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

[Math Amateur]

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:View attachment 9443
View attachment 9444
After the proof of Proposition 8.21, Browder makes the following remarks:

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

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

where $$D_jf$$ is the column vector $$[D_jf_1, \ldots , D_jf_m ]^t$$, that is $$( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }$$ where the left-hand side denotes the entry in the ith row and jth column of the matrix $$f'(p)$$ ...

... ... ... "

My questions are as follows ...How/why exactly do we know that

$$f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]$$ ... ... and further ...How/why exactly do we know that

$$( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }$$ ...
Help will be much appreciated ...

Peter

 

[Math Amateur]

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 $$f'(p)$$ of the linear transformation $$\text{df}_p$$ associated with the map $$f = ( f_1, \ldots , f_m )$$ namely

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

where $$D_jf$$ is the column vector $$[D_jf_1, \ldots , D_jf_m ]^t$$, that is$$( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }$$where the left-hand side denotes the entry in the ith row and jth column of the matrix $$f'(p)$$ ...

... ... ... "

My questions are as follows ...How/why exactly do we know that

$$f'(p) = [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]$$ ... ...and further ...How/why exactly do we know that

$$( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j }$$ ...
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 $$f: \mathbb{R}^2 \to \mathbb{R}^3$$

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

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

and also $$e_1 = (1, 0)$$ , $$e_2 = (0, 1)$$ and $$p = (a, b)$$ ... ...Now calculate $$D_1 f(p)$$ and $$D_2 f(p)$$ from first principles

First calculate $$D_1 f(p) = \lim_{ h^1 \to 0 } \frac{ f(p + h^1e_1) - f(p) }{ h^1}$$Now $$f(p + h^1e_1) = f ( (a, b) + h^1(1,0) )$$

$$= f( a + h^1, b )$$

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

Also we have $$f(p) = f(a, b) = ( ab, a + b, a^2 )$$... so then ...$$\frac{ f(p + h^1e_1) - f(p) }{ h} = \frac{ ( h^1b , h^1, 2ah^1 + h^{1 \ 2} ) }{ h^1} = \begin{pmatrix}b \\ 1 \\ 2a + h^1 \end{pmatrix}$$Therefore $$D_1 f(p) = \lim_{ h^1 \to 0 } \frac{ f(p + h^1e_1) - f(p) }{ h^1} = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}$$Similarly we can determine $$D_2 f(p) = = \begin{pmatrix}a \\ 1 \\ 0 \end{pmatrix}$$
Now the Proposition shows that $$D_j f(p) = Le_j$$ so in terms of the example$$Le_1 = D_1 f(p) = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}$$

and

$$Le_2 = D_2 f(p) = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}$$Now if we take $$L = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix}$$ ( BUT! can we legitimately do this at this point ... ?)

... then ...

$$Le_1 = L \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix}$$

and

$$Le_2 = L \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} b & a \\ 1 & 1 \\ 2a & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}$$
Now we can illustrate the last line of the proof of Proposition 8.21 in terms of the example ...We have ...$$Lh = L ( h^1, h^2 ) = L( \sum_{ j = 1}^2 h^j e_j ) = L (h^1 e_1 + h^2 e_2)$$$$\Longrightarrow Lh = h^1 Le_1 + h^2 Le_2$$$$\Longrightarrow Lh = h^1 \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix} + h^2 \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}$$$$= \begin{pmatrix} bh^1 \\ h^1 \\ 2ah^1 \end{pmatrix} + \begin{pmatrix} ah^2 \\ h^2 \\ 0 \end{pmatrix}$$$$= \begin{pmatrix} bh^1 + ah^2 \\ h^1 + h^2 \\ 2ah^1 \end{pmatrix}$$... or in terms of partial derivatives ...$$\Longrightarrow Lh = \sum_{ j = 1}^2 h^j D_j f(p)$$$$= h^1 D_1 f(p) + h^2 D_2 f(p)$$$$= h^1 \begin{pmatrix} b \\ 1 \\ 2a \end{pmatrix} + h^2 \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix}$$$$= \begin{pmatrix} bh^1 + ah^2 \\ h^1 + h^2 \\ 2ah^1 \end{pmatrix}$$
Is the above example essentially correct ...

Any comments ...Peter

 

[soloenergy]

Great question, Peter. Let's break down these remarks and try to understand them better.

First, let's recall that the map f: U \to \mathbb{R}^m is differentiable at a point p \in U if there exists a linear transformation \text{df}_p : \mathbb{R}^n \to \mathbb{R}^m such that

\lim_{h\to 0} \frac{ \| f(p+h) - f(p) - \text{df}_p(h) \| }{ \| h \| } = 0

In other words, the map f is differentiable at p if it can be approximated by a linear transformation at that point.

Now, let's look at the matrix f'(p). This is a m \times n matrix, where the (i,j)th entry is given by ( f'(p) )_j^i. In other words, this entry tells us how the jth component of f varies with respect to the ith variable at the point p. We can also interpret this as the jth partial derivative of the ith component of f at p.

Now, the column vector D_jf(p) is defined as [D_jf_1(p), \ldots , D_jf_m(p)]^t. This vector represents the partial derivatives of the jth component of f at p with respect to all the variables. So, for example, D_1f(p) represents the partial derivatives of the first component of f at p with respect to all the variables.

Therefore, the matrix f'(p) can be written as [ \ D_1 f(p) \ \ D_2 f(p) \ \ldots \ D_n f(p)]. This is because each column represents the partial derivatives of a different component of f at p.

Finally, we can see that ( f'(p) )_j^i = D_jf_i(p) = \frac{ \partial f_i }{ \partial x_j } by comparing the (i,j)th entry of the matrix f'(p) with the (i,j)th entry of the vector D_jf(p). They are both representing the partial derivative of the ith component of f at p with respect to the jth variable.

I hope this helps clarify the remarks made by Browder. Let me know if you have any further questions. Happy reading