Check if 1-x is completely Monotonic functiona problem on uniform convergence of monotonic increasing...
Is this saw blade faulty?
Reasons for having MCU pin-states default to pull-up/down out of reset
Has the laser at Magurele, Romania reached a tenth of the Sun's power?
Started in 1987 vs. Starting in 1987
Why would five hundred and five same as one?
"Marked down as someone wanting to sell shares." What does that mean?
Air travel with refrigerated insulin
Offset in split text content
What is the period/term used describe Giuseppe Arcimboldo's style of painting?
Asserting that Atheism and Theism are both faith based positions
I keep switching characters, how do I stop?
Mortal danger in mid-grade literature
Capacitor electron flow
Do native speakers use "ultima" and "proxima" frequently in spoken English?
A seasonal riddle
Error in master's thesis, I do not know what to do
Is there any common country to visit for persons holding UK and Schengen visas?
Put the phone down / Put down the phone
Strange behavior in TikZ draw command
How to test the sharpness of a knife?
How do I prevent inappropriate ads from appearing in my game?
Walter Rudin's mathematical analysis: theorem 2.43. Why proof can't work under the perfect set is uncountable.
How can I, as DM, avoid the Conga Line of Death occurring when implementing some form of flanking rule?
How do I lift the insulation blower into the attic?
Check if 1-x is completely Monotonic function
a problem on uniform convergence of monotonic increasing functionDifferentiability of monotonic functionsHow can I tell if a function is a monotonic transformation?Technical name for an almost-monotonic functionShow function is monotonic and if it has maximum or minimum without second derivativeAre there cases where using the definition is easier than using the derivative to study the monotonic behaviour of a function?Monotonicity of $(f circ f)$ and $(f circ f circ f)$ for strictly monotonic $f$Monotonic function is rectificiableProve monotonic function on $mathbb{R}$ under given conditionWhy does a monotonic function always have a positive rate of change?
$begingroup$
I had a question about how can I check whether $f(x)=1-x$ is completely monotonic. Could somebody provide a simple example based on this function.
real-analysis functions derivatives monotone-functions
asked Mar 12 at 15:23
AlexAlex
115
$endgroup$
add a comment |
$begingroup$
I had a question about how can I check whether $f(x)=1-x$ is completely monotonic. Could somebody provide a simple example based on this function.
real-analysis functions derivatives monotone-functions
asked Mar 12 at 15:23
AlexAlex
115
$endgroup$
add a comment |
$begingroup$
I had a question about how can I check whether $f(x)=1-x$ is completely monotonic. Could somebody provide a simple example based on this function.
real-analysis functions derivatives monotone-functions
asked Mar 12 at 15:23
AlexAlex
115
$endgroup$
I had a question about how can I check whether $f(x)=1-x$ is completely monotonic. Could somebody provide a simple example based on this function.
real-analysis functions derivatives monotone-functions
real-analysis functions derivatives monotone-functions
asked Mar 12 at 15:23
AlexAlex
115
asked Mar 12 at 15:23
AlexAlex
115
asked Mar 12 at 15:23
AlexAlex
115
asked Mar 12 at 15:23
AlexAlex
115
asked Mar 12 at 15:23
AlexAlex
115
115
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
According to this, a completely monotonic function must satisty $$ (-1)^n frac {d^n}{dt^n}f(t)ge0$$ for all nonnegative $n$ and $t>0$; the special case $n=0$ reduces to requiring $f$ non-negative on $(0,infty)$. Which your $f$ clearly does not satisfy.
The characterization of completely monotone functions as Laplace transforms of non-negative measures is also a tip-off: if $x=2$, say, you would be representing $-1$ as the integral of a non-negative function.
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3145214%2fcheck-if-1-x-is-completely-monotonic-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to this, a completely monotonic function must satisty $$ (-1)^n frac {d^n}{dt^n}f(t)ge0$$ for all nonnegative $n$ and $t>0$; the special case $n=0$ reduces to requiring $f$ non-negative on $(0,infty)$. Which your $f$ clearly does not satisfy.
The characterization of completely monotone functions as Laplace transforms of non-negative measures is also a tip-off: if $x=2$, say, you would be representing $-1$ as the integral of a non-negative function.
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
add a comment |
$begingroup$
According to this, a completely monotonic function must satisty $$ (-1)^n frac {d^n}{dt^n}f(t)ge0$$ for all nonnegative $n$ and $t>0$; the special case $n=0$ reduces to requiring $f$ non-negative on $(0,infty)$. Which your $f$ clearly does not satisfy.
The characterization of completely monotone functions as Laplace transforms of non-negative measures is also a tip-off: if $x=2$, say, you would be representing $-1$ as the integral of a non-negative function.
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
$endgroup$
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
add a comment |
$begingroup$
According to this, a completely monotonic function must satisty $$ (-1)^n frac {d^n}{dt^n}f(t)ge0$$ for all nonnegative $n$ and $t>0$; the special case $n=0$ reduces to requiring $f$ non-negative on $(0,infty)$. Which your $f$ clearly does not satisfy.
The characterization of completely monotone functions as Laplace transforms of non-negative measures is also a tip-off: if $x=2$, say, you would be representing $-1$ as the integral of a non-negative function.
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
$endgroup$
According to this, a completely monotonic function must satisty $$ (-1)^n frac {d^n}{dt^n}f(t)ge0$$ for all nonnegative $n$ and $t>0$; the special case $n=0$ reduces to requiring $f$ non-negative on $(0,infty)$. Which your $f$ clearly does not satisfy.
The characterization of completely monotone functions as Laplace transforms of non-negative measures is also a tip-off: if $x=2$, say, you would be representing $-1$ as the integral of a non-negative function.
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
edited Mar 12 at 16:00
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
answered Mar 12 at 15:40
kimchi loverkimchi lover
11.3k31229
11.3k31229
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
add a comment |
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Thanks @kimchi lover. But the above function satisfies $ ( - 1 ) ^ { n } frac { d ^ { n } } { d t ^ { n } } f ( t ) geq 0 $: $(-1)f'(x) = -1 (-1) =1geq0$, $(-1)^2f''(x)=(-1)^2cdot0 =0 geq0$ etc.
$endgroup$
– Alex
Mar 12 at 15:43
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
$begingroup$
Brilliant! Many thanks @kimchi lover, this explanation is much more clear!
$endgroup$
– Alex
Mar 12 at 16:20
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3145214%2fcheck-if-1-x-is-completely-monotonic-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
