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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add a comment |
$begingroup$
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
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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
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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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
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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 $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
add a comment |
$begingroup$
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.
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
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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 $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
add a comment |
$begingroup$
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 $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
add a comment |
$begingroup$
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 $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
$endgroup$
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 $f,f',cdots,f^{(n-1)},g,g',cdots, g^{(n-1)}$.
edited Dec 14 '18 at 9:53
answered Dec 14 '18 at 9:43
Kavi Rama MurthyKavi Rama Murthy
64.5k42765
64.5k42765
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Dec 14 '18 at 10:08
Martund
1,667213
1,667213
answered Dec 14 '18 at 9:55
Jermain McDermottJermain McDermott
11
11
add a comment |
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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