Why is there an “implication” rather than and “and” in this definition of the derivative?
I am readig Pugh's Analysis book:
Definition
Let $f:U to mathbb{R}^m$ be given where $U$ is an open subset of $mathbb{R}^n$. The function $f$ is differentiable a $p in U$ with derivative $(Df)_p = T$ if $T:mathbb{R}^n to mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) implies lim_{|v| to 0} dfrac {R(v)}{|v|}=0$.
Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$implies$" there rather than a "$wedge$". Isn't it more natural to say
"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$"?
I'm having trouble seeing what the impact of changing these would be.
real-analysis analysis logic frechet-derivative
asked Dec 18 at 0:24
Ovi
12.3k1038110
add a comment |
I am readig Pugh's Analysis book:
Definition
Let $f:U to mathbb{R}^m$ be given where $U$ is an open subset of $mathbb{R}^n$. The function $f$ is differentiable a $p in U$ with derivative $(Df)_p = T$ if $T:mathbb{R}^n to mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) implies lim_{|v| to 0} dfrac {R(v)}{|v|}=0$.
Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$implies$" there rather than a "$wedge$". Isn't it more natural to say
"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$"?
I'm having trouble seeing what the impact of changing these would be.
real-analysis analysis logic frechet-derivative
asked Dec 18 at 0:24
Ovi
12.3k1038110
1
It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35
add a comment |
I am readig Pugh's Analysis book:
Definition
Let $f:U to mathbb{R}^m$ be given where $U$ is an open subset of $mathbb{R}^n$. The function $f$ is differentiable a $p in U$ with derivative $(Df)_p = T$ if $T:mathbb{R}^n to mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) implies lim_{|v| to 0} dfrac {R(v)}{|v|}=0$.
Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$implies$" there rather than a "$wedge$". Isn't it more natural to say
"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$"?
I'm having trouble seeing what the impact of changing these would be.
real-analysis analysis logic frechet-derivative
asked Dec 18 at 0:24
Ovi
12.3k1038110
I am readig Pugh's Analysis book:
Definition
Let $f:U to mathbb{R}^m$ be given where $U$ is an open subset of $mathbb{R}^n$. The function $f$ is differentiable a $p in U$ with derivative $(Df)_p = T$ if $T:mathbb{R}^n to mathbb{R}^m$ is a linear transformation and $f(p+v) = f(p)+T(v)+R(v) implies lim_{|v| to 0} dfrac {R(v)}{|v|}=0$.
Partly due to the missing quantifiers, I'm having trouble understanding why there is a "$implies$" there rather than a "$wedge$". Isn't it more natural to say
"T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$"?
I'm having trouble seeing what the impact of changing these would be.
real-analysis analysis logic frechet-derivative
real-analysis analysis logic frechet-derivative
asked Dec 18 at 0:24
Ovi
12.3k1038110
asked Dec 18 at 0:24
Ovi
12.3k1038110
edited Dec 18 at 0:32
asked Dec 18 at 0:24
Ovi
12.3k1038110
asked Dec 18 at 0:24
Ovi
12.3k1038110
asked Dec 18 at 0:24
Ovi
12.3k1038110
12.3k1038110
1
It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35
add a comment |
1
It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35
1
1
It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35
add a comment |
2 Answers
2
active
oldest
votes
The author means
$$forall R left((forall v ~ f(p+v) = f(p) + T(v) + R(v)) implies lim_{|v| to 0} frac{R(v)}{|v|} = 0right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ bigg(R = v mapsto f(p + v) - f(p) - T(v)bigg) land bigg(lim_{|v| to 0} frac{R(v)}{|v|} = 0bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$begin{cases} text{define } R text{ as } v mapsto f(p + v) - f(p) - T(v) \
lim_{|v| to 0} frac{R(v)}{|v|} = 0 end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.
answered Dec 18 at 1:00
DanielV
17.8k42754
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
|
show 3 more comments
I can't speak for Pugh, but your construction
T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$
is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.
Look at it this way, would you write the following?
T is the derivative if $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$
answered Dec 18 at 1:02
Chris Culter
20k43482
Thank you for the response!
– Ovi
Dec 18 at 1:17
add a comment |
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2 Answers
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2 Answers
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The author means
$$forall R left((forall v ~ f(p+v) = f(p) + T(v) + R(v)) implies lim_{|v| to 0} frac{R(v)}{|v|} = 0right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ bigg(R = v mapsto f(p + v) - f(p) - T(v)bigg) land bigg(lim_{|v| to 0} frac{R(v)}{|v|} = 0bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$begin{cases} text{define } R text{ as } v mapsto f(p + v) - f(p) - T(v) \
lim_{|v| to 0} frac{R(v)}{|v|} = 0 end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.
answered Dec 18 at 1:00
DanielV
17.8k42754
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
|
show 3 more comments
The author means
$$forall R left((forall v ~ f(p+v) = f(p) + T(v) + R(v)) implies lim_{|v| to 0} frac{R(v)}{|v|} = 0right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ bigg(R = v mapsto f(p + v) - f(p) - T(v)bigg) land bigg(lim_{|v| to 0} frac{R(v)}{|v|} = 0bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$begin{cases} text{define } R text{ as } v mapsto f(p + v) - f(p) - T(v) \
lim_{|v| to 0} frac{R(v)}{|v|} = 0 end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.
answered Dec 18 at 1:00
DanielV
17.8k42754
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
|
show 3 more comments
The author means
$$forall R left((forall v ~ f(p+v) = f(p) + T(v) + R(v)) implies lim_{|v| to 0} frac{R(v)}{|v|} = 0right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ bigg(R = v mapsto f(p + v) - f(p) - T(v)bigg) land bigg(lim_{|v| to 0} frac{R(v)}{|v|} = 0bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$begin{cases} text{define } R text{ as } v mapsto f(p + v) - f(p) - T(v) \
lim_{|v| to 0} frac{R(v)}{|v|} = 0 end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.
answered Dec 18 at 1:00
DanielV
17.8k42754
The author means
$$forall R left((forall v ~ f(p+v) = f(p) + T(v) + R(v)) implies lim_{|v| to 0} frac{R(v)}{|v|} = 0right)$$
which is taking advantage of how only one function for $R$ satisfies the condition. You are maybe thinking something like
$$ bigg(R = v mapsto f(p + v) - f(p) - T(v)bigg) land bigg(lim_{|v| to 0} frac{R(v)}{|v|} = 0bigg)$$
The problem with the second equation is that it isn't defining $R$ as $v mapsto f(p + v) - f(p) - T(v)$, it is saying 'if' $R$ is defined as such. Alternatively,
$$begin{cases} text{define } R text{ as } v mapsto f(p + v) - f(p) - T(v) \
lim_{|v| to 0} frac{R(v)}{|v|} = 0 end{cases}$$
is an option, in that case sort of using 'and' in the casual sequential sense, like 'crack the egg and put the egg in the bowl and beat the egg and pour the egg into the pan...'.
answered Dec 18 at 1:00
DanielV
17.8k42754
answered Dec 18 at 1:00
DanielV
17.8k42754
answered Dec 18 at 1:00
DanielV
17.8k42754
answered Dec 18 at 1:00
DanielV
17.8k42754
17.8k42754
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
|
show 3 more comments
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
Thanks for the response! I think I understand it but it seems really really weird; your third approach seems the most natural and I don't see anybody would use anything else; will have to digest it more.
– Ovi
Dec 18 at 1:12
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
It seems like the author is using the first route just so he can avoid making a definition for $R$ before the actual definition of differentiability?
– Ovi
Dec 18 at 1:14
1
1
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
Yeah, it is odd he didn't bother directly using a definition, and also didn't just write $$lim_{|v| to 0} frac{f(p + v) - f(p) - T(v)}{|v|} = 0$$
– DanielV
Dec 18 at 1:36
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
If I could ask you another thing; can I formulate it like this? $f: mathbb{R}^n to mathbb{R}^m$ is differentiable at $p$ if $exists T exists R forall v : left( dfrac{}{} f(p+v) = f(p)+T(v)+R(v) right) wedge left( lim_{v to 0} dfrac {|R(v)|}{|v|} = 0 right) $?
– Ovi
Dec 20 at 16:55
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
@Ovi You should take the $exists T$ off because $T$ is being defined so it must be left free in the definition, but otherwise yes that is fine. It is a common trick when encoding things in First Order Logic to convert $a = f(g(b))$ to $exists y~(a = f(y) land y = g(b))$
– DanielV
Dec 20 at 19:43
|
show 3 more comments
I can't speak for Pugh, but your construction
T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$
is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.
Look at it this way, would you write the following?
T is the derivative if $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$
answered Dec 18 at 1:02
Chris Culter
20k43482
Thank you for the response!
– Ovi
Dec 18 at 1:17
add a comment |
I can't speak for Pugh, but your construction
T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$
is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.
Look at it this way, would you write the following?
T is the derivative if $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$
answered Dec 18 at 1:02
Chris Culter
20k43482
Thank you for the response!
– Ovi
Dec 18 at 1:17
add a comment |
I can't speak for Pugh, but your construction
T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$
is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.
Look at it this way, would you write the following?
T is the derivative if $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$
answered Dec 18 at 1:02
Chris Culter
20k43482
I can't speak for Pugh, but your construction
T is the derivative if we can write $f(p+v) = f(p)+T(v)+R(v)$ AND $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$
is awkward because neither the left nor the right side of the "AND" is a property of $T$ that needs checking. The left side is always true; you can always do that. And the right side makes no sense unless $R$ is constrained.
Look at it this way, would you write the following?
T is the derivative if $lim_{|v| to 0} dfrac {R(v)}{|v|}=0$ AND we can write $f(p+v) = f(p)+T(v)+R(v)$
answered Dec 18 at 1:02
Chris Culter
20k43482
answered Dec 18 at 1:02
Chris Culter
20k43482
answered Dec 18 at 1:02
Chris Culter
20k43482
answered Dec 18 at 1:02
Chris Culter
20k43482
20k43482
Thank you for the response!
– Ovi
Dec 18 at 1:17
add a comment |
Thank you for the response!
– Ovi
Dec 18 at 1:17
Thank you for the response!
– Ovi
Dec 18 at 1:17
Thank you for the response!
– Ovi
Dec 18 at 1:17
add a comment |
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It would probably be clearer to write "...is a linear transformation and $lim_{|v| to 0} R(V)/|v| = 0$, where $R(v) = f(p+v) - f(p) - T(v)$.
– Bungo
Dec 18 at 0:29
@Bungo Thanks for the reply. I've seen that definition in another book as well. I'm having trouble understanding how that definition related to Pugh's; is it equivalent to the definition with $implies$? Or to the one with $wedge$? Or is the one with $wedge$ nonsense? I have a vague feeling that $implies$ and $wedge$ won't matter because the derivative is unique, but I don't really understand it.
– Ovi
Dec 18 at 0:35