How to prove that a function obeys a certain functional equation












0












$begingroup$


I saw this equation in a book and It was used as part of calculating something else in here and here Can I get any help about this equation



the statement is this:



suppose that:
$$f: Rtimes R rightarrow R$$
$$k: Rrightarrow R$$
we assume function G exits such that:
$$f(x,k(x)+z) = f(x,k(x)) + G(x,z)z$$
prove that G is calculated by this formula:
$G(x,z) = int_{0}^{1} D_x(f(x,k(x)+yz)dy$



My problem is that the writers of the links I gave above use this equation and there is no proof or reference for that I want to know how this can be proved






calculus integration definite-integrals dynamical-systems







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $k$? What is $f$?
    $endgroup$
    – 5xum
    Jan 14 at 14:35






  • 1




    $begingroup$
    What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
    $endgroup$
    – Gerry Myerson
    Jan 14 at 15:02








  • 1




    $begingroup$
    F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
    $endgroup$
    – Glyphack
    Jan 14 at 15:05












  • $begingroup$
    I have updated the question @gerry-myerson
    $endgroup$
    – Glyphack
    Jan 14 at 16:48






  • 1




    $begingroup$
    hope it's more clear now @GerryMyerson
    $endgroup$
    – Glyphack
    Jan 15 at 4:37


















0












$begingroup$


I saw this equation in a book and It was used as part of calculating something else in here and here Can I get any help about this equation



the statement is this:



suppose that:
$$f: Rtimes R rightarrow R$$
$$k: Rrightarrow R$$
we assume function G exits such that:
$$f(x,k(x)+z) = f(x,k(x)) + G(x,z)z$$
prove that G is calculated by this formula:
$G(x,z) = int_{0}^{1} D_x(f(x,k(x)+yz)dy$



My problem is that the writers of the links I gave above use this equation and there is no proof or reference for that I want to know how this can be proved






calculus integration definite-integrals dynamical-systems







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is $k$? What is $f$?
    $endgroup$
    – 5xum
    Jan 14 at 14:35






  • 1




    $begingroup$
    What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
    $endgroup$
    – Gerry Myerson
    Jan 14 at 15:02








  • 1




    $begingroup$
    F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
    $endgroup$
    – Glyphack
    Jan 14 at 15:05












  • $begingroup$
    I have updated the question @gerry-myerson
    $endgroup$
    – Glyphack
    Jan 14 at 16:48






  • 1




    $begingroup$
    hope it's more clear now @GerryMyerson
    $endgroup$
    – Glyphack
    Jan 15 at 4:37
















0












0








0





$begingroup$


I saw this equation in a book and It was used as part of calculating something else in here and here Can I get any help about this equation



the statement is this:



suppose that:
$$f: Rtimes R rightarrow R$$
$$k: Rrightarrow R$$
we assume function G exits such that:
$$f(x,k(x)+z) = f(x,k(x)) + G(x,z)z$$
prove that G is calculated by this formula:
$G(x,z) = int_{0}^{1} D_x(f(x,k(x)+yz)dy$



My problem is that the writers of the links I gave above use this equation and there is no proof or reference for that I want to know how this can be proved






calculus integration definite-integrals dynamical-systems







share|cite|improve this question











$endgroup$




I saw this equation in a book and It was used as part of calculating something else in here and here Can I get any help about this equation



the statement is this:



suppose that:
$$f: Rtimes R rightarrow R$$
$$k: Rrightarrow R$$
we assume function G exits such that:
$$f(x,k(x)+z) = f(x,k(x)) + G(x,z)z$$
prove that G is calculated by this formula:
$G(x,z) = int_{0}^{1} D_x(f(x,k(x)+yz)dy$



My problem is that the writers of the links I gave above use this equation and there is no proof or reference for that I want to know how this can be proved






calculus integration definite-integrals dynamical-systems




calculus integration definite-integrals dynamical-systems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 4:32







Glyphack

















asked Jan 14 at 14:33









GlyphackGlyphack

63




63








  • 1




    $begingroup$
    What is $k$? What is $f$?
    $endgroup$
    – 5xum
    Jan 14 at 14:35






  • 1




    $begingroup$
    What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
    $endgroup$
    – Gerry Myerson
    Jan 14 at 15:02








  • 1




    $begingroup$
    F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
    $endgroup$
    – Glyphack
    Jan 14 at 15:05












  • $begingroup$
    I have updated the question @gerry-myerson
    $endgroup$
    – Glyphack
    Jan 14 at 16:48






  • 1




    $begingroup$
    hope it's more clear now @GerryMyerson
    $endgroup$
    – Glyphack
    Jan 15 at 4:37
















  • 1




    $begingroup$
    What is $k$? What is $f$?
    $endgroup$
    – 5xum
    Jan 14 at 14:35






  • 1




    $begingroup$
    What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
    $endgroup$
    – Gerry Myerson
    Jan 14 at 15:02








  • 1




    $begingroup$
    F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
    $endgroup$
    – Glyphack
    Jan 14 at 15:05












  • $begingroup$
    I have updated the question @gerry-myerson
    $endgroup$
    – Glyphack
    Jan 14 at 16:48






  • 1




    $begingroup$
    hope it's more clear now @GerryMyerson
    $endgroup$
    – Glyphack
    Jan 15 at 4:37










1




1




$begingroup$
What is $k$? What is $f$?
$endgroup$
– 5xum
Jan 14 at 14:35




$begingroup$
What is $k$? What is $f$?
$endgroup$
– 5xum
Jan 14 at 14:35




1




1




$begingroup$
What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
$endgroup$
– Gerry Myerson
Jan 14 at 15:02






$begingroup$
What does $K:Rtimes R$ mean? You have to put a lot more thought into your presentation, if you expect anyone to be able to understand, much less answer, your question. In fact, I don't even see a question.
$endgroup$
– Gerry Myerson
Jan 14 at 15:02






1




1




$begingroup$
F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
$endgroup$
– Glyphack
Jan 14 at 15:05






$begingroup$
F should be infinitely differentiable for example you can suppose that $f(x,z) = sin(x+z^2)$ and $k(x)=exp(x)$
$endgroup$
– Glyphack
Jan 14 at 15:05














$begingroup$
I have updated the question @gerry-myerson
$endgroup$
– Glyphack
Jan 14 at 16:48




$begingroup$
I have updated the question @gerry-myerson
$endgroup$
– Glyphack
Jan 14 at 16:48




1




1




$begingroup$
hope it's more clear now @GerryMyerson
$endgroup$
– Glyphack
Jan 15 at 4:37






$begingroup$
hope it's more clear now @GerryMyerson
$endgroup$
– Glyphack
Jan 15 at 4:37












1 Answer
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Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is
$$
h'(s)=(D_yF)(x,k(x)+sz)cdot z,~~ text{ which means } ~~ h'(s)=left.frac{partial F(x,y)}{partial y}right|_{y=k(x)+sz}.
$$
Now integrate this equation over $[0,1]$ and use that $z$ is constant to get
$$
F(x,k(x)+z)=h(1)=h(0)+int_0^1h'(s),ds=F(x,k(x))+left[int_0^1 (D_yF)(x,k(x)+sz),dsright]cdot z.
$$
The notation in your claim seems slightly skewed in this light.






share|cite|improve this answer









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    0












    $begingroup$

    Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is
    $$
    h'(s)=(D_yF)(x,k(x)+sz)cdot z,~~ text{ which means } ~~ h'(s)=left.frac{partial F(x,y)}{partial y}right|_{y=k(x)+sz}.
    $$
    Now integrate this equation over $[0,1]$ and use that $z$ is constant to get
    $$
    F(x,k(x)+z)=h(1)=h(0)+int_0^1h'(s),ds=F(x,k(x))+left[int_0^1 (D_yF)(x,k(x)+sz),dsright]cdot z.
    $$
    The notation in your claim seems slightly skewed in this light.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is
      $$
      h'(s)=(D_yF)(x,k(x)+sz)cdot z,~~ text{ which means } ~~ h'(s)=left.frac{partial F(x,y)}{partial y}right|_{y=k(x)+sz}.
      $$
      Now integrate this equation over $[0,1]$ and use that $z$ is constant to get
      $$
      F(x,k(x)+z)=h(1)=h(0)+int_0^1h'(s),ds=F(x,k(x))+left[int_0^1 (D_yF)(x,k(x)+sz),dsright]cdot z.
      $$
      The notation in your claim seems slightly skewed in this light.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is
        $$
        h'(s)=(D_yF)(x,k(x)+sz)cdot z,~~ text{ which means } ~~ h'(s)=left.frac{partial F(x,y)}{partial y}right|_{y=k(x)+sz}.
        $$
        Now integrate this equation over $[0,1]$ and use that $z$ is constant to get
        $$
        F(x,k(x)+z)=h(1)=h(0)+int_0^1h'(s),ds=F(x,k(x))+left[int_0^1 (D_yF)(x,k(x)+sz),dsright]cdot z.
        $$
        The notation in your claim seems slightly skewed in this light.






        share|cite|improve this answer









        $endgroup$



        Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is
        $$
        h'(s)=(D_yF)(x,k(x)+sz)cdot z,~~ text{ which means } ~~ h'(s)=left.frac{partial F(x,y)}{partial y}right|_{y=k(x)+sz}.
        $$
        Now integrate this equation over $[0,1]$ and use that $z$ is constant to get
        $$
        F(x,k(x)+z)=h(1)=h(0)+int_0^1h'(s),ds=F(x,k(x))+left[int_0^1 (D_yF)(x,k(x)+sz),dsright]cdot z.
        $$
        The notation in your claim seems slightly skewed in this light.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 21:07









        LutzLLutzL

        57.7k42054




        57.7k42054






























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