first order Taylor expansion term of a function multiplied by a dot product of gradients.












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$begingroup$


I need to do the following Taylor expansion.
Let's suppose we have a function $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$, where by $(nabla (|nabla n|^2),nabla n)$ I mean the dot product and $n$ is a real-valued three-variable function $n(bf {r})=$ $n(x,y,z)$. I would like to find the first order Taylor expansion term of it around the point $n_0$ if the perturbation is $n_1$, so that $n=n_0+n_1$.
I am doing it in two ways.





First way



I suppose, that the mathematically CORRECT is to treat $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$ as a whole and calculate the derivatives of it considering that it is a product of two functions. But if I will try to do it in that way I will encounter with the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$, which I don´t know how to calculate.





Second way



$f(n_0+n_1,|nabla (n_0+n_1)|^2)(nabla (|nabla (n_0+n_1)|^2),nabla (n_0+n_1))=[f(n_0,|nabla n_0|^2)+(n_1+n_0-n_0)f_{n}|_0+(|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)-|nabla n_0|^2)f_{|nabla n|^2}|_0][nabla (|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)), (nabla n_0+nabla n_1)]=[f(n_0,|nabla n_0|^2)+n_1f_{n}|_0+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0][(nabla(|nabla n_0|^2)+2nabla(nabla n_0,nabla n_1), nabla n_0+nabla n_1)]$



($f_{n}|_0$ means partial derivative with respect to $n$ at $n_0$ and $f_{|nabla n|^2}|_0$ means partial derivative with respect to $|nabla n|^2$ at $|nabla n_0|^2$).



So the higher order terms, like $|nabla n_1|^2$ and zero order terms, like $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_0)$ should not be considered and finally I have the following expression:



$f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_1)+2f(n_0,|nabla n_0|^2)(nabla (nabla n_0,nabla n_1),nabla n_0)+n_1f_{n}|_0(nabla(|nabla n_0|^2),nabla n_0)+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0(nabla(|nabla n_0|^2),nabla n_0)$



In this way, I am actually expanding $f(n,|nabla n|^2)$ and $(nabla (|nabla n|^2),nabla n)$ functions SEPARATELY.





So, could you please let me know, what is the problem with my calcs?
And how the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$ can be calculated.



Best Regards



Henrikh










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I need to do the following Taylor expansion.
    Let's suppose we have a function $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$, where by $(nabla (|nabla n|^2),nabla n)$ I mean the dot product and $n$ is a real-valued three-variable function $n(bf {r})=$ $n(x,y,z)$. I would like to find the first order Taylor expansion term of it around the point $n_0$ if the perturbation is $n_1$, so that $n=n_0+n_1$.
    I am doing it in two ways.





    First way



    I suppose, that the mathematically CORRECT is to treat $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$ as a whole and calculate the derivatives of it considering that it is a product of two functions. But if I will try to do it in that way I will encounter with the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$, which I don´t know how to calculate.





    Second way



    $f(n_0+n_1,|nabla (n_0+n_1)|^2)(nabla (|nabla (n_0+n_1)|^2),nabla (n_0+n_1))=[f(n_0,|nabla n_0|^2)+(n_1+n_0-n_0)f_{n}|_0+(|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)-|nabla n_0|^2)f_{|nabla n|^2}|_0][nabla (|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)), (nabla n_0+nabla n_1)]=[f(n_0,|nabla n_0|^2)+n_1f_{n}|_0+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0][(nabla(|nabla n_0|^2)+2nabla(nabla n_0,nabla n_1), nabla n_0+nabla n_1)]$



    ($f_{n}|_0$ means partial derivative with respect to $n$ at $n_0$ and $f_{|nabla n|^2}|_0$ means partial derivative with respect to $|nabla n|^2$ at $|nabla n_0|^2$).



    So the higher order terms, like $|nabla n_1|^2$ and zero order terms, like $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_0)$ should not be considered and finally I have the following expression:



    $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_1)+2f(n_0,|nabla n_0|^2)(nabla (nabla n_0,nabla n_1),nabla n_0)+n_1f_{n}|_0(nabla(|nabla n_0|^2),nabla n_0)+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0(nabla(|nabla n_0|^2),nabla n_0)$



    In this way, I am actually expanding $f(n,|nabla n|^2)$ and $(nabla (|nabla n|^2),nabla n)$ functions SEPARATELY.





    So, could you please let me know, what is the problem with my calcs?
    And how the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$ can be calculated.



    Best Regards



    Henrikh










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I need to do the following Taylor expansion.
      Let's suppose we have a function $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$, where by $(nabla (|nabla n|^2),nabla n)$ I mean the dot product and $n$ is a real-valued three-variable function $n(bf {r})=$ $n(x,y,z)$. I would like to find the first order Taylor expansion term of it around the point $n_0$ if the perturbation is $n_1$, so that $n=n_0+n_1$.
      I am doing it in two ways.





      First way



      I suppose, that the mathematically CORRECT is to treat $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$ as a whole and calculate the derivatives of it considering that it is a product of two functions. But if I will try to do it in that way I will encounter with the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$, which I don´t know how to calculate.





      Second way



      $f(n_0+n_1,|nabla (n_0+n_1)|^2)(nabla (|nabla (n_0+n_1)|^2),nabla (n_0+n_1))=[f(n_0,|nabla n_0|^2)+(n_1+n_0-n_0)f_{n}|_0+(|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)-|nabla n_0|^2)f_{|nabla n|^2}|_0][nabla (|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)), (nabla n_0+nabla n_1)]=[f(n_0,|nabla n_0|^2)+n_1f_{n}|_0+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0][(nabla(|nabla n_0|^2)+2nabla(nabla n_0,nabla n_1), nabla n_0+nabla n_1)]$



      ($f_{n}|_0$ means partial derivative with respect to $n$ at $n_0$ and $f_{|nabla n|^2}|_0$ means partial derivative with respect to $|nabla n|^2$ at $|nabla n_0|^2$).



      So the higher order terms, like $|nabla n_1|^2$ and zero order terms, like $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_0)$ should not be considered and finally I have the following expression:



      $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_1)+2f(n_0,|nabla n_0|^2)(nabla (nabla n_0,nabla n_1),nabla n_0)+n_1f_{n}|_0(nabla(|nabla n_0|^2),nabla n_0)+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0(nabla(|nabla n_0|^2),nabla n_0)$



      In this way, I am actually expanding $f(n,|nabla n|^2)$ and $(nabla (|nabla n|^2),nabla n)$ functions SEPARATELY.





      So, could you please let me know, what is the problem with my calcs?
      And how the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$ can be calculated.



      Best Regards



      Henrikh










      share|cite|improve this question









      $endgroup$




      I need to do the following Taylor expansion.
      Let's suppose we have a function $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$, where by $(nabla (|nabla n|^2),nabla n)$ I mean the dot product and $n$ is a real-valued three-variable function $n(bf {r})=$ $n(x,y,z)$. I would like to find the first order Taylor expansion term of it around the point $n_0$ if the perturbation is $n_1$, so that $n=n_0+n_1$.
      I am doing it in two ways.





      First way



      I suppose, that the mathematically CORRECT is to treat $f(n,|nabla n|^2)(nabla (|nabla n|^2),nabla n)$ as a whole and calculate the derivatives of it considering that it is a product of two functions. But if I will try to do it in that way I will encounter with the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$, which I don´t know how to calculate.





      Second way



      $f(n_0+n_1,|nabla (n_0+n_1)|^2)(nabla (|nabla (n_0+n_1)|^2),nabla (n_0+n_1))=[f(n_0,|nabla n_0|^2)+(n_1+n_0-n_0)f_{n}|_0+(|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)-|nabla n_0|^2)f_{|nabla n|^2}|_0][nabla (|nabla n_0|^2+|nabla n_1|^2+2(nabla n_0,nabla n_1)), (nabla n_0+nabla n_1)]=[f(n_0,|nabla n_0|^2)+n_1f_{n}|_0+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0][(nabla(|nabla n_0|^2)+2nabla(nabla n_0,nabla n_1), nabla n_0+nabla n_1)]$



      ($f_{n}|_0$ means partial derivative with respect to $n$ at $n_0$ and $f_{|nabla n|^2}|_0$ means partial derivative with respect to $|nabla n|^2$ at $|nabla n_0|^2$).



      So the higher order terms, like $|nabla n_1|^2$ and zero order terms, like $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_0)$ should not be considered and finally I have the following expression:



      $f(n_0,|nabla n_0|^2) (nabla(|nabla n_0|^2),nabla n_1)+2f(n_0,|nabla n_0|^2)(nabla (nabla n_0,nabla n_1),nabla n_0)+n_1f_{n}|_0(nabla(|nabla n_0|^2),nabla n_0)+2(nabla n_0,nabla n_1)f_{|nabla n|^2}|_0(nabla(|nabla n_0|^2),nabla n_0)$



      In this way, I am actually expanding $f(n,|nabla n|^2)$ and $(nabla (|nabla n|^2),nabla n)$ functions SEPARATELY.





      So, could you please let me know, what is the problem with my calcs?
      And how the derivatives like $frac{d(nabla (|nabla n|^2),nabla n)}{dn}$ can be calculated.



      Best Regards



      Henrikh







      sequences-and-series taylor-expansion perturbation-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 20 '18 at 23:24









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