Prove derivative of order $n$ at neighbourhood $x_0$












0












$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







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  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19
















0












$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19














0












0








0





$begingroup$



If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity







share|cite|improve this question









$endgroup$





If
$$f(x)-f(x_0)=g(x)(x-x_0)$$
and $g in C^{(n-1)}(U(x_0))$, where $U(x_0)$ is a neighbourhood of $x_0$, then show that $f(x)$ has a derivative $f^{(n)}(x_0)$ of order $n$ at $x_0$.




I think since $f(x)$ is continuous on its derivative of order $n-1$, I think I just need to verify that $f^{(n-1)}(x)-f^{(n-1)}(x_0)=g^{(n-1)}(x)(x-x_0)$ works and completes the proof. How do I continue it? Thanks in advance.






derivatives proof-verification continuity




derivatives proof-verification continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 17:11









weilam06weilam06

16011




16011












  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19


















  • $begingroup$
    Did induction failed?
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:14










  • $begingroup$
    I didn't try induction and I think there are some ways can prove it directly.
    $endgroup$
    – weilam06
    Dec 7 '18 at 17:16






  • 1




    $begingroup$
    Sure, you can find $n$th the derivative of a product... by induction.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:17










  • $begingroup$
    By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
    $endgroup$
    – Will M.
    Dec 7 '18 at 17:19
















$begingroup$
Did induction failed?
$endgroup$
– Will M.
Dec 7 '18 at 17:14




$begingroup$
Did induction failed?
$endgroup$
– Will M.
Dec 7 '18 at 17:14












$begingroup$
I didn't try induction and I think there are some ways can prove it directly.
$endgroup$
– weilam06
Dec 7 '18 at 17:16




$begingroup$
I didn't try induction and I think there are some ways can prove it directly.
$endgroup$
– weilam06
Dec 7 '18 at 17:16




1




1




$begingroup$
Sure, you can find $n$th the derivative of a product... by induction.
$endgroup$
– Will M.
Dec 7 '18 at 17:17




$begingroup$
Sure, you can find $n$th the derivative of a product... by induction.
$endgroup$
– Will M.
Dec 7 '18 at 17:17












$begingroup$
By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
$endgroup$
– Will M.
Dec 7 '18 at 17:19




$begingroup$
By the way, $f^{(n-1)}(x) - f^{(n-1)}(x_0) = g^{(n-1)}(x)(x - x_0)$ is not true in general. I suspect virtually any pair of functions will NOT satisfy that equality given the original equation.
$endgroup$
– Will M.
Dec 7 '18 at 17:19










1 Answer
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$begingroup$

Hint: Note that
begin{align}
f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
end{align}
which means
begin{align}
frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
end{align}






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    $begingroup$

    Hint: Note that
    begin{align}
    f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
    end{align}
    which means
    begin{align}
    frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
    end{align}






    share|cite|improve this answer









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      0












      $begingroup$

      Hint: Note that
      begin{align}
      f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
      end{align}
      which means
      begin{align}
      frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint: Note that
        begin{align}
        f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
        end{align}
        which means
        begin{align}
        frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
        end{align}






        share|cite|improve this answer









        $endgroup$



        Hint: Note that
        begin{align}
        f(x) = f(x_0)+g(x)(x-x_0) implies f^{(n-1)}(x) = g^{(n-1)}(x)(x-x_0)+(n-1)g^{(n-2)}(x)
        end{align}
        which means
        begin{align}
        frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0} = g^{(n-1)}(x)+(n-1)frac{g^{(n-2)}(x)-g^{(n-2)}(x_0)}{x-x_0}.
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 17:37









        Jacky ChongJacky Chong

        17.9k21128




        17.9k21128






























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