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PF Gold
P: 162
 Quote by TranscendArcu 1. The problem statement, all variables and given/known data 3. The attempt at a solution So I observed: T(B) = λB T-1(B) = λ'B Also, T-1(T(B)) = λ'λB = B This implies, λ'λ = 1 And so, there should be a relation $λ = \frac{1}{λ'}$. Is that right?
Your conclusion is correct but I'm not sure I agree with your proof. In particular this
T-1(B) = λ'B
is only true if B is an eigenvector of T, and thats not really clear a priori. Just say that B is an eigenvector of T so

T(B) = λB

Apply T-1 to both sides
T-1(T(B)) = T-1(λB)

T is a linear operator (I assume) ... I'll let you finish the argument. It's pretty trivial.

Also as to the diagonalization comment, while I agree that's a good way to think about it intuitively it's not a good technique for a proof. Obviously there are issues with some finite dimensional space maps being non-diagonalizible; but even more fundamentally for a Hilbert space I think you would need the operator to be compact to have any hope of such an approach working.


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# Do the vectors (1,3,-2),(2,1,1) belong to R(T)?

• Dec 12th 2009, 08:38 PM
Undefdisfigure
Do the vectors (1,3,-2),(2,1,1) belong to R(T)?
Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

My first approach was to make a column vector (a) = (a+b)
(b) (b-2c)
(c) (a+2c)

and then remove the coefficients like this:

a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
(0) (1) (-2)
(1) (0) (2)

The second thing I tried was to compute two vectors for R(T) like this:

T(1,3,-2) = (4,7,-3)

T(2,1,1) = (3,-1,4)

And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.
• Dec 13th 2009, 05:53 AM
tonio
Quote:

Originally Posted by Undefdisfigure
Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

My first approach was to make a column vector (a) = (a+b)
(b) (b-2c)
(c) (a+2c)

and then remove the coefficients like this:

a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
(0) (1) (-2)
(1) (0) (2)

The second thing I tried was to compute two vectors for R(T) like this:

T(1,3,-2) = (4,7,-3)

T(2,1,1) = (3,-1,4)

And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.

Since $T\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatri x}a+b\\b-2c\\a+2c\end{pmatrix}$, you only have to ask whether there exist $a,b,c,\in\mathbb{R}\,\,\,s.t.\,\,\,\begin{pmatrix} a+b\\b-2c\\a+2c\end{pmatrix}=\begin{pmatrix}\;\;1\\\;\;3\ \-2\end{pmatrix}\,\,or\,\,\begin{pmatrix}2\\1\\1\end {pmatrix}$.

For example, for the first one, we'd get:

** $a+b=1$

** $b-2c=3$

** $a+2c=-2$

Substracting the third eq. from the first one we get $b-2c=3=$ second eq., so the system seems to be congruent (i.e., has at least one solution), for example: $a=2,b=-1,c=-2$, so this first vector does belong to the range of $T$. Now you continue with the other one.

Tonio
• Dec 13th 2009, 01:39 PM
Undefdisfigure
how do I thank you for this reply??
• Dec 13th 2009, 02:00 PM
tonio
Quote:

Originally Posted by Undefdisfigure
how do I thank you for this reply??

You just have...any time!(Wink)

Tonio

format: plain


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# Joint Rain Detection and Removal from a Single Image

Figure 1. An example result of our proposed rain removal method that removes heavy rain streaks and enhances the visibility significantly. Top: The raw image with rain streaks (left) and the output image of our method (right). Bottom: closer looks at specific regions (blue and red) for more details.

## Abstract

In this paper, we address a rain removal problem from a single image, even in the presence of heavy rain and rain streak accumulation. Our core ideas lie in the new rain image models and a novel deep learning architecture. We first modify an existing model comprising a rain streak layer and a background layer, by adding a binary map that locates rain streak regions. Second, we create a new model consisting of a component representing rain streak accumulation (where individual streaks cannot be seen, and thus visually similar to mist or fog), and another component representing various shapes and directions of overlapping rain streaks, which usually happen in heavy rain. Based on the first model, we develop a multi-task deep learning architecture that learns the binary rain streak map, the appearance of rain streaks, and the clean background, which is our ultimate output. The additional binary map is critically beneficial, since its loss function can provide additional strong information to the network. To handle rain streak accumulation (again, a phenomenon visually similar to mist or fog) and various shapes and directions of overlapping rain streaks, we propose a recurrent rain detection and removal network that removes rain streaks and clears up the rain accumulation iteratively and progressively. In each recurrence of our method, a new contextualized dilated network is developed to exploit regional contextual information and outputs better representation for rain detection. The evaluation on real images, particularly on heavy rain, shows the effectiveness of our novel models and architecture, outperforming the state-of-the-art methods significantly. Our codes and data sets will be publicly available.

Figure. 2. The architecture of our proposed recurrent rain detection and removal. Each recurrence is a multi-task network to perform a joint rain detection and removal (in the blue dash box). In such a network, a contextualized dilated network (in the gray region) extracts rain features $$\mathbf{F}_t$$ from the input rain image $$\mathbf{O}_t$$. Then, $$\mathbf{R}_t$$, $$\mathbf{S}_t$$ and $$\mathbf{B}_t$$ are predicted to perform joint rain detection, estimation and removal. The contextualized dilated network has two features: 1) it takes a recurrent structure, which refines the extracted features progressively; 2) for each recurrence, the output features are aggregated from three convolution paths (P1, P2 and P3) with different dilated factors (DF) and receptive fields.

## Results

 Rain images DSC [2] LP[1] The proposed

• Paper: arXiv CVPR


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# Heston Model Closed-Form Solution¶

## Overview¶

The Heston Model , published by Steven Heston in paper “A Closed-Form Solution for Options with Stochastic Volatility with Applications to Bond and Currency Options” in 1993 [HEST1993] , extends the well-known Black-Scholes options pricing model by adding a stochastic process for the stock volatility.

The stochastic equations of the model, and the partial differential equation (PDE) derived from them, are shown in the section on the Heston Model under “Models”. There are a number of ways in which the equations may be solved, including Monte-Carlo and Finite-Difference methods. Here, we show the closed-form solution originally obtained in [HEST1993] - sometimes referred to as the “semi-analytic” solution since an analytic solution of its integrals is not known.

The expression for a European Call option, derived from the Heston PDE, is shown below, (in the form presented by Crisostomo in [CHRSO2014] ):

$C_0 = S_0.\Pi_1 - \mathrm{e}^{-rT}K.\Pi_2$

where here and in the following: $$T$$ (and $$t$$) = Time to Expiration; $$K$$ = Option Strike Price; $$S_0$$ = stock price at $$t = 0$$; $$r$$ = interest-rate; $$V_0$$ = stock-price variance at $$t = 0$$; $$\eta$$ = “volatility-of-volatility” (elsewhere $$\sigma$$); $$a$$ = rate-of-reversion (elsewhere $$\kappa$$); $$\tilde{V}$$ = long-term average variance (elsewhere $$\theta$$); $$\rho$$ = correlation of $$z_1(t)$$ and $$z_2(t)$$ processes.

Using a solution based on Characteristic functions, the values of probabilities $$\Pi_1$$ and $$\Pi_2$$ are given by:

$\Pi_1 = \frac{1}{2} + \frac{1}{\pi} \int_0^\infty Re\left[\frac{\mathrm{e}^{-i.w.ln(K)}.\Psi_{lnS_T}(w - i)}{i.w.\Psi_{lnS_T}(-i)} \right]\mathrm{d}w$
$\Pi_2 = \frac{1}{2} + \frac{1}{\pi} \int_0^\infty Re\left[\frac{\mathrm{e}^{-i.w.ln(K)}.\Psi_{lnS_T}(w)}{i.w} \right]\mathrm{d}w$

where the Characteristic function ψ is:

$\Psi_{lnS_T}(w) = \mathrm{e}^{[C(t,w).\tilde{V} + D(t,w).V_0 + i.w.ln(S_0.\mathrm{e}^{rt})]}$

where:

\begin{align}\begin{aligned}C(t,w) = a.\left[r_-.t - \frac{2}{\eta^2}.ln\left(\frac{1 - g.\mathrm{e}^{-ht}}{1 - g}\right)\right]\\D(t,w) = r_- .\frac{1 -\mathrm{e}^{-ht}}{1 - g.\mathrm{e}^{-ht}}\\r_{\pm} = \frac{\beta \pm h}{\eta^2}; h = \sqrt{\beta^2 - 4.\alpha.\gamma}\\g = \frac{r_-}{r_+}\\\alpha = -\frac{w^2}{2} - \frac{iw}{2} ; \beta = \alpha - \rho.\eta. i. w ; \gamma = \frac{\eta^2}{2}\end{aligned}\end{align}

To obtain a solution for Call $$C_0$$ the integrands in the $$\Pi_1$$ and $$\Pi_2$$ terms must be evaluated using a selected numeric integration technique suited to integration from 0 to ∞, and the internal characteristic function terms must be obtained using complex-number computation.

## References¶

 [HEST1993] (1, 2) Heston, S. L., “A Closed-Form Solution for Options with Stochastic Volatility with Applications to Bond and Currency Options”, The Review of Financial Studies (1993)
 [CHRSO2014] Crisostomo, R., “An Analysis of the Heston Stochastic Volatility Model: Implementation and Calibration using Matlab”, CNMV Working Paper 58: 1-46, (2014).


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## G = C3×D4⋊6D10order 480 = 25·3·5

### Direct product of C3 and D4⋊6D10

Series: Derived Chief Lower central Upper central

 Derived series C1 — C10 — C3×D4⋊6D10
 Chief series C1 — C5 — C10 — C30 — C6×D5 — D5×C2×C6 — C3×D4×D5 — C3×D4⋊6D10
 Lower central C5 — C10 — C3×D4⋊6D10
 Upper central C1 — C6 — C6×D4

Generators and relations for C3×D46D10
G = < a,b,c,d,e | a3=b4=c2=d10=e2=1, ab=ba, ac=ca, ad=da, ae=ea, cbc=dbd-1=b-1, be=eb, dcd-1=ece=b2c, ede=d-1 >

Subgroups: 1040 in 332 conjugacy classes, 170 normal (22 characteristic)
C1, C2, C2, C3, C4, C4, C22, C22, C22, C5, C6, C6, C2×C4, C2×C4, D4, D4, Q8, C23, C23, D5, C10, C10, C12, C12, C2×C6, C2×C6, C2×C6, C15, C2×D4, C2×D4, C4○D4, Dic5, C20, D10, D10, C2×C10, C2×C10, C2×C10, C2×C12, C2×C12, C3×D4, C3×D4, C3×Q8, C22×C6, C22×C6, C3×D5, C30, C30, 2+ 1+4, Dic10, C4×D5, D20, C2×Dic5, C5⋊D4, C2×C20, C5×D4, C22×D5, C22×C10, C6×D4, C6×D4, C3×C4○D4, C3×Dic5, C60, C6×D5, C6×D5, C2×C30, C2×C30, C2×C30, C4○D20, D4×D5, D42D5, C2×C5⋊D4, D4×C10, C3×2+ 1+4, C3×Dic10, D5×C12, C3×D20, C6×Dic5, C3×C5⋊D4, C2×C60, D4×C15, D5×C2×C6, C22×C30, D46D10, C3×C4○D20, C3×D4×D5, C3×D42D5, C6×C5⋊D4, D4×C30, C3×D46D10
Quotients: C1, C2, C3, C22, C6, C23, D5, C2×C6, C24, D10, C22×C6, C3×D5, 2+ 1+4, C22×D5, C23×C6, C6×D5, C23×D5, C3×2+ 1+4, D5×C2×C6, D46D10, D5×C22×C6, C3×D46D10

Smallest permutation representation of C3×D46D10
On 120 points
Generators in S120
(1 22 12)(2 23 13)(3 24 14)(4 25 15)(5 21 11)(6 28 18)(7 29 19)(8 30 20)(9 26 16)(10 27 17)(31 55 45)(32 51 41)(33 52 42)(34 53 43)(35 54 44)(36 60 50)(37 56 46)(38 57 47)(39 58 48)(40 59 49)(61 87 71)(62 88 72)(63 89 73)(64 90 74)(65 81 75)(66 82 76)(67 83 77)(68 84 78)(69 85 79)(70 86 80)(91 111 101)(92 112 102)(93 113 103)(94 114 104)(95 115 105)(96 116 106)(97 117 107)(98 118 108)(99 119 109)(100 120 110)
(1 92 10 97)(2 98 6 93)(3 94 7 99)(4 100 8 95)(5 96 9 91)(11 106 16 101)(12 102 17 107)(13 108 18 103)(14 104 19 109)(15 110 20 105)(21 116 26 111)(22 112 27 117)(23 118 28 113)(24 114 29 119)(25 120 30 115)(31 70 36 65)(32 66 37 61)(33 62 38 67)(34 68 39 63)(35 64 40 69)(41 76 46 71)(42 72 47 77)(43 78 48 73)(44 74 49 79)(45 80 50 75)(51 82 56 87)(52 88 57 83)(53 84 58 89)(54 90 59 85)(55 86 60 81)
(1 61)(2 67)(3 63)(4 69)(5 65)(6 62)(7 68)(8 64)(9 70)(10 66)(11 75)(12 71)(13 77)(14 73)(15 79)(16 80)(17 76)(18 72)(19 78)(20 74)(21 81)(22 87)(23 83)(24 89)(25 85)(26 86)(27 82)(28 88)(29 84)(30 90)(31 91)(32 97)(33 93)(34 99)(35 95)(36 96)(37 92)(38 98)(39 94)(40 100)(41 107)(42 103)(43 109)(44 105)(45 101)(46 102)(47 108)(48 104)(49 110)(50 106)(51 117)(52 113)(53 119)(54 115)(55 111)(56 112)(57 118)(58 114)(59 120)(60 116)
(1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30)(31 32 33 34 35)(36 37 38 39 40)(41 42 43 44 45)(46 47 48 49 50)(51 52 53 54 55)(56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70)(71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110)(111 112 113 114 115 116 117 118 119 120)
(1 5)(2 4)(6 8)(9 10)(11 12)(13 15)(16 17)(18 20)(21 22)(23 25)(26 27)(28 30)(31 37)(32 36)(33 40)(34 39)(35 38)(41 50)(42 49)(43 48)(44 47)(45 46)(51 60)(52 59)(53 58)(54 57)(55 56)(61 70)(62 69)(63 68)(64 67)(65 66)(71 80)(72 79)(73 78)(74 77)(75 76)(81 82)(83 90)(84 89)(85 88)(86 87)(91 97)(92 96)(93 95)(98 100)(101 107)(102 106)(103 105)(108 110)(111 117)(112 116)(113 115)(118 120)

G:=sub<Sym(120)| (1,22,12)(2,23,13)(3,24,14)(4,25,15)(5,21,11)(6,28,18)(7,29,19)(8,30,20)(9,26,16)(10,27,17)(31,55,45)(32,51,41)(33,52,42)(34,53,43)(35,54,44)(36,60,50)(37,56,46)(38,57,47)(39,58,48)(40,59,49)(61,87,71)(62,88,72)(63,89,73)(64,90,74)(65,81,75)(66,82,76)(67,83,77)(68,84,78)(69,85,79)(70,86,80)(91,111,101)(92,112,102)(93,113,103)(94,114,104)(95,115,105)(96,116,106)(97,117,107)(98,118,108)(99,119,109)(100,120,110), (1,92,10,97)(2,98,6,93)(3,94,7,99)(4,100,8,95)(5,96,9,91)(11,106,16,101)(12,102,17,107)(13,108,18,103)(14,104,19,109)(15,110,20,105)(21,116,26,111)(22,112,27,117)(23,118,28,113)(24,114,29,119)(25,120,30,115)(31,70,36,65)(32,66,37,61)(33,62,38,67)(34,68,39,63)(35,64,40,69)(41,76,46,71)(42,72,47,77)(43,78,48,73)(44,74,49,79)(45,80,50,75)(51,82,56,87)(52,88,57,83)(53,84,58,89)(54,90,59,85)(55,86,60,81), (1,61)(2,67)(3,63)(4,69)(5,65)(6,62)(7,68)(8,64)(9,70)(10,66)(11,75)(12,71)(13,77)(14,73)(15,79)(16,80)(17,76)(18,72)(19,78)(20,74)(21,81)(22,87)(23,83)(24,89)(25,85)(26,86)(27,82)(28,88)(29,84)(30,90)(31,91)(32,97)(33,93)(34,99)(35,95)(36,96)(37,92)(38,98)(39,94)(40,100)(41,107)(42,103)(43,109)(44,105)(45,101)(46,102)(47,108)(48,104)(49,110)(50,106)(51,117)(52,113)(53,119)(54,115)(55,111)(56,112)(57,118)(58,114)(59,120)(60,116), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30)(31,32,33,34,35)(36,37,38,39,40)(41,42,43,44,45)(46,47,48,49,50)(51,52,53,54,55)(56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120), (1,5)(2,4)(6,8)(9,10)(11,12)(13,15)(16,17)(18,20)(21,22)(23,25)(26,27)(28,30)(31,37)(32,36)(33,40)(34,39)(35,38)(41,50)(42,49)(43,48)(44,47)(45,46)(51,60)(52,59)(53,58)(54,57)(55,56)(61,70)(62,69)(63,68)(64,67)(65,66)(71,80)(72,79)(73,78)(74,77)(75,76)(81,82)(83,90)(84,89)(85,88)(86,87)(91,97)(92,96)(93,95)(98,100)(101,107)(102,106)(103,105)(108,110)(111,117)(112,116)(113,115)(118,120)>;

G:=Group( (1,22,12)(2,23,13)(3,24,14)(4,25,15)(5,21,11)(6,28,18)(7,29,19)(8,30,20)(9,26,16)(10,27,17)(31,55,45)(32,51,41)(33,52,42)(34,53,43)(35,54,44)(36,60,50)(37,56,46)(38,57,47)(39,58,48)(40,59,49)(61,87,71)(62,88,72)(63,89,73)(64,90,74)(65,81,75)(66,82,76)(67,83,77)(68,84,78)(69,85,79)(70,86,80)(91,111,101)(92,112,102)(93,113,103)(94,114,104)(95,115,105)(96,116,106)(97,117,107)(98,118,108)(99,119,109)(100,120,110), (1,92,10,97)(2,98,6,93)(3,94,7,99)(4,100,8,95)(5,96,9,91)(11,106,16,101)(12,102,17,107)(13,108,18,103)(14,104,19,109)(15,110,20,105)(21,116,26,111)(22,112,27,117)(23,118,28,113)(24,114,29,119)(25,120,30,115)(31,70,36,65)(32,66,37,61)(33,62,38,67)(34,68,39,63)(35,64,40,69)(41,76,46,71)(42,72,47,77)(43,78,48,73)(44,74,49,79)(45,80,50,75)(51,82,56,87)(52,88,57,83)(53,84,58,89)(54,90,59,85)(55,86,60,81), (1,61)(2,67)(3,63)(4,69)(5,65)(6,62)(7,68)(8,64)(9,70)(10,66)(11,75)(12,71)(13,77)(14,73)(15,79)(16,80)(17,76)(18,72)(19,78)(20,74)(21,81)(22,87)(23,83)(24,89)(25,85)(26,86)(27,82)(28,88)(29,84)(30,90)(31,91)(32,97)(33,93)(34,99)(35,95)(36,96)(37,92)(38,98)(39,94)(40,100)(41,107)(42,103)(43,109)(44,105)(45,101)(46,102)(47,108)(48,104)(49,110)(50,106)(51,117)(52,113)(53,119)(54,115)(55,111)(56,112)(57,118)(58,114)(59,120)(60,116), (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30)(31,32,33,34,35)(36,37,38,39,40)(41,42,43,44,45)(46,47,48,49,50)(51,52,53,54,55)(56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110)(111,112,113,114,115,116,117,118,119,120), (1,5)(2,4)(6,8)(9,10)(11,12)(13,15)(16,17)(18,20)(21,22)(23,25)(26,27)(28,30)(31,37)(32,36)(33,40)(34,39)(35,38)(41,50)(42,49)(43,48)(44,47)(45,46)(51,60)(52,59)(53,58)(54,57)(55,56)(61,70)(62,69)(63,68)(64,67)(65,66)(71,80)(72,79)(73,78)(74,77)(75,76)(81,82)(83,90)(84,89)(85,88)(86,87)(91,97)(92,96)(93,95)(98,100)(101,107)(102,106)(103,105)(108,110)(111,117)(112,116)(113,115)(118,120) );

G=PermutationGroup([[(1,22,12),(2,23,13),(3,24,14),(4,25,15),(5,21,11),(6,28,18),(7,29,19),(8,30,20),(9,26,16),(10,27,17),(31,55,45),(32,51,41),(33,52,42),(34,53,43),(35,54,44),(36,60,50),(37,56,46),(38,57,47),(39,58,48),(40,59,49),(61,87,71),(62,88,72),(63,89,73),(64,90,74),(65,81,75),(66,82,76),(67,83,77),(68,84,78),(69,85,79),(70,86,80),(91,111,101),(92,112,102),(93,113,103),(94,114,104),(95,115,105),(96,116,106),(97,117,107),(98,118,108),(99,119,109),(100,120,110)], [(1,92,10,97),(2,98,6,93),(3,94,7,99),(4,100,8,95),(5,96,9,91),(11,106,16,101),(12,102,17,107),(13,108,18,103),(14,104,19,109),(15,110,20,105),(21,116,26,111),(22,112,27,117),(23,118,28,113),(24,114,29,119),(25,120,30,115),(31,70,36,65),(32,66,37,61),(33,62,38,67),(34,68,39,63),(35,64,40,69),(41,76,46,71),(42,72,47,77),(43,78,48,73),(44,74,49,79),(45,80,50,75),(51,82,56,87),(52,88,57,83),(53,84,58,89),(54,90,59,85),(55,86,60,81)], [(1,61),(2,67),(3,63),(4,69),(5,65),(6,62),(7,68),(8,64),(9,70),(10,66),(11,75),(12,71),(13,77),(14,73),(15,79),(16,80),(17,76),(18,72),(19,78),(20,74),(21,81),(22,87),(23,83),(24,89),(25,85),(26,86),(27,82),(28,88),(29,84),(30,90),(31,91),(32,97),(33,93),(34,99),(35,95),(36,96),(37,92),(38,98),(39,94),(40,100),(41,107),(42,103),(43,109),(44,105),(45,101),(46,102),(47,108),(48,104),(49,110),(50,106),(51,117),(52,113),(53,119),(54,115),(55,111),(56,112),(57,118),(58,114),(59,120),(60,116)], [(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30),(31,32,33,34,35),(36,37,38,39,40),(41,42,43,44,45),(46,47,48,49,50),(51,52,53,54,55),(56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70),(71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110),(111,112,113,114,115,116,117,118,119,120)], [(1,5),(2,4),(6,8),(9,10),(11,12),(13,15),(16,17),(18,20),(21,22),(23,25),(26,27),(28,30),(31,37),(32,36),(33,40),(34,39),(35,38),(41,50),(42,49),(43,48),(44,47),(45,46),(51,60),(52,59),(53,58),(54,57),(55,56),(61,70),(62,69),(63,68),(64,67),(65,66),(71,80),(72,79),(73,78),(74,77),(75,76),(81,82),(83,90),(84,89),(85,88),(86,87),(91,97),(92,96),(93,95),(98,100),(101,107),(102,106),(103,105),(108,110),(111,117),(112,116),(113,115),(118,120)]])

111 conjugacy classes

 class 1 2A 2B ··· 2F 2G 2H 2I 2J 3A 3B 4A 4B 4C 4D 4E 4F 5A 5B 6A 6B 6C ··· 6L 6M ··· 6T 10A ··· 10F 10G ··· 10N 12A 12B 12C 12D 12E ··· 12L 15A 15B 15C 15D 20A 20B 20C 20D 30A ··· 30L 30M ··· 30AB 60A ··· 60H order 1 2 2 ··· 2 2 2 2 2 3 3 4 4 4 4 4 4 5 5 6 6 6 ··· 6 6 ··· 6 10 ··· 10 10 ··· 10 12 12 12 12 12 ··· 12 15 15 15 15 20 20 20 20 30 ··· 30 30 ··· 30 60 ··· 60 size 1 1 2 ··· 2 10 10 10 10 1 1 2 2 10 10 10 10 2 2 1 1 2 ··· 2 10 ··· 10 2 ··· 2 4 ··· 4 2 2 2 2 10 ··· 10 2 2 2 2 4 4 4 4 2 ··· 2 4 ··· 4 4 ··· 4

111 irreducible representations

 dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 4 4 4 4 type + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C3 C6 C6 C6 C6 C6 D5 D10 D10 D10 C3×D5 C6×D5 C6×D5 C6×D5 2+ 1+4 C3×2+ 1+4 D4⋊6D10 C3×D4⋊6D10 kernel C3×D4⋊6D10 C3×C4○D20 C3×D4×D5 C3×D4⋊2D5 C6×C5⋊D4 D4×C30 D4⋊6D10 C4○D20 D4×D5 D4⋊2D5 C2×C5⋊D4 D4×C10 C6×D4 C2×C12 C3×D4 C22×C6 C2×D4 C2×C4 D4 C23 C15 C5 C3 C1 # reps 1 2 4 4 4 1 2 4 8 8 8 2 2 2 8 4 4 4 16 8 1 2 4 8

Matrix representation of C3×D46D10 in GL6(𝔽61)

 1 0 0 0 0 0 0 1 0 0 0 0 0 0 47 0 0 0 0 0 0 47 0 0 0 0 0 0 47 0 0 0 0 0 0 47
,
 60 0 0 0 0 0 0 60 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 60 0 0 0 0 60 0 0 0
,
 60 0 0 0 0 0 0 60 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0
,
 0 43 0 0 0 0 17 18 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 60 0 0 0 0 0 0 60
,
 1 0 0 0 0 0 60 60 0 0 0 0 0 0 1 0 0 0 0 0 0 60 0 0 0 0 0 0 60 0 0 0 0 0 0 1

G:=sub<GL(6,GF(61))| [1,0,0,0,0,0,0,1,0,0,0,0,0,0,47,0,0,0,0,0,0,47,0,0,0,0,0,0,47,0,0,0,0,0,0,47],[60,0,0,0,0,0,0,60,0,0,0,0,0,0,0,0,0,60,0,0,0,0,60,0,0,0,0,1,0,0,0,0,1,0,0,0],[60,0,0,0,0,0,0,60,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,1,0,0],[0,17,0,0,0,0,43,18,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,60,0,0,0,0,0,0,60],[1,60,0,0,0,0,0,60,0,0,0,0,0,0,1,0,0,0,0,0,0,60,0,0,0,0,0,0,60,0,0,0,0,0,0,1] >;

C3×D46D10 in GAP, Magma, Sage, TeX

C_3\times D_4\rtimes_6D_{10}
% in TeX

G:=Group("C3xD4:6D10");
// GroupNames label

G:=SmallGroup(480,1141);
// by ID

G=gap.SmallGroup(480,1141);
# by ID

G:=PCGroup([7,-2,-2,-2,-2,-3,-2,-5,555,1571,18822]);
// Polycyclic

G:=Group<a,b,c,d,e|a^3=b^4=c^2=d^10=e^2=1,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c=d*b*d^-1=b^-1,b*e=e*b,d*c*d^-1=e*c*e=b^2*c,e*d*e=d^-1>;
// generators/relations

׿
×
𝔽


text:
# Positive Solutions for the Neumann p-Laplacian with Superdiffusive Reaction

Bulletin of the Malaysian Mathematical Sciences Society, Aug 2015

We consider a generalized logistic equation driven by the Neumann p-Laplacian and with a reaction that exhibits a superdiffusive kind of behavior. Using variational methods based on the critical point theory, together with truncation and comparison techniques, we show that there exists a critical value $\lambda _*>0$ of the parameter, such that if $\lambda >\lambda _*$, the problem has at least two positive solutions, if $\lambda =\lambda _*$, the problem has at least one positive solution and it has no positive solution if $\lambda \in (0,\lambda _*)$. Finally, we show that for all $\lambda \geqslant \lambda _*$, the problem has a smallest positive solution.

This is a preview of a remote PDF: https://link.springer.com/content/pdf/10.1007%2Fs40840-015-0212-3.pdf

Leszek Gasiński, Nikolaos S. Papageorgiou. Positive Solutions for the Neumann p-Laplacian with Superdiffusive Reaction, Bulletin of the Malaysian Mathematical Sciences Society, 2017, 1711-1731, DOI: 10.1007/s40840-015-0212-3


text:
# How do you solve -( - n + 7) = 25- 3n?

Jun 9, 2018

$n = 8$

#### Explanation:

First, distribute the negative sign (it flips the existing sign)
$n - 7 = 25 - 3 n$

Add 7 to both sides
$n = 32 - 3 n$

Add 3n to both sides
$4 n = 32$

Divide both sides by 4
$n = 8$

The key is to use the INVERSE operation (ex: addition/subtraction, multiplication/division)


text:
• Volume 109, Issue 2

May 1999,   pages  117-230

• Jacobian of meromorphic curves

The contact structure of two meromorphic curves gives a factorization of their jacobian.

• Degenerations of the moduli spaces of vector bundles on curves II (generalized Gieseker moduli spaces)

LetX0 be a projective curve whose singularity is one ordinary double point. We construct a birational modelG(n, d) of the moduli spaceU(n, d) of stable torsion free sheaves in the case (n, d)= 1, such that G(n, d) has normal crossing singularities and behaves well under specialization i.e. if a smooth projective curve specializes toX0, then the moduli space of stable vector bundles of rankn and degreed onX specializes toG(n, d). This generalizes an earlier work of Gieseker in the rank two case.

• On a new mean and functions of bounded deviation

The main purpose of the present paper is to introduce a new type of mean of a function and to study some of its special properties. In the last section we make use of this mean to show that a function of bounded deviation is not necessarily a function of bounded variation.

• Nonlinear boundary value problems

In this paper we consider two quasilinear boundary value problems. The first is vector valued and has periodic boundary conditions. The second is scalar valued with nonlinear boundary conditions determined by multivalued maximal monotone maps. Using the theory of maximal monotone operators for reflexive Banach spaces and the Leray-Schauder principle we establish the existence of solutions for both problems.

• # Proceedings – Mathematical Sciences

Current Issue
Volume 129 | Issue 3
June 2019


text:
# Tag Info

2

While I haven't managed to beat @Jaap's result I think I gained some good intuition on the problem that is worth while sharing: MAIN IDEA The centre piece of the advanced strategies we have seen is the following neat little trick: Assume for the sake of argument that the events each player can see form a fixed recurrent sequence and only the phase is random....

1

My friend Adam came up with a new solution while we were hiking up a steep mountain. I want to post it here, because it attacks the problem from a different angle.

2

There is a simple strategy that is pretty good: Probability calculation: There is however a better strategy in which outcomes are grouped together in triplets.

10

In this paper about Levine's hat puzzle there is a better strategy with a winning probability of $0.7$. Let $a_i$ be the coin toss outcomes that are told to $A$, and $b_i$ the ones that are given to $B$. This is a bit easier than having the tosses interleaved as a single sequence. The strategy is as follows: Now let's calculate the probability of winning.

Top 50 recent answers are included


text:
True or False

Number Theory Level 1

True or false:

$$\quad$$ $$3n$$ is divisible by 6 for all positive integer $$n$$.

×


text:
Comment
Share
Q)

# What is wavelength of sound with frequency 330hz.Take the speed in air 330

What is wavelength of sound with frequency 330hz.Take the speed in air 330 m/s


text:
# Fitting a theoretical density function with log-return sample

I am working on a exercise to fit a theoretical density function,presented below, with log-return series of a stock.

$\frac{1}{2}a e^{a(x-\mu)} \quad \text{if x } < \mu$

or

$\frac{1}{2}be^{-b(x-\mu)} \quad \text{if x } \geq \mu \\$

I am now stuck on estimating the parameters for the function. I have tried different approaches, such as creating a probability distribution and use FindDist.Param. and also tried to fit it with a ExponentialDistr(or Laplace), multiplying neg. returns with -1 for case $(x < \mu)$, however without any success. Given I have the moments of the sample and at least think I have worked out the correct moments (mean, var, skew, kurt) I would also like to try a method of moments, but not sure how to do proceed with the above function? As I am relatively new to Mathematica and keep getting lost but enjoying the endless possibilities, any suggestions or recommendations on procedure is greatly appreciated.

For importing my log-returns I have used ImportString["...", "List"].

Many thanks for the feedback guys! Alas, my original post was written 4 am in the morning. Anyway, I'll be more thorough in coming posts. The following is the procedure I have tried,

1. logretun = ImportString["r1, r2, .., rT", "List"] Where $r_i$ are daily log-returns of a stock, copied from an .csv spreadsheet.
2. Define the above density as a function,

f[x_] = 1/2*a*Exp[a*(x - m)]*Boole[x < m] +
1/2*b*Exp[-b*(x - m)]*Boole[x >= m]

(is the use of Boole correct?)

3. Define a function from (2) as distribution

fdist = ProbabilityDistribution[f, {x, -Infinity, Infinity}]

4. Run

FindDistributionParameter[logreturn, fdist]

With no luck. (Just runs and runs with no answer - even if I take a smaller sample, so I do not think its me just being impatient..).

I have gotten some values for $a$ and $b$ (87.76922562 and -82.00007391, respectively) through the following method; however I am not sure it is justifiable.

Step (3) piecewise, for a ExponentialDistribution[λ]:

1. Sort the return series, select the ones that are $x < \mu$ and multiply by -1 and import as "minuslogreturns"

2. Run step 3 as

FindDistributionParameter[minuslogreturns, ExponentialDistribution[a]]

3. Run again for the positive part, with $b$ as the parameter for ExponentialDistribution.

I am not confident that this is a feasible approach. I have tested my $a$ and $b$ by moments ($\mu$ = sample median), getting relative close mean, variance and skewness, however I cannot seem to get the kurtosis close. (sample kurtosis 9.494758128 vs estimated 15.33348985). On a side note, it might be very well that the expressions for expected return, variance, skewness and kurtosis also be off, as they are quite long expressions. I can post these if deemed necessary?

Hope this make my questions a bit more clearer.

-
Welcome to Mathematica.SE, Geoffrey!. FindDist.Param does not look like Mathematica syntax. Are you sure you are not getting your software mixed up? Are you using RLink? Please add some more information about what you have tried for estimating this model, i.e. all your estimation code, not just the specific functions used. –  Verbeia Mar 15 '13 at 5:39
@Verbeia I think the OP is a bit lazy and is arbitrarily shortening function names. He is referring to FindDistributionParameters, ExponentialDistribution, LaplaceDistribution. @Geoffrey: You have to be less sloppy with your questions if you like to get first class answers (or any answers at all). –  Sjoerd C. de Vries Mar 15 '13 at 7:17
Please post the exact code you used and describe your problems you encounter more accurately than "without any success". If you don't improve your question I will have to vote to close. –  Sjoerd C. de Vries Mar 15 '13 at 11:51


text:
# How do you solve lim_x->pi/4 (tanx-cotx)/(x-pi/4) ?

Jan 21, 2018

L'Hospital Rule yields:

${\lim}_{x \to \frac{\pi}{4}} \frac{\tan x - \cot x}{x - \frac{\pi}{4}} = 4$

#### Explanation:

by L'Hospital rule, which states that: when you get an indeterminate form after assuming that ${\lim}_{x \to c} f \left(x\right) = f \left(c\right)$, then ${\lim}_{x \to c} \frac{g \left(x\right)}{h \left(x\right)} = \frac{g ' \left(c\right)}{h ' \left(c\right)}$

$\implies {\lim}_{x \to \frac{\pi}{4}} \frac{{\left(\sec x\right)}^{2} + {\left(\csc x\right)}^{2}}{1}$

placing values of $x = \frac{\pi}{4}$

$= \frac{{\left(\sqrt{2}\right)}^{2} + {\left(\sqrt{2}\right)}^{2}}{1} = \frac{4}{1} = 4$

you can find the value of the limit as now it is not in the indertiminate form

hope you find it helpful :)

Aug 14, 2018

Here is a Solution without use of the L'Hospital's Rule.

#### Explanation:

Let, $x = \frac{\pi}{4} + y$.

$\therefore \text{ As } x \to \frac{\pi}{4} , y \to 0$.

Now, $\tan x - \cot x = \sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x$,

$= \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x}$,

$= \frac{- 2 \left({\cos}^{2} x - {\sin}^{2} x\right)}{2 \sin x \cos x}$,

$= \frac{- 2 \cos 2 x}{\sin 2 x}$,

$= - 2 \cot 2 x$.

$\therefore \text{ The Reqd. Lim.} = {\lim}_{x \to \frac{\pi}{4}} \frac{- 2 \cot 2 x}{x - \frac{\pi}{4}}$,

$= {\lim}_{y \to 0} \frac{- 2 \cot \left(2 \left(\frac{\pi}{4} + y\right)\right)}{y}$,

$= {\lim}_{y \to 0} \frac{- 2 \cot \left(\frac{\pi}{2} + 2 y\right)}{y}$,

$= {\lim}_{y \to 0} \frac{- 2 \left(- \tan 2 y\right)}{y}$,

$= {\lim}_{y \to 0} \left\{2 \cdot \frac{\tan 2 y}{2 y} \cdot 2\right\}$.

Knowing that, ${\lim}_{\theta \to 0} \tan \frac{\theta}{\theta} = 1$,

$\text{The Lim.} = 4 \cdot 1 = 4$.

Aug 14, 2018

#### Explanation:

Solution without L'Hospital Rule :

Let,

$L = {\lim}_{x \to \frac{\pi}{4}} \frac{\tan x - \cot x}{x - \frac{\pi}{4}}$

Here,

$\tan x - \cot x = \sin \frac{x}{\cos} x - \cos \frac{x}{\sin} x = \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x}$

$\therefore \tan x - \cot x = - \frac{{\cos}^{2} x - {\sin}^{2} x}{2 \sin x \cos x} \times 2$

$\therefore \tan x - \cot x = - 2 \frac{\cos 2 x}{\sin 2 x} = - 2 \cot 2 x$

So,

$L = {\lim}_{x \to \frac{\pi}{4}} \frac{- 2 \cot 2 x}{\left(x - \frac{\pi}{4}\right)} = - 2 {\lim}_{x \to \frac{\pi}{4}} \frac{\cot 2 x}{\left(x - \frac{\pi}{4}\right)}$

Let , $x - \frac{\pi}{4} = \theta \implies x = \frac{\pi}{4} + \theta \implies 2 x = \frac{\pi}{2} + 2 \theta$

$\mathmr{and} x \to \frac{\pi}{4} \implies \theta \to 0$

$\therefore L = - 2 {\lim}_{\theta \to 0} \frac{\cot \left(\frac{\pi}{2} + 2 \theta\right)}{\theta} = - 2 {\lim}_{\theta \to 0} \frac{- \tan 2 \theta}{\theta}$

$\therefore L = 2 {\lim}_{\theta \to 0} \frac{\tan 2 \theta}{2 \theta} \times 2 = 2 \left(1\right) \times 2 = 4$


text:
Version 7 (modified by diatchki, 10 months ago) (diff)

--

# Literate Markdown

Markdown has grown in popularity since github started encouraging people to write their documentation with it. Github highlights the source according to how it's labeled, so the haskell code blocks look nice, as do the HTML blocks. As an example, here is a blog post written in markdown and rendered by github's source view:

And here is the source:

As you can see, the source is a haskell module, and if GHC accepted .md or .markdown as input, this blog post would be executable.

## Current Literate Processing

Using "bird-tracks":

This is a comment.  Lines starting with '>' are the actual code.

> average xs = sum xs / length xs

Or, using the LaTeX compatible notation:

This is a comment.

\begin{code}
average = sum xs / length xs
\end{code}

Unfortunately, neither of this is compatible with mark-down: in mark-down the bird-tracks signify quoting (just like in e-mail clients) and, of course, \begin{code} is LaTeX.

## The Proposal

The idea is to extend Haskell's literate notation so that it is compatible with markdown, in the same way that \begin{code} makes it work with LaTeX. This is great for two reasons:

1. markdown is a simple language that is used by many programmers
2. there are many existing tools that know how to process the markdown notation (e.g., github, pandoc, etc.)

To support literate Haskell written in markdown we need two changes:

1. A new way to indicate what are the code parts in a literate Haskell file
2. (Optional, but nice.) Disable bird-tracks style Haskell blocks in markdown files, so that GHC does not accidentally interpret quotes as code.

### How to Recognize Code Blocks

A new code block would be started by  or haskell at the beginning of a line, the code starts on the next line. The code block is terminated by a line that starts with . For example:

# Heading

This is a comment, and next there will be some code:

haskell
x :: Int
x  = 10

`

Markdown supports snippets of code in different languages, which is why there is haskelll after the ticks. If a language is not explicitly specified, then we assume that the block contains Haskell---after all, we are working with literate Haskell files.

### When to Use Literate Markdown

Currently, GHC supports mixing various styles code blocks---one may have a literate Haskell file that contains both bird-tracks and LaTeX style literate Haskell. It is very unclear if this generality is ever useful.

For literate markdown we propose a different approach: the style of the literate Haskell file will be determined by the file's extension. So, if the compiler is handed a file ending in .md or .markdown, the common suffixes for markdown files, then it will unlit the file using the markdown convention. Ordinary .lhs files will be processed as usual, so this is fully backward compatible.

## Implementation

The above design has been implemented in https://github.com/elliottt/ghc/tree/literate-markdown .

The implementation involved the following:

1. Modify unlit to support the new form of code block
2. Don't get confused with CPP handline (see bug #4836)
3. Make GHC look for modules in .md and .markdown files.
4. TODO when unliting markdown, do not recognize bird-tracks as code blocks.


text:
# Al Khwarizmi’s Identity and the Quadratic Formula

The quadratic formula is a main tool for understanding quadratic relations. It is based on al Khwarizmi's technique of completing the square.
Al Khwarizmi is often considered the father of algebra, due to an influential text he wrote, and his name is the origin of the term algorithm. His `completing-the-square’ technique lies at the heart of a beautiful formula that we call al Khwarizmi’s identity. The usual quadratic formula is a consequence.
• will see how the completing-the-square leads to al Khwarizmi’s identity
• how the quadratic formula follows from al Khwarizmi’s identity
On this page is a very important derivation. Make sure you proceed slowly and carefully, please check all the steps by writing things out line by line, and then apply your understanding to work out the answers to the questions. This is how we learn mathematics.

## Solving a quadratic equation using completing the square

Suppose we want to solve the quadratic equation
[Large{x^2-14x-1887=0}.]
Half of the coefficient of (normalsize{x}) is (normalsize{-7}), so we take the (normalsize{1887}) to the other side, and add the square of (normalsize{-7}) to both sides. This gives
[Large{x^2-14x+49=1887+49=1936}.]
Now we rewrite the left-hand side as a perfect square:
[Large{(x-7)^2=1936}.]
At this stage we have to “take the square root” of (normalsize{1936}). What does this mean? It means finding a number (normalsize{r}) with the property that (normalsize{r^2=1936}). In this case such a number actually exists: it is (normalsize{r=44}). But otherwise we would just write (normalsize{pm sqrt{1936}}) to represent an approximate square root, and its negative. We can’t forget about the negative, since we want two solutions!
So in our case (normalsize{x-7=44}) or (normalsize{x-7=-44}). Thus we do get two solutions, namely (normalsize{x=51}) or (normalsize{x=-37}).

## Deriving al Khwarizmi’s identity

Now let’s apply this to transform the general quadratic polynomial
[Large{ax^2+bx+c}]
where (aneq 0). First step, factor out the (normalsize{a}) to get
[Large{ax^2+bx+c = a left( x^2+frac{b}{a}x+frac{c}{a}right) }.]
Add and subtract (normalsize{left(frac{b}{2a}right)^2}) inside the right hand expression:
[Large{ax^2+bx+c = a left( x^2+frac{b}{a}x + left(frac{b}{2a}right)^2 +frac{c}{a} -left(frac{b}{2a}right)^2right)}.]
Now there is a perfect square inside these brackets:
[Large{ax^2+bx+c = a left( left(x+frac{b}{2a}right)^2 +frac{c}{a} -left(frac{b}{2a}right)^2right)}.]
Since
[{Large frac{c}{a} -left(frac{b}{2a}right)^2 = frac{4ac-b^2}{(2a)^2}}]
we get
[Large{ax^2+bx+c = a left( left(x+frac{b}{2a}right)^2 +frac{4ac-b^2}{(2a)^2}right)}.]
Now for the last step, we multiply through by (normalsize{a}) to get
[Large{ax^2+bx+c=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}. label{1} tag 1}]
This is one of the really great derivations in mathematics, resulting in an essential formula, which we call al Khwarizmi’s identity. This step has lots of questions to give you practice with this formula!
Q1 (E): Apply the steps above to rewrite (normalsize{x^2+12x-85}) using al Khwarizmi’s identity.
Q2 (M): Apply the steps above to rewrite (normalsize{3x^2+5x-22}) using al Khwarizmi’s identity.

## Solving quadratic equations using al Khwarizmi’s identity

Once we have written a quadratic function in the form of al Khwarizmi’s identity, solving a quadratic equation involving it is relatively easy.
For example suppose that we want to solve (normalsize{x^2+12x-85=0}). Having already solved Q1, we rewrite this equation as
[Large{x^2+12x-85=(x+6)^2-121=0}]
and then as
[Large{(x+6)^2=121}.]
Now taking square roots, (normalsize{x+6=11}) or (normalsize{x+6=-11}) from which we deduce that (normalsize{x=5}) or (normalsize{x=-17}). These are the two solutions.
Q3 (M): i) Solve (normalsize{x^2+12x-85=-21}).
Q4 (M): i) Solve (normalsize{3x^2+5x-22=0}) using al Khwarizmi’s identity.

The familiar quadratic formula follows from al Khwarizmi’s identity.
If (normalsize{ax^2+bx+c = 0}) then
[Large{ax^2+bx+c=aleft(x+frac{b}{2a}right)^2+frac{4ac-b^2}{4a}=0}.]
So
[Large{ left(x+frac{b}{2a}right)^2 = frac{b^2-4ac}{(2a)^2}}.]
Taking square roots
[Large{ x+frac{b}{2a}=pmfrac{sqrt{b^2-4ac}}{2a}}.]
And finally, we isolate (x)
[Large{ x= frac{- b pmsqrt{b^2-4ac}}{2a}}.]
This is the famous equation that all students memorise.
Q5 (E): Use the quadratic formula to solve (normalsize{x^2+4x+3=0}).

Once we know how to solve quadratics, Descartes theorem allows us a simpler way to factor a quadratic expression: just find its zeroes first, and then each of these will correspond to a linear factor!
Q6 (E) Use the quadratic formula and Descartes’ theorem to factor (normalsize{x^2+x-132}).

A1. We transform the quadratic expression (normalsize{x^2+12x-85}) to
[Large{begin{align}x^2+12x-85&=x^2+12x+6^2-85-6^2\&=(x+6)^2-121.end{align}}]
A2. We transform
[Large{begin{align}3x^2+5x-22&=3Bigg(x^2+frac{5}{3}x+Bigg(frac{5}{6}Bigg)^2-frac{22}{3}-Bigg(frac{5}{6}Bigg)^2Bigg)\&=3Bigg(Bigg(x+frac{5}{6}Bigg)^2-frac{22}{3}-frac{25}{36}Bigg)\&=3Bigg(x+frac{5}{6}Bigg)^2-22-frac{25}{12}\&=3Bigg(x+frac{5}{6}Bigg)^2-frac{289}{12}.end{align}}]
A3.
We use the working of A1. to rewrite (normalsize{x^2+12x-85=-21}) as
[Large{x^2+12x-85=(x+6)^2-121=-21}.]
This gives
[Large{(x+6)^2=100}.]
In this case we can take square roots, giving (normalsize{x+6=10}) or (normalsize{x+6=-10}), from which we deduce that (normalsize{x=4}) or (normalsize{x=-16}). These are the two solutions.
A4.
We use the working of A2. to write (normalsize{3x^2+5x-22=0}) as
[Large{3x^2+5x-22=3Bigg(x+frac{5}{6}Bigg)^2-frac{289}{12}=0}.]
This gives
[Large{Bigg(x+frac{5}{6}Bigg)^2=frac{289}{36}}.]
So taking square roots, (normalsize{x+5/6=17/6}) or (normalsize{x+5/6=-17/6}) from which we deduce that (normalsize{x=12/6=2}) or (normalsize{x=-22/6=-11/3}). These are the two solutions.
A5. Using the quadratic formula we have:
[Large{x=frac{-4+sqrt{16-12}}{2}=-frac{2}{2}=-1}]
and
[Large{x=frac{-4-sqrt{16-12}}{2}=-frac{6}{2}=-3}.]
A6. Solving equation (normalsize{x^2+x-132=0}) we have zeroes
[Large{x=frac{-1+sqrt{1+528}}{2}=11}]
and
[Large{x=frac{-1-sqrt{1+528}}{2}=-12}]
hence
[Large{x^2+x-132=(x-11)(x+12)=0}.]
You are now in a position to systematically solve all quadratic factorisation problems, without any guess work. Thank you al Khwarizmi!

Created by


text:
# Density of polyhedral partitions

created by braidesa on 12 Sep 2016
modified by garroni on 18 Sep 2020

[BibTeX]

Published Paper

Inserted: 12 sep 2016
Last Updated: 18 sep 2020

Journal: Calc. Var. Partial Diff. Equations
Volume: 56
Number: 2
Pages: 10
Year: 2017
Doi: 10.1007/s00526-017-1108-x

Abstract:

We prove the density of polyhedral partitions in the set of finite Caccioppoli partitions. Precisely, we consider a decomposition $u$ of a bounded Lipschitz set $\Omega\subset{\mathbb R}^n$ into finitely many subsets of finite perimeter, which can be identified with a function in $SBV_{\rm loc}(\Omega;{\cal Z})$ with ${\cal Z}\subset {\mathbb R}^N$ a finite set of parameters. For all $\varepsilon>0$ we prove that such a $u$ is $\varepsilon$-close to a small deformation of a polyhedral decomposition $v_\varepsilon$, in the sense that there is a $C^1$ diffeomorphism $f_\varepsilon:{\mathbb R}^n\to{\mathbb R}^n$ which is $\varepsilon$-close to the identity and such that $u\circ f_\varepsilon-v_\varepsilon$ is $\varepsilon$-small in the strong $BV$ norm. This implies that the energy of $u$ is close to that of $v_\varepsilon$ for a large class of energies defined on partitions. Such types of approximations are very useful in order to simplify computations in the estimates of $\Gamma$-limits.


text:
# $\lim_{n\to\infty}\int f_n d\mu=0$ implies that $f_n \to 0$ a.e $\mu$

Let $(X,F,\mu)$ be finite measure space.

Let $f_n$ be sequence of measurable function from $X$ and $f_n\geq 0$ almost everywhere $\mu$.

Claim.


text:
Chapter 10 - Compositions, Inverses, and Combinations of Functions - 10.1 Composition of Functions - Exercises and Problems for Section 10.1 - Exercises and Problems - Page 401: 19

$$x^3+3x+1$$

Work Step by Step

We know that $f(x)$ is the function that is plugged into the original equation for x. Thus, it follows that $f(x)$ is: $$x^3+3x+1$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Has anyone ever used IdeaPaint?

Someone sent me the link to IdeaPaint which appears to be a paint that turns any surface into a whiteboard. It looks pretty appealing, but I was wondering:

1. Whether anyone here has ever used it and if so whether they are happy with it?
2. What if I want my wall to be a magnetic whiteboard? What might I do to the wall (and how) before applying the whiteboard paint, to make magnets stick to it?
-

I'm not familiar with this product, but it seem like the thing that you're looking for...

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Be fantastic if it can be used under dry erase paint! –  shirlock homes Mar 28 '11 at 10:40
My wife used something similar to this (it was put on with a brush, not in a spray can) under regular paint, and it worked fine. –  KeithB Mar 29 '11 at 15:37

I haven't used Ideapaint yet, but my wife is a teacher and I have used other similar products before to make dry erase boards for her. I did notice a display at Lowes last week in the paint dept of a dry erase paint. I had to play with it, and it seems to work great. So my point is, you can stop by a Lowes and actually try it out and check competitive prices etc.

If you want a magnetic board, you have to use a ferrous back board. Something made of or has a layer of steel foil. Aluminum or stainless steel is not magnetic and won't work, so ask for a ferrous material.

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I just saw Asaf's answer about Kyrlon magnetic paint. What a great idea! If this krylon paint can be used under the dry erase paint, you got a perfect combination. Guess we would have to read the krylon instructions carefully first. –  shirlock homes Mar 28 '11 at 10:39
hey, i've just googled a bit and found this... can't be any simpler then this :) –  Asaf Chertkoff Mar 28 '11 at 10:49
Will have to get ahold of some of these paints and try out the combo. May take me a while since I don't live in the US...unless someone beats me to it and posts their results:) –  Zippy Mar 28 '11 at 12:12
From my experience, the magnetic paints don't work. You need so many layers that at some point, it's cheaper/easier just to get some sheet metal and paint that. –  DA01 Nov 2 '11 at 3:28
I agree with DA01. After a long day of putting up magnetic primer, only the large neodymium magnets were strong enough to hold up papers. –  Arelius Feb 1 '12 at 8:07

I haven't used whiteboard paint, but I have used chalkboard paint. One note from this experience: The texture in any previous layers of paint is very significant if you're trying to write on the wall. The texture in this case was from a standard paint roller; it was quite palpable even after the chalkboard paint was applied. You'll want to apply primer to the surface in thin layers with a sprayer if you have access to one, or sand thoroughly if your surface is already painted and not smooth.

Also, drywall and layers of paint aren't very durable for day-to-day use. My low-use chalkboard is weathering fairly well. After 8 months of writing the last time someone took the dog out, there are wear patterns where it doesn't erase or take the chalk as well as it used to, but it's not chipped or worn through anywhere. I erase the times once a week. If you intend to use this frequently, consider choosing a different product. Most real whiteboards are on hardboard or steel backing layers.

The different option I'd suggest is to use tileboard. Here's a 1/8" thick 4x8 panel from Home Depot for $13. It's prohibitively expensive to coat your walls in commercial whiteboards, which run from$150 to \$300 for a 4x8 sheet, but panelboard or tileboard is cheap enough to coat the walls in. (Some would argue that IdeaPaint is also prohibitively expensive, but that could be debated). You can mount the panels with screws, but it will bubble. You should glue it to the wall or to a thicker piece of MDF or hardboard that won't bubble. You can also get plastic strips that clip across the gaps to make the joints less obnoxious. After 14 months of fairly heavy use as a daily whiteboard in my home office, I can say that I'm quite impressed with this stuff and will never pay the ridiculous sums that others charge to buy real whiteboards.

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I really like the chalkboard surface I painted on my basement refridgerator, but I might have use the chalkboard contact-paper had I known about it at the time. –  skiggety Sep 5 '14 at 17:21

I once worked at an office that had a very large IdeaPaint wall. I didn't do any of the painting, but I did use the wall as a whiteboard. I was pretty impressed; it erases approximately as well as a real whiteboard, and has a nice bright white color.

When they first put it in, there were a lot of signs warning us not to use it yet -- apparently it takes a few days to cure, and if you use it before it's cured, you can stain it permanently. So, have fun but read the directions!

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Rust-Oleum offers these two magnetic primers: here and here. I belive they can be used together with IdeaPaint.

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I have seen systems like "Magic Whiteboard" used in the UK and have been very inpressed with them, not a paint but does the same job.

Magic Whiteboard is made from statically charged polypropylene film. It allows you to create a whiteboard from a roll anywhere, in seconds.

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I've seen it in the US. Not for sale specifically, but my neighbor had brought a pack in from his office that he had found when cleaning up that they were no longer using. –  Joe Nov 2 '11 at 13:46

protected by Niall C.♦Oct 18 '12 at 15:21

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.


text:
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The numerator of a fraction is $4$less than the denominator. If the numerator is decreased by $2$ and the denominator is increased by $1$, then the denominator is eight times the numerator. The fraction is also written as $\dfrac{m}{{14}}$. What is the value of $m$?

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Hint: First make linear equations from the given information and then solve those equations in order to find the value of two numbers.

Let the numerator be $x$and denominator be$y$.
According to the question:
${\text{Case1: }}x = y - 4{\text{ }} \ldots \ldots \left( 1 \right) \\ {\text{Case2: }}8\left( {x - 2} \right) = y + 1 \\ \Rightarrow 8x - 16 = y + 1 \\ \Rightarrow 8x - y = 17{\text{ }} \ldots \ldots \left( 2 \right) \\$
Put the value of $x$from equation $\left( 1 \right)$in equation $\left( 2 \right)$and solve for$y$, we get:
$\Rightarrow 8\left( {y - 4} \right) - y = 17 \\ \Rightarrow 8y - 32 - y = 17 \\ \Rightarrow 7y = 49 \\ \Rightarrow y = 7 \\$
Now, put the value of$y$in equation$\left( 1 \right)$and solve for$x$, we get:
$\Rightarrow x = y - 4 \\ \Rightarrow x = 7 - 4 \\ \Rightarrow x = 3 \\$
Hence, the fraction $\dfrac{x}{y}$is equal to $\dfrac{3}{7}.$
Now, the given fraction is of the form $\dfrac{m}{{14}}$. So, we will also convert our fraction of the same form.
Therefore, multiplying and dividing the fraction by$2$, we get:
$\dfrac{x}{y}{\text{ = }}\dfrac{3}{7} \times \dfrac{2}{2} \\ \dfrac{x}{y} = \dfrac{6}{{14}}{\text{ }} \ldots \ldots \left( 3 \right) \\$
Now, the equation$\left( 3 \right)$is of the form $\dfrac{m}{{14}}$. By comparison, we get:
$m = 6$

Note- Whenever you see a problem like this, always try to identify the number of variables and try to make that many equations. Also, in order to compare two fractions, always make their denominator or numerator equal.


text:
ShowBoundsEffect¶

from panda3d.core import ShowBoundsEffect

class ShowBoundsEffect

Bases: RenderEffect

Applied to a GeomNode to cause a visible bounding volume to be drawn for this node. This is generally used only during development to help identify bounding volume issues.

Inheritance diagram

static getClassType()TypeHandle
getTight()bool

Returns true if the “tight” flag was set, meaning the effect should compute and draw the tight bounding-box of the node’s vertices every frame.

static make(tight: bool)RenderEffect

Constructs a new ShowBoundsEffect object.


text:
## Renormalization group and critical phenomena. II: Phase-space cell analysis of critical behavior.(English)Zbl 1236.82016

Summary: A generalization of the Ising model is solved, qualitatively, for its critical behavior. In the generalization the spin $$s_{\vec n}$$ at a lattice site $$\vec n$$ can take on any value from $$-\infty$$ to $$\infty$$. The interaction contains a quartic term in order not to be pure Gaussian. The interaction is investigated by making a change of variable $$s_{\vec n}=\sum_m\psi_m(\vec n)s_m'$$, where the functions $$\psi_m(\vec n)$$ are localized wavepacket functions. There are a set of orthogonal wave-packet functions for each order-of-magnitude range of the momentum $$\vec k$$. An effective interaction is defined by integrating out the wave-packet variables with momentum of order 1, leaving unintegrated the variables with momentum $$<0.5$$. Then the variables with momentum between 0.25 and 0.5 are integrated, etc. The integrals are computed qualitatively. The result is to give a recursion formula for a sequence of effective Landau-Ginsberg-type interactions. Solution of the recursion formula gives the following exponents: $$\eta=0, \gamma=1.22, \nu=0.61$$ for three dimensions. In five dimensions or higher one gets $$\eta=0, \gamma=1$$, and $$\nu=1/2$$, as in the Gaussian model (at least for a small quartic term). Small corrections neglected in the analysis may make changes (probably small) in the exponents for three dimensions.
For part I, cf. Phys. Rev. B (3) 4, No. 9, 3174–3183 (1971; Zbl 1236.82017).

### MSC:

 82B20 Lattice systems (Ising, dimer, Potts, etc.) and systems on graphs arising in equilibrium statistical mechanics 82B28 Renormalization group methods in equilibrium statistical mechanics

Zbl 1236.82017
Full Text:

### References:

 [1] K. Wilson, Phys. Rev. B 4 pp 3174– (1971) · Zbl 1236.82017 [2] L. P. Kadanoff, Physics 2 pp 263– (1966) [3] L. P. Kadanoff, Rev. Mod. Phys. 39 pp 395– (1967) [4] M. E. Fisher, Rept. Progr. Phys. 30 pp 731– (1967) [5] V. L. Ginsberg, Zh. Eksperim. i Teor. Fiz. 20 pp 1064– (1950) [6] J. Schrieffer, in: Superconductivity (1964) [7] K. Wilson, Phys. Rev. 140 pp B445– (1965) [8] K. Wilson, Phys. Rev. D 2 pp 1439– (1970) [9] A. A. Migdal, Zh. Eksperim. i Teor. Fiz. 59 pp 1015– (1970) [10] A. M. Polyakov, Zh. Eksperim. i. Teor. Fiz. 59 pp 542– (1970) [11] T. H. Berlin, Phys. Rev. 86 pp 821– (1952) · Zbl 0047.45703 [12] J. Langer, Phys. Rev. 137 pp A– (1965) [13] L. Landau, in: Quantum Mechanics (1958) [14] F. Hildebrand, in: Introduction to Numerical Analysis (1956) · Zbl 0070.12401
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Points of Authority

Geometry Level pending

Consider the rectangle $$ABCD$$ and points $$M, N, O, P$$ on the segments $$\overline{AB}$$, $$\overline{BC}$$, $$\overline{CD}$$ and $$\overline{AD}$$, respectively.

$$MO$$ and $$NP$$ are perpendicular to each, intersecting at the point $$X$$.

If $$\left [ AMXP \right ] = 6, \left [ PXOD \right ]= 8$$ and $$\left [ MBNX \right ] = 18$$ , evaluate $$\left [ MNOP \right ] + \left [ ABCD \right ]$$.

×


text:
# Sum of roots of a trigonometric equation

The equation $$\sin x + \sqrt{3} \cos x = 1,$$ where $x \in [-2\pi, 2\pi]$ has four roots $-\dfrac{3\pi}{2}$, $\dfrac{\pi}{2}$, $\dfrac{\pi}{2}$ and $\dfrac{11\pi}{6}$. Sum of roots of the given equation is $\dfrac{2\pi}{3}$. How do I tell Mathematica to do that?

I tried

Solve[{Sin[x] + Sqrt[3]*Cos[x] == 1, -2*\[Pi] <= x <= 2*\[Pi]}, {x}]

edit from comment

I repaired the given equation to

Solve[{Sin[x] + Sqrt[3]Cos[x] == 1, -2[Pi] <= x <= 2*[Pi]}, {x}]

and used

Simplify@Total[ x /. Solve[{Sin[x] + Sqrt[3] Cos[x] == 2, -2 Pi <= x <= 2 Pi}, {x}]],

-4 [Pi] + 8 ArcTan[1/(2 + Sqrt[3])].

How do I get the answer -((5 [Pi])/3)?

-

One way :

Total[Reduce[Sin[x] + Sqrt[3] Cos[x] == 1 && -2 \[Pi] <= x <= 2 \[Pi], x] /. Or -> List /. x == rhs_ -> rhs] // Simplify

(* (2 \[Pi])/3 *)
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Thank Artes very much. –  minthao_2011 Sep 28 '12 at 8:18

FullSimplify @ Total[ DeleteDuplicates[ x
/. Solve[{Sin[x] + Sqrt[3] Cos[x] == 2, -2 Pi <= x <= 2 Pi}, {x}]]]

returns

$\dfrac{-5 \pi}{3}$

Like Artes says you need DeleteDuplicates[] because of the multiple roots.

From the documentation:

FullSimplify[expr]
tries a wide range of transformations on expr involving elementary and special functions, and returns the simplest form it finds.

-


text:
# How do you multiply (- 5k ^ { 2} j ^ { 4} ) ^ { 3} ( k ^ { 7} j ^ { 7} ) ^ { 3}?

Feb 4, 2017

See the entire solution process below:

#### Explanation:

First, expand the terms in parenthesis using this rule for exponents:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left(- 5 {k}^{\textcolor{red}{2}} {j}^{\textcolor{red}{4}}\right)}^{\textcolor{b l u e}{3}} {\left({k}^{\textcolor{red}{7}} {j}^{\textcolor{red}{7}}\right)}^{\textcolor{b l u e}{3}} = \left(- {5}^{\textcolor{b l u e}{3}} {k}^{\textcolor{red}{2} \times \textcolor{b l u e}{3}} {j}^{\textcolor{red}{4} \times \textcolor{b l u e}{3}}\right) \left({k}^{\textcolor{red}{7} \times \textcolor{b l u e}{3}} {j}^{\textcolor{red}{7} \times \textcolor{b l u e}{3}}\right) =$

$\left(- 125 {k}^{6} {j}^{12}\right) \left({k}^{21} {j}^{21}\right)$

Next, rearrange the expression grouping like terms:

$- 125 \left({k}^{6} {k}^{21}\right) \left({j}^{12} {j}^{21}\right)$

We can now use this rule for exponents to complete the multiplication: ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$- 125 \left({k}^{\textcolor{red}{6}} \times {k}^{\textcolor{b l u e}{21}}\right) \left({j}^{\textcolor{red}{12}} \times {j}^{\textcolor{b l u e}{21}}\right) = - 125 {k}^{\textcolor{red}{6} + \textcolor{b l u e}{21}} {j}^{\textcolor{red}{12} + \textcolor{b l u e}{21}} =$

$- 125 {k}^{27} {j}^{33}$


text:
model category

for ∞-groupoids

∞-Lie theory

# Contents

## Idea

There exist various model category structures which present the homotopy theory of L-∞ algebras.

By definition L-∞ algebras are the ∞-algebras in the category of chain complexes over the Lie operad. As such they carry a model structure on algebras over an operad. There is a strictification which leads equivalently to a model structure on dg-Lie algebras.

A more geometric way is to think of L-∞ algebras as being the tangent spaces to connected smooth ∞-groupoids, hence to the delooping/moduli ∞-stacks $BG$ of smooth ∞-groups, hence as the first order infinitesimal neighbourhood

$B𝔤↪BG$\mathbf{B}\mathfrak{g} \hookrightarrow \mathbf{B}G

of the essentially unique point $*\to BG$ (see at Lie differentiation). Since this is equivalently the first order neighbourhood of the formal neighbourhood (the jets) and the formal neighbourhood can be described purely in terms of associative algebra/coalgebra (see at smooth algebra) many models for ${L}_{\infty }$-algebras are formulated in terms of such data.

In particular, one succinct way to present L-∞ algebras (as discussed there) is as dg-coalgebras:

###### Proposition/Definition

An L-∞ algebra $\left(𝔤,\left[-\right],\left[-,-\right],\left[-,-,-\right],\cdots \right)$ structure on a graded vector space $𝔤$ is equivalently a dg-coalgebra structure on the graded-commutative cofree coalgebra over $𝔤$.

Conversely, the category of L-∞ algebras (and general “weak” morphisms between them) is the full subcategory of that of counital cocommutative dg-coalgebras on those whose underlying bare graded-commutative coalgebra (forgetting the codifferential) is free

${L}_{\infty }\mathrm{Alg}↪\mathrm{dgCoCAlg}\phantom{\rule{thinmathspace}{0ex}}.$L_\infty Alg \hookrightarrow dgCoCAlg \,.

Accordingly it is of interest to have also model structure on dg-coalgebras or dually (on pro-objects of) dg-algebras which presents ${L}_{\infty }$-algebras. Prop. 8 below identifies the category ${L}_{\infty }\mathrm{Alg}$ as the (full sub-)category of fibrant objects inside such a model structure of “differential graded formal spaces”, which in turn is related by a zig-zag of Quillen equivalence to various other models

###### Remark/Warning

So we write ”${L}_{\infty }\mathrm{Alg}$” here for the 1-category of ${L}_{\infty }$-algebras and general morphisms between them, since this is an entry on model category presentations. If we want to refer to the (∞,1)-category of ${L}_{\infty }$-algebras we here write explicitly ”${L}_{W}\left({L}_{\infty }\mathrm{Alg}\right)$”, referring to the simplicial localization of this 1-category.

###### Remark/Warning

All gradings in the following are $ℤ$-gradings, unless explicitly stated otherwise. In terms of the underlying geometry this means that we are dealing with derived geometry (see below the section Simplicial sheaves over comsimplicial formal spaces for details): the algebra elements in positive degree correspond to categorical/simplicial/∞-groupoid/∞-stack-degree, and those in negative degree to the cosimplicial degree of the derived site of cosimplicial formal spaces.

Technically this affects for instance the nature of fibrations: for instance the model structure on dg-Lie algebras below is transferred from a model structure on chain complexes. For unbounded chain complexes this is the ”Categorical projective class structure” whose fibrations are the chain maps that are surjective in every degree. This appears for instance in prop. 2 and prop. 9 below.

On the other hand, if one considered chain complexes in non-negative degree (for tangent complexes in “higher but non-derived geometry”), then one would use the Projective structure on chain complexes in non-negative degree. This has as fibrations precisely the chain maps that are surjective in every positive degree. This case is (currently) not discussed in the following.

###### Remark/Warning

Some of the model structures below are on the category of ${L}_{\infty }$-algebras with “strict” morphisms between them, namely for those morphisms which are morphisms of algebras over an operad for an ${L}_{\infty }$-algebra regarded as an algebra over a cofibrant resolution of the Lie operad. We write

${L}_{\infty }{\mathrm{Alg}}_{\mathrm{str}}\to {L}_{\infty }\mathrm{Alg}$L_\infty Alg_{str} \to L_\infty Alg

for the wide subcategory on the strict ${L}_{\infty }$-morphisms.

## Definition as algebras over an operad

As their name indicates, L-∞ algebras are the homotopy algebras over the Lie operad (in a category of chain complexes). As such, the general theory of model structures on algebras over an operad provides a model category structure on the category of ${L}_{\infty }$-algebras.

This we discuss here. But there is also a natural identification of ${L}_{\infty }$-algebras with infinitesimal derived ∞-stacks. For expressing this a host of other, Quillen equivalent model structures are available. These we discuss below in Definitions as formal/infinitesimal ∞-stacks.

By the general discussion at model structure on dg-algebras over an operad, if $k$ is a field which contains the field of rational numbers, then for every symmetric operad (uncolored) $𝒪$ in the category of chain complexes (unbounded) ${\mathrm{Ch}}_{•}\left(k\right)$, the free-forgetful adjunction

$\mathrm{Alg}\left(𝒪\right)\stackrel{\stackrel{F}{←}}{\underset{U}{\to }}{\mathrm{Ch}}_{•}\left(k\right)$Alg(\mathcal{O}) \stackrel{\overset{F}{\leftarrow}}{\underset{U}{\to}} Ch_\bullet(k)

between the algebras over an operad and the underlying chain complexes induces a transferred model structure from the projective unbounded model structure on chain complexes where hence on both sides

• the weak equivalences are the morphisms that are quasi-isomorphisms on the (underlying) chain complexes;

• the fibrations are the morphisms that are degreewise surjections on the (underlying) chain complexes.

Hence in particular $\left(F⊢U\right)$ is Quillen adjunction between these model structures.

So this is in particular true for $𝒪=\stackrel{^}{\mathrm{Lie}}$ the standard cofibrant resolution of the Lie operad. In this case $\mathrm{Alg}\left(\stackrel{^}{\mathrm{Lie}}\right)\simeq {L}_{\infty }{\mathrm{Alg}}_{\mathrm{str}}$ is the category of (unbounded) ${L}_{\infty }$-algebras (with strict ${L}_{\infty }$-maps between them as in remark 3 above) and hence is equipped with a transferred model structure this way

${L}_{\infty }{\mathrm{Alg}}_{\mathrm{str}}\left(k\right)\stackrel{\stackrel{F}{←}}{\underset{U}{\to }}{\mathrm{Ch}}_{•}\left(k\right)\phantom{\rule{thinmathspace}{0ex}}.$L_\infty Alg_{str}(k) \stackrel{\overset{F}{\leftarrow}}{\underset{U}{\to}} Ch_\bullet(k) \,.

Moreover, by the rectification result discussed at model structure on dg-algebras over an operad, the resolution map $\stackrel{^}{\mathrm{Lie}}\stackrel{\simeq }{\to }\mathrm{Lie}$ induces a Quillen equivalence

${L}_{\infty }{\mathrm{Alg}}_{\mathrm{str}}\left(k\right)\stackrel{\simeq }{\to }\mathrm{dgLieAlg}\left(k\right)$L_\infty Alg_{str}(k) \stackrel{\simeq}{\to} dgLieAlg(k)

with the model structure on dg-Lie algebras, similarly transferred from the model structure on chain complexes.

## Definitions as formal/infinitesimal $\infty$-stacks

We list here definitions of various further model category structures that all present the (∞,1)-category of L-∞ algebras and describe a web of zig-zags of Quillen equivalences between them. These Quillen equivalences may be thought of as presenting an equivalence between the (∞,1)-category of ${L}_{\infty }$-algebras and that of infinitesimal derived ∞-stacks (“formal moduli problems”).

### Summary

The following tabulates the main categories considered below, the functors relating them and their homotopy theoretic nature. The last row points to the relevant definitions and propositions of the following text.

L-∞ algebrasform Chevalley-Eilenberg algebrapro-objects in commutative Artin dg-algebrasdualizecommutative dg-coalgebraform tangentsdg-Lie algebras
${L}_{\infty }\mathrm{Alg}$$\stackrel{\mathrm{CE}}{↪}$$\mathrm{Pro}\left(\mathrm{dgArtinCAlg}{\right)}^{\mathrm{op}}$$\stackrel{\left(-{\right)}^{*}}{↪}$$\mathrm{dgCoCAlg}$$\stackrel{ℒ}{\to }$$\mathrm{dgLieAlg}$
$=:\mathrm{dgFormalSpace}$
category of fibrant objectsequivalence of (∞,1)-categories under simplicial localizationopposite model structure of cofibrantly generated model categoryleft Quillen equivalencemodel categoryleft Quillen equivalencecofibrantly generated model category
prop. 1def. 8def. 7prop. 12def. 3prop. 3def. 1

Here we are trying to use suggestive names of the categories involved. The notation used here corresponds to that in (Pridham) by the following dictionary

(handle with care, may still need attention)

notation used herenotation in Pridham
$\mathrm{DerivedFormalSpace}$, def. 6$\mathrm{scSp}$, def. 1.32
$\mathrm{dgFormalSpace}$, def. 7${\mathrm{DG}}_{ℤ}\mathrm{Sp}$. def. 3.1
${\mathrm{FormalSpace}}^{{\Delta }^{\mathrm{op}}}$$\mathrm{sDGSp}$, def. 4.6

### On dg-Lie algebras

Let $k$ be a field of characteristic 0.

###### Definition

Write ${\mathrm{dgLieAlg}}_{k}\in \mathrm{Cat}$ for the category of dg-Lie algebras over $k$.

###### Proposition

The category ${\mathrm{dgLieAlg}}_{k}$ carries a cofibrantly generated model category structure in which

on the underlying chain complexes. This is the transferred model structure of the corresponding model structure on chain complexes along the forgetful functor to the category of chain complexes.

We call this the model structure on dg-Lie algebras.

###### Definition

Write

$𝒞\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}{\mathrm{dgLieAlg}}_{k}\to {\mathrm{dgCoCAlg}}_{k}$\mathcal{C} \;\colon\; dgLieAlg_k \to dgCoCAlg_k

for the functor which sends a dg-Lie algebra $\left(𝔤,d,\left[-,-\right]\right)$ to the dg-coalgebra whose underlying coalgebra is free on the underlying graded vector space $𝔤$ and whose coderivation is given by

$\delta :{v}_{1}↦d{v}_{1}$\delta \colon v_1 \mapsto d v_1
$\delta :\left({v}_{1},{v}_{2}\right)↦\left[{v}_{1},{v}_{2}\right]$\delta \colon (v_1, v_2) \mapsto [v_1, v_2]

and then extended as a coderivation.

###### Proposition

The functor from def. 2 has a left adjoint

$ℒ\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}{\mathrm{dgCoCAlg}}_{k}\to {\mathrm{dgLieAlg}}_{k}\phantom{\rule{thinmathspace}{0ex}}.$\mathcal{L} \;\colon\; dgCoCAlg_k \to dgLieAlg_k \,.

(Quillen, App. B6) (Hinich98, 1.2.1, 2.2.5) See also (Pridham, def. 3.23).

###### Remark

If one thinks of a dg-coalgebra as presenting a a derived formal space, as discuss below then its image under $ℒ$, prop. 3, may be thought of as its tangent dg-Lie algebra. Therefore $ℒ$ is also called the tangent Lie algebra functor.

### On dg-coalgebras

Let $k$ be a field of characteristic 0.

###### Definition

Write ${\mathrm{dgCoCAlg}}_{k}\in \mathrm{Cat}$ for the category of co-commutative counital $ℤ$-graded dg-coalgebras over $k$.

###### Proposition

There exists a model category structure on ${\mathrm{dgCoCAlg}}_{k}$ for which

(Hinich98, theorem 3.1) See also (Pridham, lemma 3.25).

We call this the model structure on dg-coalgebras.

###### Proposition

The pair of adjoint functors

$\left(ℒ⊣𝒞\right)\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}{\mathrm{dgLie}}_{k}\stackrel{\stackrel{ℒ}{←}}{\underset{𝒞}{\to }}{\mathrm{dgCoCAlg}}_{k}$(\mathcal{L} \dashv \mathcal{C}) \;\colon\; dgLie_k \stackrel{\overset{\mathcal{L}}{\leftarrow}}{\underset{\mathcal{C}}{\to}} dgCoCAlg_k

from prop. 3 constitutes a Quillen equivalence between the model structure on dg-Lie algebras, prop. 2, and the model structure on dg-coalgebras, prop. 4.

###### Remark

Since every object in ${\mathrm{dgCoCAlg}}_{k}$ is a cofibrant object and every object in ${\mathrm{dgLie}}_{k}$ is a fibrant object, the composite

$𝒞ℒ:{\mathrm{dgCoCAlg}}_{k}\to {\mathrm{dgCocAlg}}_{k}$\mathcal{C}\mathcal{L} \colon dgCoCAlg_k \to dgCocAlg_k

is already its derived functor and the unit

$𝔤\to 𝒞ℒ𝔤$\mathfrak{g} \to \mathcal{C}\mathcal{L}\mathfrak{g}

is a weak equivalence that exhibits $𝒞ℒ𝔤$ as a fibrant resolution and moreover, if $𝔤$ was already fibrant, hence by prop. 8 below an L-∞ algebra, as a strictification of $𝔤$: because a dg-Lie algebra is an ${L}_{\infty }$-algebra in which the Lie bracket satisfies its Jacobi identity strictly (not just up to a homotopy measured by the trinary bracket) and in which the “Jacobiator identity” holds strictly, etc.

### On simplicial presheaves over cosimplicial formal spaces

###### Definition

Write

$\mathrm{InfThPoint}↪{\mathrm{Alg}}^{\mathrm{op}}$InfThPoint \hookrightarrow Alg^{op}

for the category of infinitesimally thickened points, the full subcategory of the opposite category of Artin algebras (“Weil alebras” in the language of synthetic differential geometry). The category of infinitesimally thickened points.

Write

$\mathrm{cInfThPoint}↪{\mathrm{sAlg}}^{\mathrm{op}}\simeq \left({\mathrm{Alg}}^{\mathrm{op}}{\right)}^{\Delta }$cInfThPoint \hookrightarrow sAlg^{op} \simeq (Alg^{op})^{\Delta}

for the full subcategory of the opposite category on simplicial algebras on those which are Artinian (or “Weil” ): the category of cosimplicial infinitesimally thickened points.

###### Definition

Write

$\mathrm{FormalSpace}≔{\mathrm{Fun}}^{\mathrm{lex}}\left({\mathrm{InfThPoints}}^{\mathrm{op}},\mathrm{Set}\right)$FormalSpace \coloneqq Fun^{lex}(InfThPoints^{op}, Set)

for the full subcategory of the category of presheaves over infinitesimally thickened points on those given by left exact functors.

In (Pridham) this is def. 1.18.

###### Definition

Write

$\mathrm{DerivedFormalSpace}≔{\mathrm{Fun}}^{\mathrm{lex}}\left({\mathrm{cInfThPoints}}^{\mathrm{op}},\mathrm{sSet}\right)$DerivedFormalSpace \coloneqq Fun^{lex}(cInfThPoints^{op}, sSet)

for the full subcategory of the category of simplicial presheaves over cosimplicial infinitesimally thickened points on those given by left exact functors.

In (Pridham) this is def. 1.32.

###### Theorem

There exists a cofibrantly generated model category structure on $\mathrm{DerivedFormalSpace}$ whose

(…)

This is (Pridham, def. 2.7, theorem 2.14)

###### Proposition

Between quasi-smooth objects in $\mathrm{DerivedFormalSpace}$ the weak equivalences are precisely the morphisms which are weak homotopy equivalences of simplicial sets over each object in $\mathrm{cInfThPoint}$.

This is (Pridham, cor. 2.16).

### On dg formal spaces

###### Definition

Write

${\mathrm{dgArtCAlg}}_{k}\in \mathrm{Cat}$dgArtCAlg_k \in Cat

for the category of graded commutative Artin dg-algebras over $k$.

Write

$\mathrm{Pro}\left({\mathrm{dgArtCAlg}}_{k}\right)\in \mathrm{Cat}$Pro(dgArtCAlg_k) \in Cat

for its category of pro-objects and write

$\mathrm{dgFormalSpace}≔\mathrm{Pro}\left({\mathrm{dgArtCAlg}}_{k}{\right)}^{\mathrm{op}}$dgFormalSpace \coloneqq Pro(dgArtCAlg_k)^{op}

for the opposite category of that.

This is (Pridham, def. 3.1).

###### Proposition

There is a cofibrantly generated model category structure on $\mathrm{Pro}\left({\mathrm{dgArtinCAlg}}_{k}\right)$, def. 7 – hence an opposite model structure on $\mathrm{dgFormalSpaces}$ – whose weak equivalences are those morphisms that are local morphisms relative to quasi-smooth maps in the homotopy category of the slice category over their codomain.

This is (Pridham, prop. 4.36).

###### Definition

Write

$\mathrm{CE}\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}{L}_{\infty }\mathrm{Alg}\stackrel{}{\to }\mathrm{Pro}\left({\mathrm{dgArtinCAlg}}_{k}{\right)}^{\mathrm{op}}$CE \;\colon\; L_\infty Alg \stackrel{}{\to} Pro(dgArtinCAlg_k)^{op}

which regards an ${L}_{\infty }$-algebra $𝔤$ as a dg-coalgebra by prop. 1 and then forms the linear dual dg-algebra, the Chevalley-Eilenberg algebra $\mathrm{CE}\left(𝔤\right)$ of $𝔤$.

###### Proposition

${L}_{\infty }$-algebras are precisely the fibrant objects in $\mathrm{dgFormalSpace}$: the Chevalley-Eilenberg algebra functor of def. 8,

${L}_{\infty }\mathrm{Alg}\stackrel{\mathrm{CE}}{\to }\mathrm{Pro}\left({\mathrm{dgArtinCAlg}}_{k}{\right)}^{\mathrm{op}}\simeq \mathrm{dgFormalSpace}$L_\infty Alg \stackrel{CE}{\to} Pro(dgArtinCAlg_k)^{op} \simeq dgFormalSpace

is an equivalence of categories onto its essential image, which are the fibrant objects of $\mathrm{dgFormalSpace}$:

$\mathrm{CE}\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}{L}_{\infty }\mathrm{Alg}\stackrel{\simeq }{\to }{\mathrm{dgFormalSpace}}_{\mathrm{fib}}↪\mathrm{dgFormalSpace}\phantom{\rule{thinmathspace}{0ex}}.$CE \;\colon\; L_\infty Alg \stackrel{\simeq}{\to} dgFormalSpace_{fib} \hookrightarrow dgFormalSpace \,.

This is proven inside the proof of (Pridham, prop. 4.42).

###### Remark

In particular this shows that the category ${L}_{\infty }\mathrm{Alg}$ of prop/def. 1 carries the structure of a category of fibrant objects that presents the homotopy theory of ${L}_{\infty }$-algebras. Notice that, of course, passing to the full subcategory of fibrant objects does not change the homotopy theory presented by the underlying category with weak equivalences in that we have an equivalence of (∞,1)-categories between the simplicial localizations

${L}_{W}\left(\mathrm{dgFormalSpace}\right)\simeq {L}_{W}\left({L}_{\infty }\mathrm{Alg}\right)\phantom{\rule{thinmathspace}{0ex}}.$L_W(dgFormalSpace) \simeq L_W(L_\infty Alg) \,.

The following proposition characterizes the structure of this category of fibrant objects.

###### Proposition

The induced structure of a category of fibrant objects on ${L}_{\infty }\mathrm{Alg}$ under the inclusion of prop. 8 has

1. weak equivalences are precisely the maps that are quasi-isomorphisms on the underlying chain complexes;

2. fibrations include in particular the maps that are surjections on the underlying chain complexes.

###### Proof

The first statement is proven in the proof of (Pridham, prop. 4.42), the second follows by (Pridham, def. 4.34) with the existence of the model structure on $\mathrm{dgFormalSpaces}$.

###### Proposition

If $k$ is of characteristic 0 then there is a zig-zag of Quillen equivalences between $\mathrm{DerivedFormalSpace}$, def. 6 and $\mathrm{dgFormalSpace}$, def. 7, hence an equivalence of (∞,1)-categories between their simplicial localizations

${L}_{W}\mathrm{DerivedFormalSpace}\simeq {L}_{W}\mathrm{dgFormalSpace}\phantom{\rule{thinmathspace}{0ex}}.$L_W DerivedFormalSpace \simeq L_W dgFormalSpace \,.

This is (Pridham, cor. 4.49).

###### Proposition

For arbitrary $k$, there is a Quillen equivalence

${\mathrm{dgLie}}_{k}\stackrel{\simeq }{\to }\mathrm{dgFormalSpace}\phantom{\rule{thinmathspace}{0ex}}.$dgLie_k \stackrel{\simeq}{\to} dgFormalSpace \,.
###### Proposition

The inclusion

$\mathrm{dgFormalSpace}\to {\mathrm{dgCoCAlg}}_{k}$dgFormalSpace \to dgCoCAlg_k

given by sending an object in $\mathrm{Pro}\left({\mathrm{dgArticCAlg}}_{k}{\right)}^{\mathrm{op}}≔\mathrm{dgFormalSpace}$, hence an dg-algebra $A$, to its dual dg-coalgebra ${A}^{*}$, is the left adjoint part of a Quillen equivalence between the model structure on $\mathrm{dgFormalSpace}$, prop. 7, and the model structure on dg-coalgebras, prop. 4.

### On cosimplicial algebras (and dual Dold-Kan correspondence)

Also a version of the ”dual monoidal Dold-Kan correspondence” gives a Quillen equivalence between two model structures for ${L}_{\infty }$-algebras. This is (Pridham, section 4.4). This we discuss now

###### Remark

This equivalence has the nice property that starting with the Chevalley-Eilenberg algebra and then “denormalizing” it under dual monoidal Dold-Kan to a cosimplicial nilpotent algebra yields manifestly an incarnation of the ${L}_{\infty }$-algebra in terms of simplicial complexes of infinitesimal simplices as is implicit in the work of Anders Kock in synthetic differential geometry. This is spelled out further in dcct, section 4.5.1.

###### Definition

Write

$\left(\mathrm{dg}\stackrel{^}{𝒞}{\right)}^{\Delta }$(dg\hat {\mathcal{C}})^{\Delta}

for cosimplicial pro-objects of dg-Artin algebras ($ℕ$-graded).

###### Proposition

The category $\left(\mathrm{dg}\stackrel{^}{𝒞}{\right)}^{\Delta }$ of def. 9 carries a model category structure where

(…)

This is (Pridham, def. 4.11, prop. 4.12).

###### Definition

Write

$\mathrm{DGdg}\stackrel{^}{𝒞}$DGdg\hat \mathcal{C}

for pro-objects in dg-algebras ($ℕ$-graded) in dg-Artin algebras ($ℕ$-graded).

###### Proposition

The dual monoidal Dold-Kan correspondence functor from dg-algebras to cosimplicial algebras? (the inverse equivalence to the normalized cochain complex functor)

$D\phantom{\rule{thickmathspace}{0ex}}:\phantom{\rule{thickmathspace}{0ex}}\mathrm{DGdg}\stackrel{^}{𝒞}\to \left(\mathrm{dg}\stackrel{^}{𝒞}{\right)}^{\Delta }$D \;\colon\; DGdg\hat \mathcal{C} \to (dg\hat \mathcal{C})^{\Delta}

induces on $\mathrm{DGdg}\stackrel{^}{𝒞}$ the transferred model structure from that of prop. 13 and is the right adjoint of a Quillen equivalence with respect to these model structures

This is (Pridham, theorem 4.26).

## Properties

### General

We discuss some further properties of the above model category structures.

###### Proposition

The model category $\mathrm{dgFormalSpace}$, def. 7, is a right proper model category.

This observation has been communicated privately by Jonathan Pridham

###### Proof

We need to show that the pullback of a weak equivalence $w$ along a fibration $f$ is again a weak equivalence. If $w$ is a fibration, this is automatic, so by factorisation we reduce to the case where $w$ is a cofibration. Now, every trivial cofibration is $\mathrm{Spf}$ of a composition of acyclic small extensions, so we may take $w$ to be $\mathrm{Spf}$ of an acyclic small extension $A\to B$ with kernel $I$. Then $f$ is $\mathrm{Spf}$ of a quasi-free map $A\to R$, so the pullback is $\mathrm{Spf}$ of $R\to R/\mathrm{IR}$, and $\mathrm{IR}=I{\stackrel{^}{\otimes }}_{A}R$, so $R\to R/\mathrm{IR}$ is also an acyclic small extension.

### Homotopies and derived hom spaces

In any model category we have a notion of homotopy between 1-morphisms. In any category of fibrant objects we still have a notion of right homotopy, given by maps into a path space object. So all of the above model category/fibrant object category structures yield models for homotopies between morphisms of ${L}_{\infty }$-algebras.

A discussion of path space objects of and hence of right homotopies between ${L}_{\infty }$-algebras (in the category of def. 1) is for instance in (Dolgushev 07, section 5).

More generally, a description of the full derived hom space between two ${L}_{\infty }$-algebras is obtained via remark 6 from the description of derived hom-spaces in categories of fibrant objects.

### Homotopy fiber products

Recognizing homotopy fiber products in any of the model structure above can be a bit subtle. A recognition principle of homotopy fibers over abelian ${L}_{\infty }$-algebras , hence useful for discussion of ∞-Lie algebra extensions), is described in (Fiorenza-Rogers-Schreiber 13, theorem 3.1.13).

## References

Precursors for 2-reduced dg-algebras are dicussed in

• Dan Quillen, Rational homotopy theory, Annals of Math., 90(1969), 205–295.

The homotopy-theoretic nature of ${L}_{\infty }$-algebras and their relation to deformation problems was then notably amplified in

• Maxim Kontsevich, Topics in algebra — deformation theory Lecture Notes (1994)

Model structurs on algebras over operads in chain complexes were discussed generally in

The full model structure on dg-coalgebras (in characteristic 0) and the Quillen equivalence of dg-Lie algebras as well as the interpretation in terms of formal $\infty$-stacks is due to

In

the relation to $\infty$-stacks is discussed more in detail.

More model category theoretic developments relating various of the previous approaches and generalizing to arbitrary characteristic are in

A useful summary of that paper is given in the notes, by Stefano Maggiolo.

A discussion of path space objects for ${L}_{\infty }$-algebras is in section 5 of

• Vasiliy A. Dolgushev, Erratum to: “A Proof of Tsygan’s Formality Conjecture for an Arbitrary Smooth Manifold” (arXiv:math/0703113)

A discussion of homotopy fibers of morphusms to abelian ${L}_{\infty }$-algebras and hence ∞-Lie algebra extensions) is in section 3.1 of

Revised on August 29, 2013 18:45:51 by Urs Schreiber (89.204.155.82)


text:
## Basic College Mathematics (10th Edition)

Published by Pearson

# Chapter 8 - Geometry - 8.6 Circles - 8.6 Exercises - Page 573: 2

#### Answer

To use the formula for finding the area of a circle, you need to know the length of the radius. If you are given the diameter of a circle, what should you do to find the radius. Divide by two.

#### Work Step by Step

Area of a Circle: $A$ = $\pi$$r$$^{2}$ Therefore, to find the area the radius is required. Diameter = 2$r$ Therefore: $r$ = Diameter / 2

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.