Are there continuous functions who are the same in an interval but differ in at least one other point? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Function which is continuous everywhere in its domain, but differentiable only at one pointAre there non-periodic continuous functions with this property?Derivative defined at some point but not continuous there?Are the two statements about continuous functions equivalent?Prove or disprove: for any two given functions, one must be upper bounding the otherIf a function is derivable in a point then there exists an open interval around the point in which the function is continuousIs there a function on a compact interval that is differentiable but not Lipschitz continuous?Are there continuous functions for which the epsilon-delta property doesn't hold?Show that two continuous functions that are surjective over the same interval intersectProve a non-constant continuous function on a compact interval must admit at least one non-local extremum
How to make Illustrator type tool selection automatically adapt with text length
Using dividends to reduce short term capital gains?
Match Roman Numerals
Can withdrawing asylum be illegal?
One-dimensional Japanese puzzle
What force causes entropy to increase?
Why are PDP-7-style microprogrammed instructions out of vogue?
What can I do if neighbor is blocking my solar panels intentionally?
Can I visit the Trinity College (Cambridge) library and see some of their rare books
Does Parliament need to approve the new Brexit delay to 31 October 2019?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
Homework question about an engine pulling a train
Can a flute soloist sit?
Can the DM override racial traits?
Working through the single responsibility principle (SRP) in Python when calls are expensive
Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)
How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?
"... to apply for a visa" or "... and applied for a visa"?
Is there a writing software that you can sort scenes like slides in PowerPoint?
Why doesn't a hydraulic lever violate conservation of energy?
How to handle characters who are more educated than the author?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
What aspect of planet Earth must be changed to prevent the industrial revolution?
Store Dynamic-accessible hidden metadata in a cell
Are there continuous functions who are the same in an interval but differ in at least one other point?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Function which is continuous everywhere in its domain, but differentiable only at one pointAre there non-periodic continuous functions with this property?Derivative defined at some point but not continuous there?Are the two statements about continuous functions equivalent?Prove or disprove: for any two given functions, one must be upper bounding the otherIf a function is derivable in a point then there exists an open interval around the point in which the function is continuousIs there a function on a compact interval that is differentiable but not Lipschitz continuous?Are there continuous functions for which the epsilon-delta property doesn't hold?Show that two continuous functions that are surjective over the same interval intersectProve a non-constant continuous function on a compact interval must admit at least one non-local extremum
$begingroup$
You are given a function $f: mathbbRrightarrow mathbbR$. Every derivative $fracd^ndx^n(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbbRrightarrow mathbbR$, for which every derivative $fracd^ndx^n(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
$endgroup$
add a comment |
$begingroup$
You are given a function $f: mathbbRrightarrow mathbbR$. Every derivative $fracd^ndx^n(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbbRrightarrow mathbbR$, for which every derivative $fracd^ndx^n(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
$endgroup$
add a comment |
$begingroup$
You are given a function $f: mathbbRrightarrow mathbbR$. Every derivative $fracd^ndx^n(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbbRrightarrow mathbbR$, for which every derivative $fracd^ndx^n(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
$endgroup$
You are given a function $f: mathbbRrightarrow mathbbR$. Every derivative $fracd^ndx^n(f(x)), ,n >0$ of the function is continuous.
Is there a function $g: mathbbRrightarrow mathbbR$, for which every derivative $fracd^ndx^n(g(x)), ,n >0$ is also continuous, such that:
$$forall xin[a,b]: , g(x) = f(x)land , exists x notin [a,b]: f(x) neq g(x),, a neq b$$
Thanks!
real-analysis calculus
real-analysis calculus
edited 1 hour ago
ZeroXLR
1,528519
1,528519
asked 3 hours ago
TVSuchtyTVSuchty
325
325
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
$endgroup$
add a comment |
Your Answer
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%2f3185630%2fare-there-continuous-functions-who-are-the-same-in-an-interval-but-differ-in-at%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
$endgroup$
Define the real functions $f$ and $g$ thus:
$$
f(x) = begincases expBig(-frac1(x - 1)^2Big) &textif x > 1 \
0 &textif x in [-1, 1] \
expBig(-frac1(x + 1)^2Big) &textif x < -1
endcases
$$ and
$g(x) = 0$. $f$ and $g$ are both $0$ on $[-1, 1]$ but they differ in value everywhere else.
Obviously $g$ is continuously differentiable infinitely many times as it is a constant function. You can also check that $f$ is continuously differentiable infinitely many times at $x = -1$ and $x = 1$ by applying L'Hôpital's rule inductively. Checking this is a fine exercise in Real Analysis; you should try it. Here is a first taste of it:
beginalign*
lim_x to 1^+fracdf(x)dx &= limlimits_x to 1^+frac2expbig(- frac1(x - 1)^2big)(x - 1)^3 \
&= 2lim_x to 1^+fracfrac1(x - 1)^3expbig(frac1(x - 1)^2big) quadtextthis limit is of the form fracinftyinfty text so L'Hôpital applies \
&= 2 lim_x to 1^+fracfracddx(x - 1)^-3fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 2 lim_x to 1^+frac-3(x - 1)^-4-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= 3lim_x to 1^+frac(x - 1)^-1expbig(frac1(x - 1)^2big) quadtextagain, this has the form fracinftyinfty text so L'Hôpital applies \
&= 3 lim_x to 1^+fracfracddx(x - 1)^-1fracddxexpbig(frac1(x - 1)^2big) text by L'Hôpital \
&= 3 lim_x to 1^+frac-(x - 1)^-2-2expbig(frac1(x - 1)^2big)(x - 1)^-3 \
&= frac32 lim_x to 1^+fracx - 1expbig(frac1(x - 1)^2big) \
&= frac32 lim_x to 1^+ Big[(x - 1)expBig(-frac1(x - 1)^2Big)Big] \
&= frac32 Big[lim_x to 1^+ (x - 1)Big] Big[lim_x to 1^+ expBig(-frac1(x - 1)^2Big)Big] = frac32 times 0 times 0 = 0
endalign* That was a long calculation but take my word: it can be repeated inductively to show that $limlimits_x to 1+fracd^nfdx^n = 0$ for all $n in mathbbZ_+!$ At all other points i.e. on $(-infty, -1) cup (-1, 1) cup (1, infty)$, $f$ is infinitely differentiable because exponentials and constant functions are infinitely differentiable.
Bonus Fact:
Both $fracd^n f(x)dx^n$ and $fracd^n g(x)dx^n$ also have the same value $0$ on $[-1, 1]$ for all positive integers $n$!
edited 2 hours ago
answered 3 hours ago
ZeroXLRZeroXLR
1,528519
1,528519
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
$begingroup$
Well Done! Unfortunately, I am a high school student and never heard of L'Hôpitals Rule. EDIT: This function is actually amazing, never saw something like this before.
$endgroup$
– TVSuchty
3 hours ago
1
1
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
It basically says under certain conditions, $limlimits_x to a(f(x) / g(x)) = limlimits_x to a(fracd f(x)dx / fracd g(x)dx)$. en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule
$endgroup$
– ZeroXLR
3 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
I am stunned. Do you know of more complex solutions?
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
@TVSuchty If you are asking questions like this at high school and are studying math seriously, you will very soon learn about this rule (and a whole host of other rules from Calculus). Take a re-look at that function afterwards.
$endgroup$
– ZeroXLR
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
$begingroup$
I look forward to. Thank you for your assistance.
$endgroup$
– TVSuchty
2 hours ago
|
show 2 more comments
$begingroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
$endgroup$
add a comment |
$begingroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
$endgroup$
add a comment |
$begingroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
$endgroup$
Taylors theorem implies that if two functions are the same in one interval, they must be the same everywhere. This is because when you look at one point in the interval, the nth derivatives of both will be equal. Thus, their Taylor series centered at that point will be equal. Then you can move away from the center and find the Taylor series of both centered around another point to get more information about the function, and they will still be equal. So anywhere you look, the two functions will be equal. (This applies for all analytic functions, not so much for piecewise functions)
answered 1 hour ago
uhhhhidkuhhhhidk
1266
1266
add a comment |
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%2f3185630%2fare-there-continuous-functions-who-are-the-same-in-an-interval-but-differ-in-at%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