgia , Peter and his group are pulled over by the local sheriff , and Peter makes every effort to talk as annoyingly and rudely to him as possible . The Sheriff is more offended by Cleveland 's attempt to calmly explain themselves , prompting him to punch out one of their headli... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 58,479 | text | {} |
normal-00028763 | In 2005 , he returned to Ramsbottom as their professional for a second season , with more effect . He took 83 wickets at 9 @.@ 19 and scored 519 runs at 37 @.@ 07 in 20 matches . He scored five fifties including a best of 66 , and took eight five @-@ wicket innings hauls including a best of 9 / 47 against Church as wel... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 68,951 | text | {} |
mixed-00021292 | Combinatorial interpretation of sum of squares, cubes Consider the sum of the first $n$ integers:
$$\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
This makes the following bit of combinatorial sense. Imagine the set $\{*,1,2,\ldots,n\}$. We can choose two from this set, order them in decreasing order and thereby obt... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 12,280 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/95047", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "23", "answer_count": 4, "answer_id": 0}} |
normal-00028499 | Since uranium metal had been so scarce before the war , little was known about its metallurgy , but with uranium being used in the reactors , the Manhattan Project became keenly interested in its properties . In particular , with water being used for cooling , there was speculation about alloys with high thermal conduc... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 68,234 | text | {} |
latex-00023263 | \zeta_{p}(1-s)=\frac{2 \cos(\frac{\pi s}{2})\Gamma(s)}{(2\pi)^{s}}\eta_{p}(s). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,432 | latex_formula | {"original_latex": "\\begin{align*}\\zeta_{p}(1-s)=\\frac{2\\,\\cos\\left(\\frac{\\pi s}{2}\\right)\\Gamma(s)}{(2\\pi)^{s}}\\eta_{p}(s).\\end{align*}"} |
latex-00040317 | 2T(s,s-1,t+1)=T(s,s,t) . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 40,824 | latex_formula | {"original_latex": "\\begin{align*}2T(s,s-1,t+1)=T(s,s,t)\\,.\\end{align*}"} |
normal-00016412 | The program featured an interview with Morales 's former attorney Robert Walton , who quoted him as having said , " I was in Dallas when we got the son of a bitch and I was in Los Angeles when we got the little bastard . " O 'Sullivan reported that the CIA declined to comment on the officers in question . It was also a... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 39,228 | text | {} |
normal-00041130 | Flight 736 , a four @-@ engined DC @-@ 7 propliner with registration N6328C , departed Los Angeles International Airport at 7 : 37 a.m. on a flight to New York City with stops in Denver , Kansas City and Washington , D.C. On board were 42 passengers and five crew members ; Captain Duane M. Ward , 44 , First Officer Arl... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 99,705 | text | {} |
mixed-00009427 | HINT
By Taylor's expansion
$$ \frac{\tan(x)-\sin(x)}{x-\sin(x)}= \frac{x+\frac{x^3}3-x+\frac{x^3}6+o(x^3)}{x-x+\frac{x^3}6+o(x^3)}=\frac{\frac{x^3}2+o(x^3)}{\frac{x^3}6+o(x^3)}=\frac{\frac{1}2+o(1)}{\frac{1}6+o(1)}\to 3$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,083 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2717908", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
normal-00046256 | The Air Force changed the program to full weapon development and awarded a contract for an XB @-@ 70 prototype and 11 YB @-@ 70s in August 1960 . In November 1960 , the B @-@ 70 program received a $ 265 million appropriation from Congress for FY 1961 . Nixon , trailing in his home state of California , also publicly en... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 112,364 | text | {} |
normal-00015209 | Alkan 's Op. 31 set of Préludes includes a number of pieces based on Jewish subjects , including some titled Prière ( Prayer ) , one preceded by a quote from the Song of Songs , and another titled Ancienne mélodie de la synagogue ( Old synagogue melody ) . The collection is believed to be " the first publication of art... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 36,455 | text | {} |
latex-00005901 | \sqrt{a} = - \int_0^{\infty} \frac {d s} {\sqrt{ \pi s}} \frac {d} {d s} e^{- a s} . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,909 | latex_formula | {"original_latex": "\\begin{align*}\\sqrt{a} = - \\int_0^{\\infty} \\frac {d s} {\\sqrt{ \\pi s}} \\; \\frac {d} {d s} \\; e^{- a s} \\; .\\end{align*}"} |
latex-00010524 | M_{N}^2 (a) = m_{0}^2 +\frac{(n_{1}-a_{1})^2}{L_{1}^2} + \frac{(n_{2}-a_{2})^2}{L_{2}^2}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 10,546 | latex_formula | {"original_latex": "\\begin{align*}M_{N}^2 (a) = m_{0}^2 +\\frac{(n_{1}-a_{1})^2}{L_{1}^2} + \\frac{(n_{2}-a_{2})^2}{L_{2}^2},\\end{align*}"} |
latex-00020077 | \hskip-36pt \begin{array}{lcp{10.7cm}} \hline\noalign{\smallskip}P_{an} &=& $(x-1)^2 \partial^2 -(x-1)(3a+3b+4c+2g+2-(a+c)/x) \partial$ \ && $+ (2a+2b+2c+g+2)(a+b+2c+g+1-c/x)$,\\Q_{an} &=& $ (x-1)^2 \partial^2 -(x-1)(1+2g+3b+4c+3a-(1+a+c)/x)\partial$ \\&& $+(2a+2b+2c+g+1)(a+b+2c+g+1)$\\&& $-(c+1)(2a+2b+2c+g+2)/x +(a+c+... | latex | OleehyO/latex-formulas | cleaned_formulas | train | 20,166 | latex_formula | {"original_latex": "\\begin{align*} \\hskip-36pt \\begin{array}{lcp{10.7cm}} \\hline\\noalign{\\smallskip}P_{an} &=& $(x-1)^2 \\partial^2 -(x-1)(3a+3b+4c+2g+2-(a+c)/x) \\partial$ \\\\ && $+ (2a+2b+2c+g+2)(a+b+2c+g+1-c/x)$,\\\\Q_{an} &=& $ (x-1)^2 \\partial^2 -(x-1)(1+2g+3b+4c+3a-(1+a+c)/x)\\partial$ \\\\&& $+(2a+2b+2c+... |
normal-00030380 | " 2 Become 1 " was released in the United Kingdom on 16 December 1996 , in three single versions . The first one , a maxi single that included the single version of the track , an orchestral take , the Junior Vasquez remix of " Wannabe " , and " One of These Girls " , a song written by the group with Paul Wilson and An... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 72,787 | text | {} |
mixed-00023410 | Smallest number whose $\sin(x)$ in radian and degrees is equal Question:
What is the smallest positive real number $x$ with the property that the sine of $x$ degrees is equal to the sine of $x$ radians?
My try: 0. But zero isn't a positive number. How do I even begin to solve it? I tried taking inverse on both sides... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 13,561 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1846912", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
normal-00018962 | The Canadian federal election of 1957 was held June 10 , 1957 , to select the 265 members of the House of Commons of Canada . In one of the great upsets in Canadian political history , the Progressive Conservative Party ( also known as " PCs " or " Tories " ) , led by John Diefenbaker , brought an end to 22 years of Li... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 45,539 | text | {} |
latex-00042345 | d(a-a^n) &= \nu(n)c_a \mbox{ for all } a\in A \ c_{a+b}-c_a-c_b &= d\frac{a^n + b^n - (a+b)^n}{\nu(n)} \mbox{ for all } a,b\in A \ ac_b + b^nc_a &= c_{ab} \mbox{ for all } a,b\in A . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 42,952 | latex_formula | {"original_latex": "\\begin{align*} d(a-a^n) &= \\nu(n)c_a \\mbox{\\ \\ for\\ all\\ } a\\in A \\\\ c_{a+b}-c_a-c_b &= d\\frac{a^n + b^n - (a+b)^n}{\\nu(n)} \\mbox{\\ \\ for\\ all\\ } a,b\\in A \\\\ ac_b + b^nc_a &= c_{ab} \\mbox{\\ \\ for\\ all\\ } a,b\\in A .\\end{align*}"} |
mixed-00000890 | Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 454 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/616639", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 5, "answer_id": 1}} |
mixed-00042219 | I am assuming that $X$, $Y$, and $Z$ are mutually independent, standard Gaussian.
The expression of your expectation can be simplified.
Let us denote your expectation as $E$.
First, notice that $X$ and $Y$ are zero-mean, so
$$
E = \frac{1}{2}\mathsf{E}\left[\sqrt{X^2 + 4Y^2 - 2XZ + Z^2}\right]
$$
Grouping terms, we hav... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,084 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/777573", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0}} |
mixed-00010053 | what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,452 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3309188", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 6, "answer_id": 0}} |
mixed-00001647 | Alternatively, let $D, S$ and $L$ be the values, and let $L=D+1+L_0$. Then you want non-negative integer solutions to $1+D+S+L_0=100$, or $2D+S+L_0=99$.
You are doing $$\sum_{D=0}^{49} \sum_{S=0}^{99-2D} 1=\sum_{D=0}^{49} (100-2D)$$ which is a correct way to count this value.
A generating function solution would be to ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 837 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1144328", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1}} |
normal-00038569 | Ragnarök is briefly referenced in stanza 40 of the poem Helgakviða Hundingsbana II . Here , the valkyrie Sigrún 's unnamed maid is passing the deceased hero Helgi Hundingsbane 's burial mound . Helgi is there with a retinue of men , surprising the maid . The maid asks if she is witnessing a delusion since she sees dead... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 93,400 | text | {} |
normal-00040021 | When it came time to animate Simba during the " I Just Can 't Wait to Be King " musical sequence , Henn felt it essential that the character remain on all fours at all times , despite the fact that he is meant to be dancing . In terms of personality , Henn aimed to depict Simba as a " cocky , confident character " at t... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 96,985 | text | {} |
mixed-00038421 | The approach shown by Elliot G is perfectly fine, but if you like a viable alternative, you may consider that:
$$ \frac{1}{x^2+1} = \frac{1}{(x+i)(x-i)} = \frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right) $$
hence by squaring both sides:
$$ \frac{1}{(x^2+1)^2} = -\frac{1}{4}\left(\frac{1}{(x-i)^2}+\frac{1}{(x+i)^2}-... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,742 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1749007", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 3, "answer_id": 0}} |
mixed-00002581 | Induction on the power of $9$ dividing the number. Begin with any number not divisible by $9,$ although it is allowed to be divisible by $3.$ The hypothesis at this stage is just that this number is the sum of three squares, say $n = a^2 + b^2 + c^2.$ Since this $n$ is not divisible by $9,$ it follows that at least one... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,311 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1777521", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 1}} |
latex-00023578 | .\begin{array}{ll}\xi_{rr}(0)=-\dfrac{ 2m_1Q_1(0)}{N},\quad\eta_{rr}(0)=-\dfrac{2m_2Q_2(0)}{N}.\end{array}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,748 | latex_formula | {"original_latex": "\\begin{align*}\\left.\\begin{array}{ll}\\displaystyle\\xi_{rr}(0)=-\\dfrac{ 2m_1Q_1(0)}{N},\\quad\\eta_{rr}(0)=-\\dfrac{2m_2Q_2(0)}{N}.\\end{array}\\right.\\end{align*}"} |
latex-00028678 | W^{s,p}(\mathbb{R}^n) := (1 - \Delta)^{-s/2} (L^p(\mathbb{R}^n) ), ||f||_{W^{s,p}} := ||(1 - \Delta)^{s/2} f ||_{L^p}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 28,940 | latex_formula | {"original_latex": "\\begin{align*} W^{s,p}(\\mathbb{R}^n) := (1 - \\Delta)^{-s/2} \\left(L^p(\\mathbb{R}^n) \\right), ||f||_{W^{s,p}} := ||(1 - \\Delta)^{s/2} f ||_{L^p}.\\end{align*}"} |
latex-00035595 | [\alpha (u)\ge 0,\alpha (-u)\ge 0]\Rightarrow u=0. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 35,959 | latex_formula | {"original_latex": "\\begin{align*} [\\alpha (u)\\ge 0,\\alpha (-u)\\ge 0]\\Rightarrow u=0.\\end{align*}"} |
normal-00032366 | Fleury was born on June 29 , 1968 , in Oxbow , Saskatchewan , the first of Wally and Donna Fleury 's three sons . Wally was a hockey player whose dreams of a professional career ended when he broke his leg playing baseball in the summer of 1963 ; the injury helped fuel a drinking problem . Donna was a quiet , religious... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 77,680 | text | {} |
normal-00049783 | New to the series is the Autovista mode , a mode in which players can view precise details such as engine parts and interior gauges on a select number of cars . It features a partnership with BBC 's Top Gear as well as its American counterpart . Jeremy Clarkson , presenter for Top Gear , provides commentary in the game... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 121,190 | text | {} |
mixed-00042379 | Draw a Venn diagram!
Let probabilities be as follows:
$a$=Blue only
$b$=Blue and Left
$c$=Left only
$d$=None
From probabilities given,
$$\begin{align}
\frac b{a+b}=\frac 17 \quad \Rightarrow a&=6b\\
\frac b{b+c}=\frac 13 \quad \Rightarrow c&=2b\\
d&=\frac 45\\
\end{align}$$
As the probabilites sum to 1,
$$\begin{alig... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,179 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/898539", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3}} |
normal-00012876 | However , in dinoflagellates , the nuclear envelope remains intact , the centrosomes are located in the cytoplasm , and the microtubules come in contact with chromosomes , whose centromeric regions are incorporated into the nuclear envelope ( the so @-@ called closed mitosis with extranuclear spindle ) . In many other ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 31,235 | text | {} |
mixed-00009935 | $y_0 \ge 2$, $y_n = y_{n-1}^2 -2$ $\Rightarrow$ $\frac{1}{y_0}+\frac{1}{y_0y_1}+\cdots = \frac{y_0 - \sqrt{y_0^2 - 4}}{2}$ $y_0 \ge 2$ and $y_n = y_{n-1}^2 - 2$. Let $S_n = \frac{1}{y_0} + \frac{1}{y_0 y_1}+\cdots + \frac{1}{y_0 y_1 \cdots y_{n}}$. Prove that
$$\lim_{n \rightarrow \infty} S_n = \frac{y_0 - \sqrt{y_0^2 ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,383 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3192721", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00025697 | Prior to its merger with Swiss Bank Corporation , UBS operated as a full @-@ service bank and a provider of wholesale financial services through its retail banking , commercial banking , investment banking , asset management and wealth management businesses . In 1997 , prior to its merger with Swiss Bank Corporation , ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 61,644 | text | {} |
mixed-00034048 | If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 20,051 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2238320", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 0}} |
normal-00024735 | On March 22 , 1834 , he published the first issue of The New @-@ Yorker in partnership with Jonas Winchester . It was less expensive than other literary magazines of the time and published both contemporary ditties and political commentary . Circulation reached 9 @,@ 000 , then a sizable number , yet it was ill @-@ man... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 59,336 | text | {} |
mixed-00011404 | Suppose $m>0$.
If $2^k = 9^m + 7^n$, then $2^k \equiv 1 \pmod 3$, thus $k \equiv 0 \pmod 2$.
Put $k = 2l$ : $7^n = 4^l - 9^m = (2^l - 3^m)(2^l + 3^m)$.
If $(2^l - 3^m) > 1$, then both factors are multiples of $7$, and thus so is $3^m$, which is impossible.
Thus $2^l - 3^m = 1$. The only solution to this is $l=2$ and $... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,261 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/120151", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 2, "answer_id": 0}} |
latex-00037824 | &\widehat{B}_{r,i} = \sum_{l=1}^R B_l \widehat{\lambda}_{l,r,i} \widehat{f}_r = \sum_{l=1}^L f_l \widehat{\gamma}_{l,r}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 38,309 | latex_formula | {"original_latex": "\\begin{align*}&\\widehat{B}_{r,i} = \\sum_{l=1}^R B_l \\widehat{\\lambda}_{l,r,i} \\widehat{f}_r = \\sum_{l=1}^L f_l \\widehat{\\gamma}_{l,r},\\end{align*}"} |
mixed-00037542 | The limit: $\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$ $$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)}$$
My steps
$\lim_{x\rightarrow0}\frac{1-\cos2x}{\sin3x\cdot\ln(1+\sin4x)} = \lim_{x\rightarrow0}\frac{1-\cos2x}{2x}\cdot2x\cdot\frac{3x}{\sin3x}\cdot\frac{1}{3x}\cdot\frac{1}{\ln(1+... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,195 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1005961", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
mixed-00009183 | $f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$
$f(x)=\sqrt{\dfrac{x-\sqrt {20}}{x-3}}$, $g(x)=\sqrt{x^2-4x-12}$, find the domain of $f(g(x))$.
For some reason, I cannot get the correct answer. Here is what I tried.
$D_g:$
$$x^2-4x-12\ge0$$
$$(x-6)(x+2)\ge0$$
$$\boxed{(-... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 4,941 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2488746", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0}} |
normal-00046480 | The steel track of Dæmonen is approximately 1 @,@ 850 @.@ 4 feet ( 564 @.@ 0 m ) long , and the height of the lift is 92 feet ( 28 m ) high . The track was designed by Bolliger & Mabillard and is filled with sand to reduce the noise made by the trains . Also , the track is painted red while the supports are silver . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 112,923 | text | {} |
normal-00017658 | " Oubliette " is the only X @-@ Files screenplay written by Craig , who exited the writing staff before the entry was produced . The extensive outdoor filming lead to several difficulties for the production crew . Amy was 12 years old in the original screenplay . The Fox network was concerned her situation was an uncom... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 42,308 | text | {} |
mixed-00028753 | sum of series $\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms sum of series $\displaystyle \frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n$ terms
assuming $\displaystyle S_{n} =\frac{n}{1\cdot 2 \cdot 3}+\fra... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 16,824 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2079081", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0}} |
latex-00017689 | UL(x)&= \begin{bmatrix} x&0\\0&1\end{bmatrix}& LR(x) &= \begin{bmatrix} 1&0\\0&x\end{bmatrix}&D(x)&= \begin{bmatrix} x&0\\0&x\end{bmatrix}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 17,736 | latex_formula | {"original_latex": "\\begin{align*}UL(x)&= \\begin{bmatrix} x&0\\\\0&1\\end{bmatrix}& LR(x) &= \\begin{bmatrix} 1&0\\\\0&x\\end{bmatrix}&D(x)&= \\begin{bmatrix} x&0\\\\0&x\\end{bmatrix}.\\end{align*}"} |
normal-00000652 | The composition of the main facade , with its dominant drawn @-@ out and elevated projection and the two symmetrical lower lateral parts , reflects the internal division into three naves . At ground @-@ floor level , the front was distinguished by the three @-@ arch entrance and bifora , whereas the first @-@ floor lev... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 1,596 | text | {} |
mixed-00009842 | Let $\arctan x=y\implies x=\tan y,x^2-1=\dfrac{\sin^2y}{\cos^2y}-1=-(1+x^2)\cos2y$
$$2x\cos(2x)+(x^2-1)\sin2x=(1+x^2)\sin2(\arctan x-x)$$
$$I=\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} =\int\dfrac{2x^2}{(1+x^2)\sin2(\arctan x-x)}$$
Now set $u=\arctan x-x,du=-\dfrac{x^2\ dx}{1+x^2}$
$$I=-2\int\dfrac{du}{\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,332 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3097078", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 5, "answer_id": 3}} |
normal-00035747 | During the First Indochina War ( 1945 to 1954 ) , both Viet Minh and French forces considered the Central Highlands to be their ‘ home ’ , as it was considered the key to domination in Indochina . Both sides recognised that in order to occupy the Central Highlands , they had to possess a sufficient reserve of manpower ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 86,082 | text | {} |
normal-00007453 | Wiśniowiecki was widely popular among the noble class , who saw in him a defender of tradition , a patriot and an able military commander . He was praised by many of his contemporaries , including a poet , Samuel Twardowski , as well as numerous diary writers and early historians . For his protection of civilian popula... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 18,200 | text | {} |
latex-00031877 | \mathring{E}_p^2&=M^{k_0}(\mathring{E}_p^2)=M^{k_0}(\{M\frac{e_1}{p},M^2\frac{e_1}{p},\cdots,M^{p^2-1}\frac{e_1}{p}\}) ({\rm mod} \mathbb{Z}^2)\\&=\{M(\lambda+v_0),M^{2}(\lambda+v_0),\cdots,M^{p^2-1}(\lambda+v_0)\} ({\rm mod} \mathbb{Z}^2)\\&=\{M\lambda,M^{2}\lambda,\cdots,M^{p^2-1}\lambda\} ({\rm mod} \mathbb{Z}^2). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 32,175 | latex_formula | {"original_latex": "\\begin{align*}\\mathring{E}_p^2&=M^{k_0}(\\mathring{E}_p^2)=M^{k_0}(\\{M\\frac{e_1}{p},M^2\\frac{e_1}{p},\\cdots,M^{p^2-1}\\frac{e_1}{p}\\}) \\;({\\rm mod} \\;\\mathbb{Z}^2)\\\\&=\\{M(\\lambda+v_0),M^{2}(\\lambda+v_0),\\cdots,M^{p^2-1}(\\lambda+v_0)\\} \\;({\\rm mod} \\;\\mathbb{Z}^2)\\\\&=\\{M\\la... |
mixed-00036532 | Approximation for $\pi$ I just stumbled upon
$$ \pi \approx \sqrt{ \frac{9}{5} } + \frac{9}{5} = 3.141640786 $$
which is $\delta = 0.0000481330$ different from $\pi$. Although this is a rather crude approximation I wonder if it has been every used in past times (historically). Note that the above might also be related... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 21,573 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/146831", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "25", "answer_count": 3, "answer_id": 0}} |
normal-00042499 | Plutarch mentions that the Athenians saw the phantom of King Theseus , the mythical hero of Athens , leading the army in full battle gear in the charge against the Persians , and indeed he was depicted in the mural of the Stoa Poikile fighting for the Athenians , along with the twelve Olympian gods and other heroes . P... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 102,898 | text | {} |
mixed-00010575 | $$
\frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} = \frac{x^2}{\left( \dfrac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \right)} = \frac{x^2}{\text{a positive number}}
$$
It is not generally true that
$$ \require{cancel}
\xcancel{\sigma_1 \sigma_2 = \frac{\sigma_1\sigma_2}{\sqrt{\sigma_2^2 + \sigma_2^2}},}
$$
as may... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,764 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3836236", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00027145 | The Leopard 2E replaced the Leopard 2A4 in Spanish mechanized units , which in turn replaced M60s in cavalry units . Both versions of the Leopard 2 are expected to remain in service with the Spanish army until 2025 . In terms of industrial scale , the production and development of the Leopard 2E represents a total of 2... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 64,935 | text | {} |
normal-00021646 | Grasshoppers are occasionally depicted in artworks , such as the Dutch Golden Age painter Balthasar van der Ast 's still life oil painting , Flowers in a Vase with Shells and Insects , c . 1630 , now in the National Gallery , London , though the insect may be a bush @-@ cricket . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 51,892 | text | {} |
latex-00006626 | G_{ab} = \frac{\partial^2}{\partial \xi^a\partial \xi^b}\log( \kappa_{cde}\xi^c\xi^d\xi^e). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 6,634 | latex_formula | {"original_latex": "\\begin{align*}G_{ab} = \\frac{\\partial^2}{\\partial \\xi^a\\partial \\xi^b}\\log\\left( \\kappa_{cde}\\xi^c\\xi^d\\xi^e\\right).\\end{align*}"} |
latex-00025971 | T_6 =-\frac{s}{2} \iint_Q \sigma_{t t}|u|^2dx d t \geq-C s \iint_Q \xi^{3 / 2}|u|^2dx d t. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 26,172 | latex_formula | {"original_latex": "\\begin{align*}T_6 =-\\frac{s}{2} \\iint_Q \\sigma_{t t}|u|^2dx d t \\geq-C s \\iint_Q \\xi^{3 / 2}|u|^2dx d t.\\end{align*}"} |
mixed-00044168 | I'm not sure about how obvious this seems, but you can just make the direct substitution $z=x+\sqrt{1+x^2}$ in$$\int\frac {\mathrm dx}{\sqrt{1+x^2}}$$
Because$$\frac {\mathrm dz}{\mathrm dx}=1+\frac {x}{\sqrt{1+x^2}}=\frac {z}{\sqrt{1+x^2}}$$
This can be noted because the derivative of $\sqrt{1+x^2}$ is $x/\sqrt{1+x^2}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 26,280 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2445250", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 2}} |
normal-00010260 | This paradigm led to innovative work in machine vision by Gerald Sussman ( who led the team ) , Adolfo Guzman , David Waltz ( who invented " constraint propagation " ) , and especially Patrick Winston . At the same time , Minsky and Papert built a robot arm that could stack blocks , bringing the blocks world to life . ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 25,051 | text | {} |
mixed-00009691 | Eqn 1
$$(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$$
$$(-\frac{6}{x^2})+(-\frac{6}{x^3})+(\frac{6}{x^3})(1+\frac{1}{x^2}+\frac{2}{x})$$
$$(-\frac{6}{x^2})+ (-\frac{6}{x^3}) + \frac{6}{x^3}+\frac{6}{x^5}+\frac{12}{x^4}$$
$$\frac6{x^2}[\frac{1}{x^3}+\frac{2}{x^2}-1]$$
Eqn 2
$$\frac{6}{x^3}(1+\frac{1... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,242 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2935936", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
latex-00008201 | {\cal M} = e^{2\phi}(\begin{array}{cc}|\lambda |^{2} & a \\& \\a & 1 \\\end{array}) ,\hspace{1cm}{\cal M}^{-1} = e^{2\phi}(\begin{array}{cc}1 & -a \\& \\-a & |\lambda |^{2} \\\end{array}) , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 8,211 | latex_formula | {"original_latex": "\\begin{align*}{\\cal M} = e^{2\\phi}\\left(\\begin{array}{cc}|\\lambda |^{2} & a \\\\& \\\\a & 1 \\\\\\end{array}\\right)\\, ,\\hspace{1cm}{\\cal M}^{-1} = e^{2\\phi}\\left(\\begin{array}{cc}1 & -a \\\\& \\\\-a & |\\lambda |^{2} \\\\\\end{array}\\right)\\, ,\\end{align*}"} |
normal-00003124 | Codium fragile ssp. atlanticum has been established to be native , although for many years it was regarded as an alien species . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 7,464 | text | {} |
latex-00015594 | \hbox{detGr}_{n,0}\propto\alpha^{\frac{1}{2}C_{2n}^n}\prod_{i=1}^{n}(\alpha-P_i^2)^{C_{2n}^{n-i}} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 15,626 | latex_formula | {"original_latex": "\\begin{align*}\\hbox{detGr}_{n,0}\\propto\\alpha^{\\frac{1}{2}C_{2n}^n}\\prod_{i=1}^{n}\\left(\\alpha-P_i^2\\right)^{C_{2n}^{n-i}}\\end{align*}"} |
normal-00009826 | Alabama also played four non @-@ conference games . The game against Penn State was originally scheduled as part of the 2004 season , however the series was moved back at the request of Alabama due to fallout from NCAA sanctions being levied on the program . The non @-@ conference schedule also included games against S... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 24,086 | text | {} |
latex-00003538 | {\cal S}_\mathrm{E} \alpha(y,t) = (\frac{\hbar}{2m})^{3/2} \frac{\partial^3}{\partial y^3} \psi(y,t) . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 3,539 | latex_formula | {"original_latex": "\\begin{align*}{\\cal S}_\\mathrm{E} \\alpha(y,t) = \\left(\\frac{\\hbar}{2m}\\right)^{3/2} \\frac{\\partial^3}{\\partial y^3} \\psi(y,t) \\, .\\end{align*}"} |
normal-00016479 | In 2009 , Brain Scan Studios released the parody Watchmensch , a comic in which writer Rich Johnston chronicled " the debate surrounding Watchmen , the original contracts , the current legal suits over the Fox contract " . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 39,394 | text | {} |
latex-00007302 | z_i~\rightarrow~(-1)^{\epsilon_i}z_i+{1\over 2}\delta_i, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 7,311 | latex_formula | {"original_latex": "\\begin{align*}z_i~\\rightarrow~(-1)^{\\epsilon_i}z_i+{1\\over 2}\\delta_i,\\end{align*}"} |
mixed-00048486 | Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = (x-y)/(x+y)$ Solve
$$\frac { { d }y }{ { d }x } =\frac { x-y }{ x+y } $$
It is homogeneous, thus let $y = vx$. From this, $\frac{\mathrm{d}y}{\mathrm{d}x} = x\frac{\mathrm{d}v}{\mathrm{d}x} + v$
Thus,
$v'x + v = (1-v)/(1+v)$ thus,
$\frac{2}{1-v} + \ln(v - 1) = \ln(x) + C$.
W... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 28,910 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1852208", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0}} |
latex-00036567 | {\bf A}^uV_{\bf \hat{u}}(t,x) = {\bf A}^uf_{\bf \hat{u}}(t,x,x)+ {\bf A}^uG\diamond g_{\bf \hat{u}}(t,x), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 36,936 | latex_formula | {"original_latex": "\\begin{align*}{\\bf A}^uV_{\\bf \\hat{u}}(t,x) = {\\bf A}^uf_{\\bf \\hat{u}}(t,x,x)+ {\\bf A}^uG\\diamond g_{\\bf \\hat{u}}(t,x),\\end{align*}"} |
normal-00023049 | Lamarckism was popularised in the English @-@ speaking world by the speculative Vestiges of the Natural History of Creation , published anonymously by Robert Chambers in 1844 . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 55,403 | text | {} |
mixed-00012005 | Let's evaluate each $(p^2 + 1)$ expression, counting up the factors of $2$ and $5$, stopping when they both are at least $6$:
$n = 1$
$p = 2$, and $(p^2 + 1) = 5$. 2s so far = $0$, 5s so far = $1$.
$n = 2$
$p = 3$, and $(p^2 + 1) = 10$. 2s so far = $1$, 5s so far = $2$.
$n = 3$
$p = 5$, and $(p^2 + 1) = 26$. 2s so far ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,624 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/557255", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 1}} |
normal-00024957 | The traditional standard for height of a horse or a pony at maturity is 14 @.@ 2 hands ( 58 inches , 147 cm ) . An animal 14 @.@ 2 h or over is usually considered to be a horse and one less than 14 @.@ 2 h a pony , but there are many exceptions to the traditional standard . In Australia , ponies are considered to be th... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 59,806 | text | {} |
normal-00035865 | The 1999 – 2000 season , the first under new chairman George Reynolds , was marked by Darlington becoming the first team to lose an FA Cup tie and still qualify for the next round . Manchester United 's involvement in the FIFA Club World Championship meant they did not enter the FA Cup . To decide who took their place ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 86,445 | text | {} |
latex-00038812 | r^i_j&:=a^{im}r_{mj}, s^i_j:=a^{im}s_{mj}, r_j:=b^mr_{mj}, s_j:=b^ms_{mj}, r:=b^ir_i, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 39,302 | latex_formula | {"original_latex": "\\begin{align*}r^i_j&:=a^{im}r_{mj},\\ \\ s^i_j:=a^{im}s_{mj}, \\ \\ r_j:=b^mr_{mj},\\ \\ s_j:=b^ms_{mj}, \\ \\ r:=b^ir_i,\\end{align*}"} |
normal-00016868 | ESPN 's Chantel Jennings tweeted a picture of a Flag of Canada redone in Michigan 's team colors of maize and blue at Crisler Arena on December 4 . One of her followers noted that the big version on the wall was a general Stauskas fan flag and that a little version of the flag was added next to it for each three @-@ po... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 40,422 | text | {} |
latex-00021683 | C_r^1 = \widehat\psi_r(1) = \inf_{x\in[0,1]} \int_{[0,1]} \d y \log\frac{1}{r(x,y)}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 21,832 | latex_formula | {"original_latex": "\\begin{align*}C_r^1 = \\widehat\\psi_r(1) = \\inf_{x\\in[0,1]} \\int_{[0,1]} \\d y\\,\\log\\frac{1}{r(x,y)}.\\end{align*}"} |
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