The implication of zero mixed partial derivatives for multivariate function's minimizationIf the second mixed partial is identically zero, then the function can be written as a sum $f(x,y) = f_1(x) + f_2(y)$$dfracpartial^2 fpartial x partial y = 0 nRightarrow f(x,y) = g(x) + h(y)$If a nonnegative function of $x_1,dots,x_n$ can be written as $sum g_k(x_k)$, then the summands can be taken nonnegative$f(x,y) = f_1(x) + f_2(x)$ with continuous differential real-valued functionsShow that both mixed partial derivatives exist at the origin but are not equalFunction on $mathbbR^2-0$.Let $F_1,F_2:bf R^2 to R$ be functions such that…How many n-th Order Partial Derivatives Exist for a Function of k Variables?Maximize $f(textbfx,textbfy) = f_1(textbfx) + f_2(textbfx,textbfy)$Symmetry of higher order mixed partial derivatives under weaker assumptionsWhy does the partial of $f: Delta to mathbbR^2$ fail to exist?To find function satisfying given partial derivatives2 variable, 2 valued function $f(x_1,,x_2)=(x_1,,x_2)$What will be $frac partial F_i partial x_j$?

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The implication of zero mixed partial derivatives for multivariate function's minimization


If the second mixed partial is identically zero, then the function can be written as a sum $f(x,y) = f_1(x) + f_2(y)$$dfracpartial^2 fpartial x partial y = 0 nRightarrow f(x,y) = g(x) + h(y)$If a nonnegative function of $x_1,dots,x_n$ can be written as $sum g_k(x_k)$, then the summands can be taken nonnegative$f(x,y) = f_1(x) + f_2(x)$ with continuous differential real-valued functionsShow that both mixed partial derivatives exist at the origin but are not equalFunction on $mathbbR^2-0$.Let $F_1,F_2:bf R^2 to R$ be functions such that…How many n-th Order Partial Derivatives Exist for a Function of k Variables?Maximize $f(textbfx,textbfy) = f_1(textbfx) + f_2(textbfx,textbfy)$Symmetry of higher order mixed partial derivatives under weaker assumptionsWhy does the partial of $f: Delta to mathbbR^2$ fail to exist?To find function satisfying given partial derivatives2 variable, 2 valued function $f(x_1,,x_2)=(x_1,,x_2)$What will be $frac partial F_i partial x_j$?













2












$begingroup$


Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$



A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
    $endgroup$
    – user147263
    Dec 29 '15 at 0:24















2












$begingroup$


Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$



A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
    $endgroup$
    – user147263
    Dec 29 '15 at 0:24













2












2








2





$begingroup$


Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$



A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.










share|cite|improve this question











$endgroup$




Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_12=f''_21=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_textbf x f(textbf x)equiv min_x_1f_1(x_1)+ min_x_2f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$



A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.







multivariable-calculus optimization partial-derivative






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 1 '15 at 20:04







user147263

















asked Jun 24 '13 at 22:43









jorter.jijorter.ji

258




258











  • $begingroup$
    Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
    $endgroup$
    – user147263
    Dec 29 '15 at 0:24
















  • $begingroup$
    Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
    $endgroup$
    – user147263
    Dec 29 '15 at 0:24















$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24




$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_textbf x f(textbf x) = min_x_1f_1(x_1)+ min_x_2f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24










2 Answers
2






active

oldest

votes


















1












$begingroup$

For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_xy ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_yx$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 6:13











  • $begingroup$
    @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 12:32










  • $begingroup$
    Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 16:49










  • $begingroup$
    @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 16:52











  • $begingroup$
    OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 17:41


















0












$begingroup$

The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.



Here I try to provide a proof without that assumption.



Restatement



I can restate the conjecture with little weaker conditions:



If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.



Proof



From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.



$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$



Annotation



The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.






share|cite|improve this answer









$endgroup$












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
    $$
    f_x(x,y) = int f_xy ,dy + h(x).
    $$
    Integrating with respect to $x$:
    $$
    f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
    $$
    Similar result yields if we start from $f_yx$, now this implies
    $$
    f(x,y) = f_1(x) + f_2(y),
    $$
    and there goes your conclusion in the question.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 6:13











    • $begingroup$
      @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 12:32










    • $begingroup$
      Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 16:49










    • $begingroup$
      @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 16:52











    • $begingroup$
      OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 17:41















    1












    $begingroup$

    For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
    $$
    f_x(x,y) = int f_xy ,dy + h(x).
    $$
    Integrating with respect to $x$:
    $$
    f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
    $$
    Similar result yields if we start from $f_yx$, now this implies
    $$
    f(x,y) = f_1(x) + f_2(y),
    $$
    and there goes your conclusion in the question.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 6:13











    • $begingroup$
      @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 12:32










    • $begingroup$
      Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 16:49










    • $begingroup$
      @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 16:52











    • $begingroup$
      OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 17:41













    1












    1








    1





    $begingroup$

    For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
    $$
    f_x(x,y) = int f_xy ,dy + h(x).
    $$
    Integrating with respect to $x$:
    $$
    f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
    $$
    Similar result yields if we start from $f_yx$, now this implies
    $$
    f(x,y) = f_1(x) + f_2(y),
    $$
    and there goes your conclusion in the question.






    share|cite|improve this answer









    $endgroup$



    For a mixed derivative $f_xy = 0$, integrating with respect to $y$ gives:
    $$
    f_x(x,y) = int f_xy ,dy + h(x).
    $$
    Integrating with respect to $x$:
    $$
    f(x,y) = iint f_xy ,dydx + int h(x)dx + g(y).
    $$
    Similar result yields if we start from $f_yx$, now this implies
    $$
    f(x,y) = f_1(x) + f_2(y),
    $$
    and there goes your conclusion in the question.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 24 '13 at 23:14









    Shuhao CaoShuhao Cao

    16.1k34292




    16.1k34292











    • $begingroup$
      Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 6:13











    • $begingroup$
      @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 12:32










    • $begingroup$
      Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 16:49










    • $begingroup$
      @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 16:52











    • $begingroup$
      OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 17:41
















    • $begingroup$
      Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 6:13











    • $begingroup$
      @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 12:32










    • $begingroup$
      Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 16:49










    • $begingroup$
      @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
      $endgroup$
      – Shuhao Cao
      Jun 26 '13 at 16:52











    • $begingroup$
      OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
      $endgroup$
      – jorter.ji
      Jun 26 '13 at 17:41















    $begingroup$
    Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 6:13





    $begingroup$
    Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 6:13













    $begingroup$
    @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 12:32




    $begingroup$
    @jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 12:32












    $begingroup$
    Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 16:49




    $begingroup$
    Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 16:49












    $begingroup$
    @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 16:52





    $begingroup$
    @jorter.ji I meant to say that provided only this differential relation $f_xy = f_yx = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
    $endgroup$
    – Shuhao Cao
    Jun 26 '13 at 16:52













    $begingroup$
    OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 17:41




    $begingroup$
    OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
    $endgroup$
    – jorter.ji
    Jun 26 '13 at 17:41











    0












    $begingroup$

    The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.



    Here I try to provide a proof without that assumption.



    Restatement



    I can restate the conjecture with little weaker conditions:



    If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.



    Proof



    From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.



    $forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$



    Annotation



    The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.



      Here I try to provide a proof without that assumption.



      Restatement



      I can restate the conjecture with little weaker conditions:



      If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.



      Proof



      From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.



      $forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$



      Annotation



      The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.



        Here I try to provide a proof without that assumption.



        Restatement



        I can restate the conjecture with little weaker conditions:



        If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.



        Proof



        From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.



        $forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$



        Annotation



        The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.






        share|cite|improve this answer









        $endgroup$



        The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.



        Here I try to provide a proof without that assumption.



        Restatement



        I can restate the conjecture with little weaker conditions:



        If $f(x, y)$ has $f_yx = 0$, then $z(x, y) = f(x) + g(y)$.



        Proof



        From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_yx(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.



        $forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$



        Annotation



        The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 16 at 8:52









        TA123TA123

        958




        958



























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