Prove that function is smooth
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There is a function:
$V(n) = sum_i=1^infty f(b_i(n))b_i^'(n)- f(a_i(n))a_i^'(n)$,
where $f$,$b$ and $a$ are $C^infty$ smooth functions.
I know that we can prove that the function $V(n)$ is also smooth under some assumptions.
I SUGGEST THE NEXT SOLUTION
As I understand we should take
$V^'(n) = sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)$,
where $f^'$ exists, $b^''$ and $a^''$ should also be smooth.
MY QUESTION: Can somebody check this logic and help me?
I will really appreciate your help!
calculus real-analysis integration derivatives
asked Sep 10 at 8:58
Caim
10810
add a comment |Â
up vote
0
down vote
favorite
There is a function:
$V(n) = sum_i=1^infty f(b_i(n))b_i^'(n)- f(a_i(n))a_i^'(n)$,
where $f$,$b$ and $a$ are $C^infty$ smooth functions.
I know that we can prove that the function $V(n)$ is also smooth under some assumptions.
I SUGGEST THE NEXT SOLUTION
As I understand we should take
$V^'(n) = sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)$,
where $f^'$ exists, $b^''$ and $a^''$ should also be smooth.
MY QUESTION: Can somebody check this logic and help me?
I will really appreciate your help!
calculus real-analysis integration derivatives
asked Sep 10 at 8:58
Caim
10810
2
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
There is a function:
$V(n) = sum_i=1^infty f(b_i(n))b_i^'(n)- f(a_i(n))a_i^'(n)$,
where $f$,$b$ and $a$ are $C^infty$ smooth functions.
I know that we can prove that the function $V(n)$ is also smooth under some assumptions.
I SUGGEST THE NEXT SOLUTION
As I understand we should take
$V^'(n) = sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)$,
where $f^'$ exists, $b^''$ and $a^''$ should also be smooth.
MY QUESTION: Can somebody check this logic and help me?
I will really appreciate your help!
calculus real-analysis integration derivatives
asked Sep 10 at 8:58
Caim
10810
There is a function:
$V(n) = sum_i=1^infty f(b_i(n))b_i^'(n)- f(a_i(n))a_i^'(n)$,
where $f$,$b$ and $a$ are $C^infty$ smooth functions.
I know that we can prove that the function $V(n)$ is also smooth under some assumptions.
I SUGGEST THE NEXT SOLUTION
As I understand we should take
$V^'(n) = sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)$,
where $f^'$ exists, $b^''$ and $a^''$ should also be smooth.
MY QUESTION: Can somebody check this logic and help me?
I will really appreciate your help!
calculus real-analysis integration derivatives
calculus real-analysis integration derivatives
asked Sep 10 at 8:58
Caim
10810
asked Sep 10 at 8:58
Caim
10810
edited Sep 10 at 9:46
asked Sep 10 at 8:58
Caim
10810
asked Sep 10 at 8:58
Caim
10810
asked Sep 10 at 8:58
Caim
10810
10810
2
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54
add a comment |Â
2
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54
2
2
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54
add a comment |Â
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2
You need more information on the functions $f,b_i,a_i$ to justify this result. You don't even know that the series for $V'(n)$ converges (even assuming that the series for $V(n)$ does). By the way, you calculated the derivatives of the terms incorrectly, for instance the first term should be $f'(b_i(n))b'_i(n)^2$.
– Lorenzo Quarisa
Sep 10 at 9:13
Thank you, you are right, but in my case the problem is that I don'thave a lot of information about $f,b_i$ and$ a_i$ so I can do only assumptions... It seems that we can conclude that $V^'$ converges uniformly if $sum_i=1^infty f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)=s(n)$ and $forall epsilon>0$, $kgeq N(epsilon)$ so that $|sum_i=1^k f^'(b_i(n))b_i^'(n)^2+f(b_i(n))b_i^''(n)- f^'(a_i(n))a_i^'(n)^2-f(a_i(n))a_i^''(n)-s(n)|<epsilon$
– Caim
Sep 10 at 9:54