MHB The Chain Rule for Multivariable Vector-Valued Functions .... ....

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The discussion centers on understanding Theorem 12.7 from Apostol's "Mathematical Analysis," specifically regarding the Chain Rule for multivariable vector-valued functions. Participants seek clarification on how Equation (16) derives from the earlier equation, the definition and significance of the function E(0) and E(y), and the formal reasoning behind these definitions. It is established that E(y) is defined piecewise, with E(0) equating to zero and E(y) for non-zero y incorporating terms from E_a(y) and E_b(v). The conversation emphasizes the importance of these definitions in comprehending the theorem's proof.
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I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...I am focused on Chapter 12: Multivariable Differential Calculus ... and in particular on Section 12.9: The Chain Rule ... ...I need help in order to fully understand Theorem 12.7, Section 12.9 ...Theorem 12.7 (including its proof) reads as follows:
View attachment 8523
View attachment 8524
In the proof of Theorem 12.7 we read the following:

" ... ... Using (14) in (15) we find$$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + f'(b) [ \| y \| E_a(y) ] + \|v \| E_b(v) $$$$= f'(b) [ g'(a) (y) ] + \| y \| E(y)$$ ... ... ... (16)Where $$E(0) = 0$$ and $$E(y) = f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v) \ \ \ \ \text{ if } y\neq 0$$ ... ... ... (17)... ... ... "

My questions are as follows:Question 1

Can someone show how Equation (16) follows ... that is ...

... how exactly does $$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + \| y \| E(y)$$

follow from

$$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + f'(b) [ \| y \| E_a(y) ] + \|v \| E_b(v)$$?

Question 2

What is $$E(0)$$ ... I know what $$E_a$$ and $$E_b$$ are ... but what is $$E$$?

Similarly ... what is $$E(y)$$ in (16) and in (17) ... shouldn't it be $$E_a(y)$$ ... ?

Further ... why (formally and rigorously) does $$E(0) = 0$$
Question 3

Can someone please demonstrate how/why

$$E(y) = f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v)$$
Help will be appreciated ...

Peter

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It may help MHB readers of the above post to have access to Apostol's section on the Total Derivative ... so I am providing the same ... as follows:
View attachment 8525
View attachment 8526
Hope that helps ...

Peter
 

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  • Apostol - 1 - Section 12.4 - PART 1 ... .png
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    Apostol - 2 - Section 12.4 - PART 2 ... .png
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Last edited:
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Peter said:
In the proof of Theorem 12.7 we read the following:

" ... ... Using (14) in (15) we find$$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + f'(b) [ \| y \| E_a(y) ] + \|v \| E_b(v) $$$$= f'(b) [ g'(a) (y) ] + \| y \| E(y)$$ ... ... ... (16)Where $$E(0) = 0$$ and $$E(y) = f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v) \ \ \ \ \text{ if } y\neq 0$$ ... ... ... (17)... ... ... "

My questions are as follows:Question 1

Can someone show how Equation (16) follows ... that is ...

... how exactly does $$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + \| y \| E(y)$$

follow from

$$f(b+v) - f(b) = f'(b) [ g'(a) (y) ] + f'(b) [ \| y \| E_a(y) ] + \|v \| E_b(v)$$?

Question 2

What is $$E(0)$$ ... I know what $$E_a$$ and $$E_b$$ are ... but what is $$E$$?

Similarly ... what is $$E(y)$$ in (16) and in (17) ... shouldn't it be $$E_a(y)$$ ... ?

Further ... why (formally and rigorously) does $$E(0) = 0$$
Question 3

Can someone please demonstrate how/why

$$E(y) = f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v)$$
The answer to all three of those questions is that the equations in (17) are meant to be the definition of $E(y)$. In other words, if you define \[E(y) = \begin{cases}0&\text{if }y=0,\\ f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v)&\text{if }y\ne0,\end{cases}\] then the second line of (16) follows immediately from the first line.
 
Opalg said:
The answer to all three of those questions is that the equations in (17) are meant to be the definition of $E(y)$. In other words, if you define \[E(y) = \begin{cases}0&\text{if }y=0,\\ f'(b) [ E_a(y) ] + \frac{ \| v \| }{ \| y \| } E_b (v)&\text{if }y\ne0,\end{cases}\] then the second line of (16) follows immediately from the first line.
Thanks Opalg ...

Indeed ... you are right, of course!

Should have seen that ...

Peter
 

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