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Differentials in R^n ... Remark by Brwder, Section 8.2 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.2 Differentials ... ...

I need some further help in fully understanding a remark by Browder after Definition 8.9 ...

The relevant text from Browder reads as follows:



Browder - Definition 8.9  ... Differentials ....png




At the end of the above text from Browder, we read the following:


"... ... Thus Definition 8.9 says roughly that a function is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by an affine function ... "



My question is as follows:


Can someone please demonstrate formally and rigorously that Definition 8.9 implies that a function is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by an affine function ... ...



Help will be much appreciated ... ...

Peter
 

steep

Member
Dec 17, 2018
51
Peter, what are you looking for here? Can you prove this in the special case of $\mathbf c =\mathbf 0$? (Then can you show why this implies the general case?)

I know this is technically a different problem, but it feels in spirit to be basically the same as this recent one
https://mathhelpboards.com/analysis...ariable-markushevich-theorem-7-1-a-26712.html

the point in both cases is that you can make $\Big \vert \epsilon ( z, z_0 ) \Big \vert$ arbitrarily small -- or changing notation, I'd say $\Big \vert \epsilon ( \mathbf x, \mathbf x_0 ) \Big \vert$ may be made arbitrarily small
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
That says explicitely that, near p, f(x) can be approximated by f(p)- L(x) where L is linear. That is the "affine function" approximating f(x).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
That says explicitely that, near p, f(x) can be approximated by f(p)- L(x) where L is linear. That is the "affine function" approximating f(x).


My thanks to steep and HallsofIvy for the help ...


I will try to make clear my dilemma/issues ...


As I see it ...


... if \(\displaystyle f\) is approximated near \(\displaystyle p\) by \(\displaystyle c + Lx\) then


\(\displaystyle f(p) \approx c + Lp\) ... ... ... ... ... (1)



Now I think ... ??? ... that (1) is equivalent to


\(\displaystyle f(p + h) - f(p)\) is approximated by \(\displaystyle c + Lh\) ... ... ... (2) ... ... ... (is that correct?)


so that ... ... ...


\(\displaystyle \lim_{ h \to 0 } \frac{1}{ \| h \| } ( [ f(p + h) - f(p) ] - [ c + Lh ] ) = 0\) ... ... ... (3)



BUT ... how does (3) reduce (exactly) to Browders 8.4 ...


Must say that not even sure that (1) \(\displaystyle \Longrightarrow\) (2) ... ... let alone how (3) \(\displaystyle \Longrightarrow\) 8.4 ...





Can someone please comment on and clarify the above ....

Peter
 

steep

Member
Dec 17, 2018
51
My thanks to steep and HallsofIvy for the help ...


I will try to make clear my dilemma/issues ...


As I see it ...


... if \(\displaystyle f\) is approximated near \(\displaystyle p\) by \(\displaystyle c + Lx\) then


\(\displaystyle f(p) \approx c + Lp\) ... ... ... ... ... (1)



Now I think ... ??? ... that (1) is equivalent to


\(\displaystyle f(p + h) - f(p)\) is approximated by \(\displaystyle c + Lh\) ... ... ... (2) ... ... ... (is that correct?)


so that ... ... ...


\(\displaystyle \lim_{ h \to 0 } \frac{1}{ \| h \| } ( [ f(p + h) - f(p) ] - [ c + Lh ] ) = 0\) ... ... ... (3)



BUT ... how does (3) reduce (exactly) to Browders 8.4 ...


Must say that not even sure that (1) \(\displaystyle \Longrightarrow\) (2) ... ... let alone how (3) \(\displaystyle \Longrightarrow\) 8.4 ...





Can someone please comment on and clarify the above ....

Peter
you didn't do the estimation correctly-- it should be

$\lim_{ h \to 0 } \frac{1}{ \| h \| } \Big \vert \big( f(p + h) - f(p) \big) - \big( c + L(p+h) - \{c - L(p)\} \big) \Big \vert = \lim_{ h \to 0 } \frac{1}{ \| h \| } \big \vert \big( f(p + h) - f(p) \big) - L(h) \big \vert = 0$
i.e. the $c$'s net -- what you had was a common bug with affine functions which is why I suggested first considering c = 0. Equivalently, your statement (2) is false
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
you didn't do the estimation correctly-- it should be

$\lim_{ h \to 0 } \frac{1}{ \| h \| } \Big \vert \big( f(p + h) - f(p) \big) - \big( c + L(p+h) - \{c - L(p)\} \big) \Big \vert = \lim_{ h \to 0 } \frac{1}{ \| h \| } \big \vert \big( f(p + h) - f(p) \big) - L(h) \big \vert = 0$
i.e. the $c$'s net -- what you had was a common bug with affine functions which is why I suggested first considering c = 0. Equivalently, your statement (2) is false



Hi steep ...

... thanks for a very helpful post indeed!!! ...


But ... sorry to belabour the point but I'm trying to make sure I understand the issue ...

... it seems to challenge my (possibly wrongheaded ...) interpretation of the differential/derivative ...


I have previously thought that the differential/derivative at \(\displaystyle p\) was the linear transformation \(\displaystyle L\)

such that \(\displaystyle L\) was a good approximation to \(\displaystyle f\) near \(\displaystyle p\) ...

... that is ... \(\displaystyle f\) is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by a linear transformation/function \(\displaystyle L\) ...


Is the above correct?


I suspect the above interpretation is wrong in some sense ...

... because Browder is asserting ... and you have proved that a function \(\displaystyle f\) is differentiable at \(\displaystyle p\)

if it can be approximated near p by an affine function ...



Can you clarify...

Peter
 

steep

Member
Dec 17, 2018
51
Hi steep ...

... thanks for a very helpful post indeed!!! ...


But ... sorry to belabour the point but I'm trying to make sure I understand the issue ...

... it seems to challenge my (possibly wrongheaded ...) interpretation of the differential/derivative ...


I have previously thought that the differential/derivative at \(\displaystyle p\) was the linear transformation \(\displaystyle L\)

such that \(\displaystyle L\) was a good approximation to \(\displaystyle f\) near \(\displaystyle p\) ...

... that is ... \(\displaystyle f\) is differentiable at \(\displaystyle p\) if it can be approximated near \(\displaystyle p\) by a linear transformation/function \(\displaystyle L\) ...


Is the above correct?
sounds good to me



I suspect the above interpretation is wrong in some sense ...

... because Browder is asserting ... and you have proved that a function \(\displaystyle f\) is differentiable at \(\displaystyle p\)

if it can be approximated near p by an affine function ...
Unless I'm misunderstanding you, you are hung up on whether the function approximation (called a derivative) is linear or affine... is that right?
Let me try to run this forward and backward.

Forward case: geometrically, if you consider the derivative at a point x and orient yourself so f(x) is your origin of the space that is the image of your domain under your function, then the derivative is a linear approximation. Of course when I say "orient yourself to f(x) is your origin" all I'm talking about is an affine translation (i.e. by a constant), -- the addition of an affine translation and a linear map is an affine map.

this is another way of saying "a function is differentiable if it is locally approximately linear... a function is differentiable at a point x if it is locally approximately linear, with an error which decreases to zero faster than linearly, as we approach x"

Backward case: I think "we all know" that if a derivative exists and has magnitude zero everywhere in a path connected domain then the function is constant in said domain. (Good proof: use mean value inequality,) This implies that if two functions have the same derivative in such a domain, then they are the same up to a constant, i.e. f(x) = g(x) +b. So the derivative is the linear map, and the constant is really just a tweak used to specify a particular detail of the function.

- - - -
your book's layout and formatting remind me of Koerner's A Companion to Analysis (where I got that quote from) --- I think it is a nice "companion" to another book on analysis.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,915
sounds good to me





Unless I'm misunderstanding you, you are hung up on whether the function approximation (called a derivative) is linear or affine... is that right?
Let me try to run this forward and backward.

Forward case: geometrically, if you consider the derivative at a point x and orient yourself so f(x) is your origin of the space that is the image of your domain under your function, then the derivative is a linear approximation. Of course when I say "orient yourself to f(x) is your origin" all I'm talking about is an affine translation (i.e. by a constant), -- the addition of an affine translation and a linear map is an affine map.

this is another way of saying "a function is differentiable if it is locally approximately linear... a function is differentiable at a point x if it is locally approximately linear, with an error which decreases to zero faster than linearly, as we approach x"

Backward case: I think "we all know" that if a derivative exists and has magnitude zero everywhere in a path connected domain then the function is constant in said domain. (Good proof: use mean value inequality,) This implies that if two functions have the same derivative in such a domain, then they are the same up to a constant, i.e. f(x) = g(x) +b. So the derivative is the linear map, and the constant is really just a tweak used to specify a particular detail of the function.

- - - -
your book's layout and formatting remind me of Koerner's A Companion to Analysis (where I got that quote from) --- I think it is a nice "companion" to another book on analysis.




Thanks so much for the above post, steep ...

... ... just now working through what you have written and reflecting on it ...

Thanks again,

Peter