Differentiable maps and be unique linear map












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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







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    0















    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







    share|cite|improve this question

























      0












      0








      0


      1






      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







      share|cite|improve this question














      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






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 26 '18 at 22:44









      Almot1960

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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, $r(x)= F(x)-F(a)-df_a(x-a)$.






          share|cite|improve this answer





















          • Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
            – Almot1960
            Dec 26 '18 at 23:55











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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, $r(x)= F(x)-F(a)-df_a(x-a)$.






          share|cite|improve this answer





















          • Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
            – Almot1960
            Dec 26 '18 at 23:55
















          1














          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, $r(x)= F(x)-F(a)-df_a(x-a)$.






          share|cite|improve this answer





















          • Thanks . so function $r$ must be $r :mathbb{R}^n to mathbb{R}^m$ .that's right ?
            – Almot1960
            Dec 26 '18 at 23:55














          1












          1








          1






          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, $r(x)= F(x)-F(a)-df_a(x-a)$.






          share|cite|improve this answer












          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, $r(x)= F(x)-F(a)-df_a(x-a)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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