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-00005837 | \{ \Gamma_+^{-1} , C_3\} = C + \frac{1}{2} \frac{\alpha_1 \alpha_2}{1 + \mu \alpha k} C U_3^{-1} Y Y^{\rm T} CU_3^{-1}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,845 | latex_formula | {"original_latex": "\\begin{align*}\\{ \\Gamma_+^{-1} , C_3\\} = C + \\frac{1}{2} \\frac{\\alpha_1 \\alpha_2}{1 + \\mu \\alpha k} C U_3^{-1} Y Y^{\\rm T} CU_3^{-1}.\\end{align*}"} |
mixed-00046915 | After using dark magic to make an initial guess $y_0$, the fast inverse square root algoritm essentially approximates $y=1/\sqrt{x}$, $x > 0$, by using the iteration
$$
y_{n+1} = \frac{3}{2}y_n - \frac{1}{2}xy_n^3.
$$
There is no division here, unless you count the multiplication by the constants $3/2$ and $1/2$.
If th... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 27,972 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/550177", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0}} |
normal-00021319 | In 756 , Óengus is found campaigning alongside Eadberht of Northumbria . The campaign is reported as follows : | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 51,031 | text | {} |
normal-00000022 | Perhaps because Abraham Lincoln had not yet been inaugurated as President , Captain Totten received no instructions from his superiors and was forced to withdraw his troops . He agreed to surrender the arsenal as long as the governor agreed to three provisions : | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 67 | text | {} |
latex-00037550 | \partial_1^2\Phi(x,\lambda)+\frac{A'(x)}{A(x)} \partial_1\Phi(x,\lambda)=\lambda \Phi(x,\lambda) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 38,029 | latex_formula | {"original_latex": "\\begin{align*}\\partial_1^2\\Phi(x,\\lambda)+\\frac{A'(x)}{A(x)} \\,\\partial_1\\Phi(x,\\lambda)=\\lambda\\,\\Phi(x,\\lambda)\\end{align*}"} |
latex-00020254 | Y_{n+1}=(A\otimes I)Y_n+h\Phi(t_n,Y_n,h)\quad(n\in\mathbb{N}), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 20,359 | latex_formula | {"original_latex": "\\begin{align*} Y_{n+1}=(A\\otimes I)Y_n+h\\Phi(t_n,Y_n,h)\\quad(n\\in\\mathbb{N}),\\end{align*}"} |
latex-00043084 | G_N(\mathbf{x}) = \sum_{\substack{u \in \mathcal{U}_N \ \ell < \delta_4^{dN}}} c_{u,\ell} \cdot u \Pi_N^{\ell}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 43,697 | latex_formula | {"original_latex": "\\begin{align*} G_N(\\mathbf{x}) = \\sum_{\\substack{u \\in \\mathcal{U}_N \\\\ \\ell < \\delta_4^{dN}}} c_{u,\\ell} \\cdot u \\Pi_N^{\\ell},\\end{align*}"} |
mixed-00041244 | Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 24,489 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4579291", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 3, "answer_id": 0}} |
normal-00003172 | Food and cuisine in Ireland takes its influence from the crops grown and animals farmed in the island 's temperate climate and from the social and political circumstances of Irish history . For example , whilst from the Middle Ages until the arrival of the potato in the 16th century the dominant feature of the Irish ec... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 7,553 | text | {} |
latex-00022362 | a=g_1g_2\cdots g_s | latex | OleehyO/latex-formulas | cleaned_formulas | train | 22,522 | latex_formula | {"original_latex": "\\begin{align*}a=g_1g_2\\cdots g_s\\end{align*}"} |
latex-00014082 | V:=& \{v_0\}\cup \bigcup_{i=1}^k (V_i\backslash \{v_0^i\}) \\E:=& \bigcup_{i=1}^k ( \{ab: ab\in E_i, a,b\neq v^i_0\} \cup \{v_0b: v_0^ib\in E_i \}\cup \{av_0: av_0^i\in E_i \} ). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 14,110 | latex_formula | {"original_latex": "\\begin{align*}V:=& \\left\\{v_0\\right\\}\\cup \\bigcup_{i=1}^k \\left(V_i\\backslash \\left\\{v_0^i\\right\\}\\right) \\\\E:=& \\bigcup_{i=1}^k \\left(\\, \\left\\{ab:\\ ab\\in E_i, a,b\\neq v^i_0\\right\\} \\cup \\left\\{v_0b: v_0^ib\\in E_i \\right\\}\\cup \\left\\{av_0: av_0^i\\in E_i \\right\\... |
mixed-00048573 | Simplifying a rational function with infinite series in numerator and denominator We're working with Taylor Series and I have to simplify the rational expression
$$ \frac{x-2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2\frac{x^6}{6!} + \cdots}{x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}.$$
I'm ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 28,966 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1913795", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0}} |
normal-00004334 | Following General of the Army Douglas MacArthur 's dismissal as Commander @-@ in @-@ Chief of UN forces in Korea , he was replaced by General Matthew B. Ridgway . Consequently , on 14 April 1951 , General James Van Fleet replaced Ridgway as commander of the US Eighth Army and the United Nations forces in Korea . The Ch... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 10,268 | text | {} |
latex-00040641 | i^{\bullet}_B([c])=[\rho] | latex | OleehyO/latex-formulas | cleaned_formulas | train | 41,153 | latex_formula | {"original_latex": "\\begin{align*} i^{\\bullet}_B([c])=[\\rho] \\end{align*}"} |
mixed-00034905 | Evaluate the indefinite integral $\int {(1-x^2)^{-3/2}}dx$ Evaluate the indefinite integral
$$\int \frac {dx}{(1-x^2)^{3/2}} .$$
My answer: I have taken $u = 1-x^2$ but I arrived at the integral of $\frac{-1}{(u^3 - u^4)^{1/2}}$. What should I do next? | mixed | math-ai/StackMathQA | stackmathqa100k | train | 20,577 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2999549", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0}} |
latex-00028368 | \widetilde{L}=(L+1)/2=[1,1,0,0,1,1,0,1,1] | latex | OleehyO/latex-formulas | cleaned_formulas | train | 28,630 | latex_formula | {"original_latex": "\\begin{align*}\\widetilde{L}=(L+1)/2=[1,1,0,0,1,1,0,1,1]\\end{align*}"} |
mixed-00043954 | Let $h(x)=x^{-1}$. Then
\begin{align}
f_Y(y)&=f_X(x)(h^{-1}(y))\times\left\lvert \frac{dh^{-1}(y)}{dy}\right\rvert \\
&=\frac{1}{\pi}\times\frac{1}{1+(1/y-\theta)^2}\times\frac{1}{y^2} \\
&=\frac{1}{\pi}\times\frac{1}{\gamma\left(1+[\gamma^{-1}(y-\tau)]^2\right)},
\end{align}
where $\gamma\equiv (1+\theta^2)^{-1}$ and ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 26,147 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2278050", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
latex-00022349 | c_x=\prod_{p\le\exp(\log^\beta x)}(1-p^{-1}). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 22,508 | latex_formula | {"original_latex": "\\begin{align*}c_x=\\prod_{p\\le\\exp(\\log^\\beta x)}(1-p^{-1}).\\end{align*}"} |
mixed-00015515 | How to derive the condition? (I am not sure it is correct.) Given a polynomial with real coefficients $\alpha x^2 + \beta x + a^2 + b^2 + c^2 - ab- bc - ca $ has imaginary roots, how do we prove $ 2(\alpha - \beta) + ((a - b)^2 + (b - c)^2 + (c - a)^2) > 0 $?
Given all coefficients are real, roots are $ z $ and $\overl... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 8,758 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3673235", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00044902 | " R & B Junkie " received positive reviews from music critics , who described it as " infectious " and one of the best tracks on Damita Jo . The song peaked at number one on the Bubbling Under R & B / Hip @-@ Hop Singles , as it received a limited release . " R & B Junkie " was performed by Jackson during the 2004 BET ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 108,969 | text | {} |
latex-00040358 | \mathbf{N}_E:=\mathbb{Z}^E/ \langle\mathbf{e}_E\rangle, \mathbf{M}_E:=\mathbf{e}_E^\perp \subset \mathbb{Z}^E, \langle -,- \rangle:\mathbf{N}_E \times \mathbf{M}_E \longrightarrow \mathbb{Z}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 40,865 | latex_formula | {"original_latex": "\\begin{align*}\\mathbf{N}_E:=\\mathbb{Z}^E/ \\langle\\mathbf{e}_E\\rangle, \\mathbf{M}_E:=\\mathbf{e}_E^\\perp \\subset \\mathbb{Z}^E, \\langle -,- \\rangle:\\mathbf{N}_E \\times \\mathbf{M}_E \\longrightarrow \\mathbb{Z}.\\end{align*}"} |
mixed-00012105 | The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares. The sum of areas of 2 squares is 400and the difference between their perimeters is 16cm. Find the sides of both squares.
I HAVE TRIED IT AS BELOW BUT ANSWER IS NOT CORRECT.......CHECK - HELP!
Let sid... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,683 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/639271", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2}} |
normal-00028539 | Credits are taken from Pink Friday liner notes . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 68,340 | text | {} |
mixed-00025173 | Example 1:
When trying to find the area between curves $f(x)$ and $g(x)$ you can achieve this by integrating the function $H(x) = |f(x) - g(x)|$. However integration of an absolute value function is piecewise. In this case though $f(x) \ge g(x)$ for $-1\le x \le 2$ so integrating $\int_{-1}^2 (f(x) - g(x)) dx$ is vali... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,644 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3512517", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1}} |
mixed-00026242 | Suppose $xy\mid x^2+y^2+1$ and let $t=\displaystyle\frac{x^2+y^2+1}{xy}$ such that $t\in\mathbb{N}$.
Construct the set,
$$S=\left\{(x,y)\in\mathbb{N}\times\mathbb{N} : \frac{x^2+y^2+1}{xy}=t\in\mathbb{N}\right\}$$
We deduce that $\displaystyle\frac{x^2+y^2+1}{xy} \ge 3$ because $\displaystyle\frac{x^2+y^2+1}{xy}<3$ imp... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 15,299 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/115272", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "19", "answer_count": 5, "answer_id": 0}} |
latex-00020767 | \begin{aligned}([1, g],1) ([\epsilon, 1], 1)&=([\epsilon, g^{s(\epsilon)}], \widetilde{C}_{X^{\ast}}([1, g], [\epsilon, 1]))\\&=([\epsilon, g^{s(\epsilon)}], \nu( \epsilon,g)\widetilde{c}_{X^{\ast}}(g^{s(\epsilon)}, 1))\\&=([\epsilon, g^{s(\epsilon)}], \nu( \epsilon,g))\\&=([\epsilon, 1], 1)([1, g^{s(\epsilon)}],\nu( \... | latex | OleehyO/latex-formulas | cleaned_formulas | train | 20,904 | latex_formula | {"original_latex": "\\begin{align*}\\begin{aligned}([1, g],1) ([\\epsilon, 1], 1)&=([\\epsilon, g^{s(\\epsilon)}], \\widetilde{C}_{X^{\\ast}}([1, g], [\\epsilon, 1]))\\\\&=([\\epsilon, g^{s(\\epsilon)}], \\nu( \\epsilon,g)\\widetilde{c}_{X^{\\ast}}(g^{s(\\epsilon)}, 1))\\\\&=([\\epsilon, g^{s(\\epsilon)}], \\nu( \\epsi... |
normal-00049203 | In 2000 , the Hardy Boyz found a new manager in their real @-@ life friend Lita . Together , the three became known as " Team Xtreme " . They continued their feud with Edge and Christian throughout 2000 , defeating them for the WWF Tag Team Championship on two occasions . At SummerSlam , the Hardy Boyz competed in the ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 119,790 | text | {} |
mixed-00033483 | Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a p... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 19,704 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1775572", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "124", "answer_count": 8, "answer_id": 2}} |
latex-00026679 | E_{11}^T=-E_{11},E_{22}^T=-E_{22},0=\dot E_{11},0=A_{12}+\dot E_{12},A_{22}^T=A_{22}+\dot E_{22}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 26,910 | latex_formula | {"original_latex": "\\begin{align*}E_{11}^T=-E_{11},E_{22}^T=-E_{22},0=\\dot E_{11},0=A_{12}+\\dot E_{12},A_{22}^T=A_{22}+\\dot E_{22}.\\end{align*}"} |
latex-00008435 | {\cal D}_rg_{\ell}(r,x)=-\frac{2}{\pi}\int_0^{\infty}j_{\ell}({\alpha}r)j_{\ell}({\alpha}x){\alpha}^2d{\alpha}=-\frac{1}{r^2}{\delta}(r-x), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 8,445 | latex_formula | {"original_latex": "\\begin{align*}{\\cal D}_rg_{\\ell}(r,x)=-\\frac{2}{\\pi}\\int_0^{\\infty}j_{\\ell}({\\alpha}r)j_{\\ell}({\\alpha}x){\\alpha}^2d{\\alpha}=-\\frac{1}{r^2}{\\delta}(r-x),\\end{align*}"} |
latex-00041245 | \xi'_\alpha:=-J_{\alpha}N | latex | OleehyO/latex-formulas | cleaned_formulas | train | 41,842 | latex_formula | {"original_latex": "\\begin{align*}\\xi'_\\alpha:=-J_{\\alpha}N\\end{align*}"} |
normal-00007002 | A weak disturbance was first observed near Grand Cayman on August 23 , gaining tropical storm strength by 0600 UTC that day . Moving towards the north @-@ northwest , the storm passed over western Cuba on August 24 , without much change in intensity . Once in the Gulf of Mexico , the tropical storm marginally intensifi... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 17,145 | text | {} |
latex-00029320 | \theta=t^{2}H+t^{2}u\sigma+tO(t^{\frac{3}{2}}\rho^{3}) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 29,595 | latex_formula | {"original_latex": "\\begin{align*}\\theta=t^{2}H+t^{2}u\\sigma+tO(t^{\\frac{3}{2}}\\rho^{3})\\end{align*}"} |
mixed-00043311 | $$I=\int \frac{4t}{1-t^4}dt$$
Let $t=\sqrt{\sin\theta}$. Then, $dt=\frac{1}{2\sqrt{\sin\theta}}\cos\theta d\theta$.
Then,
$$I=\int \frac{4\sqrt{\sin\theta}\cos\theta d\theta}{2\sqrt{\sin\theta}\cos^2\theta}=2\int\sec\theta d\theta=2\ln|\sec\theta+\tan\theta|+c=2\ln\left|\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,756 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1690210", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 8, "answer_id": 5}} |
latex-00031977 | \sum_{ h=-\infty }^{\infty} T_h =\sum_{h=-\infty}^{\infty}\gamma(h)\gamma(h-p+q)+\gamma(h+q)\gamma(h-p). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 32,285 | latex_formula | {"original_latex": "\\begin{align*}\\sum_{ h=-\\infty }^{\\infty} T_h =\\sum_{h=-\\infty}^{\\infty}\\gamma(h)\\gamma(h-p+q)+\\gamma(h+q)\\gamma(h-p).\\end{align*}"} |
latex-00025710 | F(p) = \frac{1}{2}||p||^2 + \sum_{i = 1}^{n}\frac{m_i}{\vert \vert p - x_i \vert \vert}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 25,906 | latex_formula | {"original_latex": "\\begin{align*}F(p) = \\frac{1}{2}||p||^2 + \\sum_{i = 1}^{n}\\frac{m_i}{\\vert \\vert p - x_i \\vert \\vert}.\\end{align*}"} |
mixed-00002500 | $$\sum_{i=1}^\infty \frac{2^n + 3^n}{6^n}=\sum_{i=1}^\infty (\frac{2}{6})^n + \sum_{i=1}^\infty (\frac{3}{6})^n=\frac{1}{2}+1=\frac{3}{2}$$
sums from forumla of geometric series $1/3^n$ and $1/2^n$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,270 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1712458", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
mixed-00042835 | I have differentiated your answer and the book's answer. They are both equal to the integrand. So your work is indeed correct.
In fact, the book's answer can be considered wrong since it lacks an absolute value around the argument to $ln$, whereas your answer does not need one there since the argument to $ln$ in your e... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,461 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1311398", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "14", "answer_count": 3, "answer_id": 2}} |
mixed-00045414 | Just for the fun of it !
The problem of the convergence being solved, there are analytical solution for this kind of integrals (and antiderivatives; have a look here.
Since @Von Neumann wrote an answer where complex numbers do appear, I wondered what would give the $1,400$ years old approximation
$$\sin(x) \simeq \frac... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 27,054 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3726225", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 5, "answer_id": 0}} |
normal-00008265 | As Chen was born in a culture other than the culture he now lives in , he tries to make games that appeal universally to all people . His goal with his games is to help video games mature as a medium by making games that inspire emotional responses in the player that other games are lacking . Although he and Thatgameco... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 20,236 | text | {} |
latex-00010519 | A^{(n)} C=C G^{(n)}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 10,541 | latex_formula | {"original_latex": "\\begin{align*}A^{(n)} C=C G^{(n)}. \\end{align*}"} |
latex-00040228 | \sigma_+(\xi)\tilde w_+(\xi)+\sigma_-^{-1}(\xi)\tilde w_-(\xi)=Q\Pi_+(Q^{-1}\tilde h)+Q\Pi_-(Q^{-1}\tilde h) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 40,735 | latex_formula | {"original_latex": "\\begin{align*}\\sigma_+(\\xi)\\tilde w_+(\\xi)+\\sigma_-^{-1}(\\xi)\\tilde w_-(\\xi)=Q\\Pi_+(Q^{-1}\\tilde h)+Q\\Pi_-(Q^{-1}\\tilde h)\\end{align*}"} |
mixed-00002053 | How to calculate the closed form of the Euler Sums We know that the closed form of the series
$$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}\left( {\begin{array}{*{20}{c}}
{2n} \\
n \\
\end{array}} \right)}}} = \frac{1}{3}\zeta \left( 2 \right).$$
but how to evaluate the following series
$$\sum\limits_{n = 1}^\infty {... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 1,044 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1415824", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 1, "answer_id": 0}} |
latex-00018136 | \frac{P_{m}}{p_m}\Delta_{10}u_{mn}=O_L(1) \frac{Q_{n}}{q_n}\Delta_{01}u_{mn}=O_L(1), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 18,195 | latex_formula | {"original_latex": "\\begin{align*}\\frac{P_{m}}{p_m}\\Delta_{10}u_{mn}=O_L(1)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{Q_{n}}{q_n}\\Delta_{01}u_{mn}=O_L(1),\\end{align*}"} |
latex-00032497 | \frac{H_A(\zeta_+, b)}{H_A(\zeta_-,b)} = \frac{\Im F(\zeta_+)}{\Im F(\zeta_-)} (1+O(R^{-u}) ) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 32,812 | latex_formula | {"original_latex": "\\begin{align*} \\frac{H_A(\\zeta_+, b)}{H_A(\\zeta_-,b)} = \\frac{\\Im F(\\zeta_+)}{\\Im F(\\zeta_-)} \\left(1+O(R^{-u}) \\right) \\end{align*}"} |
mixed-00026971 | If it is to be shown by induction that $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
then $f\left(0\right)=g\left(0\right)$ must be proved (the base case) and
secondly it must be shown that: $$\sum_{i=0}^{n+1}f\left(i\right)=g\left(n+1\right)$$
on base of: $$\sum_{i=0}^{n}f\left(i\right)=g\left(n\right)$$
That amou... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 15,745 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/649054", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00006980 | Wimbledon 2007 Final : Federer vs. Nadal ( 2007 ) . Kultur White Star , DVD release date : 30 October 2007 , run time : 180 minutes , ASIN B000V02CU0 . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 17,059 | text | {} |
mixed-00007781 | Express both $\sin\alpha$ and $\cos\alpha$ in terms of $t=\tan\frac{\alpha}{2}$:
\begin{align*}
\sin\alpha&=\frac{2t}{1+t^2},\\
\cos\alpha&=\frac{1-t^2}{1+t^2}.
\end{align*} | mixed | math-ai/StackMathQA | stackmathqa100k | train | 4,101 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1387753", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 5, "answer_id": 3}} |
latex-00027479 | b' = (\begin{tabular}{cccccc}0 &0& 0& 0& $(1+s)$ & $3+s$\end{tabular})\textrm{and}b = \sqrt{32} \frac{b'}{||b'||_2}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 27,727 | latex_formula | {"original_latex": "\\begin{align*}b' = \\left(\\begin{tabular}{cccccc}0 &0& 0& 0& $(1+s)$ & $3+s$\\end{tabular}\\right)\\textrm{and}b = \\sqrt{32} \\frac{b'}{||b'||_2}.\\end{align*}"} |
normal-00008364 | Bertin is presented as strong , energetic and warm @-@ hearted . His hair is grey verging on white , his fingers spread across his knees . Bertin 's fingers were described in 1959 by artist Henry de Waroquier as " crab @-@ like claws ... emerging from the tenebrous caverns that are the sleeves of his coat . " The bulk ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 20,509 | text | {} |
mixed-00012131 | Method of moments on uniform distributions I need help on how to find the estimates $a$ and $b$ in the uniform distribution $\mathcal U[a,b]$ using the method of moments.
This is where I am at:
I have found $U_1=\overline X$ and $m_1=\frac{a+b}2$ Also, $m_2=\frac1{n(E(X_i^2)}$ and $u_2=E(X_i^2)$ when I equate $m_1$ to ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 6,697 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/664811", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
normal-00002416 | The conclave began on February 15 , 1769 . Initially only 27 cardinals participated . Zelanti , taking advantage of the small number of the electors and the absence of the French and Spanish cardinals , tried to achieve a quick election of Cardinal Flavio Chigi . In one ballot he was only two votes short of being elect... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 5,831 | text | {} |
latex-00005556 | S_{BF}[A,\phi] = \int d^{2}x tr( \phi^{0}F^{0} + \phi^{a}F^{b} ) = \int d^{2}x ( - \phi^{0}F^{0}_{\alpha\beta}\epsilon^{\alpha\beta} + \Lambda \delta_{ab}\phi^{a}F^{b}_{\alpha\beta}\epsilon^{\alpha\beta}). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,562 | latex_formula | {"original_latex": "\\begin{align*}S_{BF}[A,\\phi] = \\int d^{2}x\\, tr\\left( \\phi^{0}F^{0} + \\phi^{a}F^{b} \\right) = \\int d^{2}x\\, \\left( - \\phi^{0}F^{0}_{\\alpha\\beta}\\epsilon^{\\alpha\\beta} + \\Lambda \\delta_{ab}\\phi^{a}F^{b}_{\\alpha\\beta}\\epsilon^{\\alpha\\beta}\\right). \\end{align*}"} |
mixed-00012691 | Using taylor series you have
$$\color{#05f}{\frac{\sin x}{x}} = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n}}{(2n+1)!} = 1 - \frac{x^2}{6} + \frac{x^4}{120} + O(x^6)$$
and $$\ln \Bigg(\color{#05f}{\frac{\sin x}{x}}\Bigg) = - \frac{x^2}{6} - O(x^4)$$
$$\lim _{x\to 0 } \frac{- \frac{x^2}{6} - O(x^4)}{x^2}= \lim_{x\to 0} -\fr... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 7,039 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1147074", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 7, "answer_id": 3}} |
normal-00017812 | As an early affluent Spokane neighborhood , the Browne 's Addition neighborhood and residences contain the largest variety of residential architecture in the city . These residences are lavish and personalized , featuring many architecture styles that were popular and trendy in the Pacific Northwest from the late 19th ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 42,725 | text | {} |
latex-00041579 | c_H(x) = \sum_{p=1}^m (u^1_pD_1+\cdots+u^m_pD_m) x x^\top (u^1_pD_1+\cdots+u^m_pD_m)^\top, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 42,180 | latex_formula | {"original_latex": "\\begin{align*}c_H(x) = \\sum_{p=1}^m (u^1_pD_1+\\cdots+u^m_pD_m)\\, x x^\\top (u^1_pD_1+\\cdots+u^m_pD_m)^\\top,\\end{align*}"} |
mixed-00037527 | Complete the square of the exponent $x^2+y^2-2\rho xy = \left(x-\rho y\right)^2+y^2(1-\rho^2)=:u^2+y^2(1-\rho^2)$
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-\frac{x^2+y^2-2 \rho x y}{2(1-\rho^2)}} dx dy=\\
=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2 \pi \sqrt{1-\rh... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 22,186 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/985552", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0}} |
mixed-00049700 | how to prove that this sequence converges to $0$? $0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$
When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to find its limit. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 29,656 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2936952", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0}} |
normal-00000405 | The Type 2 and 3 gold dollars depict Liberty as a Native American princess , with a fanciful feathered headdress not resembling any worn by any Indian tribe . This image is an inexact copy of the design Longacre had made for the three @-@ dollar piece , and is one of a number of versions of Liberty Longacre created bas... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 1,016 | text | {} |
mixed-00039333 | Evaluate $\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$ Evaluate $$\displaystyle\int_0^\infty \frac {(\log x)^4dx}{(1+x)(1+x^2)}$$
This is a past final term exam problem of a complex analysis course at my university. I am studying for this year’s exam and I found this problem. The examiner assumes us to use residue... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 23,318 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2571670", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0}} |
mixed-00024497 | There is a mistake in my presentation that eludes me, but the essence is the following.
Using
$$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$
then let $S_{n}$ be the desired summation to be evaluated to obtain:
\begin{align}
S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\
&= ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 14,229 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2789309", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "10", "answer_count": 3, "answer_id": 2}} |
latex-00005319 | &\bullet \quad(f=g)[h,k](x):=\begin{cases} h(x)&\mbox{ if }f(x)=g(x),\\k(x)&\mbox{ if }f(x)\neq g(x).\end{cases}& | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,324 | latex_formula | {"original_latex": "\\begin{align*}&\\bullet \\quad(f=g)[h,k](x):=\\begin{cases} h(x)&\\mbox{ if }f(x)=g(x),\\\\k(x)&\\mbox{ if }f(x)\\neq g(x).\\end{cases}&\\end{align*}"} |
normal-00008382 | Critics were divided with the Harajuku Lovers Tour . Patrick MacDonald of the Seattle Times , while applauding Stefani 's song @-@ writing efforts and the show 's " frothy fun " antics , reprimanded the singer 's dancing and limited material , given that she performed only twelve songs from Love . Angel . Music . Baby.... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 20,549 | text | {} |
normal-00041929 | The total programme cost , as of FY2013 , was around € 45 @.@ 9 billion , which translated to a unit programme cost of approximately € 160 @.@ 5 million . This figure takes in account improved hardware of the F3 standard , and which includes development costs over a period of 40 years , including inflation . The unit f... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 101,730 | text | {} |
normal-00013508 | Iguanodon teeth are , as the name suggests , like those of an iguana , but larger . Unlike hadrosaurids , which had columns of replacement teeth , Iguanodon only had one replacement tooth at a time for each position . The upper jaw held up to 29 teeth per side , with none at the front of the jaw , and the lower jaw 25 ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 32,719 | text | {} |
mixed-00001902 | HINT: Split
$$\frac{n+1}{2(n-1)!}= \frac{1}{2(n-2)!}+\frac{1}{(n-1)!}$$
and use the fact that $\sum_{n=0}^{\infty} \frac{1}{n!}=e$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 966 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1334661", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "10", "answer_count": 1, "answer_id": 0}} |
latex-00005708 | G_{\mu\nu\rho\sigma}=-\bar{g}_{\mu\rho}\bar{g}_{\nu\sigma}\nabla_{\beta}\nabla^{\beta}-2\bar{R}_{\mu\rho\nu\sigma}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 5,715 | latex_formula | {"original_latex": "\\begin{align*}G_{\\mu\\nu\\rho\\sigma}=-\\bar{g}_{\\mu\\rho}\\bar{g}_{\\nu\\sigma}\\nabla_{\\beta}\\nabla^{\\beta}-2\\bar{R}_{\\mu\\rho\\nu\\sigma}.\\end{align*}"} |
mixed-00041334 | Can we make an integral domain with any number of members? Is it true that rings without zero divisors (integral domains) can have any number of members except for 4,6? and if this is true then what would the multiplication operator be? | mixed | math-ai/StackMathQA | stackmathqa100k | train | 24,547 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/43788", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 2}} |
normal-00022493 | Picnic areas include a group pavilion , tennis courts , a children ’ s playground and a disc golf course . The park has an interpretive center near the main parking lot , adjacent to the canyon . Annual events hosted in the park include a Wildflower Program in April , Adventure Weekend ( also in April ) , and a Kids ' ... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 54,079 | text | {} |
normal-00039136 | Although found in many types of woodland , the pied currawong prefers to breed in mature forests . It builds a nest of thin sticks lined with grass and bark high in trees in spring ; generally eucalypts are chosen and never isolated ones . It produces a clutch of three eggs ; they are a light pinkish @-@ brown colour (... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 94,725 | text | {} |
mixed-00020135 | Yes, the improper integral is convergent, but your computation is not correct. Note that
$$\begin{align}
\int \frac{x \ln(x)}{(x^2+1)^2} dx& = -\frac{\ln(x)}{2(x^2+1)}+ \int\frac{1}{2x(x^2+1)}dx\\
&=
-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx\\
&=-\frac{\ln(x)}{2(x^2+1)}+\frac{\... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 11,573 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3409650", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 6, "answer_id": 1}} |
latex-00025433 | a\chi(f\otimes m) &= af(m) = \chi(af\otimes m), \mbox{ and} \ (\chi(f\otimes m))a &= \lambda(a)\chi(f\otimes m) = \chi(f\otimes \lambda(a)m) = \chi((f\otimes m)a). | latex | OleehyO/latex-formulas | cleaned_formulas | train | 25,627 | latex_formula | {"original_latex": "\\begin{align*} a\\chi(f\\otimes m) &= af(m) = \\chi(af\\otimes m), \\mbox{\\ \\ \\ and} \\\\ (\\chi(f\\otimes m))a &= \\lambda(a)\\chi(f\\otimes m) = \\chi(f\\otimes \\lambda(a)m) = \\chi((f\\otimes m)a).\\end{align*}"} |
mixed-00016085 | A better approach would be to treat the given expression as a quadratic in $x$ and, so we can re-write the expression as $ 4x^2 -2x(y+2) +3(y-1)$. Set it equal to zero, and solve for $x$ using the quadratic formula.
$x = \frac{ 2(y+2) \pm \sqrt{4(y+2)^2-48(y-1)}} {8}$
This gives us the solutions $x=\frac{3}{2} $ and $x... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 9,103 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4290758", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 8, "answer_id": 0}} |
mixed-00009779 | This double integral is first transformed into a single integral of a complete elliptic integral:
\begin{align}
I&=\displaystyle\int_{0}^{1}\displaystyle \int_{0}^{1}\dfrac{dxdy}{\sqrt{1-x^2}{\sqrt{1-y^2}}{\sqrt{4x^2+y^2-x^2y^2}}}\\
&\stackrel{x=\cos t}{=}\int_0^1\frac{dy}{\sqrt{1-y^2}}\int_{0}^{\pi/2}\frac{dt}{\sqrt... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 5,294 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3029929", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 1, "answer_id": 0}} |
mixed-00040895 | Just compute
$$
(u-v)^2=(b+b^4-b^2-b^3)^2=-(b^4+b^3+b^2+b+1)+5=5.
$$
We have used $b^5=1$ in the second step. | mixed | math-ai/StackMathQA | stackmathqa100k | train | 24,274 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/4172762", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
normal-00030691 | The third disc from Integral was released as a stand @-@ alone game in North America under the title of Metal Gear Solid : VR Missions on September 23 , 1999 . While the content of VR Missions are virtually identical to the VR Disc , the unlocking requirements for the Ninja missions and the photoshoot mode were changed... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 73,579 | text | {} |
mixed-00014405 | We can compute the integral by Fourier series. Consider
\begin{align}
f(x) = \frac{\sin^2(4x)}{\sin x}
\end{align}
then the sine series expansion of $f$ is given by
\begin{align}
f(x) = \sin x + \sin 3x + \sin 5x + \sin 7 x.
\end{align}
In particular, we see
\begin{align}
\int^{2\pi}_0 f^2(x)\ dx = \int^{2\pi}_0 \sin^2... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 8,080 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2539108", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0}} |
latex-00002392 | (ij(n+1)) \circ (i_1 i_2\ldots i_{l-1} i)(j_1 j_2\ldots j_{l'-1} j)(n+1) = (i_1 i_2\ldots i_{l-1} i j_1 j_2\ldots j_{l'-1} j (n+1)) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 2,393 | latex_formula | {"original_latex": "\\begin{align*}(ij(n+1)) \\circ (i_1 i_2\\ldots i_{l-1} i)(j_1 j_2\\ldots j_{l'-1} j)(n+1) = (i_1 i_2\\ldots i_{l-1} i j_1 j_2\\ldots j_{l'-1} j (n+1))\\end{align*}"} |
latex-00049854 | \frac{k^2+u^2}{k} = k+\frac{u^2}{k}, | latex | OleehyO/latex-formulas | cleaned_formulas | train | 50,550 | latex_formula | {"original_latex": "\\begin{align*}\\frac{k^2+u^2}{k} = k+\\frac{u^2}{k},\\end{align*}"} |
latex-00019816 | \Pr(X_t \le k) = ( 1 - \frac{1}{n} )^t \sum_{j=0}^k \Pr(\widetilde{X}_t = k \vert \widetilde{X}_0 = j), | latex | OleehyO/latex-formulas | cleaned_formulas | train | 19,903 | latex_formula | {"original_latex": "\\begin{align*}\\Pr(X_t \\le k) = ( 1 - \\frac{1}{n} )^t \\sum_{j=0}^k \\Pr(\\widetilde{X}_t = k \\,\\vert\\, \\widetilde{X}_0 = j),\\end{align*}"} |
latex-00016388 | Tr_w K_{H_2}={e^{-{\bar{s}/4}}\over (4\pi \bar{s})^{1/2}}\int_0^\infty {dy \cosh y e^{-{y^2/ \bar{s}}}\over (\sinh^2y+\sin^2{w\over 2})}~~. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 16,426 | latex_formula | {"original_latex": "\\begin{align*}Tr_w K_{H_2}={e^{-{\\bar{s}/4}}\\over (4\\pi \\bar{s})^{1/2}}\\int_0^\\infty {dy \\cosh y e^{-{y^2/ \\bar{s}}}\\over (\\sinh^2y+\\sin^2{w\\over 2})}~~.\\end{align*}"} |
mixed-00019155 | Where the object that moves along the intersection of $x^2+y^2=1$ and $y+z=1$ needs to be if we want the sum $x+2y+z$ to be max/min?
An object is moving along the curve which is derived from the intersection of the cylinder $x^2+y^2=1$ and the plane $y+z=1$. Where does the object need to be located if we want to maxim... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 10,974 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2393232", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0}} |
normal-00036024 | Aaron Sorkin ( February 2004 ) . The West Wing Seasons 3 & 4 : The Shooting Scripts : Eight Teleplays . Newmarket Press . ISBN 978 @-@ 1 @-@ 55704 @-@ 612 @-@ 3 . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 86,927 | text | {} |
normal-00001449 | The garage has two doors and two windows . Both the doors and windows have two rowlock brick arches over them . The car entrance is on the west side ; a passage door is on the north side . A clear @-@ glass window with 16 panes is on the east side . On the west side , north of the car entrance , is a window with bevele... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 3,714 | text | {} |
latex-00013829 | \mbox{\boldmath$S$} = \kappa^{\frac{1}{2}} ~(\begin{array}{cc}m_q^{-1} {\mbox{\boldmath$S$}}_q~I^{(6)} & 0 \\0 & m_D^{-1} {\mbox{\boldmath$S$}}_D~I^{(21)}\end{array} ) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 13,857 | latex_formula | {"original_latex": "\\begin{align*}\\mbox{\\boldmath$S$} = \\kappa^{\\frac{1}{2}} ~\\left(\\begin{array}{cc}m_q^{-1} {\\mbox{\\boldmath$S$}}_q~I^{(6)} & 0 \\\\0 & m_D^{-1} {\\mbox{\\boldmath$S$}}_D~I^{(21)}\\end{array} \\right)\\end{align*}"} |
mixed-00013833 | The brute force method :
See it as a polynomial of degree $3$ of $K[X]$ with coefficients in $K = \mathbb{C}(y)$. If it is not irreducible then $$X^3+y^3+3X^2y+3Xy^2-X^2-y^2=( X +a)(X^2+bX+c)=X^3+bX^2 +cX+aX^2+abX+ac$$
$c = \frac{y^3-y^2}{a}$, $b = 3y-1-a$
$$= X^3+(3y-1-a)X^2 +\frac{y^3-y^2}{a}X+aX^2+a(3y-1-a)X+1$$
I ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 7,728 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2095132", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0}} |
mixed-00030605 | Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$
In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 17,953 | Q | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/3901672", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 3, "answer_id": 2}} |
normal-00023638 | Performed by Thirty Seconds to Mars | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 56,667 | text | {} |
mixed-00047105 | To solve the recurrence that you have given,
$$
\begin{align}
f(x)
&=\sum_{k=0}^\infty a_kx^k\\
&=1+x+\sum_{k=2}^\infty (\color{#C00000}{a_{k-1}}+\color{#00A000}{(k-1)a_{k-2}})x^k\\
&=1+x+\color{#C00000}{x(f(x)-1)}+\color{#00A000}{x^2(xf'(x)+f(x))}\\[10pt]
-1
&=x^3f'(x)+f(x)(x^2+x-1)
\end{align}
$$
This equation can be... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 28,083 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/717679", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0}} |
latex-00009788 | ( - \frac{1+h_q^2}{2h_p^2} + B - g \rho (h-d) )_p + ( \frac{h_q}{h_p} )_q + g\rho h_p & = 0,\textrm{in } \bigcup_i \mathcal{R}_i, \\-\frac{1+h_q^2}{2h_p^2} - g\rho h + \frac{Q}{2} & = 0, \textrm{on } \{ p = 0 \}, \ h & = 0, \textrm{on } \{ p = p_0\}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 9,807 | latex_formula | {"original_latex": "\\begin{align*}\\left( - \\frac{1+h_q^2}{2h_p^2} + B - g \\rho (h-d) \\right)_p + \\left( \\frac{h_q}{h_p} \\right)_q + g\\rho h_p & = 0,\\textrm{in } \\bigcup_i \\mathcal{R}_i, \\\\-\\frac{1+h_q^2}{2h_p^2} - g\\rho h + \\frac{Q}{2} & = 0, \\textrm{on } \\{ p = 0 \\}, \\\\ h & = 0, \\textrm{on } \\{... |
latex-00008376 | \frac{1}{2}\widehat{\chi }_{p}( \tau ) ( \widehat{\chi }_{p}( \frac{\tau }{2}) -\widehat{\chi }_{p}( \frac{\tau +1}{2}) ) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 8,386 | latex_formula | {"original_latex": "\\begin{align*}\\frac{1}{2}\\widehat{\\chi }_{p}\\left( \\tau \\right) \\left( \\widehat{\\chi }_{p}\\left( \\frac{\\tau }{2}\\right) -\\widehat{\\chi }_{p}\\left( \\frac{\\tau +1}{2}\\right) \\right)\\end{align*}"} |
latex-00048281 | d^{\rm sup}(10^{-1}) \le \{\begin{array}{ll}1&\mbox{for }a=4,\\1&\mbox{for }a=3,\\2&\mbox{for }a=2, \end{array}. | latex | OleehyO/latex-formulas | cleaned_formulas | train | 48,950 | latex_formula | {"original_latex": "\\begin{align*}d^{\\rm sup}(10^{-1})\\,\\le\\,\\left\\{\\begin{array}{ll}1&\\mbox{for\\ }a=4,\\\\1&\\mbox{for\\ }a=3,\\\\2&\\mbox{for\\ }a=2, \\end{array}\\right.\\end{align*}"} |
normal-00020142 | Pennock received only one start apiece in the months of April and May , as the 1919 Red Sox relied on George Dumont , Bill James , and Bullet Joe Bush , leading Pennock to threaten to quit in late @-@ May unless Barrow fulfilled his earlier promise to Pennock . Barrow continued to use Pennock regularly after Memorial D... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 48,246 | text | {} |
mixed-00043374 | How about this:
\begin{align}
u & = \sqrt{x^2-1} \\
u^2 & = x^2 - 1 \\
u^2+1 & = x^2 \\
2u\,du & = 2x\,dx \\
\end{align}
\begin{align}
\int x^3 \sqrt{x^2-1} \, dx & = \int x^2 \sqrt{x^2-1} \Big( x\,dx\Big) \\
& = \int (u^2+1) u \left( \frac 1 2 \, du \right) \\
& = \frac 1 2 \int (u^4 + u^2) \, du \\
& \phantom{{}={}} ... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 25,793 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/1768356", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1}} |
mixed-00026317 | $$\tan(\theta) + \cot(\theta) = {\sin(\theta)\over \cos(\theta)} + {\cos(\theta)\over \sin(\theta)} = {\sin^2(\theta) + \cos^2(\theta) \over\cos(\theta)\sin(\theta)}
= {1\over\sin(\theta)\cos(\theta)}.$$
Now avail yourself of the fact that $$\sin(2\theta) = 2\cos(
\theta)\sin(\theta).$$ | mixed | math-ai/StackMathQA | stackmathqa100k | train | 15,347 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/170951", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1}} |
latex-00044085 | \Omega_n(D)=([D,L]\times C_D(L))\cap \Omega_n(D)=[\Omega_n(D),L]\times C_{\Omega_n(D)}(L) | latex | OleehyO/latex-formulas | cleaned_formulas | train | 44,711 | latex_formula | {"original_latex": "\\begin{align*}\\Omega_n(D)=([D,L]\\times C_D(L))\\cap \\Omega_n(D)=[\\Omega_n(D),L]\\times C_{\\Omega_n(D)}(L)\\end{align*}"} |
normal-00035064 | The only documented predator of the silky sifaka , other than humans , is the fossa , a cat @-@ like carnivore found only on Madagascar . Although no aerial predators are known , the silky sifaka often watch the sky and emit loud " aerial disturbance " roars at the sight of the large Madagascar buzzard ( Buteo brachypt... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 84,371 | text | {} |
normal-00022483 | Until 1939 the only access to the area ( and much of Dade County , Georgia ) was through Tennessee or Alabama . That year Georgia began work on Highway 136 to connect U.S. 41 to the recently established park . The Civilian Conservation Corps did much of the early work to construct the state park and access roads . | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 54,060 | text | {} |
mixed-00024025 | $$\sum_{n=1}^\infty\frac{\sin (n\pi/k)}{n} = \sum_{n=1}^\infty\frac{\pi}{n\pi}\sin2\pi \frac{n}{2}\frac1{k}$$
The right hand side is a sawtooth wave of period $2$, at $t = 1/k$:
$$f(t) = \frac{\pi}{2}-\frac{\pi}2t, \quad 0<t<2\\
f(t+2) = f(t)\\
f(2n) = 0, \quad n\in\mathbb Z$$
So specifically,
$$\sum_{n=1}^\infty\frac{... | mixed | math-ai/StackMathQA | stackmathqa100k | train | 13,941 | A | {"meta": {"language": "en", "url": "https://math.stackexchange.com/questions/2346681", "timestamp": "2023-03-29 00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1}} |
normal-00026086 | Joaquim Dos Santos won the " Best Directing in an Animated Television Production " caption in the 2008 Annie Awards for his directing in " Into the Inferno " . Additionally , music editor and composer Jeremy Zuckerman and the sound editing team were nominated a Golden Reel award for " Best Sound Editing in a Television... | normal | Salesforce/wikitext | wikitext-103-raw-v1 | train | 62,527 | text | {} |
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