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Meaning of the nth order derivative
Clash Royale CLAN TAG#URR8PPP
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We say that the number zero is infinitely differentiable(because every higher order derivative exists and is identically zero). But then if that is the case , shouldn’t every function be infinitely differentiable ? Suppose we differentiate a differentiable function $n$ number of times and we get zero , we can still differentiate it infinitely right ? Then why do many textbooks call a function twice differentiable , thrice differentiable etc. ? I’m sorry if this sounds really stupid but this was something me and my friends had a huge debate on so I wanted to clear it once and for all ! Please correct me if I am mistaken somewhere Thanks for your help
derivatives soft-question
asked Jul 20 at 18:18
Aditi
694314
 |Â
show 4 more comments
up vote
2
down vote
favorite
We say that the number zero is infinitely differentiable(because every higher order derivative exists and is identically zero). But then if that is the case , shouldn’t every function be infinitely differentiable ? Suppose we differentiate a differentiable function $n$ number of times and we get zero , we can still differentiate it infinitely right ? Then why do many textbooks call a function twice differentiable , thrice differentiable etc. ? I’m sorry if this sounds really stupid but this was something me and my friends had a huge debate on so I wanted to clear it once and for all ! Please correct me if I am mistaken somewhere Thanks for your help
derivatives soft-question
asked Jul 20 at 18:18
Aditi
694314
When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
1
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
2
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
1
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
1
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30
 |Â
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We say that the number zero is infinitely differentiable(because every higher order derivative exists and is identically zero). But then if that is the case , shouldn’t every function be infinitely differentiable ? Suppose we differentiate a differentiable function $n$ number of times and we get zero , we can still differentiate it infinitely right ? Then why do many textbooks call a function twice differentiable , thrice differentiable etc. ? I’m sorry if this sounds really stupid but this was something me and my friends had a huge debate on so I wanted to clear it once and for all ! Please correct me if I am mistaken somewhere Thanks for your help
derivatives soft-question
asked Jul 20 at 18:18
Aditi
694314
We say that the number zero is infinitely differentiable(because every higher order derivative exists and is identically zero). But then if that is the case , shouldn’t every function be infinitely differentiable ? Suppose we differentiate a differentiable function $n$ number of times and we get zero , we can still differentiate it infinitely right ? Then why do many textbooks call a function twice differentiable , thrice differentiable etc. ? I’m sorry if this sounds really stupid but this was something me and my friends had a huge debate on so I wanted to clear it once and for all ! Please correct me if I am mistaken somewhere Thanks for your help
derivatives soft-question
asked Jul 20 at 18:18
Aditi
694314
asked Jul 20 at 18:18
Aditi
694314
asked Jul 20 at 18:18
Aditi
694314
asked Jul 20 at 18:18
Aditi
694314
694314
When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
1
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
2
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
1
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
1
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30
 |Â
show 4 more comments
When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
1
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
2
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
1
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
1
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30
When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
1
1
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
2
2
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
1
1
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
1
1
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
First, if you differentiate a function $n$ times and get zero, then yes the function in question is infinitely differentiable because the zero function is itself infinitely differentiable. Of course in this case the function was necessarily a polynomial of degree at most $n-1$, so you sort of already knew they were infinitely differentiable.
Examples of functions which are finitely many times differentiable are $x|x|^n$, which is $n$ times differentiable at $0$ for positive integers $n$.
answered Jul 20 at 18:25
Ian
65k24681
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
add a comment |Â
up vote
1
down vote
I'm guessing that by saying that the number $0$ is differentiable you mean that the funcion $f(x)=0$ is infinitely differentiable, which is true because of your argument.
Anyways, if you assume that the $n$-th derivative of a funcion is 0, you can integrate $n$ times and conclude that the funcion is a polynomial, which most funcions aren't.
If you want a concrete example or a funcion which is $n$ tines differentiable, integrate $f(x)=|x|$ that amount of time. The result is a function which is $n$ times differentiable over $mathbbR$, but it isn't at the origin.
answered Jul 20 at 18:29
Darth Lubinus
113
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First, if you differentiate a function $n$ times and get zero, then yes the function in question is infinitely differentiable because the zero function is itself infinitely differentiable. Of course in this case the function was necessarily a polynomial of degree at most $n-1$, so you sort of already knew they were infinitely differentiable.
Examples of functions which are finitely many times differentiable are $x|x|^n$, which is $n$ times differentiable at $0$ for positive integers $n$.
answered Jul 20 at 18:25
Ian
65k24681
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
add a comment |Â
up vote
3
down vote
accepted
First, if you differentiate a function $n$ times and get zero, then yes the function in question is infinitely differentiable because the zero function is itself infinitely differentiable. Of course in this case the function was necessarily a polynomial of degree at most $n-1$, so you sort of already knew they were infinitely differentiable.
Examples of functions which are finitely many times differentiable are $x|x|^n$, which is $n$ times differentiable at $0$ for positive integers $n$.
answered Jul 20 at 18:25
Ian
65k24681
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First, if you differentiate a function $n$ times and get zero, then yes the function in question is infinitely differentiable because the zero function is itself infinitely differentiable. Of course in this case the function was necessarily a polynomial of degree at most $n-1$, so you sort of already knew they were infinitely differentiable.
Examples of functions which are finitely many times differentiable are $x|x|^n$, which is $n$ times differentiable at $0$ for positive integers $n$.
answered Jul 20 at 18:25
Ian
65k24681
First, if you differentiate a function $n$ times and get zero, then yes the function in question is infinitely differentiable because the zero function is itself infinitely differentiable. Of course in this case the function was necessarily a polynomial of degree at most $n-1$, so you sort of already knew they were infinitely differentiable.
Examples of functions which are finitely many times differentiable are $x|x|^n$, which is $n$ times differentiable at $0$ for positive integers $n$.
answered Jul 20 at 18:25
Ian
65k24681
answered Jul 20 at 18:25
Ian
65k24681
answered Jul 20 at 18:25
Ian
65k24681
answered Jul 20 at 18:25
Ian
65k24681
65k24681
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
add a comment |Â
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
Ohh ! I’d thought a function stopped being differentiable at that point where it’s derivative becomes zero but now I understand that’s not the case . Thanks for clearing my doubt !
– Aditi
Jul 20 at 18:27
1
1
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
@Aditi Nope, zero derivatives, even identically zero derivatives, don't break differentiability. However, both points of nondifferentiability and points of zero derivative are critical points (i.e. candidates for local extrema). That might be the source of your confusion.
– Ian
Jul 20 at 18:28
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
Yes you’re correct . Thanks for helping !
– Aditi
Jul 20 at 18:33
add a comment |Â
up vote
1
down vote
I'm guessing that by saying that the number $0$ is differentiable you mean that the funcion $f(x)=0$ is infinitely differentiable, which is true because of your argument.
Anyways, if you assume that the $n$-th derivative of a funcion is 0, you can integrate $n$ times and conclude that the funcion is a polynomial, which most funcions aren't.
If you want a concrete example or a funcion which is $n$ tines differentiable, integrate $f(x)=|x|$ that amount of time. The result is a function which is $n$ times differentiable over $mathbbR$, but it isn't at the origin.
answered Jul 20 at 18:29
Darth Lubinus
113
add a comment |Â
up vote
1
down vote
I'm guessing that by saying that the number $0$ is differentiable you mean that the funcion $f(x)=0$ is infinitely differentiable, which is true because of your argument.
Anyways, if you assume that the $n$-th derivative of a funcion is 0, you can integrate $n$ times and conclude that the funcion is a polynomial, which most funcions aren't.
If you want a concrete example or a funcion which is $n$ tines differentiable, integrate $f(x)=|x|$ that amount of time. The result is a function which is $n$ times differentiable over $mathbbR$, but it isn't at the origin.
answered Jul 20 at 18:29
Darth Lubinus
113
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'm guessing that by saying that the number $0$ is differentiable you mean that the funcion $f(x)=0$ is infinitely differentiable, which is true because of your argument.
Anyways, if you assume that the $n$-th derivative of a funcion is 0, you can integrate $n$ times and conclude that the funcion is a polynomial, which most funcions aren't.
If you want a concrete example or a funcion which is $n$ tines differentiable, integrate $f(x)=|x|$ that amount of time. The result is a function which is $n$ times differentiable over $mathbbR$, but it isn't at the origin.
answered Jul 20 at 18:29
Darth Lubinus
113
I'm guessing that by saying that the number $0$ is differentiable you mean that the funcion $f(x)=0$ is infinitely differentiable, which is true because of your argument.
Anyways, if you assume that the $n$-th derivative of a funcion is 0, you can integrate $n$ times and conclude that the funcion is a polynomial, which most funcions aren't.
If you want a concrete example or a funcion which is $n$ tines differentiable, integrate $f(x)=|x|$ that amount of time. The result is a function which is $n$ times differentiable over $mathbbR$, but it isn't at the origin.
answered Jul 20 at 18:29
Darth Lubinus
113
answered Jul 20 at 18:29
Darth Lubinus
113
answered Jul 20 at 18:29
Darth Lubinus
113
answered Jul 20 at 18:29
Darth Lubinus
113
113
add a comment |Â
add a comment |Â
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When we differentiate $0$, we actually mean function $x mapsto 0$, not the number $0$
– Rumpelstiltskin
Jul 20 at 18:22
1
Numbers aren't differentiable, functions are. Sometimes some $n$th derivative of a function will happen to have a zero value. That tells you something about the function. Sometimes a function can be differentiated twice but not three times.
– Ethan Bolker
Jul 20 at 18:22
2
Polynomials are infinitely differentiable. But the function $f(x)=x|x|$ is differentiable once, but not twice.
– quasi
Jul 20 at 18:23
1
@Aditi $e^x$ is infinitely differentiable and its derivative is never $0$.
– saulspatz
Jul 20 at 18:27
1
@Aditi: There are lots of functions other than polynomials which are infinitely differentiable. For example:$;sin(x)$,$;e^x$,$;frac1x^2+1$.
– quasi
Jul 20 at 18:30