How to detect sharp corners in graphs
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I have drawn a graph for this function
$$logleft(1+|x|right)$$
It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".
calculus
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up vote
0
down vote
favorite
I have drawn a graph for this function
$$logleft(1+|x|right)$$
It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".
calculus
Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have drawn a graph for this function
$$logleft(1+|x|right)$$
It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".
calculus
I have drawn a graph for this function
$$logleft(1+|x|right)$$
It looks like its graph has a sharp corner in $x=0$. In fact, for $xrightarrow0^pm$, $f(x)sim pm x$. The right and left derivates in $x=0$ are $+1$ and $-1$ respectively. This should mean that the function has a sharp corner because the right and left derivates differ for the same point. Is there are tool to confirm this? I see that Wolfram Alpha does not report this information and I don't want to "guess".
calculus
asked Aug 16 at 9:04
Cesare
53129
53129
Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10
add a comment |Â
Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10
Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :
$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$
So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :
add a comment |Â
up vote
2
down vote
Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.
Then:
$ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$
and
$ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.
add a comment |Â
up vote
1
down vote
On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
add a comment |Â
up vote
0
down vote
Basic properties of functions:
Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.
Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.
Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.
For more information, read about Smoothness.
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :
$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$
So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :
add a comment |Â
up vote
2
down vote
accepted
You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :
$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$
So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :
$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$
So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :
You can be sure that there's a sharp corner when you look at the value of the derivatives near $0$. Indeed, general theorems say that $f$ is continuous, but as you said the value of the left derivative in $0$ and the value of the right derivative in $0$ are different. We have :
$$ undersetx rightarrow 0^+lim f(x) = 1 quad textand quad undersetx rightarrow 0^-lim f(x) = -1 $$
So it means that near $0$ the function will "look like" $x mapsto x$ on the right side and $x mapsto -x$ on the left side (see the definition of a derivative), exactly like the absolute value function :
answered Aug 16 at 9:21
tmaths
1,332113
1,332113
add a comment |Â
add a comment |Â
up vote
2
down vote
Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.
Then:
$ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$
and
$ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.
add a comment |Â
up vote
2
down vote
Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.
Then:
$ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$
and
$ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.
Then:
$ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$
and
$ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.
Let $f(x)=logleft(1+|x|right)$ for $x in mathbb R$ and $g(x)=logleft(1+xright)$ for $x>-1$.
Then:
$ lim_x to 0+fracf(x)-f(0)x-0=lim_x to 0+fracg(x)-g(0)x-0=g'(0)=1$
and
$ lim_x to 0-fracf(x)-f(0)x-0=lim_x to 0-fracg(-x)-g(0)x-0=-lim_x to 0-fracg(-x)-g(0)-x-0=-g'(0)=-1$.
answered Aug 16 at 9:28
Fred
38k1238
38k1238
add a comment |Â
add a comment |Â
up vote
1
down vote
On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
add a comment |Â
up vote
1
down vote
On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
add a comment |Â
up vote
1
down vote
up vote
1
down vote
On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point
On "sharp corners" first derivatives do funky stuff: they can either don't exist at all or, if your function has a cusp, they can go to infinity approaching that point
answered Aug 16 at 9:13
Davide Morgante
2,357322
2,357322
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
add a comment |Â
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Thanks, does my function has a sharp corner?
â Cesare
Aug 16 at 9:14
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Yes, probably it has! (by sketching the graph out you clearly see that it has) It's well known that $|x|$ is non differentiable at $x=0$. Try to differentiate the function $y = log(sqrtx^2+1)$
â Davide Morgante
Aug 16 at 9:22
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
Ne approfitto per farti i complimenti per i tuoi lavori, invidio le tue grandi capacità ad una così giovane età ! Keep up the good work
â Davide Morgante
Aug 16 at 9:32
add a comment |Â
up vote
0
down vote
Basic properties of functions:
Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.
Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.
Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.
For more information, read about Smoothness.
add a comment |Â
up vote
0
down vote
Basic properties of functions:
Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.
Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.
Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.
For more information, read about Smoothness.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Basic properties of functions:
Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.
Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.
Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.
For more information, read about Smoothness.
Basic properties of functions:
Continuous: This is the property that if the function maps to some values a and b, that it passes through every value between a and b too. A discontinuity is a point where the derivative is positive or negative infinity. Some functions are continuous everywhere, some are continuous but with specific discontinuities, some are discontinuous everywhere.
Smooth (1st order): This is the property that there are no kinks or corners as you describe. We can rewrite it as a requirement that the first derivative is continuous (as above). In this instance, your derivative is has a discontinuity at $x=0$; the derivative is a function like any other, and in this case it is one with a discontinuity. So yes, we can analyse the function of the derivative to see any discontinuities.
Smoothness (general): A function is said to be 'smooth' mathematically if we can keep differentiating it and obtain continuous functions. So while your function can be differentiated once, the result is not continuous so the smoothness is limited. We can classify functions according to how many times they can be differentiated before the result has a discontinuity. Some functions (e.g. a polynomial, or $sin(x)$) can be differentiated forever and remain continuous.
For more information, read about Smoothness.
answered Aug 16 at 9:38
Phil H
9531513
9531513
add a comment |Â
add a comment |Â
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Hi Cesare, what do you mean by tool to confirm this ? Are you unsure about the value of the derivatives or you don't understand why there's a "sharp corner" ?
â tmaths
Aug 16 at 9:09
How can I be sure that a sharp corner exists? Do I need to check the derivates?
â Cesare
Aug 16 at 9:10