Taylor Expansion and Derivative Formulas for Matrix Logarithms
I give the derivation of formulas for the Taylor expansion and derivative of a matrix logarithm. log(x + y) - log(x) - log(x + y + U) + log(x + U) =.
matrixlog tex( )
“Covariates impacts in compositional models and simplicial
2 dic 2019 logarithm is on the right hand side of the regression equation) or symmetrically the partial derivative of E(log(Y )) with respect to X in ...
wp tse
“Covariates impacts in compositional models and simplicial
2 dic 2019 logarithm is on the right hand side of the regression equation) or symmetrically the partial derivative of E(log(Y )) with respect to X in ...
wp tse
DIFFERENTIAL EQUATIONS
Now substituting x = 1 in the above
leep
Impact of covariates in compositional models and simplicial derivatives
25 mar 2021 with respect to log(X) (if the logarithm is on the right hand side of the regression equation) or symmetrically the partial derivative of ...
Impact of Covariates in Compositional Models and S
On a Linear Differential Equation of the Second Order
zero and consequently the fundamental integrals of the equation in the domain of x = O are Yi = ii (x)
CONTINUITY AND DIFFERENTIABILITY
Then we say logarithm of a to base b is x if bx=a log. = x b. 6. logb b = 1 and logb 1 = 0. (iv) The derivative of ex w.r.t.
leep
Differential Equations Assignment #1: answers.
One quickly checks that x(t) = log(t) · et2 is a solution of the equation and satisfies the initial condition; it is even unique as such by the Cauchy-
A S
Dimensions of Logarithmic Quantities
note that eq 1 may be written in a differential form: (x + d log (x) = log (x + dje) - tion multiplying each equation by the stoichiometric num-.
A REFINEMENT OF SELBERG'S ASYMPTOTIC EQUATION
ψ(x) - Σ Λ{n) - Σ log p R{x)= ψ(x) - x we have the two equivalent forms of Selberg's asymptotic equation. (1). R(x)log x + J*Λ(-
) df(t) = O(x)
Stephen L. Adler
Institute for Advanced Study, Einstein Drive, Princeton, NJ 08540, USA. I give the derivation of formulas for the Taylor expansion and derivative of a matrix logarithm. They may well be in the literature, but I have not found derivations by using standard search tools, so thought it useful to document them as a memo.I. INTRODUCTION
A query from \Backpacker" on Physics Forum says \A paper I'm reading states that: for positive hermitian matrices A and B, the Taylor expansion oflog(A+tB) att= 0 is log(A+tB) =log(A) +t∫ 1 01B+zIA1
B+zIdz+O(t2):(1)
However, there is no source or proof given, and I cannot seem to nd a derivation of this identity anywhere!" (No citation for the paper containing this formula was given in the query.) Derivations of this and related formulas are given in the following sections, with no attempt at mathematical rigor.II. TAYLOR EXPANSION OF THE MATRIX LOG
Letxandybe noncommuting matrices or operators. Then the expansion 1 x+y=1 x 1 x y1 x +1 x y1 x y1 x :::(2) is easily veried by multiplying through from the left (or from the right) byx+y. Replacingxby x+a1 and integrating the left hand side with respect toafrom 0 to an upper limitUgives U 0 da1 x+y+a1=log(x+y+U1)log(x+y); U 0 da1 x+a1=log(x+U1)log(x); (3)Electronic address:adler@ias.edu
2 so that subtracting and substituting Eq. (2) gives log(x+y)log(x)log(x+y+U) + log(x+U) =∫ U 0 da(1 x+a11 x+y+a1) U 0 da(1 x+a1y1 x+a11 x+a1y1 x+a1y1 x+a1+:::) (4) Since log(x+y+U)log(x+U) = log(1 + (x+y)=U)log(1 +x=U)(5) vanishes asU! 1, taking this limit gives the Taylor expansion formula log(x+y)log(x) =∫ 1 0 da(1 x+a1y1 x+a11 x+a1y1 x+a1y1 x+a1+:::) :(6) The rst term in this expansion is the equation given in the query.III. DERIVATIVE OF THE MATRIX LOG
Lettingy=dx(t) and dividing bydt, one gets
d dt log(x(t))=∫ 1 0 da1 x(t) +a1dx(t) dt 1 x(t) +a1;(7) even whenx(t) anddx(t)=dtdo not commute. Making the substitutiona=b=(1b), a related formula can be given as an integral ∫1 0db, d dt log(x(t))=∫ 1 0 db1 (1b)x(t) +b1dx(t) dt 1 (1b)x(t) +b1:(8) These formulas are analogs of the well-known formula for the derivative of the matrix exponen- tial, d dt exp(x(t))=∫ 1 0 daexp(ax(t))dx(t) dt exp((1a)x(t));(9) which is given in the Wikipedia article on the matrix exponential. Taylor Expansion and Derivative Formulas for Matrix LogarithmsStephen L. Adler
Institute for Advanced Study, Einstein Drive, Princeton, NJ 08540, USA. I give the derivation of formulas for the Taylor expansion and derivative of a matrix logarithm. They may well be in the literature, but I have not found derivations by using standard search tools, so thought it useful to document them as a memo.I. INTRODUCTION
A query from \Backpacker" on Physics Forum says \A paper I'm reading states that: for positive hermitian matrices A and B, the Taylor expansion oflog(A+tB) att= 0 is log(A+tB) =log(A) +t∫ 1 01B+zIA1
B+zIdz+O(t2):(1)
However, there is no source or proof given, and I cannot seem to nd a derivation of this identity anywhere!" (No citation for the paper containing this formula was given in the query.) Derivations of this and related formulas are given in the following sections, with no attempt at mathematical rigor.II. TAYLOR EXPANSION OF THE MATRIX LOG
Letxandybe noncommuting matrices or operators. Then the expansion 1 x+y=1 x 1 x y1 x +1 x y1 x y1 x :::(2) is easily veried by multiplying through from the left (or from the right) byx+y. Replacingxby x+a1 and integrating the left hand side with respect toafrom 0 to an upper limitUgives U 0 da1 x+y+a1=log(x+y+U1)log(x+y); U 0 da1 x+a1=log(x+U1)log(x); (3)Electronic address:adler@ias.edu
2 so that subtracting and substituting Eq. (2) gives log(x+y)log(x)log(x+y+U) + log(x+U) =∫ U 0 da(1 x+a11 x+y+a1) U 0 da(1 x+a1y1 x+a11 x+a1y1 x+a1y1 x+a1+:::) (4) Since log(x+y+U)log(x+U) = log(1 + (x+y)=U)log(1 +x=U)(5) vanishes asU! 1, taking this limit gives the Taylor expansion formula log(x+y)log(x) =∫ 1 0 da(1 x+a1y1 x+a11 x+a1y1 x+a1y1 x+a1+:::) :(6) The rst term in this expansion is the equation given in the query.III. DERIVATIVE OF THE MATRIX LOG
Lettingy=dx(t) and dividing bydt, one gets
d dt log(x(t))=∫ 1 0 da1 x(t) +a1dx(t) dt 1 x(t) +a1;(7) even whenx(t) anddx(t)=dtdo not commute. Making the substitutiona=b=(1b), a related formula can be given as an integral ∫1 0db, d dt log(x(t))=∫ 1 0 db1 (1b)x(t) +b1dx(t) dt 1 (1b)x(t) +b1:(8) These formulas are analogs of the well-known formula for the derivative of the matrix exponen- tial, d dt exp(x(t))=∫ 1 0 daexp(ax(t))dx(t) dt exp((1a)x(t));(9) which is given in the Wikipedia article on the matrix exponential.- log x differentiation formula