Verification on proof of $sigma(kU)=k^{2n+1}sigma(U)$?












1












$begingroup$


Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let



$$sigma(U)=iint_U d(x,y)dxdy$$



Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.



I think I have a proof of $sigma(kU)=k^{2n+1}sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:




Each of its points of $U$ will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations



$$x_i'=kx_i$$



$$dx_i'=kdx_i$$



Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x'=kx}=U$:



begin{align}
sigma(kU) &= iint_{kU}d(x',y')dx'dy' \
&= iint_U d(kx,ky)kdx kdy \
&= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(kx_1-ky_1)^2+cdots+(kx_n-ky_n)^2}kdx_1ldots kdx_nkdy_1ldots kdy_n \
&= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} ksqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}k^{2n}dx_1ldots dx_ndy_1ldots dy_n \
&= k^{2n+1}int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}dx_1ldots dx_ndy_1ldots dy_n \
&= k^{2n+1}iint_Ud(x,y)dxdy \
sigma(kU)&= k^{2n+1}sigma(U)
end{align}



Is this correct? Thanks.






calculus proof-verification metric-spaces







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let



    $$sigma(U)=iint_U d(x,y)dxdy$$



    Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.



    I think I have a proof of $sigma(kU)=k^{2n+1}sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:




    Each of its points of $U$ will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations



    $$x_i'=kx_i$$



    $$dx_i'=kdx_i$$



    Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x'=kx}=U$:



    begin{align}
    sigma(kU) &= iint_{kU}d(x',y')dx'dy' \
    &= iint_U d(kx,ky)kdx kdy \
    &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(kx_1-ky_1)^2+cdots+(kx_n-ky_n)^2}kdx_1ldots kdx_nkdy_1ldots kdy_n \
    &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} ksqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}k^{2n}dx_1ldots dx_ndy_1ldots dy_n \
    &= k^{2n+1}int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}dx_1ldots dx_ndy_1ldots dy_n \
    &= k^{2n+1}iint_Ud(x,y)dxdy \
    sigma(kU)&= k^{2n+1}sigma(U)
    end{align}



    Is this correct? Thanks.






    calculus proof-verification metric-spaces







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let



      $$sigma(U)=iint_U d(x,y)dxdy$$



      Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.



      I think I have a proof of $sigma(kU)=k^{2n+1}sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:




      Each of its points of $U$ will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations



      $$x_i'=kx_i$$



      $$dx_i'=kdx_i$$



      Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x'=kx}=U$:



      begin{align}
      sigma(kU) &= iint_{kU}d(x',y')dx'dy' \
      &= iint_U d(kx,ky)kdx kdy \
      &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(kx_1-ky_1)^2+cdots+(kx_n-ky_n)^2}kdx_1ldots kdx_nkdy_1ldots kdy_n \
      &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} ksqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}k^{2n}dx_1ldots dx_ndy_1ldots dy_n \
      &= k^{2n+1}int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}dx_1ldots dx_ndy_1ldots dy_n \
      &= k^{2n+1}iint_Ud(x,y)dxdy \
      sigma(kU)&= k^{2n+1}sigma(U)
      end{align}



      Is this correct? Thanks.






      calculus proof-verification metric-spaces







      share|cite|improve this question











      $endgroup$




      Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let



      $$sigma(U)=iint_U d(x,y)dxdy$$



      Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.



      I think I have a proof of $sigma(kU)=k^{2n+1}sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:




      Each of its points of $U$ will have coordinates $x=(x_1,ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',ldots,x_n')$ with the relations



      $$x_i'=kx_i$$



      $$dx_i'=kdx_i$$



      Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to ${xmid exists x'in kU:x'=kx}=U$:



      begin{align}
      sigma(kU) &= iint_{kU}d(x',y')dx'dy' \
      &= iint_U d(kx,ky)kdx kdy \
      &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(kx_1-ky_1)^2+cdots+(kx_n-ky_n)^2}kdx_1ldots kdx_nkdy_1ldots kdy_n \
      &= int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} ksqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}k^{2n}dx_1ldots dx_ndy_1ldots dy_n \
      &= k^{2n+1}int_{x_1^0}^{x_1^1}cdotsint_{x_n^0}^{x_n^1}int_{y_1^0}^{y_1^1}cdotsint_{y_n^0}^{y_n^1} sqrt{(x_1-y_1)^2+cdots+(x_n-y_n)^2}dx_1ldots dx_ndy_1ldots dy_n \
      &= k^{2n+1}iint_Ud(x,y)dxdy \
      sigma(kU)&= k^{2n+1}sigma(U)
      end{align}



      Is this correct? Thanks.






      calculus proof-verification metric-spaces




      calculus proof-verification metric-spaces






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      share|cite|improve this question













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      edited Jan 15 at 0:10







      Garmekain

















      asked Jan 14 at 21:25









      GarmekainGarmekain

      1,372720




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          $begingroup$

          One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$int_{x_1^0}^{x_1^1}...int_{x_n^0}^{x_n^1}int_{y_0^0}^{y_0^1}...int_{y_n^0}^{y_n^1}d(x,y),dx,dy.$$



          But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,yin U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.






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            $begingroup$

            One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$int_{x_1^0}^{x_1^1}...int_{x_n^0}^{x_n^1}int_{y_0^0}^{y_0^1}...int_{y_n^0}^{y_n^1}d(x,y),dx,dy.$$



            But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,yin U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$int_{x_1^0}^{x_1^1}...int_{x_n^0}^{x_n^1}int_{y_0^0}^{y_0^1}...int_{y_n^0}^{y_n^1}d(x,y),dx,dy.$$



              But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,yin U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$int_{x_1^0}^{x_1^1}...int_{x_n^0}^{x_n^1}int_{y_0^0}^{y_0^1}...int_{y_n^0}^{y_n^1}d(x,y),dx,dy.$$



                But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,yin U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.






                share|cite|improve this answer









                $endgroup$



                One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$int_{x_1^0}^{x_1^1}...int_{x_n^0}^{x_n^1}int_{y_0^0}^{y_0^1}...int_{y_n^0}^{y_n^1}d(x,y),dx,dy.$$



                But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,yin U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 0:25









                AyoubAyoub

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