Differentiable maps and be unique linear map
Let $pmb F : U subseteq mathbb{R}^n to mathbb{R}^m$
is differentiable at $pmb a in U$ and let $pmb a $ is an cluster point of $U$ if there exists a linear map $T : mathbb{R}^n to mathbb{R}^m $ satisfying:
$$big(forall epsilon > 0 in mathbb{R} big) big(exists delta_{(epsilon,pmb a)}>0 in mathbb{R}big) $$
$$text{such that} :$$
$$parallel big(pmb x-pmb a big) parallel leq delta text{for every } pmb x in mathbb{R}^n Longrightarrow parallel pmb F(pmb x)- pmb F(pmb a) - T(pmb x -pmb a)parallel leq epsilon parallel pmb x -pmb aparallel$$
The map $T$ is unique, denoted $df_{pmb a} $ and called ‘differential (or ‘derivative’) of $f$ at $pmb a$ . Another notation for this is:
$$pmb F(pmb x)= pmb F(pmb a) + df_{pmb a}(pmb x -pmb a)+pmb r(pmb x)$$
I have two questions :
$1-)$ How prove that $T$ is unique ?
$2-)$ What is function of $pmb r$ ? What is domain and range of function $pmb r$ ?
multivariable-calculus
asked Dec 26 '18 at 22:44
Almot1960
2,503723
add a comment |
Let $pmb F : U subseteq mathbb{R}^n to mathbb{R}^m$
is differentiable at $pmb a in U$ and let $pmb a $ is an cluster point of $U$ if there exists a linear map $T : mathbb{R}^n to mathbb{R}^m $ satisfying:
$$big(forall epsilon > 0 in mathbb{R} big) big(exists delta_{(epsilon,pmb a)}>0 in mathbb{R}big) $$
$$text{such that} :$$
$$parallel big(pmb x-pmb a big) parallel leq delta text{for every } pmb x in mathbb{R}^n Longrightarrow parallel pmb F(pmb x)- pmb F(pmb a) - T(pmb x -pmb a)parallel leq epsilon parallel pmb x -pmb aparallel$$
The map $T$ is unique, denoted $df_{pmb a} $ and called ‘differential (or ‘derivative’) of $f$ at $pmb a$ . Another notation for this is:
$$pmb F(pmb x)= pmb F(pmb a) + df_{pmb a}(pmb x -pmb a)+pmb r(pmb x)$$
I have two questions :
$1-)$ How prove that $T$ is unique ?
$2-)$ What is function of $pmb r$ ? What is domain and range of function $pmb r$ ?
multivariable-calculus
asked Dec 26 '18 at 22:44
Almot1960
2,503723
add a comment |
Let $pmb F : U subseteq mathbb{R}^n to mathbb{R}^m$
is differentiable at $pmb a in U$ and let $pmb a $ is an cluster point of $U$ if there exists a linear map $T : mathbb{R}^n to mathbb{R}^m $ satisfying:
$$big(forall epsilon > 0 in mathbb{R} big) big(exists delta_{(epsilon,pmb a)}>0 in mathbb{R}big) $$
$$text{such that} :$$
$$parallel big(pmb x-pmb a big) parallel leq delta text{for every } pmb x in mathbb{R}^n Longrightarrow parallel pmb F(pmb x)- pmb F(pmb a) - T(pmb x -pmb a)parallel leq epsilon parallel pmb x -pmb aparallel$$
The map $T$ is unique, denoted $df_{pmb a} $ and called ‘differential (or ‘derivative’) of $f$ at $pmb a$ . Another notation for this is:
$$pmb F(pmb x)= pmb F(pmb a) + df_{pmb a}(pmb x -pmb a)+pmb r(pmb x)$$
I have two questions :
$1-)$ How prove that $T$ is unique ?
$2-)$ What is function of $pmb r$ ? What is domain and range of function $pmb r$ ?
multivariable-calculus
asked Dec 26 '18 at 22:44
Almot1960
2,503723
Let $pmb F : U subseteq mathbb{R}^n to mathbb{R}^m$
is differentiable at $pmb a in U$ and let $pmb a $ is an cluster point of $U$ if there exists a linear map $T : mathbb{R}^n to mathbb{R}^m $ satisfying:
$$big(forall epsilon > 0 in mathbb{R} big) big(exists delta_{(epsilon,pmb a)}>0 in mathbb{R}big) $$
$$text{such that} :$$
$$parallel big(pmb x-pmb a big) parallel leq delta text{for every } pmb x in mathbb{R}^n Longrightarrow parallel pmb F(pmb x)- pmb F(pmb a) - T(pmb x -pmb a)parallel leq epsilon parallel pmb x -pmb aparallel$$
The map $T$ is unique, denoted $df_{pmb a} $ and called ‘differential (or ‘derivative’) of $f$ at $pmb a$ . Another notation for this is:
$$pmb F(pmb x)= pmb F(pmb a) + df_{pmb a}(pmb x -pmb a)+pmb r(pmb x)$$
I have two questions :
$1-)$ How prove that $T$ is unique ?
$2-)$ What is function of $pmb r$ ? What is domain and range of function $pmb r$ ?
multivariable-calculus
multivariable-calculus
asked Dec 26 '18 at 22:44
Almot1960
2,503723
asked Dec 26 '18 at 22:44
Almot1960
2,503723
asked Dec 26 '18 at 22:44
Almot1960
2,503723
asked Dec 26 '18 at 22:44
Almot1960
2,503723
asked Dec 26 '18 at 22:44
Almot1960
2,503723
2,503723
add a comment |
add a comment |
1 Answer
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Let $T'$ be another differential, $|x-a|leq delta_n$ implies that $|F(x)-F(a)-T(x-a)|leq 1/n|x-a|$ and $|F(x)-F(a)-T'(x-a)|leq 1/n|x-a|$.
We deduce that $|T(x-a)-T'(x-a)|leq |(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)|leq |F(x)-F(a)-T(x-a)|+|F(x)-F(a)-T'(x-a)|leq 2/n |x-a|$.
Let $b$ such that $|x-b|=1$, we have $|delta_n(x-b)|=delta_n$ impliest that $|T(delta_n(x-b))-T'(delta_n(x-b))|leq {2over n}|delta_n(x-b)|$. This implies that $|(T-T')(x-b)|leq {2over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $aneq x$, $(T-T')({{x-a}over{|x-a|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
add a comment |
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Let $T'$ be another differential, $|x-a|leq delta_n$ implies that $|F(x)-F(a)-T(x-a)|leq 1/n|x-a|$ and $|F(x)-F(a)-T'(x-a)|leq 1/n|x-a|$.
We deduce that $|T(x-a)-T'(x-a)|leq |(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)|leq |F(x)-F(a)-T(x-a)|+|F(x)-F(a)-T'(x-a)|leq 2/n |x-a|$.
Let $b$ such that $|x-b|=1$, we have $|delta_n(x-b)|=delta_n$ impliest that $|T(delta_n(x-b))-T'(delta_n(x-b))|leq {2over n}|delta_n(x-b)|$. This implies that $|(T-T')(x-b)|leq {2over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $aneq x$, $(T-T')({{x-a}over{|x-a|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
add a comment |
Let $T'$ be another differential, $|x-a|leq delta_n$ implies that $|F(x)-F(a)-T(x-a)|leq 1/n|x-a|$ and $|F(x)-F(a)-T'(x-a)|leq 1/n|x-a|$.
We deduce that $|T(x-a)-T'(x-a)|leq |(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)|leq |F(x)-F(a)-T(x-a)|+|F(x)-F(a)-T'(x-a)|leq 2/n |x-a|$.
Let $b$ such that $|x-b|=1$, we have $|delta_n(x-b)|=delta_n$ impliest that $|T(delta_n(x-b))-T'(delta_n(x-b))|leq {2over n}|delta_n(x-b)|$. This implies that $|(T-T')(x-b)|leq {2over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $aneq x$, $(T-T')({{x-a}over{|x-a|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
add a comment |
Let $T'$ be another differential, $|x-a|leq delta_n$ implies that $|F(x)-F(a)-T(x-a)|leq 1/n|x-a|$ and $|F(x)-F(a)-T'(x-a)|leq 1/n|x-a|$.
We deduce that $|T(x-a)-T'(x-a)|leq |(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)|leq |F(x)-F(a)-T(x-a)|+|F(x)-F(a)-T'(x-a)|leq 2/n |x-a|$.
Let $b$ such that $|x-b|=1$, we have $|delta_n(x-b)|=delta_n$ impliest that $|T(delta_n(x-b))-T'(delta_n(x-b))|leq {2over n}|delta_n(x-b)|$. This implies that $|(T-T')(x-b)|leq {2over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $aneq x$, $(T-T')({{x-a}over{|x-a|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
Let $T'$ be another differential, $|x-a|leq delta_n$ implies that $|F(x)-F(a)-T(x-a)|leq 1/n|x-a|$ and $|F(x)-F(a)-T'(x-a)|leq 1/n|x-a|$.
We deduce that $|T(x-a)-T'(x-a)|leq |(T(x-a)-(F(x)-F(a))+(F(x)-F(a)-T'(x-a)|leq |F(x)-F(a)-T(x-a)|+|F(x)-F(a)-T'(x-a)|leq 2/n |x-a|$.
Let $b$ such that $|x-b|=1$, we have $|delta_n(x-b)|=delta_n$ impliest that $|T(delta_n(x-b))-T'(delta_n(x-b))|leq {2over n}|delta_n(x-b)|$. This implies that $|(T-T')(x-b)|leq {2over n}$ for every $n>0$, we deduce that $(T-T')(x-b)=0$ and $T-T'=0$. Since for every $aneq x$, $(T-T')({{x-a}over{|x-a|}})=0$.
By definition, $r(x)= F(x)-F(a)-df_a(x-a)$.
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
answered Dec 26 '18 at 23:09
Tsemo Aristide
55.7k11444
55.7k11444
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
add a comment |
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
– Almot1960
Dec 26 '18 at 23:55
add a comment |
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