Differential of scalar product












1














Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







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  • I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    – manooooh
    2 days ago












  • Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    – Mark
    2 days ago








  • 1




    Particular case of math.stackexchange.com/questions/1120430/….
    – Martín-Blas Pérez Pinilla
    2 days ago
















1














Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







share|cite|improve this question
























  • I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    – manooooh
    2 days ago












  • Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    – Mark
    2 days ago








  • 1




    Particular case of math.stackexchange.com/questions/1120430/….
    – Martín-Blas Pérez Pinilla
    2 days ago














1












1








1







Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus







share|cite|improve this question















Task from homework:




Let $f:Bbb R^ntoBbb R$, $f(x,y)=langle x, yrangle$, where $langlecdot,cdotrangle$ means the scalar product in $Bbb R^n$.
Find the differential $Df(x,y)(h,k)$.




First, the domain of $f$ is surely wrong, so with correcting it to $Bbb R^ntimes Bbb R^n$, I'm struggling to even start because every theorem we ever mentioned in class was about functions with the domain in $Bbb R^n$. How can $f$ be partially differentiated if the components $x$ and $y$ are again vectors?
Partial derivatives were my first idea, but any help would be appreciated.






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









APC89

1,920418




1,920418










asked 2 days ago









Lena

335




335












  • I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    – manooooh
    2 days ago












  • Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    – Mark
    2 days ago








  • 1




    Particular case of math.stackexchange.com/questions/1120430/….
    – Martín-Blas Pérez Pinilla
    2 days ago


















  • I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
    – manooooh
    2 days ago












  • Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
    – Mark
    2 days ago








  • 1




    Particular case of math.stackexchange.com/questions/1120430/….
    – Martín-Blas Pérez Pinilla
    2 days ago
















I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
– manooooh
2 days ago






I think $f$ can be interpreted as $f(x,y)=xcdot y$, so $D_f(h,k)=Bigl.(y,x)Bigr|_{(h,k)}=(k,h)$. Also, there is no correlation between its domain and its variables; $Bbb R^n$ but $(x,y)$?
– manooooh
2 days ago














Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
– Mark
2 days ago






Well, $mathbb{R^n}timesmathbb{R^n}$ is just $mathbb{R^{2n}}$.
– Mark
2 days ago






1




1




Particular case of math.stackexchange.com/questions/1120430/….
– Martín-Blas Pérez Pinilla
2 days ago




Particular case of math.stackexchange.com/questions/1120430/….
– Martín-Blas Pérez Pinilla
2 days ago










2 Answers
2






active

oldest

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2














The value of the differential at a point $(x,y)$ is the linear part of the difference
$;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
$$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






share|cite|improve this answer































    0














    The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






    share|cite|improve this answer























    • I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
      – Lena
      2 days ago










    • It is a bit fishy, yes, I edit the answer.
      – gangrene
      2 days ago











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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    2














    The value of the differential at a point $(x,y)$ is the linear part of the difference
    $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
    $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
    so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






    share|cite|improve this answer




























      2














      The value of the differential at a point $(x,y)$ is the linear part of the difference
      $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
      $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
      so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






      share|cite|improve this answer


























        2












        2








        2






        The value of the differential at a point $(x,y)$ is the linear part of the difference
        $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
        $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
        so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.






        share|cite|improve this answer














        The value of the differential at a point $(x,y)$ is the linear part of the difference
        $;f(x+h, y+k)-f(x,y)$. Now if $f$ is the dot product, we can use bilinearity:
        $$f(x+h, y+k)-f(x,y)=langle x+h,y+krangle -langle x, yrangle=underbrace{langle x,krangle+langle h, yrangle}_text{linear terms} +underbrace{langle h, krangle}_{o(|(h,k)|)},$$
        so $; Df_{(x,y)}(h,k)=langle x,krangle+langle h, yrangle$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        Bernard

        118k639112




        118k639112























            0














            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer























            • I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              – Lena
              2 days ago










            • It is a bit fishy, yes, I edit the answer.
              – gangrene
              2 days ago
















            0














            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer























            • I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              – Lena
              2 days ago










            • It is a bit fishy, yes, I edit the answer.
              – gangrene
              2 days ago














            0












            0








            0






            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.






            share|cite|improve this answer














            The (usual, euclidean) scalar product is defined via $$ langle x ,y rangle = x_1 y_1 + x_2 y_2 + dots + x_n y_n $$ with $x,y in mathbb{R}^n$, so your function is essentially $ f:mathbb{R}^n times mathbb{R}^n simeq mathbb{R}^{2n} to mathbb{R} $ defined indeed by $$ f(x,y) = x_1 y_1 + x_2 y_2 + dots + x_n y_n. $$From here you should be able to conclude.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            gangrene

            915514




            915514












            • I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              – Lena
              2 days ago










            • It is a bit fishy, yes, I edit the answer.
              – gangrene
              2 days ago


















            • I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
              – Lena
              2 days ago










            • It is a bit fishy, yes, I edit the answer.
              – gangrene
              2 days ago
















            I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
            – Lena
            2 days ago




            I agree that $Bbb R^n times Bbb R^n$ is isomorphic to $Bbb R^{2n}$, but is it really correct to say they're equal?
            – Lena
            2 days ago












            It is a bit fishy, yes, I edit the answer.
            – gangrene
            2 days ago




            It is a bit fishy, yes, I edit the answer.
            – gangrene
            2 days ago


















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