Intuitive reasoning that a function can't exist












7












$begingroup$


Sorry for the title; I had no idea how to make it concise yet informative.



I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!






real-analysis functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    1 hour ago










  • $begingroup$
    If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
    $endgroup$
    – Torsten Schoeneberg
    1 hour ago






  • 8




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    1 hour ago
















7












$begingroup$


Sorry for the title; I had no idea how to make it concise yet informative.



I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!






real-analysis functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    1 hour ago










  • $begingroup$
    If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
    $endgroup$
    – Torsten Schoeneberg
    1 hour ago






  • 8




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    1 hour ago














7












7








7


1



$begingroup$


Sorry for the title; I had no idea how to make it concise yet informative.



I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!






real-analysis functions







share|cite|improve this question











$endgroup$




Sorry for the title; I had no idea how to make it concise yet informative.



I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!






real-analysis functions




real-analysis functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









SvanN

1,9721422




1,9721422










asked 1 hour ago









Math-funMath-fun

7,0411425




7,0411425








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    1 hour ago










  • $begingroup$
    If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
    $endgroup$
    – Torsten Schoeneberg
    1 hour ago






  • 8




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    1 hour ago














  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    1 hour ago










  • $begingroup$
    If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
    $endgroup$
    – Torsten Schoeneberg
    1 hour ago






  • 8




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    1 hour ago








1




1




$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago




$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago












$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago




$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago




8




8




$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago




$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago










1 Answer
1






active

oldest

votes


















11












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    37 mins ago










  • $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    28 mins ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077308%2fintuitive-reasoning-that-a-function-cant-exist%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









11












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    37 mins ago










  • $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    28 mins ago
















11












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    37 mins ago










  • $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    28 mins ago














11












11








11





$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$



I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 48 mins ago

























answered 1 hour ago









HenryHenry

98.9k476164




98.9k476164








  • 1




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    37 mins ago










  • $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    28 mins ago














  • 1




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    37 mins ago










  • $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    28 mins ago








1




1




$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago




$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago












$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago




$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077308%2fintuitive-reasoning-that-a-function-cant-exist%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Magento 2 controller redirect on button click in phtml file

Image
0 I want to redirect customer using controller when button is clicked in one of my phtml files. So far I got, controller: public function execute(){ return $this->addToCart(); } public function addToCart(){ // some logic $this->_redirect('checkout/cart/index'); } The above is working fine (redirecting) when you access the controller via URL (www.mydomain.com/batchorder/index/addtocart). And here is my .phtml file: $("#addToCart").click(function(){ let controllerUrl = "<?php echo $block-getUrl('batchorder/index/addtocart'); ?>" $.post(controllerUrl, { "products": JSON.stringify(sideCart.SideCartProducts) }) The post request returns my logic as it should, however, it is not redirecting. Anyone can help? ...
Read more

Polycentropodidae

Image
Polycentropodidae Plectrocnemia conspersa Siyentipiko nga pagklasipika Ginhadi-an: Animalia Phylum: Arthropoda Ubosphylum: Hexapoda Klase: Insecta Orden: Trichoptera Labawbanay: Hydropsychoidea Banay: Polycentropodidae Binomial nga ngaran Polycentropodidae An Polycentropodidae [1] in uska familia han Insecta. An Polycentropodidae in nahilalakip ha familia nga Hydropsychoidea, ordo nga Trichoptera, classis nga Insecta, punoan nga Arthropoda, ngan regnum nga Animalia. [1] An familia nga Polycentropodidae in naglalakip hin 660 ka mga species, sumala ha Catalogue of Life [1] . An kladograma hini sumala ha Catalogue of Life [1] : Hydropsychoidea   Polycentropodidae  Adectophylax Antillopsyche Cernotina Cyrnellus Cyrnodes Cyrnopsis Cyrnus Eodipseudopsis Eoneureclipsis Holocentropus Kambaitipsyche Neucentropus ...
Read more