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#matrix-inversion

For , a pseudoinverse of is defined as a matrix satisfying all of the following four criteria:

- (
*AA*^{+}need not be the general identity matrix, but it maps all column vectors of*A*to themselves); - (
*A*^{+}is a weak inverse for the multiplicative semigroup); - (
*AA*^{+}is Hermitian); and - (
*A*^{+}*A*is also Hermitian).

Moore-Penrose Pseudo-inverse

exists for any matrix , but when the latter has full rank, can be expressed as a simple algebraic formula.

In particular, when has linearly independent columns (and thus matrix is invertible), can be computed as:

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; K ) {\displaystyle I_{n}\in \mathrm {M} (n,n;K)} denotes the n × n {\displaystyle n\times n} identity matrix. Definition[edit source] <span>For A ∈ M ( m , n ; K ) {\displaystyle A\in \mathrm {M} (m,n;K)} , a pseudoinverse of A {\displaystyle A} is defined as a matrix A + ∈ M ( n , m ; K ) {\displaystyle A^{+}\in \mathrm {M} (n,m;K)} satisfying all of the following four criteria: [8] [9] A A + A = A {\displaystyle AA^{+}A=A\,\!} (AA + need not be the general identity matrix, but it maps all column vectors of A to themselves); A + A A + = A + {\displaystyle A^{+}AA^{+}=A^{+}\,\!} (A + is a weak inverse for the multiplicative semigroup); ( A A + ) ∗ = A A + {\displaystyle (AA^{+})^{*}=AA^{+}\,\!} (AA + is Hermitian); and ( A + A ) ∗ = A + A {\displaystyle (A^{+}A)^{*}=A^{+}A\,\!} (A + A is also Hermitian). A + {\displaystyle A^{+}} exists for any matrix A {\displaystyle A} , but when the latter has full rank, A + {\displaystyle A^{+}} can be expressed as a simple algebraic formula. In particular, when A {\displaystyle A} has linearly independent columns (and thus matrix A ∗ A {\displaystyle A^{*}A} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = ( A ∗ A ) − 1 A ∗ . {\displaystyle A^{+}=(A^{*}A)^{-1}A^{*}\,.} This particular pseudoinverse constitutes a left inverse, since, in this case, A + A = I {\displaystyle A^{+}A=I} . When A {\displaystyle A} has linearly independent rows (matrix A A ∗ {\displaystyle AA^{*}} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = A ∗ ( A A ∗ ) − 1 . {\displaystyle A^{+}=A^{*}(AA^{*})^{-1}\,.} This is a right inverse, as A A + = I {\displaystyle AA^{+}=I} . Properties[edit source] Proofs for some of these facts may be found on a separate page Proofs involving the Moore–Penrose inverse. Existence and uniqueness[edit source] The pseu

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#singular-value-decomposition

The geometric content of the SVD theorem can thus be summarized as follows: for every linear map

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m , n ) , {\displaystyle T(\mathbf {V} _{i})=\sigma _{i}\mathbf {U} _{i},\qquad i=1,\ldots ,\min(m,n),} where σ i is the i-th diagonal entry of Σ, and T(V i ) = 0 for i > min(m,n). <span>The geometric content of the SVD theorem can thus be summarized as follows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with non-negative real diagonal entries. To get a more visual flavour of singular values and SVD factorization — at least when working on real vector spaces — consider the sphere S of radius one in R n . The linear map T map

#matrix-inversion

A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition.^{[1]}^{[9]}^{[15]} If is the singular value decomposition of *A* , then . For a rectangular diagonal matrix such as Σ {\displaystyle \Sigma } , we get the pseudoinverse by taking the reciprocal of each non-zero element on the diagonal, leaving the zeros in place, and then transposing the matrix. In numerical computation, only elements larger than some small tolerance are taken to be nonzero, and the others are replaced by zeros. For example, in the MATLAB, GNU Octave, or NumPy function pinv , the tolerance is taken to be *t* = ε⋅max(*m*,*n*)⋅max(Σ) , where ε is the machine epsilon.

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A {\displaystyle A} and A ∗ {\displaystyle A^{*}} . Singular value decomposition (SVD)[edit source] <span>A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition. [1] [9] [15] If A = U Σ V ∗ {\displaystyle A=U\Sigma V^{*}} is the singular value decomposition of A, then A + = V Σ + U ∗ {\displaystyle A^{+}=V\Sigma ^{+}U^{*}} . For a rectangular diagonal matrix such as Σ {\displaystyle \Sigma } , we get the pseudoinverse by taking the reciprocal of each non-zero element on the diagonal, leaving the zeros in place, and then transposing the matrix. In numerical computation, only elements larger than some small tolerance are taken to be nonzero, and the others are replaced by zeros. For example, in the MATLAB, GNU Octave, or NumPy function pinv , the tolerance is taken to be t = ε⋅max(m,n)⋅max(Σ), where ε is the machine epsilon. The computational cost of this method is dominated by the cost of computing the SVD, which is several times higher than matrix–matrix multiplication, even if a state-of-the art implem

#matrix-inversion

Moore-Penrose Pseudo-inverse

exists for any matrix , but when the latter has full rank, can be expressed as a simple algebraic formula.

In particular, when has linearly independent columns (and thus matrix is invertible), can be computed as:

This particular pseudoinverse constitutes a *left inverse*, since, in this case, .

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3; ( AA + need not be the general identity matrix, but it maps all column vectors of A to themselves); ( A + is a weak inverse for the multiplicative semigroup); ( AA + is Hermitian); and ( A + A is also Hermitian). <span>Moore-Penrose Pseudo-inverse exists for any matrix , but when the latter has full rank, can be expressed as a simple algebraic formula. In particular, when has linearly independent columns (and thus matrix is invertible), can be computed as: This particular pseudoinverse constitutes a left inverse, since, in this case, . When has linearly independent rows (matrix is invertible), can be computed as: This is a right inverse, as . <span><body><html>

; K ) {\displaystyle I_{n}\in \mathrm {M} (n,n;K)} denotes the n × n {\displaystyle n\times n} identity matrix. Definition[edit source] <span>For A ∈ M ( m , n ; K ) {\displaystyle A\in \mathrm {M} (m,n;K)} , a pseudoinverse of A {\displaystyle A} is defined as a matrix A + ∈ M ( n , m ; K ) {\displaystyle A^{+}\in \mathrm {M} (n,m;K)} satisfying all of the following four criteria: [8] [9] A A + A = A {\displaystyle AA^{+}A=A\,\!} (AA + need not be the general identity matrix, but it maps all column vectors of A to themselves); A + A A + = A + {\displaystyle A^{+}AA^{+}=A^{+}\,\!} (A + is a weak inverse for the multiplicative semigroup); ( A A + ) ∗ = A A + {\displaystyle (AA^{+})^{*}=AA^{+}\,\!} (AA + is Hermitian); and ( A + A ) ∗ = A + A {\displaystyle (A^{+}A)^{*}=A^{+}A\,\!} (A + A is also Hermitian). A + {\displaystyle A^{+}} exists for any matrix A {\displaystyle A} , but when the latter has full rank, A + {\displaystyle A^{+}} can be expressed as a simple algebraic formula. In particular, when A {\displaystyle A} has linearly independent columns (and thus matrix A ∗ A {\displaystyle A^{*}A} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = ( A ∗ A ) − 1 A ∗ . {\displaystyle A^{+}=(A^{*}A)^{-1}A^{*}\,.} This particular pseudoinverse constitutes a left inverse, since, in this case, A + A = I {\displaystyle A^{+}A=I} . When A {\displaystyle A} has linearly independent rows (matrix A A ∗ {\displaystyle AA^{*}} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = A ∗ ( A A ∗ ) − 1 . {\displaystyle A^{+}=A^{*}(AA^{*})^{-1}\,.} This is a right inverse, as A A + = I {\displaystyle AA^{+}=I} . Properties[edit source] Proofs for some of these facts may be found on a separate page Proofs involving the Moore–Penrose inverse. Existence and uniqueness[edit source] The pseu

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#matrix-inversion

Question

when has [...] the Moore-Penrose inverse is a *left inverse*

Answer

linearly independent columns

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ml> Moore-Penrose Pseudo-inverse exists for any matrix , but when the latter has full rank, can be expressed as a simple algebraic formula. In particular, when has linearly independent columns (and thus matrix is invertible), can be computed as: This particular pseudoinverse constitutes a left inverse, since, in this case, . <html>

; K ) {\displaystyle I_{n}\in \mathrm {M} (n,n;K)} denotes the n × n {\displaystyle n\times n} identity matrix. Definition[edit source] <span>For A ∈ M ( m , n ; K ) {\displaystyle A\in \mathrm {M} (m,n;K)} , a pseudoinverse of A {\displaystyle A} is defined as a matrix A + ∈ M ( n , m ; K ) {\displaystyle A^{+}\in \mathrm {M} (n,m;K)} satisfying all of the following four criteria: [8] [9] A A + A = A {\displaystyle AA^{+}A=A\,\!} (AA + need not be the general identity matrix, but it maps all column vectors of A to themselves); A + A A + = A + {\displaystyle A^{+}AA^{+}=A^{+}\,\!} (A + is a weak inverse for the multiplicative semigroup); ( A A + ) ∗ = A A + {\displaystyle (AA^{+})^{*}=AA^{+}\,\!} (AA + is Hermitian); and ( A + A ) ∗ = A + A {\displaystyle (A^{+}A)^{*}=A^{+}A\,\!} (A + A is also Hermitian). A + {\displaystyle A^{+}} exists for any matrix A {\displaystyle A} , but when the latter has full rank, A + {\displaystyle A^{+}} can be expressed as a simple algebraic formula. In particular, when A {\displaystyle A} has linearly independent columns (and thus matrix A ∗ A {\displaystyle A^{*}A} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = ( A ∗ A ) − 1 A ∗ . {\displaystyle A^{+}=(A^{*}A)^{-1}A^{*}\,.} This particular pseudoinverse constitutes a left inverse, since, in this case, A + A = I {\displaystyle A^{+}A=I} . When A {\displaystyle A} has linearly independent rows (matrix A A ∗ {\displaystyle AA^{*}} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = A ∗ ( A A ∗ ) − 1 . {\displaystyle A^{+}=A^{*}(AA^{*})^{-1}\,.} This is a right inverse, as A A + = I {\displaystyle AA^{+}=I} . Properties[edit source] Proofs for some of these facts may be found on a separate page Proofs involving the Moore–Penrose inverse. Existence and uniqueness[edit source] The pseu

Tags

#matrix-inversion

Question

The *left *Moore-Penrose Pseudo-inverse is [...]

Answer

*This is the one for linear models*

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e exists for any matrix , but when the latter has full rank, can be expressed as a simple algebraic formula. In particular, when has linearly independent columns (and thus matrix is invertible), can be computed as<span>: This particular pseudoinverse constitutes a left inverse, since, in this case, . <span><body><html>

; K ) {\displaystyle I_{n}\in \mathrm {M} (n,n;K)} denotes the n × n {\displaystyle n\times n} identity matrix. Definition[edit source] <span>For A ∈ M ( m , n ; K ) {\displaystyle A\in \mathrm {M} (m,n;K)} , a pseudoinverse of A {\displaystyle A} is defined as a matrix A + ∈ M ( n , m ; K ) {\displaystyle A^{+}\in \mathrm {M} (n,m;K)} satisfying all of the following four criteria: [8] [9] A A + A = A {\displaystyle AA^{+}A=A\,\!} (AA + need not be the general identity matrix, but it maps all column vectors of A to themselves); A + A A + = A + {\displaystyle A^{+}AA^{+}=A^{+}\,\!} (A + is a weak inverse for the multiplicative semigroup); ( A A + ) ∗ = A A + {\displaystyle (AA^{+})^{*}=AA^{+}\,\!} (AA + is Hermitian); and ( A + A ) ∗ = A + A {\displaystyle (A^{+}A)^{*}=A^{+}A\,\!} (A + A is also Hermitian). A + {\displaystyle A^{+}} exists for any matrix A {\displaystyle A} , but when the latter has full rank, A + {\displaystyle A^{+}} can be expressed as a simple algebraic formula. In particular, when A {\displaystyle A} has linearly independent columns (and thus matrix A ∗ A {\displaystyle A^{*}A} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = ( A ∗ A ) − 1 A ∗ . {\displaystyle A^{+}=(A^{*}A)^{-1}A^{*}\,.} This particular pseudoinverse constitutes a left inverse, since, in this case, A + A = I {\displaystyle A^{+}A=I} . When A {\displaystyle A} has linearly independent rows (matrix A A ∗ {\displaystyle AA^{*}} is invertible), A + {\displaystyle A^{+}} can be computed as: A + = A ∗ ( A A ∗ ) − 1 . {\displaystyle A^{+}=A^{*}(AA^{*})^{-1}\,.} This is a right inverse, as A A + = I {\displaystyle AA^{+}=I} . Properties[edit source] Proofs for some of these facts may be found on a separate page Proofs involving the Moore–Penrose inverse. Existence and uniqueness[edit source] The pseu

Tags

#singular-value-decomposition

Question

geometrically SVD finds **[...]** for every linear map *T* : *K*^{n} → *K*^{m}

Answer

orthonormal bases of K^{n} and K^{m}

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SVD as change of coordinates The geometric content of the SVD theorem can thus be summarized as follows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases,

m , n ) , {\displaystyle T(\mathbf {V} _{i})=\sigma _{i}\mathbf {U} _{i},\qquad i=1,\ldots ,\min(m,n),} where σ i is the i-th diagonal entry of Σ, and T(V i ) = 0 for i > min(m,n). <span>The geometric content of the SVD theorem can thus be summarized as follows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with non-negative real diagonal entries. To get a more visual flavour of singular values and SVD factorization — at least when working on real vector spaces — consider the sphere S of radius one in R n . The linear map T map

Tags

#singular-value-decomposition

Question

Geometrically SVD finds orthonormal bases of K^{n} and K^{m }for every linear map *T* : *K*^{n} → *K*^{m} such that T maps the i-th basis vector of K^{n} to a non-negative multiple of the i-th basis vector of K^{m} , and sends the left-over basis vectors to [...].

Answer

zero

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ollows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to <span>zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with non-negative real diagonal entries. <span><body><html>

m , n ) , {\displaystyle T(\mathbf {V} _{i})=\sigma _{i}\mathbf {U} _{i},\qquad i=1,\ldots ,\min(m,n),} where σ i is the i-th diagonal entry of Σ, and T(V i ) = 0 for i > min(m,n). <span>The geometric content of the SVD theorem can thus be summarized as follows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with non-negative real diagonal entries. To get a more visual flavour of singular values and SVD factorization — at least when working on real vector spaces — consider the sphere S of radius one in R n . The linear map T map

Tags

#singular-value-decomposition

Question

With SVD geometrically every linear map *T* : *K*^{n} → *K*^{m} is represented by a diagonal matrix with [...] entries.

Answer

non-negative real diagonal

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h that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with <span>non-negative real diagonal entries. <span><body><html>

m , n ) , {\displaystyle T(\mathbf {V} _{i})=\sigma _{i}\mathbf {U} _{i},\qquad i=1,\ldots ,\min(m,n),} where σ i is the i-th diagonal entry of Σ, and T(V i ) = 0 for i > min(m,n). <span>The geometric content of the SVD theorem can thus be summarized as follows: for every linear map T : K n → K m one can find orthonormal bases of K n and K m such that T maps the i-th basis vector of K n to a non-negative multiple of the i-th basis vector of K m , and sends the left-over basis vectors to zero. With respect to these bases, the map T is therefore represented by a diagonal matrix with non-negative real diagonal entries. To get a more visual flavour of singular values and SVD factorization — at least when working on real vector spaces — consider the sphere S of radius one in R n . The linear map T map

#matrix

In linear algebra, a square matrix *A* is called **diagonalizable** if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix *P* such that *P*^{−1}*AP* is a diagonal matrix.

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Diagonalizable matrix From Wikipedia, the free encyclopedia Jump to: navigation, search This article is about matrix diagonalisation in linear algebra. For other uses, see Diagonalisation. <span>In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P −1 AP is a diagonal matrix. If V is a finite-dimensional vector space, then a linear map T : V → V is called diagonalizable if there exists an ordered basis of V with respect to which T is represented by a diagona

Tags

#matrix

Question

Answer

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In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P −1 AP is a diagonal matrix.

Diagonalizable matrix From Wikipedia, the free encyclopedia Jump to: navigation, search This article is about matrix diagonalisation in linear algebra. For other uses, see Diagonalisation. <span>In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P −1 AP is a diagonal matrix. If V is a finite-dimensional vector space, then a linear map T : V → V is called diagonalizable if there exists an ordered basis of V with respect to which T is represented by a diagona

#english

Novelist Adam Langer skewers the publishing trade — and some of its recent trends — while digging toward something deeper.

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30+ years Terms of Service Privacy Policy YOU ARE HERE: LAT Home→Collections Book review: 'The Thieves of Manhattan' by Adam Langer <span>Novelist Adam Langer skewers the publishing trade — and some of its recent trends — while digging toward something deeper. July 18, 2010|By Ella Taylor, Special to the Los Angeles Times Email Share The Thieves of Manhattan A Novel Adam Langer Spiegel & Grau: 260 pp., $15 paper

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boost sales even further. Shades of James Frey and Margaret Seltzer, attention-seekers who thrilled a gullible public (not to mention their editors) with trumped-up accounts of drug addiction and childhood ghetto traumas they never endured. <span>Thus does Ian, a committed realist who has yet to write or, for that matter, live an adventure of his own, become embroiled in a whole series of plots. As they unfold, he gets a life, writes more than one book and falls hopelessly in love. All of which forces him to reassess the hazy borders between truth and fiction, life and art and

#logic

According to *A* *History of Formal Logic* (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries.

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the hands of thinkers such as George Boole, Gottlob Frege, Bertrand Russell, Alfred Tarski and Kurt Gödel, it’s clear that Kant was dead wrong. But he was also wrong in thinking that there had been no progress since Aristotle up to his time. <span>According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries. (Throughout this piece, the focus is on the logical traditions that emerged against the background of ancient Greek logic. So Indian and Chinese logic are not included, but medieval Ara

#reinforcement-learning

**Mountain Car**, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill. Since gravity is stronger than the car's engine, even at full throttle, the car cannot simply accelerate up the steep slope. The car is situated in a valley and must learn to leverage potential energy by driving up the opposite hill before the car is able to make it to the goal at the top of the rightmost hill. The domain has been used as a test bed in various Reinforcement Learning papers.

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[imagelink] This article is an orphan, as no other articles link to it. Please introduce links to this page from related articles; try the Find link tool for suggestions. (July 2012) [imagelink] The mountain car problem <span>Mountain Car, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill. Since gravity is stronger than the car's engine, even at full throttle, the car cannot simply accelerate up the steep slope. The car is situated in a valley and must learn to leverage potential energy by driving up the opposite hill before the car is able to make it to the goal at the top of the rightmost hill. The domain has been used as a test bed in various Reinforcement Learning papers. Contents [hide] 1 Introduction 2 History 3 Techniques used to solve mountain car 3.1 Discretization 3.2 Function approximation 3.3 Traces 4 Technical details 4.1 State v

#spanish

Usage:

- To
*ask politely*:*¿*(*Podrías*pasarme ese plato, por favor?*Could you pass me that plate, please?*). - To express
*wishes*:*¡Me*(*encantaría*ir de viaje a Australia!*I would love to go on a trip to Australia!*). - To
*suggest*:*Creo que*(*deberías*ir al médico a verte ese dolor de espalda*I think you should go to see the doctor for a check-up on that back of yours*). - To express a
*hypothesis or probability*:*Podemos posponer la salida unas horas, pero*(*llegaríamos*bastante tarde*We can postpone our trip for some hours, but we would arrive quite late*). - To express
*uncertainty*in the past:*No sabía si*(*estarías*en la oficina, por eso no te llamé*I didn’t know whether you’d be at the office. That’s why I didn’t call you*). - To
*refer to the future*from a moment*in the past*:*María me dijo que estaría en casa para las 11, pero no ha aparecido aún*(*María told me she’d be at home by 11, but she hasn’t turned up yet*).

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#spanish

Verbs: The Conditional Simple

Usage:

- To
*ask politely*:*¿*(*Podrías*pasarme ese plato, por favor?*Could you pass me that plate, please?*).

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Verbs: The Conditional Simple Usage: To ask politely: ¿Podrías pasarme ese plato, por favor? (Could you pass me that plate, please?). To express wishes: ¡Me encantaría ir de viaje a Australia! (I would love to go on a trip to Australia!). To suggest: Creo que deberías ir al médico a verte ese dolor de espalda (I think

Tags

#spanish

Question

use conditional simple to *[...]*:

*¿**Podrías* pasarme ese plato, por favor?

Answer

ask politely

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Verbs: The Conditional Simple Usage: To ask politely: ¿Podrías pasarme ese plato, por favor? (Could you pass me that plate, please?).

Verbs: The Conditional Simple Usage: To ask politely: ¿Podrías pasarme ese plato, por favor? (Could you pass me that plate, please?). To express wishes: ¡Me encantaría ir de viaje a Australia! (I would love to go on a trip to Australia!). To suggest: Creo que deberías ir al médico a verte ese dolor de espalda (I think

#spanish

- To
*refer to the future*from a moment*in the past*:*María me dijo que estaría en casa para las 11, pero no ha aparecido aún*(*María told me she’d be at home by 11, but she hasn’t turned up yet*).

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e can postpone our trip for some hours, but we would arrive quite late). To express uncertainty in the past: No sabía si estarías en la oficina, por eso no te llamé (I didn’t know whether you’d be at the office. That’s why I didn’t call you). <span>To refer to the future from a moment in the past: María me dijo que estaría en casa para las 11, pero no ha aparecido aún (María told me she’d be at home by 11, but she hasn’t turned up yet). <span><body><html>

Tags

#spanish

Question

use conditional simple to *[...]*:

*María me dijo que estaría en casa para las 11, pero no ha aparecido aún.*

Answer

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To refer to the future from a moment in the past: María me dijo que estaría en casa para las 11, pero no ha aparecido aún (María told me she’d be at home by 11, but she hasn’t turned up yet).

e can postpone our trip for some hours, but we would arrive quite late). To express uncertainty in the past: No sabía si estarías en la oficina, por eso no te llamé (I didn’t know whether you’d be at the office. That’s why I didn’t call you). <span>To refer to the future from a moment in the past: María me dijo que estaría en casa para las 11, pero no ha aparecido aún (María told me she’d be at home by 11, but she hasn’t turned up yet). <span><body><html>

Tags

#spanish

Question

como **[...]** en el acta/informe

as stated or recorded in the minutes/report

as stated or recorded in the minutes/report

Answer

consta

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repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Tags

#spanish

Question

▸ ¿te falta mucho? — no, ya casi **[...]**

do you have much to do? — no, I've nearly finished

do you have much to do? — no, I've nearly finished

Answer

acabo

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repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Tags

#spanish

Question

el tren **[...]** su salida a las 10.50

the train will depart at 10:50

the train will depart at 10:50

Answer

efectuará

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#reinforcement-learning

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Mountain Car, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill. Since gravity is stronger than the car's engine, even at full throttle, the car cannot simply accelerate up the steep slope. The car is situated in a valley and must learn to leverage p

[imagelink] This article is an orphan, as no other articles link to it. Please introduce links to this page from related articles; try the Find link tool for suggestions. (July 2012) [imagelink] The mountain car problem <span>Mountain Car, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill. Since gravity is stronger than the car's engine, even at full throttle, the car cannot simply accelerate up the steep slope. The car is situated in a valley and must learn to leverage potential energy by driving up the opposite hill before the car is able to make it to the goal at the top of the rightmost hill. The domain has been used as a test bed in various Reinforcement Learning papers. Contents [hide] 1 Introduction 2 History 3 Techniques used to solve mountain car 3.1 Discretization 3.2 Function approximation 3.3 Traces 4 Technical details 4.1 State v

Tags

#reinforcement-learning

Question

Answer

under-powered car

status | not learned | measured difficulty | 37% [default] | last interval [days] | |||
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repetition number in this series | 0 | memorised on | scheduled repetition | ||||

scheduled repetition interval | last repetition or drill |

Mountain Car, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill.

[imagelink] This article is an orphan, as no other articles link to it. Please introduce links to this page from related articles; try the Find link tool for suggestions. (July 2012) [imagelink] The mountain car problem <span>Mountain Car, a standard testing domain in Reinforcement Learning, is a problem in which an under-powered car must drive up a steep hill. Since gravity is stronger than the car's engine, even at full throttle, the car cannot simply accelerate up the steep slope. The car is situated in a valley and must learn to leverage potential energy by driving up the opposite hill before the car is able to make it to the goal at the top of the rightmost hill. The domain has been used as a test bed in various Reinforcement Learning papers. Contents [hide] 1 Introduction 2 History 3 Techniques used to solve mountain car 3.1 Discretization 3.2 Function approximation 3.3 Traces 4 Technical details 4.1 State v

Tags

#logic

Question

According to *A* *History of Formal Logic* (1961) by the distinguished J M Bocheński, the golden periods for logic were [...], the medieval scholastic period, and the mathematical period of the 19th and 20th centuries.

Answer

the ancient Greek period

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According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries.

the hands of thinkers such as George Boole, Gottlob Frege, Bertrand Russell, Alfred Tarski and Kurt Gödel, it’s clear that Kant was dead wrong. But he was also wrong in thinking that there had been no progress since Aristotle up to his time. <span>According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries. (Throughout this piece, the focus is on the logical traditions that emerged against the background of ancient Greek logic. So Indian and Chinese logic are not included, but medieval Ara

Tags

#logic

Question

According to *A* *History of Formal Logic* (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, [...], and the mathematical period of the 19th and 20th centuries.

Answer

the medieval scholastic period

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scheduled repetition interval | last repetition or drill |

According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries.

the hands of thinkers such as George Boole, Gottlob Frege, Bertrand Russell, Alfred Tarski and Kurt Gödel, it’s clear that Kant was dead wrong. But he was also wrong in thinking that there had been no progress since Aristotle up to his time. <span>According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries. (Throughout this piece, the focus is on the logical traditions that emerged against the background of ancient Greek logic. So Indian and Chinese logic are not included, but medieval Ara

Tags

#logic

Question

According to *A* *History of Formal Logic* (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and [...] of the 19th and 20th centuries.

Answer

the mathematical period

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scheduled repetition interval | last repetition or drill |

According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries.

the hands of thinkers such as George Boole, Gottlob Frege, Bertrand Russell, Alfred Tarski and Kurt Gödel, it’s clear that Kant was dead wrong. But he was also wrong in thinking that there had been no progress since Aristotle up to his time. <span>According to A History of Formal Logic (1961) by the distinguished J M Bocheński, the golden periods for logic were the ancient Greek period, the medieval scholastic period, and the mathematical period of the 19th and 20th centuries. (Throughout this piece, the focus is on the logical traditions that emerged against the background of ancient Greek logic. So Indian and Chinese logic are not included, but medieval Ara

Question

Thus does Ian, a committed realist who has yet to write or, for that matter, live an adventure of his own, become [...] in a whole series of plots.

*involve (someone) deeply in an argument, conflict, or difficult situation*

Answer

embroiled

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Thus does Ian, a committed realist who has yet to write or, for that matter, live an adventure of his own, become embroiled in a whole series of plots.

boost sales even further. Shades of James Frey and Margaret Seltzer, attention-seekers who thrilled a gullible public (not to mention their editors) with trumped-up accounts of drug addiction and childhood ghetto traumas they never endured. <span>Thus does Ian, a committed realist who has yet to write or, for that matter, live an adventure of his own, become embroiled in a whole series of plots. As they unfold, he gets a life, writes more than one book and falls hopelessly in love. All of which forces him to reassess the hazy borders between truth and fiction, life and art and

Question

Novelist Adam Langer [...] the publishing trade — and some of its recent trends — while digging toward something deeper.

*fasten together or pierce with*

•*informal* subject to sharp criticism or critical analysis

•

Answer

skewers

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Novelist Adam Langer skewers the publishing trade — and some of its recent trends — while digging toward something deeper.

30+ years Terms of Service Privacy Policy YOU ARE HERE: LAT Home→Collections Book review: 'The Thieves of Manhattan' by Adam Langer <span>Novelist Adam Langer skewers the publishing trade — and some of its recent trends — while digging toward something deeper. July 18, 2010|By Ella Taylor, Special to the Los Angeles Times Email Share The Thieves of Manhattan A Novel Adam Langer Spiegel & Grau: 260 pp., $15 paper

#matrix-inversion

A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition.^{[1]}^{[9]}^{[15]} If is the singular value decomposition of *A* , then .

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A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition. [1] [9] [15] If is the singular value decomposition of A , then . For a rectangular diagonal matrix such as Σ {\displaystyle \Sigma } , we get the pseudoinverse by taking the reciprocal of each non-zero element on the diagonal, leaving the zeros in p

A {\displaystyle A} and A ∗ {\displaystyle A^{*}} . Singular value decomposition (SVD)[edit source] <span>A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition. [1] [9] [15] If A = U Σ V ∗ {\displaystyle A=U\Sigma V^{*}} is the singular value decomposition of A, then A + = V Σ + U ∗ {\displaystyle A^{+}=V\Sigma ^{+}U^{*}} . For a rectangular diagonal matrix such as Σ {\displaystyle \Sigma } , we get the pseudoinverse by taking the reciprocal of each non-zero element on the diagonal, leaving the zeros in place, and then transposing the matrix. In numerical computation, only elements larger than some small tolerance are taken to be nonzero, and the others are replaced by zeros. For example, in the MATLAB, GNU Octave, or NumPy function pinv , the tolerance is taken to be t = ε⋅max(m,n)⋅max(Σ), where ε is the machine epsilon. The computational cost of this method is dominated by the cost of computing the SVD, which is several times higher than matrix–matrix multiplication, even if a state-of-the art implem

Tags

#matrix-inversion

Question

If is the singular value decomposition of *A* , then the pseudoinverse of *A *is [...]

Answer

.

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A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition. [1] [9] [15] If is the singular value decomposition of A , then .

A {\displaystyle A} and A ∗ {\displaystyle A^{*}} . Singular value decomposition (SVD)[edit source] <span>A computationally simple and accurate way to compute the pseudoinverse is by using the singular value decomposition. [1] [9] [15] If A = U Σ V ∗ {\displaystyle A=U\Sigma V^{*}} is the singular value decomposition of A, then A + = V Σ + U ∗ {\displaystyle A^{+}=V\Sigma ^{+}U^{*}} . For a rectangular diagonal matrix such as Σ {\displaystyle \Sigma } , we get the pseudoinverse by taking the reciprocal of each non-zero element on the diagonal, leaving the zeros in place, and then transposing the matrix. In numerical computation, only elements larger than some small tolerance are taken to be nonzero, and the others are replaced by zeros. For example, in the MATLAB, GNU Octave, or NumPy function pinv , the tolerance is taken to be t = ε⋅max(m,n)⋅max(Σ), where ε is the machine epsilon. The computational cost of this method is dominated by the cost of computing the SVD, which is several times higher than matrix–matrix multiplication, even if a state-of-the art implem

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pseudo datapoint based approximation methods for DGPs trade model complexity for a lower computational complexity of \(O(NLM^ 2 ) \) where N is the number of datapoints, L is the number of layers, and M is the number of pseudo datapoints. This complexity scales quadratically in M whereas the dependence on the number of layers L is only linear. Therefore, it can be cheaper to increase the representation power of the model

Tags

#deep-gaussian-process

Question

pseudo datapoint based approximation methods for DGPs has a computational complexity of [...]

Answer

\(O(NLM^ 2 ) \)

*where N is the number of datapoints, L is the number of layers, and M is the number of pseudo datapoints.*

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pseudo datapoint based approximation methods for DGPs trade model complexity for a lower computational complexity of \(O(NLM^ 2 ) \) where N is the number of datapoints, L is the number of layers, and M is the number of pseudo datapoints.

Tags

#deep-gaussian-process

Question

DGPs can perform [...] or dimensionality compression or expansion

Answer

input warping

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DGPs can perform input warping or dimensionality compression or expansion, and automatically learn to construct a kernel that works well for the data at hand. As a result, learning in this model provides a flexible f

Tags

#deep-gaussian-process

Question

DGPs can perform input warping or [...]

Answer

dimensionality compression or expansion

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DGPs can perform input warping or dimensionality compression or expansion, and automatically learn to construct a kernel that works well for the data at hand. As a result, learning in this model provides a flexible form of Bayesian kernel design. </sp

Tags

#deep-gaussian-process

Question

DGPs can automatically learn to [...] that works well for the data at hand.

Answer

construct a kernel

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DGPs can perform input warping or dimensionality compression or expansion, and automatically learn to construct a kernel that works well for the data at hand. As a result, learning in this model provides a flexible form of Bayesian kernel design.

Tags

#deep-gaussian-process

Question

The new method uses an [...] procedure and a novel and efficient extension of the probabilistic backpropagation algorithm for learning.

Answer

approximate Expectation Propagation

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The new method uses an approximate Expectation Propagation procedure and a novel and efficient ex- tension of the probabilistic backpropagation algorithm for learning.

Question

Deep Gaussian processes (DGPs) are [...] of Gaussian processes (GPs) and are formally equivalent to neural networks with multiple, infinitely wide hidden layers.

Answer

multi-layer hierarchical generalisations

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Deep Gaussian processes (DGPs) are multi-layer hierarchical generalisations of Gaussian pro- cesses (GPs) and are formally equivalent to neural networks with multiple, infinitely wide hidden layers.

Tags

#variational-inference

Question

Answer

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Variational Bayesian methods are a family of techniques for approximating intractable integrals arising in Bayesian inference and machine learning.

f references, but its sources remain unclear because it has insufficient inline citations. Please help to improve this article by introducing more precise citations. (September 2010) (Learn how and when to remove this template message) <span>Variational Bayesian methods are a family of techniques for approximating intractable integrals arising in Bayesian inference and machine learning. They are typically used in complex statistical models consisting of observed variables (usually termed "data") as well as unknown parameters and latent variables, with various

Question

Artículo 114.- Ningún Senador o Representante, desde el día de su elección hasta el de su cese, podrá ser acusado criminalmente, ni aun por delitos comunes que no sean de los detallados en el artículo 93, sino ante su respectiva Cámara, la cual, por dos tercios de votos del total de sus componentes, resolverá si hay lugar a la formación de causa, y, en caso afirmativo, lo declarará suspendido en sus funciones y quedará a disposición del Tribunal competente.

Answer

[default - edit me]

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