Continuously differentiable composition/convolution.
$begingroup$
My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)
Let $f$ and $g$ be continuous functions over the reals s.t.
$g(x) = int_0^x f(y)(y-x) dy$ $forall x$
and $g$ is three times continuously differentiable,
what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?
my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?
(feel free to edit the title to best fit my question)
real-analysis calculus
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
$endgroup$
add a comment |
$begingroup$
My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)
Let $f$ and $g$ be continuous functions over the reals s.t.
$g(x) = int_0^x f(y)(y-x) dy$ $forall x$
and $g$ is three times continuously differentiable,
what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?
my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?
(feel free to edit the title to best fit my question)
real-analysis calculus
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
$endgroup$
add a comment |
$begingroup$
My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)
Let $f$ and $g$ be continuous functions over the reals s.t.
$g(x) = int_0^x f(y)(y-x) dy$ $forall x$
and $g$ is three times continuously differentiable,
what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?
my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?
(feel free to edit the title to best fit my question)
real-analysis calculus
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
$endgroup$
My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)
Let $f$ and $g$ be continuous functions over the reals s.t.
$g(x) = int_0^x f(y)(y-x) dy$ $forall x$
and $g$ is three times continuously differentiable,
what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?
my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?
(feel free to edit the title to best fit my question)
real-analysis calculus
real-analysis calculus
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
asked Jan 7 at 22:30
Hossien SahebjameHossien Sahebjame
1169
1169
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
$$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
add a comment |
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1 Answer
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votes
$begingroup$
We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
$$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
add a comment |
$begingroup$
We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
$$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
$endgroup$
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
add a comment |
$begingroup$
We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
$$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
$endgroup$
We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
$$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
answered Jan 7 at 22:44
AndreiAndrei
13.2k21230
13.2k21230
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
add a comment |
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
$begingroup$
got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
$endgroup$
– Hossien Sahebjame
Jan 8 at 21:45
1
1
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
$begingroup$
You are correct.
$endgroup$
– Andrei
Jan 8 at 21:47
add a comment |
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