Why is there an “implication” rather than and “and” in this definition of the derivative?
up vote
3
down vote
favorite
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 4 hours ago
Ovi
12.2k1038109
add a comment |
up vote
3
down vote
favorite
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 4 hours ago
Ovi
12.2k1038109
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
4 hours ago
@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
4 hours ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 4 hours ago
Ovi
12.2k1038109
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 4 hours ago
Ovi
12.2k1038109
asked 4 hours ago
Ovi
12.2k1038109
edited 4 hours ago
asked 4 hours ago
Ovi
12.2k1038109
asked 4 hours ago
Ovi
12.2k1038109
asked 4 hours ago
Ovi
12.2k1038109
12.2k1038109
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
4 hours ago
@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
4 hours ago
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
4 hours ago
@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
4 hours ago
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
4 hours ago
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
4 hours ago
@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
4 hours ago
@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
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
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 4 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |
up vote
2
down vote
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 4 hours ago
Chris Culter
19.6k43381
Thank you for the response!
– Ovi
3 hours ago
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
up vote
4
down vote
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 4 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |
up vote
4
down vote
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 4 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
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 4 hours ago
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 4 hours ago
DanielV
17.8k42754
answered 4 hours ago
DanielV
17.8k42754
answered 4 hours ago
DanielV
17.8k42754
answered 4 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
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
3 hours ago
add a comment |
up vote
2
down vote
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 4 hours ago
Chris Culter
19.6k43381
Thank you for the response!
– Ovi
3 hours ago
add a comment |
up vote
2
down vote
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 4 hours ago
Chris Culter
19.6k43381
Thank you for the response!
– Ovi
3 hours ago
add a comment |
up vote
2
down vote
up vote
2
down vote
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 4 hours ago
Chris Culter
19.6k43381
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 4 hours ago
Chris Culter
19.6k43381
answered 4 hours ago
Chris Culter
19.6k43381
answered 4 hours ago
Chris Culter
19.6k43381
answered 4 hours ago
Chris Culter
19.6k43381
19.6k43381
Thank you for the response!
– Ovi
3 hours ago
add a comment |
Thank you for the response!
– Ovi
3 hours ago
Thank you for the response!
– Ovi
3 hours ago
Thank you for the response!
– Ovi
3 hours ago
add a comment |
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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
4 hours ago
@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
4 hours ago