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Multivariable Taylor's formula and approximation (big O)
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I post here because I encounter a problem to understand something. To begin, let consider $f : mathbbR^n rightarrow mathbbR$ a n times differentiable function. Then, we have, by the Taylor's formula :
$$ forall a, h in mathbbR^n quad f(a+h)=f(a)+Df_a(h) + ... + frac1n!D^nf_a(h^n) + o_o(||h||^n)$$
So, when $n=1$, it's easy to see that each term of the taylor expansion is "bigger" than the next one, i.e for $k in mathbbN$, we have : $D^kf_a(h^k)=f^(k)(a)h^k$ and then $o(h^k) = o(D^kf_a(h^k))$, but I've some trouble when $n>1$. Actually, for $n=2$ and the taylor formula at the order 2, we have :
$D^2f_a((x_1, x_2), (x_1,x_2)) = fracpartial ^2 fpartial x_1^2(a) x_1^2 + 2fracpartial ^2 fpartial x_1 partial x_2(a) x_1x_2 + fracpartial ^2 fpartial x_2^2(a) x_2^2$
And I want to show that : $o(||(x_1,x_2)||^2) = o(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. By considering (and utilizating the equivalence of the norm in finite dimension), if I consider the norm $||(x_1,x_2)|| = max (|x_1|, |x_2|)$, then I have : $D^2f_a((x_1, x_2), (x_1,x_2)) = O(||(x_1,x_2)||^2)$, but I whould like to have $||(x_1,x_2)||^2 = O(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. But, still with the norm max, if I consider $x_1x_2$, I should have : $x_1x_2$ which verify : $x_1x_2 geq C||(x_1,x_2)^2||$, but if I consider $(x_1,x_2) = (e, e^2)$, for $e<<0$, I have : $e^3 geq Ce^2$, which is not true.
So, I'm kind of lost. And here, I've only considered $mathbbR^n$, but what happen if I consider a vector space of infinite dimension ? (cause I could not use the equivalence of norm)
real-analysis multivariable-calculus derivatives
asked Jul 22 at 18:47
ChocoSavour
1017
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up vote
2
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I post here because I encounter a problem to understand something. To begin, let consider $f : mathbbR^n rightarrow mathbbR$ a n times differentiable function. Then, we have, by the Taylor's formula :
$$ forall a, h in mathbbR^n quad f(a+h)=f(a)+Df_a(h) + ... + frac1n!D^nf_a(h^n) + o_o(||h||^n)$$
So, when $n=1$, it's easy to see that each term of the taylor expansion is "bigger" than the next one, i.e for $k in mathbbN$, we have : $D^kf_a(h^k)=f^(k)(a)h^k$ and then $o(h^k) = o(D^kf_a(h^k))$, but I've some trouble when $n>1$. Actually, for $n=2$ and the taylor formula at the order 2, we have :
$D^2f_a((x_1, x_2), (x_1,x_2)) = fracpartial ^2 fpartial x_1^2(a) x_1^2 + 2fracpartial ^2 fpartial x_1 partial x_2(a) x_1x_2 + fracpartial ^2 fpartial x_2^2(a) x_2^2$
And I want to show that : $o(||(x_1,x_2)||^2) = o(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. By considering (and utilizating the equivalence of the norm in finite dimension), if I consider the norm $||(x_1,x_2)|| = max (|x_1|, |x_2|)$, then I have : $D^2f_a((x_1, x_2), (x_1,x_2)) = O(||(x_1,x_2)||^2)$, but I whould like to have $||(x_1,x_2)||^2 = O(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. But, still with the norm max, if I consider $x_1x_2$, I should have : $x_1x_2$ which verify : $x_1x_2 geq C||(x_1,x_2)^2||$, but if I consider $(x_1,x_2) = (e, e^2)$, for $e<<0$, I have : $e^3 geq Ce^2$, which is not true.
So, I'm kind of lost. And here, I've only considered $mathbbR^n$, but what happen if I consider a vector space of infinite dimension ? (cause I could not use the equivalence of norm)
real-analysis multivariable-calculus derivatives
asked Jul 22 at 18:47
ChocoSavour
1017
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I post here because I encounter a problem to understand something. To begin, let consider $f : mathbbR^n rightarrow mathbbR$ a n times differentiable function. Then, we have, by the Taylor's formula :
$$ forall a, h in mathbbR^n quad f(a+h)=f(a)+Df_a(h) + ... + frac1n!D^nf_a(h^n) + o_o(||h||^n)$$
So, when $n=1$, it's easy to see that each term of the taylor expansion is "bigger" than the next one, i.e for $k in mathbbN$, we have : $D^kf_a(h^k)=f^(k)(a)h^k$ and then $o(h^k) = o(D^kf_a(h^k))$, but I've some trouble when $n>1$. Actually, for $n=2$ and the taylor formula at the order 2, we have :
$D^2f_a((x_1, x_2), (x_1,x_2)) = fracpartial ^2 fpartial x_1^2(a) x_1^2 + 2fracpartial ^2 fpartial x_1 partial x_2(a) x_1x_2 + fracpartial ^2 fpartial x_2^2(a) x_2^2$
And I want to show that : $o(||(x_1,x_2)||^2) = o(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. By considering (and utilizating the equivalence of the norm in finite dimension), if I consider the norm $||(x_1,x_2)|| = max (|x_1|, |x_2|)$, then I have : $D^2f_a((x_1, x_2), (x_1,x_2)) = O(||(x_1,x_2)||^2)$, but I whould like to have $||(x_1,x_2)||^2 = O(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. But, still with the norm max, if I consider $x_1x_2$, I should have : $x_1x_2$ which verify : $x_1x_2 geq C||(x_1,x_2)^2||$, but if I consider $(x_1,x_2) = (e, e^2)$, for $e<<0$, I have : $e^3 geq Ce^2$, which is not true.
So, I'm kind of lost. And here, I've only considered $mathbbR^n$, but what happen if I consider a vector space of infinite dimension ? (cause I could not use the equivalence of norm)
real-analysis multivariable-calculus derivatives
asked Jul 22 at 18:47
ChocoSavour
1017
I post here because I encounter a problem to understand something. To begin, let consider $f : mathbbR^n rightarrow mathbbR$ a n times differentiable function. Then, we have, by the Taylor's formula :
$$ forall a, h in mathbbR^n quad f(a+h)=f(a)+Df_a(h) + ... + frac1n!D^nf_a(h^n) + o_o(||h||^n)$$
So, when $n=1$, it's easy to see that each term of the taylor expansion is "bigger" than the next one, i.e for $k in mathbbN$, we have : $D^kf_a(h^k)=f^(k)(a)h^k$ and then $o(h^k) = o(D^kf_a(h^k))$, but I've some trouble when $n>1$. Actually, for $n=2$ and the taylor formula at the order 2, we have :
$D^2f_a((x_1, x_2), (x_1,x_2)) = fracpartial ^2 fpartial x_1^2(a) x_1^2 + 2fracpartial ^2 fpartial x_1 partial x_2(a) x_1x_2 + fracpartial ^2 fpartial x_2^2(a) x_2^2$
And I want to show that : $o(||(x_1,x_2)||^2) = o(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. By considering (and utilizating the equivalence of the norm in finite dimension), if I consider the norm $||(x_1,x_2)|| = max (|x_1|, |x_2|)$, then I have : $D^2f_a((x_1, x_2), (x_1,x_2)) = O(||(x_1,x_2)||^2)$, but I whould like to have $||(x_1,x_2)||^2 = O(||D^2f_a((x_1, x_2), (x_1,x_2))||)$. But, still with the norm max, if I consider $x_1x_2$, I should have : $x_1x_2$ which verify : $x_1x_2 geq C||(x_1,x_2)^2||$, but if I consider $(x_1,x_2) = (e, e^2)$, for $e<<0$, I have : $e^3 geq Ce^2$, which is not true.
So, I'm kind of lost. And here, I've only considered $mathbbR^n$, but what happen if I consider a vector space of infinite dimension ? (cause I could not use the equivalence of norm)
real-analysis multivariable-calculus derivatives
asked Jul 22 at 18:47
ChocoSavour
1017
edited Jul 22 at 22:57
asked Jul 22 at 18:47
ChocoSavour
1017
asked Jul 22 at 18:47
ChocoSavour
1017
asked Jul 22 at 18:47
ChocoSavour
1017
1017
add a comment |Â
add a comment |Â
1 Answer
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Without some assumptions on $f$, the equality $|(x_1, x_2)|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = frac 12(x_1^2 - x_2^2)$ and $a = 0in mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $|(x_1, x_2)|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.
answered Jul 23 at 1:51
BindersFull
498110
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Without some assumptions on $f$, the equality $|(x_1, x_2)|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = frac 12(x_1^2 - x_2^2)$ and $a = 0in mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $|(x_1, x_2)|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.
answered Jul 23 at 1:51
BindersFull
498110
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
add a comment |Â
up vote
1
down vote
Without some assumptions on $f$, the equality $|(x_1, x_2)|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = frac 12(x_1^2 - x_2^2)$ and $a = 0in mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $|(x_1, x_2)|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.
answered Jul 23 at 1:51
BindersFull
498110
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Without some assumptions on $f$, the equality $|(x_1, x_2)|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = frac 12(x_1^2 - x_2^2)$ and $a = 0in mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $|(x_1, x_2)|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.
answered Jul 23 at 1:51
BindersFull
498110
Without some assumptions on $f$, the equality $|(x_1, x_2)|^2 = O(D^2 f_a((x_1, x_2), (x_1, x_2))$ does not hold. For example, choose $f(x_1,x_2) = frac 12(x_1^2 - x_2^2)$ and $a = 0in mathbb R^2$ so that $D^2 f_0((x_1, x_2), (x_1, x_2)) = x_1^2 - x_2^2$. For $(x_1, x_2) = (1, 1)$ we have $|(x_1, x_2)|^2 = 2$ but $D^2 f_0((x_1, x_2), (x_1, x_2)) = 0$.
answered Jul 23 at 1:51
BindersFull
498110
answered Jul 23 at 1:51
BindersFull
498110
answered Jul 23 at 1:51
BindersFull
498110
answered Jul 23 at 1:51
BindersFull
498110
498110
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
add a comment |Â
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
Thank you very much !
– ChocoSavour
Jul 25 at 22:46
add a comment |Â
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