Flaw in proof that a differentiable function has continuous derivativeProve that $f'(a)=lim_{xrightarrow...
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Flaw in proof that a differentiable function has continuous derivative
Prove that $f'(a)=lim_{xrightarrow a}f'(x)$.Composition of a continuous function with functions that converge uniformlyCounterexamples in Analysis error: everywhere continuous, nowhere differentiable functionthe derivative of a sequence of convex function covergeIf a function f: R to R has the property that $f(x)/x^n$ tends to zero as x tends to zero, does it follow that f is infinitely differentiable at 0?Is the space of almost everywhere differentiable function with bounded derivative embedded with uniform norm complete?Prove that this function is differentiableA function which is infinitely-differentiable everywhere but continuous nowhere?Proof that a lipschitz continuous function has a limitWhat is required to prove that a function $f(x)$ tends to a limit?function continuous and differentiable at exactly one point
$begingroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked 10 hours ago
user3184807user3184807
359110
$endgroup$
add a comment |
$begingroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked 10 hours ago
user3184807user3184807
359110
$endgroup$
add a comment |
$begingroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
asked 10 hours ago
user3184807user3184807
359110
$endgroup$
Let f be a function differentiable on $(a,b)$ and continuous on $cin(a,b)$. If $c+h in (a,b)$ then by the mean value theorem $frac{f(c+h)-f(c)}{h}=f'(c+theta h)$ for $theta in [0,1]$. Let $h xrightarrow{}0$, then $f'(c+theta h) xrightarrow{} f'(c)$ by the above.
My reasoning is the following: $theta$ is a function of $h$, so in fact it is not true that that $theta(h_n)h_n$ represents any arbitrary sequence that tends to 0, so using this definition of limits(the sequence definition), what we have does not follow. EDIT: I am aware of the explicit counterexample $x^2sin(1/x)$ but that doesn't capture the full meat of the question.
real-analysis limits analysis derivatives convergence
real-analysis limits analysis derivatives convergence
asked 10 hours ago
user3184807user3184807
359110
asked 10 hours ago
user3184807user3184807
359110
edited 10 hours ago
user3184807
asked 10 hours ago
user3184807user3184807
359110
asked 10 hours ago
user3184807user3184807
359110
asked 10 hours ago
user3184807user3184807
359110
359110
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered 10 hours ago
zhw.zhw.
73.7k43175
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered 10 hours ago
IanIan
68.7k25389
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
add a comment |
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
|
show 4 more comments
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3 Answers
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3 Answers
3
active
oldest
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active
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votes
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered 10 hours ago
zhw.zhw.
73.7k43175
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
add a comment |
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered 10 hours ago
zhw.zhw.
73.7k43175
$endgroup$
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
add a comment |
$begingroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered 10 hours ago
zhw.zhw.
73.7k43175
$endgroup$
It is not true that $theta$ need be a function of $h,$ because it may not be uniquely specified. There could be many $theta$'s that work for a given $h.$
We only know $f'(y)to f'(c)$ as $yto c$ within the set of $y=x+theta h$ that arise in the MVT process you described. But the set of such $y$ may not equal any deleted neighborhood of $c,$ as the example $x^2sin(1/x)$ shows.
answered 10 hours ago
zhw.zhw.
73.7k43175
edited 6 hours ago
answered 10 hours ago
zhw.zhw.
73.7k43175
answered 10 hours ago
zhw.zhw.
73.7k43175
answered 10 hours ago
zhw.zhw.
73.7k43175
73.7k43175
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
add a comment |
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
So we can make $theta$ a function by choosing a value of $h$ and then say the union of all the sequences $theta(h_n)h_n$ where $h_n xrightarrow{}0$ over all functions $theta$ need not be all sequences tending to 0?
$endgroup$
– user3184807
10 hours ago
1
1
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
$begingroup$
That's a lot of choosing, and the axiom of choice may even come in here. But yes, the sequences we get through the MVT may not be all necessary sequences. In the case of $x^2sin(1/x)$ the sequence $1/(2npi)to 0,$ but this sequence does not arise as $theta(h_n)h_n$ in the MVT process.
$endgroup$
– zhw.
9 hours ago
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered 10 hours ago
IanIan
68.7k25389
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered 10 hours ago
IanIan
68.7k25389
$endgroup$
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
add a comment |
$begingroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered 10 hours ago
IanIan
68.7k25389
$endgroup$
If you look at $f(x)=x^2 sin(1/x)$ on $(0,1]$ and $f(0)=0$, you have $f'(x)=2xsin(1/x)-cos(1/x)$ on $(0,1]$ and $f'(0)=0$. Thus MVT tells you that for every $h in (0,1)$ there exists $theta(h) in (0,1)$ with
$$hsin(1/h)=2htheta(h)sin(1/(htheta(h)))-cos(1/(htheta(h))).$$
Notice that once $h$ is small enough, $htheta(h)$ is forced to stay relatively close to the zeros of $cos(1/x)$, since
$$|cos(1/(htheta(h)))|=|hsin(1/h)-2htheta(h)sin(1/(htheta(h)))|<3h.$$
This means that the MVT is only providing you with information about $f'(x)$ approaching $f'(0)$ along sequences which stay sufficiently close to the zeros of $cos(1/x)$. If you instead look at something like $f' left ( frac{1}{pi n} right )$ you find that it doesn't go to zero. There is no contradiction because the MVT is never giving you information about those points.
answered 10 hours ago
IanIan
68.7k25389
edited 9 hours ago
answered 10 hours ago
IanIan
68.7k25389
answered 10 hours ago
IanIan
68.7k25389
answered 10 hours ago
IanIan
68.7k25389
68.7k25389
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
add a comment |
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
So does this essentially come down to the idea that $theta(h_n)h_n$ cannot always be made to be any arbitrary sequence tending to 0?
$endgroup$
– user3184807
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
$begingroup$
@user3184807 Yes, that is essentially what happens.
$endgroup$
– Ian
10 hours ago
add a comment |
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
|
show 4 more comments
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
|
show 4 more comments
$begingroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
$endgroup$
You are assuming what you want to prove: that $f’$ is continuous at $c$ when you say that $f'(c+theta h) to f'(c)$ as $hto0$.
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
answered 10 hours ago
Julián AguirreJulián Aguirre
69.2k24096
69.2k24096
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
|
show 4 more comments
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
1
1
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
No, because OP defined $theta(h)$ through the MVT in the first place. The point is that you know that $lim_{h to 0} f'(c+htheta(h))=f'(c)$, but why doesn't this imply $lim_{h to 0} f'(c+h)=f'(c)$? Presumably this is because of some very bad property of $theta(h)$.
$endgroup$
– Ian
10 hours ago
1
1
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
That has nothing to do with muy answer.
$endgroup$
– Julián Aguirre
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
Let me spell it out, then. By definition $lim_{h to 0} frac{f(c+h)-f(c)}{h}=f'(c)$, and by MVT $frac{f(c+h)-f(c)}{h}=f'(c+htheta(h))$ for some $theta(h) in (0,1)$. That's all OP assumed. Then they send $h to 0$ in the MVT relation and find $lim_{h to 0} f'(c+h theta(h))=f'(c)$. Now they're wondering why they can't conclude $lim_{h to 0} f'(c+h)=f'(c)$.
$endgroup$
– Ian
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@Ian: The problem is that the limit may not exist. If it exists, then it must equal $f'(c)$ (as has been covered many times on this site already, for example here: math.stackexchange.com/questions/257907/…).
$endgroup$
– Hans Lundmark
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
$begingroup$
@HansLundmark OP knows that already as well, the question is about intuition. The answer, for the case of something like $f(x)=x^2sin(1/x)$, is that $htheta(h)$ has numerous discontinuities in a vicinity of $h=0$, so you don't get information about entire neighborhoods of $h=0$ from the MVT.
$endgroup$
– Ian
10 hours ago
|
show 4 more comments
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