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-00020689 | \delta=\frac{v_+-v_-}{4(\beta+1)}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 20,822 | latex_formula | {"original_latex": "\\begin{align*}\\delta=\\frac{v_+-v_-}{4(\\beta+1)}.\\end{align*}"} |
latex-00023104 | .\begin{array}{ll}K=\inf\limits_{(u,v)\in\mathcal{M}}S(u,v).\end{array}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,272 | latex_formula | {"original_latex": "\\begin{align*}\\left.\\begin{array}{ll}K=\\inf\\limits_{(u,v)\\in\\mathcal{M}}S(u,v).\\end{array}\\right.\\end{align*}"} |
latex-00041973 | (\sum^{\infty}_{n,m=-\infty}q^{n^2+nm+2m^2})^2=1+\sum^{\infty}_{n=1}\sigma^{*}(n)q^n. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 42,575 | latex_formula | {"original_latex": "\\begin{align*}\\left(\\sum^{\\infty}_{n,m=-\\infty}q^{n^2+nm+2m^2}\\right)^2=1+\\sum^{\\infty}_{n=1}\\sigma^{*}(n)q^n.\\end{align*}"} |
mixed-00034178 | For a regular polygon with $n$ sides with side length $l$. The ends of each side when connected to the centre of the polygon forms a triangle with an angle of $\frac{2\pi}{n}$ at the centre. There will be $n$ such triangles. The altitude of each triangle starting from the centre of the polygon has a length of $\frac{a}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 20,127 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2326851", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "10", "answer_count": 4, "answer_id": 0}} |
normal-00033801 | Pickling salt is ultrafine to speed dissolving to make brine . Gourmet salts may be used for specific tastes . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 81,475 | text | {} |
mixed-00045296 | $$f(x)=\cos x+\cos\frac{\sqrt{3} x}{2}$$
Let $f_1(x)$ be periodic with a period of $T_1$ and $f_2(x)$ be periodic with a period of $T_2$. Then $f(x)=f_1(x)+f_2(x)$ only if $T_1/T_2$ is rational. Then the period is given by $T=LCM(T_1,T_2).$
Here $T_1=2\pi$ and $T_2=\frac{4\pi}{\sqrt{3}}$ This means $T_1/T_2=\frac{\sqrt... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 26,980 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3573254", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0}} |
latex-00037908 | \frac{\partial V_l^{PR}(c,t)}{\partial t} = - \sum_{k=1}^K \cal R_t^{kl}(A_k^{Static}, \Delta_l V_l^{PR}(c,t)) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 38,393 | latex_formula | {"original_latex": "\\begin{align*} \\frac{\\partial V_l^{PR}(c,t)}{\\partial t} = - \\sum_{k=1}^K \\cal R_t^{kl}(A_k^{Static}, \\Delta_l V_l^{PR}(c,t)) \\end{align*}"} |
normal-00010163 | The Assembly understood its mandate under the Solemn League and Covenant to have been fulfilled on 14 April 1648 when it delivered the scripture citations to Parliament , and the Scottish Commissioners had already left by the end of 1647 . The Assembly continued to meet primarily for the purpose of examination of minis... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 24,807 | text | {} |
mixed-00037731 | Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\fr... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,313 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1148203", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "12", "answer_count": 2, "answer_id": 0}} |
normal-00012095 | Shortly after his wife 's death , Rosebery left his grieving children and went alone on a tour of Spain . Following a visit to El Escorial he wrote on the sepulchral wonders of the building , but added " for the dead alone the Taj is of course supreme . " On his return home he had designed for his wife a Victorian Goth... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 29,325 | text | {} |
mixed-00029006 | By using Newton's identities, we can find
that
$$s_2:=ab+bc+ca=\frac{1}{2}(a+b+c)^2-\frac{1}{2}(a^2+b^2+c^2)=-3$$
and
$$s_3:=abc=\frac{1}{3}(a^3+b^3+c^3)-\frac{1}{3}(a+b+c)^3+(ab+bc+ca)(a+b+c)=1.$$
Then we obtain $(a,b,c)$ by solving the polynomial equation
$$x^3-s_1x^2+s_2x-s_3=x^3-x^2-3x-1=(x+1)(x^2-2x-1)\\=(x+1)... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 16,978 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2296338", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0}} |
normal-00003483 | Bosi completed the deal in June 2007 for a new site at 29 Maddox Street in London . He intended for the new Hibiscus to be open by September , and to transfer over the style of cooking he had used in Ludlow , saying , " I 'm transferring Hibiscus , not starting a new restaurant . The idea is to continue and build on wh... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 8,225 | text | {} |
mixed-00044493 | how many positive dividers that aren't multiple of 2 are there in the number 52920? i need to know how many positive dividers that aren't multiple of 2 are there in the number 52920.
How do i eliminate multiples of 2? | mixed | math-ai/StackMathQA | stackmathqa100k | train | 26,477 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2766723", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
normal-00006992 | The 1955 Atlantic hurricane season was , at the time , the costliest season ever recorded . The hurricane season officially began on June 15 , 1955 , and ended on November 15 , 1955 . It was slightly above average , with 13 recorded tropical cyclones . The first storm , Alice , had persisted since December 30 , 1954 . ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 17,108 | text | {} |
normal-00040801 | The history of vigilante justice and the Montana Vigilantes began in 1863 in what was at the time a remote part of eastern Idaho Territory . Vigilante activities continued , although somewhat sporadically , through the Montana Territorial period until the territory became the state of Montana in 1889 . Vigilantism aros... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 98,918 | text | {} |
normal-00045633 | In addition , there are two non @-@ member observer states of the United Nations General Assembly : the Holy See ( which holds sovereignty over Vatican City ) and the State of Palestine . The Cook Islands and Niue , both states in free association with New Zealand , are full members of several UN specialized agencies a... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 110,748 | text | {} |
normal-00018072 | Beatriz Michelena ( February 22 , 1890 – October 10 , 1942 ) was a Venezuelan American actress during the silent film era , known at the time for her operatic soprano voice and appearances in musical theatre . She was one of the few Latina stars visible on the silver screen in the United States in the 1910s . She was t... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 43,324 | text | {} |
normal-00047596 | A total of 15 @,@ 678 light @-@ duty plug @-@ in electric vehicles were registered in the Netherlands in 2014 , consisting of 12 @,@ 425 plug @-@ in hybrids , down 38 @.@ 4 % from 2013 , 2 @,@ 664 all @-@ electric cars , up 18 @.@ 3 % from a year earlier , and 589 vans , up 236 @.@ 6 % from 2013 . Sales in 2014 were le... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 115,534 | text | {} |
latex-00028473 | \sigma_{AB}(x,\xi)&=u_{\xi}^{-1}(x)(A(Bu_{\xi}))(x) \\&=u_{\xi}^{-1}(x)\int_{\Omega}K_{A}(x,y)\Big[\int_{\Omega} K_{B}(y,z)u_{\xi}(z)dz\Big]dy\\&=u_{\xi}^{-1}(x)\int_{\Omega}K_{A}(x,y)u_{\xi}(y)\sigma_{B}(y,\xi)dy. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 28,735 | latex_formula | {"original_latex": "\\begin{align*}\\sigma_{AB}(x,\\xi)&=u_{\\xi}^{-1}(x)(A(Bu_{\\xi}))(x) \\\\&=u_{\\xi}^{-1}(x)\\int_{\\Omega}K_{A}(x,y)\\Big[\\int_{\\Omega} K_{B}(y,z)u_{\\xi}(z)dz\\Big]dy\\\\&=u_{\\xi}^{-1}(x)\\int_{\\Omega}K_{A}(x,y)u_{\\xi}(y)\\sigma_{B}(y,\\xi)dy.\\end{align*}"} |
normal-00048881 | Freshmen were expected to build the early Bonfires to help prove their worth . For almost two decades , the students constructed Bonfire from debris and wood acquired through various , sometimes illicit , means , including appropriating lumber intended for a dormitory in 1912 . In 1935 , a farmer reported that students... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 118,932 | text | {} |
latex-00049756 | \sup_{B_{s}(R-d)} \|X_{ (\Phi^t_\chi)^\ast F - F }(\psi,\bar\psi)\|_{s} &= \sup_{B_{s}(R-d)} \|X_{ F \circ \Phi^t_\chi - F }(\psi,\bar\psi)\|_{s} \\&\stackrel{\eqref{liebrests}}{\leq} \frac{5}{d} \sup_{B_{s}(R)} \|X_\chi(\psi,\bar\psi)\|_{s} \sup_{B_{s}(R)} \|X_F(\psi,\bar\psi)\|_{s} \\&< 2 \sup_{B_{s}(R)} \|X_F(\psi,\... | latex | OleehyO/latex-formulas | cleaned_formulas | train | 50,452 | latex_formula | {"original_latex": "\\begin{align*}\\sup_{B_{s}(R-d)} \\|X_{ (\\Phi^t_\\chi)^\\ast F - F }(\\psi,\\bar\\psi)\\|_{s} &= \\sup_{B_{s}(R-d)} \\|X_{ F \\circ \\Phi^t_\\chi - F }(\\psi,\\bar\\psi)\\|_{s} \\\\&\\stackrel{\\eqref{liebrests}}{\\leq} \\frac{5}{d} \\, \\sup_{B_{s}(R)} \\|X_\\chi(\\psi,\\bar\\psi)\\|_{s} \\,\\sup... |
latex-00044401 | \tilde{\Omega}^t({\bf e}_j):=\sum_{i=1}^n\tilde{\Omega}_{ji}{\bf e}_i, T^t({\bf e}_j):=\sum_{i=1}^nT_{ji}{\bf e}_i, B_{\infty}({\bf e}_j):=\sum_{i=1}^n(B_{\infty})_{ji}{\bf e}_i | latex | OleehyO/latex-formulas | cleaned_formulas | train | 45,027 | latex_formula | {"original_latex": "\\begin{align*} \\tilde{\\Omega}^t({\\bf e}_j):=\\sum_{i=1}^n\\tilde{\\Omega}_{ji}{\\bf e}_i, T^t({\\bf e}_j):=\\sum_{i=1}^nT_{ji}{\\bf e}_i, B_{\\infty}({\\bf e}_j):=\\sum_{i=1}^n(B_{\\infty})_{ji}{\\bf e}_i\\end{align*}"} |
normal-00016835 | Consonance and dissonance # Dissonance for discussion of the nature and usage of discords in melody and harmony and similar devices in rhythm and metre | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 40,340 | text | {} |
normal-00036670 | To the north of the bunker , the Germans erected a bomb @-@ proof power station with a 2 @,@ 000 horsepower ( 1 @.@ 5 MW ) generating capacity . The site was initially powered from the main electricity grid , but it was intended that it would have its own independent power source to minimise the likelihood of disruptio... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 88,436 | text | {} |
mixed-00011511 | Consider the polynomial $f(x)=x^n-b^n.$ Then $f(b)=b^n-b^n=0.$ So $b$ is a root of $f$ and this implies $(x-b)$ divides $f(x)$. Put $x=a$, then $a-b$ divides $f(a)=a^n-b^n.$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,325 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/188657", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "45", "answer_count": 8, "answer_id": 4}} |
latex-00016090 | \lim_{S\rightarrow\infty}\rho(\mu^{N,M} (S,y), \mu^{N,M}(z)) \leq \phi(S), \lim_{S\rightarrow\infty}{\phi}(S)=0, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 16,128 | latex_formula | {"original_latex": "\\begin{align*}\\lim_{S\\rightarrow\\infty}\\rho(\\mu^{N,M} (S,y), \\mu^{N,M}(z))\\ \\leq \\ \\phi(S), \\ \\ \\ \\ \\ \\ \\lim_{S\\rightarrow\\infty}{\\phi}(S)=0,\\end{align*}"} |
mixed-00001451 | How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 738 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1023259", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0}} |
normal-00028766 | In late @-@ 2007 , Kartik was recalled to the ODI team in place of the struggling Powar mid @-@ way through the series against Australia . He played his first ODI in 18 months when he returned for the fourth match at Mohali . He took 1 / 48 , and conceded only two runs in the 48th over as Australia stumbled in a tight ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 68,957 | text | {} |
mixed-00033974 | HINT: use that we get for the sum $$\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b) (a-c) (b-c)}{(a+b) (a+c) (b+c)}$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 20,006 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2159661", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1}} |
latex-00048919 | \sum_{n=0}^{\infty}\frac{t^{n}a(D)^{n}}{n!}u | latex | OleehyO/latex-formulas | cleaned_formulas | train | 49,610 | latex_formula | {"original_latex": "\\begin{align*}\\sum_{n=0}^{\\infty}\\frac{t^{n}a(D)^{n}}{n!}u\\end{align*}"} |
normal-00040308 | Roleplaying may be seen as part of a movement in Western culture towards participatory arts , as opposed to traditional spectator arts . Participants in a LARP cast off the role of passive observer and take on new roles that are often outside of their daily life and contrary to their culture . The arrangers of a LARP a... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 97,639 | text | {} |
mixed-00024557 | Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:
Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$
Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem.
The $LHS$ is easy t... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,266 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2856589", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0}} |
latex-00020697 | |\eta(x,y)|\le e^{\gamma}(\eta_1(y_0)+\sum_{k=2}^{\infty}\xi_{k}(y_0))R(y) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 20,830 | latex_formula | {"original_latex": "\\begin{align*}|\\eta(x,y)|\\le e^{\\gamma}\\left(\\eta_1(y_0)+\\sum_{k=2}^{\\infty}\\xi_{k}(y_0)\\right)R(y)\\end{align*}"} |
normal-00008157 | Note : Only the top five positions are included for both sets of standings . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 19,987 | text | {} |
normal-00043759 | Migne , Jacques Paul , ed . ( 1865 ) . Nicephori Gregorae , Byzantinae Historiae Libri XXXVII . Patrologia Graeca , vol . 148 . Paris : Garnier . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 106,033 | text | {} |
mixed-00012228 | You shouldn't round, if you can avoid it. Your matrix is
\begin{bmatrix}
29 & 10 \\
10 & 19
\end{bmatrix}
whose characteristic polynomial is
$$
X^2 - 48X + 451
$$
The roots are given by the formula
$$
\frac{48\pm\sqrt{48^2-4\cdot 451}}{2}
$$
so they are $24+5\sqrt{5}$ and $25-5\sqrt{5}$, so you computed correctly. An e... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,758 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/759646", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00015638 | MD 36 passes through the Georges Creek Valley , which has a long history of coal mining . In recognition of this , the MDSHA has designated MD 36 as part of the Coal Heritage Scenic Byway . Coal mining was a major industry in western Maryland in the 19th century , with railroads being the major route connecting the coa... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 37,476 | text | {} |
mixed-00030678 | HINT.- If $p\ne3M+1$ then $p=3M-1$. But then
$$3M-1=X^2+3Y^2\Rightarrow X^2\equiv-1\pmod3$$ which is impossible because $\left(\mathbb Z/3\mathbb Z\right)^2= \{0,1\}$.
COMMENT.-Maybe this could help @Saegusa in his inquiring on $\mathbb Q(\sqrt{-3})$ mainly in $4p=u^2+3v^2$. This problem was conjectured by Fermat and p... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 17,996 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3985550", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
latex-00029832 | P_{\alpha \beta \gamma \delta}= P_{\beta \alpha \gamma \delta}= P_{\alpha \beta \delta \gamma}=P_{\gamma \delta \alpha \beta}, \alpha, \beta, \gamma, \delta=1,2, \hbox{in } \Omega. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 30,113 | latex_formula | {"original_latex": "\\begin{align*} P_{\\alpha \\beta \\gamma \\delta}= P_{\\beta \\alpha \\gamma \\delta}= P_{\\alpha \\beta \\delta \\gamma}=P_{\\gamma \\delta \\alpha \\beta}, \\alpha, \\beta, \\gamma, \\delta=1,2, \\ \\ \\hbox{in } \\Omega.\\end{align*}"} |
latex-00024918 | |H_{n} + \bar{c}B'|_{\delta} = |\bar{A} + \bar{c}B'|_{\delta} \geq \delta^{-\bar{\epsilon}}|\bar{A}| = \delta^{-\bar{\epsilon}}|H_{n}|. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 25,109 | latex_formula | {"original_latex": "\\begin{align*} |H_{n} + \\bar{c}B'|_{\\delta} = |\\bar{A} + \\bar{c}B'|_{\\delta} \\geq \\delta^{-\\bar{\\epsilon}}|\\bar{A}| = \\delta^{-\\bar{\\epsilon}}|H_{n}|. \\end{align*}"} |
mixed-00011840 | Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
With... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,521 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/434765", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 3, "answer_id": 0}} |
normal-00048653 | Americium is produced mostly artificially in small quantities , for research purposes . A tonne of spent nuclear fuel contains about 100 grams of various americium isotopes , mostly 241Am and 243Am . Their prolonged radioactivity is undesirable for the disposal , and therefore americium , together with other long @-@ l... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 118,329 | text | {} |
latex-00023663 | & b_1=(3+t+u)b_2-(1+2t)c_2-ub_3 | latex | OleehyO/latex-formulas | cleaned_formulas | train | 23,833 | latex_formula | {"original_latex": "\\begin{align*}& b_1=(3+t+u)b_2-(1+2t)c_2-ub_3\\end{align*}"} |
normal-00008781 | Upon returning to the United States after the war , Nicholson was placed in reduced commission in November 1919 . She was decommissioned at Philadelphia in May 1922 . She was struck from the Naval Vessel Register in January 1936 sold for scrapping in June . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 21,525 | text | {} |
mixed-00015977 | $A^{10}=(SJS^{-1})^{10}=SJ^{10}S^{-1}$ and it is easy to find $J^{10}$. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 9,036 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4172461", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0}} |
latex-00013228 | . \frac{1}{(2\pi)^2}\sum_{a<b} [X^a,X^b] [X^a,X^b]+{\rm fermions}), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 13,256 | latex_formula | {"original_latex": "\\begin{align*}\\left. \\frac{1}{(2\\pi)^2}\\sum_{a<b} [X^a,X^b] [X^a,X^b]+{\\rm fermions}\\right),\\end{align*}"} |
latex-00000203 | \phi_{i}(\mathcal{Z}(\mathcal{U}( \mathbb{Z}[G]e_{i})))=\mathcal{Z}(\mathcal{U}( \mathbb{Z}[N_{i}]\varepsilon_{i})). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 203 | latex_formula | {"original_latex": "\\begin{align*} \\phi_{i}(\\mathcal{Z}(\\mathcal{U}( \\mathbb{Z}[G]e_{i})))=\\mathcal{Z}(\\mathcal{U}( \\mathbb{Z}[N_{i}]\\varepsilon_{i})).\\end{align*}"} |
latex-00029202 | &Ric_{g_{t}}=d_{t \cdot euc}Ric(\theta)+Q_{euc}(\theta), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,472 | latex_formula | {"original_latex": "\\begin{align*}&Ric_{g_{t}}=d_{t \\cdot euc}Ric(\\theta)+Q_{euc}(\\theta),\\end{align*}"} |
mixed-00045583 | Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$ I want to prove that
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
My approach is the most logical one, rewrite as follows:
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $
But the remaining series are both... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 27,163 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3919534", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1}} |
mixed-00002221 | The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,129 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1544526", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1}} |
latex-00040074 | Z_N = e^{\zeta_N} (I_N(\Lambda) + K_N(\Lambda)). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 40,581 | latex_formula | {"original_latex": "\\begin{align*}Z_N = e^{\\zeta_N} (I_N(\\Lambda) + K_N(\\Lambda)).\\end{align*}"} |
mixed-00023342 | Another way to look at the problem is to rewrite (assuming $x\neq 0$) $$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)\implies 4 x^2 \left(\frac{\left(n-x^2\right)^2}{4 x^2}-\frac{1}{2}
\left(n-x^2\right)\right)=0$$ Expand and simplify to get $$3 x^4-4n x^2+n^2=0$$ Now, using $y=x^2$, the eq... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 13,520 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1804055", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2}} |
latex-00010898 | \int_{U\! \times Y \times Z} \!h(u,y,z) \Phi(p) (du,dy,dz) =\int_{F} \!\tilde h (\mu ,z)p(d\mu,dz) \forall h(\cdot)\in C(U\times Y\times Z), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 10,921 | latex_formula | {"original_latex": "\\begin{align*} \\int_{U\\! \\times Y \\times Z} \\!h(u,y,z) \\Phi(p) (du,dy,dz) =\\int_{F} \\!\\tilde h (\\mu ,z)p(d\\mu,dz)\\ \\ \\ \\ \\ \\forall h(\\cdot)\\in \\ C(U\\times Y\\times Z),\\end{align*}"} |
mixed-00004470 | Continuty at (3,0) : When $|(x,y)-(3,0)|=\sqrt{(x-3)^2+y^2}<\delta<1$ for some $\delta>0$ then $|f(x,y)|\leq 9-x^2-y^2=(3-x)(3+x)\leq 6 (3-x)\leq 6\sqrt{(x-3)^2+y^2}<6\delta.$ This implies that $f$ is continuous at $(3,0).$
Partial differential wrt $y$ : $f(3,y)=0$ for any $y$ and hence $f$ has a partial derivative at ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 2,265 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3197878", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0}} |
latex-00026064 | \rho(x) := \frac{\sup \mathsf{L}(x)}{\min \mathsf{L}(x)}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 26,267 | latex_formula | {"original_latex": "\\begin{align*}\\rho(x) := \\frac{\\sup \\mathsf{L}(x)}{\\min \\mathsf{L}(x)}.\\end{align*}"} |
mixed-00038670 | Notice that: $$x + \frac{1}{x} = 1 \;\implies\; x^2 - x + 1 = 0 \;\implies\; x^3 + 1 = (x+1)(x^2-x+1) = 0 \;\implies\; x^3=-1$$
Then $x^6=1$ so: $$x^5 + \frac{1}{x^5} = \frac{x^6}{x} + \frac{x}{x^6} = \frac{1}{x} + x = 1$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,902 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1976629", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 5, "answer_id": 0}} |
mixed-00010169 | Show that $\sum_{j=1}^{n-1} M_j(M_{j+1} - M_j) = \frac{1}{2}\left( \sum_{j=1}^n X_j \right)^2 - \frac{n}{2}$ The following question is taken from Steven Shreve Stochastic Calculus for Finance Volume 1, question $2.5$
Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2},$ as is the probab... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,520 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3406742", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0}} |
latex-00002260 | L'=\sum_{n>0}(\frac12|\partial\phi|^2-\frac{M^2n^2}{2}|\phi_n |^2-2\lambda \phi_n \sum_{n>0} |\phi_n|^2 ) + \frac{\lambda}{3!} \sum_{n,n'}{}' \phi_n \phi_{n'} \phi_{n+n'}^*, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 2,261 | latex_formula | {"original_latex": "\\begin{align*} L'=\\sum_{n>0}\\left(\\frac12|\\partial\\phi|^2-\\frac{M^2n^2}{2}|\\phi_n |^2-2\\lambda \\phi_n \\sum_{n>0} |\\phi_n|^2 \\right) + \\frac{\\lambda}{3!} \\sum_{n,n'}{}' \\phi_n \\phi_{n'} \\phi_{n+n'}^*,\\end{align*}"} |
normal-00007681 | Game Music Festival ~ Super Live ' 92 ~ ( 1992 ) – with Alph Lyla | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 18,821 | text | {} |
mixed-00024243 | $${ \left( \cos { x } +i\sin { x } \right) }^{ 3 }=\cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \cos ^{ 3 }{ x } +3i\cos ^{ 2 }{ x\sin { x } -3\cos { x\sin ^{ 2 }{ x } -i\sin ^{ 3 }{ x } = } } \cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \\ \cos { \left( 3x \right) =\cos ^{ 3 }{ x } -3\cos {... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,076 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2532231", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
normal-00006264 | Soon after Henry 's coronation Gerard was appointed to the recently vacant see of York , and became embroiled in the long @-@ running dispute between York and the see of Canterbury concerning which archbishopric had primacy over England . Gerard managed to secure papal recognition of York 's claim to jurisdiction over ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 15,381 | text | {} |
latex-00005019 | V(\phi =2\pi )V^{-1}(\phi =0)=-1. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,024 | latex_formula | {"original_latex": "\\begin{align*}V(\\phi =2\\pi )V^{-1}(\\phi =0)=-1.\\end{align*}"} |
mixed-00013172 | Line tangent to circle
A circle with a radius of $2$ units has its center at $(0,0)$. A circle with a radius of $7$ units has its center at $(15,0)$. A line tangent to both circles intersects the $x$-axis at $(x,0)$. What is the value of $x$? Express your answer as a common fraction.
My problem with this question is ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 7,329 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1584672", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 1}} |
normal-00040485 | Title TK came out on May 20 – 21 , 2002 , and the " Son of Three " single on September 2 of that year . The single reached No. 72 on the UK Singles Chart . The album and single versions of " Son of Three " have mostly been well received by critics . AllMusic 's Heather Phares calls the Title TK track " sweet [ and ] pl... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 98,103 | text | {} |
latex-00030443 | (u+1)z^2\sum_{k\in \mathbb{Z}/d\mathbb{Z}} x_k + (u+2)z\sum_{k\in \mathbb{Z}/d\mathbb{Z}} E^{(k)}+ \sum_{k \in \mathbb{Z}/d\mathbb{Z}}{\rm tr}(e_1^{(k)} e_2)=0. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 30,731 | latex_formula | {"original_latex": "\\begin{align*} (u+1)z^2\\sum_{k\\in \\mathbb{Z}/d\\mathbb{Z}} x_k + (u+2)z\\sum_{k\\in \\mathbb{Z}/d\\mathbb{Z}} E^{(k)}+ \\sum_{k \\in \\mathbb{Z}/d\\mathbb{Z}}{\\rm tr}(e_1^{(k)} e_2)=0. \\end{align*}"} |
latex-00045670 | \tilde\nabla_xq_a^{\varepsilon}(t,x,\xi,y)=\int_0^1 \nabla_xq_a^{\varepsilon}(t,x+\theta(y-x),\xi)d\theta. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 46,315 | latex_formula | {"original_latex": "\\begin{align*}\\tilde\\nabla_xq_a^{\\varepsilon}(t,x,\\xi,y)=\\int_0^1 \\nabla_xq_a^{\\varepsilon}(t,x+\\theta(y-x),\\xi)d\\theta.\\end{align*}"} |
mixed-00019506 | The inequality is homogenous in $(x, y, z)$, therefore we can assume
that $xyz=1$. Then
$$
\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \ge \dfrac{3}{2}
$$
as for example demonstrated in
If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ or A inequality proposed at Zhauty... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 11,182 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2723798", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0}} |
mixed-00014736 | Evaluating $\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$ I've got one integration question which I first felt was not a hard nut to crack. But, as I proceeded, difficulties arose. This is the one:
$\displaystyle\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$
I went ahead simplifying the... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 8,283 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2857711", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 2, "answer_id": 1}} |
latex-00000015 | (\begin{array}{cc} \mu_1 & 0 \ 0 & \mu_2 \ \end{array} )(\begin{array}{cc} 0 & d_1 \ d_2 & 0 \ \end{array} )(\begin{array}{cc}\bar{\mu_1}^{-1} & 0 \\0 & \bar{\mu_2}^{-1} \\\end{array}) =(\begin{array}{cc} 0 & 1 \ 1 & 0 \ \end{array} ). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 15 | latex_formula | {"original_latex": "\\begin{align*}\\left(\\begin{array}{cc} \\mu_1 & 0 \\\\ 0 & \\mu_2 \\\\ \\end{array} \\right)\\left(\\begin{array}{cc} 0 & d_1 \\\\ d_2 & 0 \\\\ \\end{array} \\right)\\left(\\begin{array}{cc}\\bar{\\mu_1}^{-1} & 0 \\\\0 & \\bar{\\mu_2}^{-1} \\\\\\end{array}\\right) =\\left(\\begin{array}{cc} 0 & 1 ... |
normal-00044250 | Beginning in the 1950s , Russian paleontologists began excavating Psittacosaurus remains at a locality near the village of Shestakovo in the oblast of Kemerovo in Siberia . Two other nearby localities were explored in the 1990s , one of which produced several complete skeletons . This species was named P. sibiricus in ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 107,150 | text | {} |
normal-00019151 | 7 @.@ 0 . The most distant observed gamma ray burst with a spectroscopic redshift measurement was GRB 090423 , which had a redshift of z | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 45,931 | text | {} |
mixed-00020860 | Finding all polynomials $P(x)$ satisfying $(P(x)+P(\frac{1}{x}))^2 =P(x^2)P(\frac{1}{x^2})$
Find all polynomials $P(x)$ satisfying
$$\left(P(x)+P\left(\frac{1}{x}\right)\right)^2 =P(x^2)P\left(\frac{1}{x^2}\right)$$
If $P'(x)\neq 0$, then $P(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$
$\Rightarrow$ ($a_nx^n+a_{n-1}x^{n-... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 12,016 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4193950", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 2, "answer_id": 0}} |
mixed-00039632 | Facing difficulty in working $\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}$ I would like to evaluate this integral,$$\int_{0}^{1}\frac{\arctan\left(\frac{ax}{1-x}\right)}{\sqrt{1-x}}\frac{dx}{x^{3/2}}\tag1$$
This is the approach I will take:
We can begin with a sub: $y=\sqrt{x}$, ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 23,498 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2834540", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 1, "answer_id": 0}} |
mixed-00013371 | A calculus approach. Let $$f(x) = \frac{x}{2}+\frac{1}{x}.$$ So $$f(x)-x = -\frac{x}{2}+\frac{1}{x}$$ is the sum of two decreasing functions on $]0,+\infty]$. Hence $f(p_{n})-p_n<0$, the sequence is decreasing.
$f(x)$ is strictly increasing on $]\sqrt{2},2]$ (easy with the derivative), thus if $x \in ]\sqrt{2},2] $, $f... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 7,449 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1710469", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3}} |
mixed-00027292 | May be not the answer you wanted given that it is of degree 8. But it has integer coefficients, so may be of interest.
If $R_n(x)=T_n(\sqrt{1-x^2})$, where $T_n$ is the Chebyshev polynomial of degree $n$, then
$$
T_n(\sin t)=\cos nt
$$
for all $t$. Because $\cos \alpha=0$, iff $\alpha$ is an odd multiple of $\pi/2$, th... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 15,937 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/905303", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 5, "answer_id": 0}} |
latex-00023970 | \lambda_{m}=\frac{\langle g,Mg \rangle}{\langle g,g \rangle}= \lambda _{k}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 24,141 | latex_formula | {"original_latex": "\\begin{align*}\\lambda_{m}=\\frac{\\langle g,Mg \\rangle}{\\langle g,g \\rangle}= \\lambda _{k}.\\end{align*}"} |
latex-00012703 | [-\frac{d^2}{dx^2}+m^2]\psi_k(x)=(\omega_k^0)^2\psi_k(x) , | latex | OleehyO/latex-formulas | cleaned_formulas | train | 12,730 | latex_formula | {"original_latex": "\\begin{align*}\\left[-\\frac{d^2}{dx^2}+m^2\\right]\\psi_k(x)=(\\omega_k^0)^2\\psi_k(x)\\;,\\end{align*}"} |
mixed-00031128 | We can considerably simplify the residue calculation when expanding numerator and denominator with $z+1$. This way we can effectively get rid of the denominator. We consider
\begin{align*}
f(z)=\frac{(z+1)\left(z^2-2cz+1\right)}{z^5+1}
\end{align*}
and obtain
\begin{align*}
\mathrm{Res}&(f,a)
=\displaystyle\lim_{z\to ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 18,272 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4489465", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0}} |
latex-00036428 | \varrho(v)=\exp\{ -v\log v-v\log\log (v+2)+v(1+O(\frac{\log\log(v+2)}{\log(v+2)})) \} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 36,797 | latex_formula | {"original_latex": "\\begin{align*} \\varrho(v)=\\exp\\left\\{ -v\\log v-v\\log\\log (v+2)+v\\left(1+O\\left(\\frac{\\log\\log(v+2)}{\\log(v+2)}\\right)\\right) \\right\\} \\end{align*}"} |
mixed-00000095 | Let the induction hypothesis be
$$ (1^3+2^3+3^3+\cdots+n^3)=(1+2+3+\cdots+n)^2$$
Now consider:
$$ (1+2+3+\cdots+n + (n+1))^2 $$
$$\begin{align}
& = \color{red}{(1+2+3+\cdots+n)^2} + (n+1)^2 + 2(n+1)\color{blue}{(1+2+3+\cdots+n)}\\
& = \color{red}{(1^3+2^3+3^3+\cdots+n^3)} + (n+1)^2 + 2(n+1)\color{blue}{(n(n+1)/2)}\\
... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 49 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/62171", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "67", "answer_count": 16, "answer_id": 2}} |
mixed-00037616 | Write $n=abc=100a+10b+c.$ Then, $S(n)=a+b+c.$ Note that $S(n)$ has two digits. Then,
if $a+b+c<10$ then $a+b+c=2$ (since $S(S(n))=a+b+c);$
if $10\le a+b+c<20$ then $a+b+c=11$ (since $S(S(n))=a+b+c-9);$
if $20\le a+b+c$ then $a+b+c=20$ (since $S(S(n))=a+b+c-18).$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,237 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1055620", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 5, "answer_id": 0}} |
mixed-00031762 | Confirm the meaning of Prime and Primitive in a Galois(2) polynomial. Here it discusses primality (or more accurately irreducibility) and primitivity of polynomials in $G(2)$. More specifically it states that $x^6 + x + 1$ is irreducible and primitive.
But here I can divide $x^7 + 1$ by $x^6 + x + 1$ and get $x$ remain... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 18,656 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/422096", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
mixed-00020207 | Homogenizing the equation, we WTS
$$\frac{(a+b+c)^3}{2} -(a+b+c)(a^2+b^2+c^2)-4abc > 0$$
Expanding and factoring, we obtain
$$(a+b-c)(a-b+c)(-a+b+c) > 0$$
This is true via the triangle inequality. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 11,617 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3468328", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 0}} |
latex-00047093 | \lambda^{\zeta}_x=\lim_{r\to0+}\frac{\lambda|_{B^T_V(x,r)}}{\lambda(B^T_V(x,r))} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 47,748 | latex_formula | {"original_latex": "\\begin{align*}\\lambda^{\\zeta}_x=\\lim_{r\\to0+}\\frac{\\lambda|_{B^T_V(x,r)}}{\\lambda(B^T_V(x,r))}\\end{align*}"} |
mixed-00049427 | Finding $\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ I am having some diffuculty finding the limit for this expression and would appreciate if anyone could give a hint, as to how to continue. I know the limit must be $e^{14}$ (trough an engine) and I can show it for
$(1+\frac{2}{x}+\frac{1}{x^2})^{7x}$ like ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 29,482 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2681037", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 5, "answer_id": 2}} |
mixed-00040351 | Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 23,939 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3552078", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 0}} |
mixed-00024484 | A functional equation problem: $ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $
Let $ \mathbb R ^ + $ denote the set of the positive real numbers. Find all functions $ f : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $$... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,222 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2776256", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0}} |
latex-00030033 | f(x,y)=\cos(x+y)\cdot(x^2+y^2)^\frac{1}{4} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 30,315 | latex_formula | {"original_latex": "\\begin{align*}f(x,y)=\\cos(x+y)\\cdot(x^2+y^2)^\\frac{1}{4}\\end{align*}"} |
latex-00037679 | E(\Theta,\chi,s):=\int_{G_n}\chi(\det(g))|\det(g)|^{s-1/2} d\mu_{\Theta}(g) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 38,163 | latex_formula | {"original_latex": "\\begin{align*}E(\\Theta,\\chi,s):=\\int_{G_n}\\chi(\\det(g))\\left|\\det(g)\\right|^{s-1/2}\\ d\\mu_{\\Theta}(g)\\end{align*}"} |
normal-00018855 | The Ripley 's Believe It or Not museum In Key West , Florida , has an exhibit recreating Elena 's body being cared for by Tanzler . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 45,265 | text | {} |
mixed-00016744 | The matrix is positive definite if and only if their eigenvalues are positive, so we calculate the characteristic polynomial $\chi_A(x)=\det(xI-A)$ and we solve for $x$ and we find:
$$\lambda_1=\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}-\frac{1}{2}\alpha+3,\quad\lambda_2=3-\frac{1}{2}\sqrt{\alpha^2+4\alpha+12}- \frac{1}{2}\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 9,512 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/353827", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1}} |
latex-00004414 | \lbrack M_{1}\sigma,M_{2}\sigma]=0 . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 4,417 | latex_formula | {"original_latex": "\\begin{align*}\\lbrack M_{1}\\sigma,M_{2}\\sigma]=0\\,\\,. \\end{align*}"} |
latex-00044799 | D_s(P\|Q) = D(P\|Q) + \frac{s-1}{2 \log e} V(P\|Q) + O(s^2) . | latex | OleehyO/latex-formulas | cleaned_formulas | train | 45,434 | latex_formula | {"original_latex": "\\begin{align*} D_s(P\\|Q) = D(P\\|Q) + \\frac{s-1}{2 \\log e} V(P\\|Q) + O(s^2) \\,. \\end{align*}"} |
latex-00039475 | R &=\frac{1}{NB}[|\mathcal{M}_2|\times|\mathcal{A}| +|\mathcal{M}_1|\times|\mathcal{A}^{*c}|+|\mathcal{M}_3|\times(|\mathcal{A}|-|\mathcal{A}^*|)]\ &=[H(p_1^*)-H(p_1)]\times q_1 +[H(p_2^*)-H(p_2)]\times q_2^*+[H(p_2)-H(p_1^*)]\times(q_1- q_1^*). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 39,978 | latex_formula | {"original_latex": "\\begin{align*}R &=\\frac{1}{NB}\\left[|\\mathcal{M}_2|\\times|\\mathcal{A}| +|\\mathcal{M}_1|\\times|\\mathcal{A}^{*c}|+|\\mathcal{M}_3|\\times(|\\mathcal{A}|-|\\mathcal{A}^*|)\\right]\\\\ &=[H(p_1^*)-H(p_1)]\\times q_1 +[H(p_2^*)-H(p_2)]\\times q_2^*+[H(p_2)-H(p_1^*)]\\times(q_1- q_1^*).\\end{alig... |
latex-00028831 | x_{i}^{H}[\lambda_{0}I_{m_{{\rm\bf x}i}}\!\!-\!\!A_{\rm\bf TT}(i) B_{\rm\bf T}(i) A_{\rm\bf TS}(i)]=0 | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,100 | latex_formula | {"original_latex": "\\begin{align*}x_{i}^{H}\\left[\\lambda_{0}I_{m_{{\\rm\\bf x}i}}\\!\\!-\\!\\!A_{\\rm\\bf TT}(i)\\;\\;B_{\\rm\\bf T}(i)\\;\\; A_{\\rm\\bf TS}(i)\\right]=0\\end{align*}"} |
normal-00037698 | NpO2 + 1 ⁄ 2 H2 + 3 HF → NpF3 + 2 H2O ( 400 ° C ) | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 91,114 | text | {} |
latex-00036941 | \lambda^{2} = \frac{\Delta t^{2}}{\Delta x^{3}} | latex | OleehyO/latex-formulas | cleaned_formulas | train | 37,419 | latex_formula | {"original_latex": "\\begin{align*}\\lambda^{2} = \\frac{\\Delta t^{2}}{\\Delta x^{3}}\\end{align*}"} |
latex-00021153 | \frac{\omega^2}{r^2} \frac{\partial z}{\partial \phi} (\phi,\omega,r;t) = \frac{\partial}{\partial s_1} F(0, p(\phi+\omega t))\cdot p'(\phi) + \frac{\partial}{\partial s_2} F(0,p(\phi+\omega t))\cdot p'(\phi+ \omega t). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 21,296 | latex_formula | {"original_latex": "\\begin{align*} \\frac{\\omega^2}{r^2}\\,\\frac{\\partial z}{\\partial \\phi} (\\phi,\\omega,r;t) = \\frac{\\partial}{\\partial s_1} F(0, p(\\phi+\\omega t))\\cdot p'(\\phi) + \\frac{\\partial}{\\partial s_2} F(0,p(\\phi+\\omega t))\\cdot p'(\\phi+ \\omega t). \\end{align*}"} |
normal-00005106 | The Guitar Hero series has made a significant cultural impact , becoming a " cultural phenomenon " . The series has helped to rekindle music education in children , influenced changes in both the video game and music industry , has found use in health and treatment of recovering patients , and has become part of the po... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 12,245 | text | {} |
normal-00025664 | Between the death of Vinculus and the gentleman with the thistledown hair , Childermass notes that tattooed upon Vinculus 's body is the last works of John Uskglass . As he tries to preserve the tattoos in memory , a man appears . Between calling Childermass his servant ( leading him to believe it is Norrell in disguis... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 61,543 | text | {} |
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