Mixing
A function that is differentiable at exactly two points
$begingroup$
As homework we were given 2 questions:
Find a function that is continuous at exactly one point and not differentiable there.
Find a function that is differentiable at exactly two points.
The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.
I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $
$ f(x) = (x-1)^2 * g(x) $
Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?
calculus derivatives
asked Jan 12 at 17:37
TegernakoTegernako
908
$endgroup$
add a comment |
$begingroup$
As homework we were given 2 questions:
Find a function that is continuous at exactly one point and not differentiable there.
Find a function that is differentiable at exactly two points.
The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.
I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $
$ f(x) = (x-1)^2 * g(x) $
Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?
calculus derivatives
asked Jan 12 at 17:37
TegernakoTegernako
908
$endgroup$
1
$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39
add a comment |
$begingroup$
As homework we were given 2 questions:
Find a function that is continuous at exactly one point and not differentiable there.
Find a function that is differentiable at exactly two points.
The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.
I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $
$ f(x) = (x-1)^2 * g(x) $
Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?
calculus derivatives
asked Jan 12 at 17:37
TegernakoTegernako
908
$endgroup$
As homework we were given 2 questions:
Find a function that is continuous at exactly one point and not differentiable there.
Find a function that is differentiable at exactly two points.
The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.
I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $
$ f(x) = (x-1)^2 * g(x) $
Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?
calculus derivatives
calculus derivatives
asked Jan 12 at 17:37
TegernakoTegernako
908
asked Jan 12 at 17:37
TegernakoTegernako
908
asked Jan 12 at 17:37
TegernakoTegernako
908
asked Jan 12 at 17:37
TegernakoTegernako
908
asked Jan 12 at 17:37
TegernakoTegernako
908
908
1
$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39
add a comment |
1
$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39
1
1
$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39
$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
add a comment |
$begingroup$
Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.
So, why is it differentiable at $0$ and $1$: well, let's just check the definition.
$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.
answered Jan 12 at 17:45
user3482749user3482749
4,266919
$endgroup$
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
$begingroup$
It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
add a comment |
$begingroup$
It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
add a comment |
$begingroup$
It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
$endgroup$
It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
answered Jan 12 at 17:41
José Carlos SantosJosé Carlos Santos
160k22127232
160k22127232
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
add a comment |
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
Good example...
$endgroup$
– Cloud JR
Jan 12 at 17:45
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
$begingroup$
An excellent example, thank you!
$endgroup$
– Tegernako
Jan 12 at 18:04
1
1
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
$begingroup$
It's your example. I just used a different notation. Anyway, I'm glad I could help.
$endgroup$
– José Carlos Santos
Jan 12 at 18:06
add a comment |
$begingroup$
Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.
So, why is it differentiable at $0$ and $1$: well, let's just check the definition.
$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.
answered Jan 12 at 17:45
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.
So, why is it differentiable at $0$ and $1$: well, let's just check the definition.
$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.
answered Jan 12 at 17:45
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.
So, why is it differentiable at $0$ and $1$: well, let's just check the definition.
$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.
answered Jan 12 at 17:45
user3482749user3482749
4,266919
$endgroup$
Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.
So, why is it differentiable at $0$ and $1$: well, let's just check the definition.
$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.
answered Jan 12 at 17:45
user3482749user3482749
4,266919
answered Jan 12 at 17:45
user3482749user3482749
4,266919
answered Jan 12 at 17:45
user3482749user3482749
4,266919
answered Jan 12 at 17:45
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
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$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39