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.










share|cite|improve this question




















  • 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















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.










share|cite|improve this question




















  • 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













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.










share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago

























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














  • 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










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...'.






share|cite|improve this answer





















  • 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


















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 $f(p+v) = f(p)+T(v)+R(v)$







share|cite|improve this answer





















  • Thank you for the response!
    – Ovi
    3 hours ago











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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...'.






share|cite|improve this answer





















  • 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















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...'.






share|cite|improve this answer





















  • 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













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...'.






share|cite|improve this answer












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...'.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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










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 $f(p+v) = f(p)+T(v)+R(v)$







share|cite|improve this answer





















  • Thank you for the response!
    – Ovi
    3 hours ago















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 $f(p+v) = f(p)+T(v)+R(v)$







share|cite|improve this answer





















  • Thank you for the response!
    – Ovi
    3 hours ago













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 $f(p+v) = f(p)+T(v)+R(v)$







share|cite|improve this answer












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 $f(p+v) = f(p)+T(v)+R(v)$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Chris Culter

19.6k43381




19.6k43381












  • 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




Thank you for the response!
– Ovi
3 hours ago


















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