Mixing
K- linear transformations and differentiable functions
Given that $T_j : mathbb{R}^n times mathbb{R}^ntimes cdots times mathbb{R}^nto mathbb{R}^m$ $j$ times $mathbb{R}^n$, such that $T_j$ is $j$-linear ,
and $f: mathbb{R}^n to mathbb{R}^m$ which is $k-1$ differentiable.
given that for $a,h in mathbb{R}^n$ we have that $f(a+h)-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(||h||^k)$
Does this imply $f$ is $k$ differentiable ?
My attempt : let $g(a) = D^{k-1} f(a)$ since $f$ is $k-1$ differentiable, i want to show that $g$ is differentiable and from this $f$ is $k$ differentiable.
$g(a+h) = D^{k-1} f(a+h) = D^{k-1} (f(a)+Df(a)h +cdots +alpha(h))$ such that $alpha(h) = o(||h||^k)$
here were i get stuck
any help for proceed or another proof would really help me.
real-analysis multivariable-calculus proof-verification
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
add a comment |
Given that $T_j : mathbb{R}^n times mathbb{R}^ntimes cdots times mathbb{R}^nto mathbb{R}^m$ $j$ times $mathbb{R}^n$, such that $T_j$ is $j$-linear ,
and $f: mathbb{R}^n to mathbb{R}^m$ which is $k-1$ differentiable.
given that for $a,h in mathbb{R}^n$ we have that $f(a+h)-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(||h||^k)$
Does this imply $f$ is $k$ differentiable ?
My attempt : let $g(a) = D^{k-1} f(a)$ since $f$ is $k-1$ differentiable, i want to show that $g$ is differentiable and from this $f$ is $k$ differentiable.
$g(a+h) = D^{k-1} f(a+h) = D^{k-1} (f(a)+Df(a)h +cdots +alpha(h))$ such that $alpha(h) = o(||h||^k)$
here were i get stuck
any help for proceed or another proof would really help me.
real-analysis multivariable-calculus proof-verification
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42
add a comment |
Given that $T_j : mathbb{R}^n times mathbb{R}^ntimes cdots times mathbb{R}^nto mathbb{R}^m$ $j$ times $mathbb{R}^n$, such that $T_j$ is $j$-linear ,
and $f: mathbb{R}^n to mathbb{R}^m$ which is $k-1$ differentiable.
given that for $a,h in mathbb{R}^n$ we have that $f(a+h)-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(||h||^k)$
Does this imply $f$ is $k$ differentiable ?
My attempt : let $g(a) = D^{k-1} f(a)$ since $f$ is $k-1$ differentiable, i want to show that $g$ is differentiable and from this $f$ is $k$ differentiable.
$g(a+h) = D^{k-1} f(a+h) = D^{k-1} (f(a)+Df(a)h +cdots +alpha(h))$ such that $alpha(h) = o(||h||^k)$
here were i get stuck
any help for proceed or another proof would really help me.
real-analysis multivariable-calculus proof-verification
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
Given that $T_j : mathbb{R}^n times mathbb{R}^ntimes cdots times mathbb{R}^nto mathbb{R}^m$ $j$ times $mathbb{R}^n$, such that $T_j$ is $j$-linear ,
and $f: mathbb{R}^n to mathbb{R}^m$ which is $k-1$ differentiable.
given that for $a,h in mathbb{R}^n$ we have that $f(a+h)-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(||h||^k)$
Does this imply $f$ is $k$ differentiable ?
My attempt : let $g(a) = D^{k-1} f(a)$ since $f$ is $k-1$ differentiable, i want to show that $g$ is differentiable and from this $f$ is $k$ differentiable.
$g(a+h) = D^{k-1} f(a+h) = D^{k-1} (f(a)+Df(a)h +cdots +alpha(h))$ such that $alpha(h) = o(||h||^k)$
here were i get stuck
any help for proceed or another proof would really help me.
real-analysis multivariable-calculus proof-verification
real-analysis multivariable-calculus proof-verification
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
asked Nov 21 '18 at 8:34
Ahmad
2,5461625
2,5461625
Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42
add a comment |
Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42
Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42
add a comment |
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Do you mean $f(a+h)color{red}{-f(a)}-sum limits_{j=0}^{k} T_j (h,cdots ,h) = o(|h|^k)$ instead?
– user1551
Nov 21 '18 at 8:39
@user1551 no since $T_0 =f(a)$ ,you could do that and start the sum from $j=1$ !
– Ahmad
Nov 21 '18 at 8:42