Confused by an unusual form of Taylor's theorem The 2019 Stack Overflow Developer Survey Results Are InTaylor's Theorem Problemprove or disprove an inequality on bounds of derivatives for radial functionsTaylor's Theorem expansionImage of Jordan measurable set under a diffeomorphism.A different form of Taylor's TheoremStrange form of Taylor's Theorem for linearizationCoherence between Hamiltonian and Lagrangian in form of an inequality (Evans PDE)A closed form for $int_0^pi lvert sin(m t) cos(n t) rvert , mathrmd t$Taylor's formula and multi-indices
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Confused by an unusual form of Taylor's theorem
The 2019 Stack Overflow Developer Survey Results Are InTaylor's Theorem Problemprove or disprove an inequality on bounds of derivatives for radial functionsTaylor's Theorem expansionImage of Jordan measurable set under a diffeomorphism.A different form of Taylor's TheoremStrange form of Taylor's Theorem for linearizationCoherence between Hamiltonian and Lagrangian in form of an inequality (Evans PDE)A closed form for $int_0^pi lvert sin(m t) cos(n t) rvert , mathrmd t$Taylor's formula and multi-indices
$begingroup$
I'm reading up on manifolds and encountered this proposition:
If $g:mathbbR^ntomathbbR$ is a $C^k$ function ($kgeq 2$) on some convex open subset $UsubseteqmathbbR^n$ about $pinmathbbR^n$, then for every $qin U$ we have
$$g(q)=g(p) + partial_ig rvert_p(q_i-p_i) + (q_i-p_i)(q_j-p_j)int_0^1partial_i,j^2 grvert_(1-t)p+tq,dt$$
in the Einstein summation notation, where the integral is a function $h:UtomathbbR$ of $q$. In particular, when $k=infty$, i.e. $g$ is smooth, so too is the integral $h$.
I'm familiar with the following form of Taylor's theorem, in multi-index notation:
beginalign
f(vecx) &= f(veca) +
sum_1leq lvert alpha rvertleq k (D^alphaf)(veca)(vecx-veca)^alpha \ &+
sum_lvertalpharvertleq k+1 dfrack+1alpha!(vecx-veca)^alpha int_0^1 (1-t)^k (D^alphaf) left( (1-t)veca + tvecx right),dt
endalign
It would seem that the first proposition is truncating the terms in the second involving derivatives of order higher than 2. How are the two results related and/or how does the first follow from the second?
calculus taylor-expansion
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
$endgroup$
add a comment |
$begingroup$
I'm reading up on manifolds and encountered this proposition:
If $g:mathbbR^ntomathbbR$ is a $C^k$ function ($kgeq 2$) on some convex open subset $UsubseteqmathbbR^n$ about $pinmathbbR^n$, then for every $qin U$ we have
$$g(q)=g(p) + partial_ig rvert_p(q_i-p_i) + (q_i-p_i)(q_j-p_j)int_0^1partial_i,j^2 grvert_(1-t)p+tq,dt$$
in the Einstein summation notation, where the integral is a function $h:UtomathbbR$ of $q$. In particular, when $k=infty$, i.e. $g$ is smooth, so too is the integral $h$.
I'm familiar with the following form of Taylor's theorem, in multi-index notation:
beginalign
f(vecx) &= f(veca) +
sum_1leq lvert alpha rvertleq k (D^alphaf)(veca)(vecx-veca)^alpha \ &+
sum_lvertalpharvertleq k+1 dfrack+1alpha!(vecx-veca)^alpha int_0^1 (1-t)^k (D^alphaf) left( (1-t)veca + tvecx right),dt
endalign
It would seem that the first proposition is truncating the terms in the second involving derivatives of order higher than 2. How are the two results related and/or how does the first follow from the second?
calculus taylor-expansion
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
$endgroup$
add a comment |
$begingroup$
I'm reading up on manifolds and encountered this proposition:
If $g:mathbbR^ntomathbbR$ is a $C^k$ function ($kgeq 2$) on some convex open subset $UsubseteqmathbbR^n$ about $pinmathbbR^n$, then for every $qin U$ we have
$$g(q)=g(p) + partial_ig rvert_p(q_i-p_i) + (q_i-p_i)(q_j-p_j)int_0^1partial_i,j^2 grvert_(1-t)p+tq,dt$$
in the Einstein summation notation, where the integral is a function $h:UtomathbbR$ of $q$. In particular, when $k=infty$, i.e. $g$ is smooth, so too is the integral $h$.
I'm familiar with the following form of Taylor's theorem, in multi-index notation:
beginalign
f(vecx) &= f(veca) +
sum_1leq lvert alpha rvertleq k (D^alphaf)(veca)(vecx-veca)^alpha \ &+
sum_lvertalpharvertleq k+1 dfrack+1alpha!(vecx-veca)^alpha int_0^1 (1-t)^k (D^alphaf) left( (1-t)veca + tvecx right),dt
endalign
It would seem that the first proposition is truncating the terms in the second involving derivatives of order higher than 2. How are the two results related and/or how does the first follow from the second?
calculus taylor-expansion
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
$endgroup$
I'm reading up on manifolds and encountered this proposition:
If $g:mathbbR^ntomathbbR$ is a $C^k$ function ($kgeq 2$) on some convex open subset $UsubseteqmathbbR^n$ about $pinmathbbR^n$, then for every $qin U$ we have
$$g(q)=g(p) + partial_ig rvert_p(q_i-p_i) + (q_i-p_i)(q_j-p_j)int_0^1partial_i,j^2 grvert_(1-t)p+tq,dt$$
in the Einstein summation notation, where the integral is a function $h:UtomathbbR$ of $q$. In particular, when $k=infty$, i.e. $g$ is smooth, so too is the integral $h$.
I'm familiar with the following form of Taylor's theorem, in multi-index notation:
beginalign
f(vecx) &= f(veca) +
sum_1leq lvert alpha rvertleq k (D^alphaf)(veca)(vecx-veca)^alpha \ &+
sum_lvertalpharvertleq k+1 dfrack+1alpha!(vecx-veca)^alpha int_0^1 (1-t)^k (D^alphaf) left( (1-t)veca + tvecx right),dt
endalign
It would seem that the first proposition is truncating the terms in the second involving derivatives of order higher than 2. How are the two results related and/or how does the first follow from the second?
calculus taylor-expansion
calculus taylor-expansion
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
asked Mar 23 at 18:05
terrygarciaterrygarcia
17211
17211
add a comment |
add a comment |
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