Prove function $g$ is infinitely differentiable for composite function












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Suppose the function $f$ is infinitely differentiable on $mathbb{R}$, and the function $g$ is differentiable on the open interval $(a,b)$ satisfying
$$g'(x)=f circ g space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space xin(a,b)$$
Prove that the function $g$ is infinitely differentiable on $(a,b)$.



I have showed that since both functions are differentiable on $(a,b)$ implies that they are continuous on $(a,b)$, thus $g'(x)$ is continuous. Besides that, $g''(x)=f' circ g cdot g'$ is also continuous. Of course, this process can be done infinitely and hence show that $g^{(n)}(x)$ is continuous. How do I show it is infinitely differentiable, because continuous does not imply differentiable.






analysis derivatives continuity







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  • $begingroup$
    Use induction on $n$ to show that the n-th derivative exists for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:45










  • $begingroup$
    So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
    $endgroup$
    – weilam06
    Dec 14 '18 at 9:47










  • $begingroup$
    For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:52
















0












$begingroup$


Suppose the function $f$ is infinitely differentiable on $mathbb{R}$, and the function $g$ is differentiable on the open interval $(a,b)$ satisfying
$$g'(x)=f circ g space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space xin(a,b)$$
Prove that the function $g$ is infinitely differentiable on $(a,b)$.



I have showed that since both functions are differentiable on $(a,b)$ implies that they are continuous on $(a,b)$, thus $g'(x)$ is continuous. Besides that, $g''(x)=f' circ g cdot g'$ is also continuous. Of course, this process can be done infinitely and hence show that $g^{(n)}(x)$ is continuous. How do I show it is infinitely differentiable, because continuous does not imply differentiable.






analysis derivatives continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use induction on $n$ to show that the n-th derivative exists for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:45










  • $begingroup$
    So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
    $endgroup$
    – weilam06
    Dec 14 '18 at 9:47










  • $begingroup$
    For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:52














0












0








0





$begingroup$


Suppose the function $f$ is infinitely differentiable on $mathbb{R}$, and the function $g$ is differentiable on the open interval $(a,b)$ satisfying
$$g'(x)=f circ g space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space xin(a,b)$$
Prove that the function $g$ is infinitely differentiable on $(a,b)$.



I have showed that since both functions are differentiable on $(a,b)$ implies that they are continuous on $(a,b)$, thus $g'(x)$ is continuous. Besides that, $g''(x)=f' circ g cdot g'$ is also continuous. Of course, this process can be done infinitely and hence show that $g^{(n)}(x)$ is continuous. How do I show it is infinitely differentiable, because continuous does not imply differentiable.






analysis derivatives continuity







share|cite|improve this question









$endgroup$




Suppose the function $f$ is infinitely differentiable on $mathbb{R}$, and the function $g$ is differentiable on the open interval $(a,b)$ satisfying
$$g'(x)=f circ g space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space xin(a,b)$$
Prove that the function $g$ is infinitely differentiable on $(a,b)$.



I have showed that since both functions are differentiable on $(a,b)$ implies that they are continuous on $(a,b)$, thus $g'(x)$ is continuous. Besides that, $g''(x)=f' circ g cdot g'$ is also continuous. Of course, this process can be done infinitely and hence show that $g^{(n)}(x)$ is continuous. How do I show it is infinitely differentiable, because continuous does not imply differentiable.






analysis derivatives continuity




analysis derivatives continuity






share|cite|improve this question













share|cite|improve this question











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asked Dec 14 '18 at 9:30









weilam06weilam06

30412




30412












  • $begingroup$
    Use induction on $n$ to show that the n-th derivative exists for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:45










  • $begingroup$
    So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
    $endgroup$
    – weilam06
    Dec 14 '18 at 9:47










  • $begingroup$
    For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:52


















  • $begingroup$
    Use induction on $n$ to show that the n-th derivative exists for each $n$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:45










  • $begingroup$
    So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
    $endgroup$
    – weilam06
    Dec 14 '18 at 9:47










  • $begingroup$
    For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 14 '18 at 9:52
















$begingroup$
Use induction on $n$ to show that the n-th derivative exists for each $n$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:45




$begingroup$
Use induction on $n$ to show that the n-th derivative exists for each $n$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:45












$begingroup$
So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
$endgroup$
– weilam06
Dec 14 '18 at 9:47




$begingroup$
So I have showed $g''$ exists so the function $g$ is twice differentiable on my post? Just use induction to derive the case to $n$-derivative?
$endgroup$
– weilam06
Dec 14 '18 at 9:47












$begingroup$
For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:52




$begingroup$
For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite product of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 9:52










2 Answers
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$begingroup$

For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite products of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.






share|cite|improve this answer











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    $begingroup$

    You know that $g'(x) = f(g)$ is differentiable on $ (a,b)$ by the chain rule ($f$ is differentiable on all reals hence on $(g(a),g(b))$, and g is differentiable on (a,b)). With the product rule, this implies that $g''= f '(g) * g'$ is differentiable, since both f '(g), g' are differentiable. Similarly, $g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      $begingroup$

      For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite products of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.






      share|cite|improve this answer











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        0












        $begingroup$

        For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite products of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.






        share|cite|improve this answer











        $endgroup$
















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          0








          0





          $begingroup$

          For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite products of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.






          share|cite|improve this answer











          $endgroup$



          For a precise argument prove the following statement by induction: for every $n$ $g^{(n)}$ exists and it is a finite sum of finite products of the functions $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 9:53

























          answered Dec 14 '18 at 9:43









          Kavi Rama MurthyKavi Rama Murthy

          64.5k42765




          64.5k42765























              0












              $begingroup$

              You know that $g'(x) = f(g)$ is differentiable on $ (a,b)$ by the chain rule ($f$ is differentiable on all reals hence on $(g(a),g(b))$, and g is differentiable on (a,b)). With the product rule, this implies that $g''= f '(g) * g'$ is differentiable, since both f '(g), g' are differentiable. Similarly, $g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You know that $g'(x) = f(g)$ is differentiable on $ (a,b)$ by the chain rule ($f$ is differentiable on all reals hence on $(g(a),g(b))$, and g is differentiable on (a,b)). With the product rule, this implies that $g''= f '(g) * g'$ is differentiable, since both f '(g), g' are differentiable. Similarly, $g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.






                share|cite|improve this answer











                $endgroup$
















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                  0





                  $begingroup$

                  You know that $g'(x) = f(g)$ is differentiable on $ (a,b)$ by the chain rule ($f$ is differentiable on all reals hence on $(g(a),g(b))$, and g is differentiable on (a,b)). With the product rule, this implies that $g''= f '(g) * g'$ is differentiable, since both f '(g), g' are differentiable. Similarly, $g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.






                  share|cite|improve this answer











                  $endgroup$



                  You know that $g'(x) = f(g)$ is differentiable on $ (a,b)$ by the chain rule ($f$ is differentiable on all reals hence on $(g(a),g(b))$, and g is differentiable on (a,b)). With the product rule, this implies that $g''= f '(g) * g'$ is differentiable, since both f '(g), g' are differentiable. Similarly, $g''' = (f '(g) )' * g' + (g')' * f '(g) = f ''(g)*(g')^2+ g'' * f '(g)$ is differentiable since it's a sum of products of differentiable functions. This pattern continues.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 14 '18 at 10:08









                  Martund

                  1,667213




                  1,667213










                  answered Dec 14 '18 at 9:55









                  Jermain McDermottJermain McDermott

                  11




                  11






























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