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Differentials and Jacobians ... Remarks After Browder Prposition 8.21 ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 help with fully understanding some remarks by Browder made after the proof of Proposition 8.21 ... ...


Proposition 8.21 (including some preliminary material and some remarks after the proof) reads as follows:


Browder - 1 -  Proposition 8.21 ... PART 1 ... .png
Browder - 2 -  Proposition 8.21 ... PART 2 ... .png



After the proof of Proposition 8.21, Browder makes the following remarks:

" ... ... This proposition can also be formulated as a description of the matrix \(\displaystyle f'(p)\) of the linear transformation \(\displaystyle \text{df}_p\) associated with the map \(\displaystyle f = ( f_1, \ldots , f_m )\) namely

\(\displaystyle f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]\)

where \(\displaystyle D_jf\) is the column vector \(\displaystyle [D_jf_1, \ldots , D_jf_m ]^t\), that is


\(\displaystyle ( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }\)


where the left-hand side denotes the entry in the ith row and jth column of the matrix \(\displaystyle f'(p)\) ...

... ... ... "




My questions are as follows ...


How/why exactly do we know that

\(\displaystyle f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]\) ... ...


and further ...


How/why exactly do we know that

\(\displaystyle ( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }\) ...



Help will be much appreciated ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 help with fully understanding some remarks by Browder made after the proof of Proposition 8.21 ... ...


Proposition 8.21 (including some preliminary material and some remarks after the proof) reads as follows:







After the proof of Proposition 8.21, Browder makes the following remarks:

" ... ... This proposition can also be formulated as a description of the matrix \(\displaystyle f'(p)\) of the linear transformation \(\displaystyle \text{df}_p\) associated with the map \(\displaystyle f = ( f_1, \ldots , f_m )\) namely

\(\displaystyle f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]\)

where \(\displaystyle D_jf\) is the column vector \(\displaystyle [D_jf_1, \ldots , D_jf_m ]^t\), that is


\(\displaystyle ( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }\)


where the left-hand side denotes the entry in the ith row and jth column of the matrix \(\displaystyle f'(p)\) ...

... ... ... "




My questions are as follows ...


How/why exactly do we know that

\(\displaystyle f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]\) ... ...


and further ...


How/why exactly do we know that

\(\displaystyle ( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }\) ...



Help will be much appreciated ...

Peter



Because my questions in the above post may be vague, I intend to illustrate the above proposition with an example and ask MHB helpers to comment on my example ...


So ... consider \(\displaystyle f: \mathbb{R}^2 \to \mathbb{R}^3\)

where \(\displaystyle f(x, y) = (xy, x + y, x^2)\)

and \(\displaystyle f_1(x, y) = xy\), \(\displaystyle f_2(x, y) = x + y\), and \(\displaystyle f_3(x, y) = x^2\)

and also \(\displaystyle e_1 = (1, 0)\) , \(\displaystyle e_2 = (0, 1)\) and \(\displaystyle p = (a, b)\) ... ...


Now calculate \(\displaystyle D_1 f(p)\) and \(\displaystyle D_2 f(p)\) from first principles

First calculate \(\displaystyle D_1 f(p) = \lim_{ h^1 \to 0 } \frac{ f(p + h^1e_1) - f(p) }{ h^1}\)


Now \(\displaystyle f(p + h^1e_1) = f ( (a, b) + h^1(1,0) )\)

\(\displaystyle = f( a + h^1, b )\)

\(\displaystyle = ( ab + h^1b, a + h^1 + b, a^2 + 2ah^1 + h^{1 \ 2} )\)

Also we have \(\displaystyle f(p) = f(a, b) = ( ab, a + b, a^2 )\)


... so then ...


\(\displaystyle \frac{ f(p + h^1e_1) - f(p) }{ h} = \frac{ ( h^1b , h^1, 2ah^1 + h^{1 \ 2} ) }{ h^1} = \begin{pmatrix}b \\ 1 \\ 2a + h^1 \end{pmatrix}\)


Therefore \(\displaystyle D_1 f(p) = \lim_{ h^1 \to 0 } \frac{ f(p + h^1e_1) - f(p) }{ h^1} = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}\)


Similarly we can determine \(\displaystyle D_2 f(p) = = \begin{pmatrix}a \\ 1 \\ 0 \end{pmatrix}\)



Now the Proposition shows that \(\displaystyle D_j f(p) = Le_j\) so in terms of the example


\(\displaystyle Le_1 = D_1 f(p) = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}\)

and

\(\displaystyle Le_2 = D_2 f(p) = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}\)


Now if we take \(\displaystyle L = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix}\) ( BUT!!! can we legitimately do this at this point ... ?)

... then ...

\(\displaystyle Le_1 = L \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}\)

and

\(\displaystyle Le_2 = L \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}\)



Now we can illustrate the last line of the proof of Proposition 8.21 in terms of the example ...


We have ...


\(\displaystyle Lh = L ( h^1, h^2 ) = L( \sum_{ j = 1}^2 h^j e_j ) = L (h^1 e_1 + h^2 e_2)\)


\(\displaystyle \Longrightarrow Lh = h^1 Le_1 + h^2 Le_2\)


\(\displaystyle \Longrightarrow Lh = h^1 \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix} + h^2 \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}\)


\(\displaystyle = \begin{pmatrix} bh^1 \\ h^1 \\ 2ah^1 \end{pmatrix} + \begin{pmatrix} ah^2 \\ h^2 \\ 0 \end{pmatrix}\)


\(\displaystyle = \begin{pmatrix} bh^1 + ah^2 \\ h^1 + h^2 \\ 2ah^1 \end{pmatrix}\)


... or in terms of partial derivatives ...


\(\displaystyle \Longrightarrow Lh = \sum_{ j = 1}^2 h^j D_j f(p)\)


\(\displaystyle = h^1 D_1 f(p) + h^2 D_2 f(p)\)


\(\displaystyle = h^1 \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix} + h^2 \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}\)


\(\displaystyle = \begin{pmatrix} bh^1 + ah^2 \\ h^1 + h^2 \\ 2ah^1 \end{pmatrix}\)



Is the above example essentially correct ...

Any comments ...


Peter
 
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