Mixing
the product of two completely monotone functions
$begingroup$
A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.
Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?
if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?
derivatives monotone-functions
asked Jan 8 at 8:12
adli farhanadli farhan
214
$endgroup$
add a comment |
$begingroup$
A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.
Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?
if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?
derivatives monotone-functions
asked Jan 8 at 8:12
adli farhanadli farhan
214
$endgroup$
$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52
add a comment |
$begingroup$
A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.
Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?
if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?
derivatives monotone-functions
asked Jan 8 at 8:12
adli farhanadli farhan
214
$endgroup$
A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.
Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?
if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?
derivatives monotone-functions
derivatives monotone-functions
asked Jan 8 at 8:12
adli farhanadli farhan
214
asked Jan 8 at 8:12
adli farhanadli farhan
214
asked Jan 8 at 8:12
adli farhanadli farhan
214
asked Jan 8 at 8:12
adli farhanadli farhan
214
asked Jan 8 at 8:12
adli farhanadli farhan
214
214
$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52
add a comment |
$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52
$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
$endgroup$
add a comment |
$begingroup$
This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
$endgroup$
add a comment |
$begingroup$
Hint:
We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
$endgroup$
add a comment |
$begingroup$
Hint:
We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
$endgroup$
Hint:
We can observe that $f$ is CM $iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,gge0,f',g'le0implies-(fg)'=-fg'-f'gge0.$$
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
answered Jan 8 at 9:17
Yves DaoustYves Daoust
126k672225
126k672225
add a comment |
add a comment |
$begingroup$
This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
$endgroup$
add a comment |
$begingroup$
This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
$endgroup$
add a comment |
$begingroup$
This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
$endgroup$
This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,infty]$, i.e. given a positive Borel measure $mu$,
$$ f in mbox{CM} Longleftrightarrow f(x) = int_0^{infty} e^{-xt} dmu(t)$$
(provided, of course that the integral converges). If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $nu$ we quickly obtain
$$fg = int_0^{infty} e^{-xt} d(mu * nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
answered Jan 8 at 9:08
postmortespostmortes
1,91721117
1,91721117
add a comment |
add a comment |
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$begingroup$
Yes, use Leibniz rule for differentiation of f(x)g(x)..
$endgroup$
– LordVader007
Jan 8 at 8:42
$begingroup$
Or, use the even simpler fact that if two functions have all of their derivatives nonnegative, then their product has the same property; and consider $f(-x)$ for your $f$ above.
$endgroup$
– Greg Martin
Jan 8 at 8:52