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

[Math Amateur]

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:

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)... ... ... "
*** EDIT ***

It now occurs to me that in fact Apostol is defining E(y) in equations (16) and (17)

I should have seen this earlier ...

Peter

================================================================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 $E$ ... 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

=========================================================================================

It may help Physics Forum readers of the above post to have access to Apostol's section on the Total Derivative ... so I am providing the same ... as follows:

Hope that helps ...

Peter

 

[fresh_42]

*** EDIT ***

It now occurs to me that in fact Apostol is defining E(y) in equations (16) and (17)

I should have seen this earlier ...

Peter

Which of the following questions does this answer already?

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

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

I guess this is the question you answered yourself: $E(y)$ is defined in (17) such that this follows.

Question 2

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

The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ with a remainder function $r()$ that has to obey $\lim_{v \to 0} \dfrac{r(v)}{||v||} = 0$. $E()$ is the remainder we get from our formulas. Now we have to show, that the limit condition is obeyed.

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

No. (17) is what defines $E(y)$, and it depends on both evaluation points. But since $b=g(a)$, in the end it depends on $a$ alone. But we do not write $E_a(y)$ because we have to distinguish our new remainder function $E()$ from the earlier one $E_a(y)$. They are two different functions, see (17).

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

We consider the general formula $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ again. Now if $v=0$, we have $r(v)=0$. Applied to the differentiations of $f(x)$ and $g(x)$ in (14) and (15), this means $r(0)=E_a(0)=0$ resp. $r(0)=E_b(0)=0$. So by (17) we have $E(0)= \text{ sth. I }\cdot 0 + \text{ sth. II }\cdot 0$. Now if $\text{ sth. II }$ is bounded as $y \to 0$, we have a continuous function $E()$ if we set $E(0)=0$. Remember, that we defined $E()$ such that it fits our needs. So if the coefficient in (17) at $E_b(0)$ will remain finite if we set $y=0$, then $E(0)=0$ is a continuous expansion of $E()$ at $y=0$.

Question 3

Can someone please demonstrate how/why

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

We set it so. $E(y) := \ldots$

 

[Math Amateur]

THanks fresh_42 ...

Appreciate your help ...

Peter

 

[Math Amateur]

Which of the following questions does this answer already?

I guess this is the question you answered yourself: $E(y)$ is defined in (17) such that this follows.

The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ with a remainder function $r()$ that has to obey $\lim_{v \to 0} \dfrac{r(v)}{||v||} = 0$. $E()$ is the remainder we get from our formulas. Now we have to show, that the limit condition is obeyed.

No. (17) is what defines $E(y)$, and it depends on both evaluation points. But since $b=g(a)$, in the end it depends on $a$ alone. But we do not write $E_a(y)$ because we have to distinguish our new remainder function $E()$ from the earlier one $E_a(y)$. They are two different functions, see (17).

We consider the general formula $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ again. Now if $v=0$, we have $r(v)=0$. Applied to the differentiations of $f(x)$ and $g(x)$ in (14) and (15), this means $r(0)=E_a(0)=0$ resp. $r(0)=E_b(0)=0$. So by (17) we have $E(0)= \text{ sth. I }\cdot 0 + \text{ sth. II }\cdot 0$. Now if $\text{ sth. II }$ is bounded as $y \to 0$, we have a continuous function $E()$ if we set $E(0)=0$. Remember, that we defined $E()$ such that it fits our needs. So if the coefficient in (17) at $E_b(0)$ will remain finite if we set $y=0$, then $E(0)=0$ is a continuous expansion of $E()$ at $y=0$.

We set it so. $E(y) := \ldots$

Hi fresh_42 ...

Thanks again for the help ...

Just a clarification ...

You write ...

" ... ... The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ ... ..."

Changing $a$ to $c$ to conform with Apostol's notation we have $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ ... ... ... (1)

Apostol's definition of differentiation (Equation (5) Section 12.4, page 346) reads as follows:

$f(c+v) = f(c) + f\, '(c)(v) + \| v \| E_c(v)$ ... ... ... (2)

Now (2) $\Longrightarrow f\, '(c)(v) = f(c+v) - f(c) - \| v \| E_c(v)$ ... ... ... (3)

So comparing (1) and (3) we find they will be equivalent if

$f\,'(a) \cdot v = f\, '(c)(v)$ ... ... ... (4)

and

$r(v) = - \| v \| E_c(v)$ ... ... ... (5)Can you please explain why (4) and (5) hold true ... ?... and just another minor issue ...


You define r in the following sentence ...

" ... ... The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ with a remainder function $r()$ that has to obey $\lim_{v \to 0} \dfrac{r(v)}{||v||} = 0$... "

and then later you assert "Now if $v=0$, we have $r(v)=0$." ... ...

Can you explain why $r(0) = 0$ ...
Thanks again for your help ...

Peter

 

[fresh_42]

Hi fresh_42 ...

Thanks again for the help ...

Just a clarification ...

You write ...

" ... ... The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ ... ..."

Changing $a$ to $c$ to conform with Apostol's notation we have $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ ... ... ... (1)

Apostol's definition of differentiation (Equation (5) Section 12.4, page 346) reads as follows:

$f(c+v) = f(c) + f\, '(c)(v) + \| v \| E_c(v)$ ... ... ... (2)

Now (2) $\Longrightarrow f\, '(c)(v) = f(c+v) - f(c) - \| v \| E_c(v)$ ... ... ... (3)

So comparing (1) and (3) we find they will be equivalent if

$f\,'(a) \cdot v = f\, '(c)(v)$ ... ... ... (4)

and

$r(v) = - \| v \| E_c(v)$ ... ... ... (5)Can you please explain why (4) and (5) hold true ... ?

(4) is just a different notation. $f\,'(c)$ for functions $f \, : \,\mathbb{R}^n \longrightarrow \mathbb{R}^m$ is the Jacobi matrix evaluated at point $x=c$. And $v$ is the direction, in which we consider the slope of the function, so $f\,'(c)(v)= \text{ matrix times vector } = f\,'(c)\cdot v\,.$

(5) is also a notational difference. I abbreviated the remainder function by $r()$ and Apostol by $E()$ - for error function I guess. It has to run faster against zero, than the vector $v$, i.e. $\lim_{v\to 0}\dfrac{r(v)}{||v||}=0\,.$ I only incorporated the $||v||$ term into the function $r()$ whereas Apostol operates with $E(v)=\dfrac{r(v)}{||v||}$, i.e. we have $\lim_{v\to 0}\dfrac{r(v)}{||v||}=\lim_{v \to 0}E(v)=0$ as condition. It's just how you write the "error".

... and just another minor issue ...


You define r in the following sentence ...

" ... ... The general formula for differentiation in direction $v$ goes: $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ with a remainder function $r()$ that has to obey $\lim_{v \to 0} \dfrac{r(v)}{||v||} = 0$... "

and then later you assert "Now if $v=0$, we have $r(v)=0$." ... ...

Can you explain why $r(0) = 0$ ...

If we take $f\,'(a) \cdot v = f(a+v)-f(a)+r(v)$ and calculate with $v=0$ we get $0=f\,'(a) \cdot 0 = f(a+0)-f(a)+r(0)=f(a)-f(a)+r(0)=r(0)$

Thanks again for your help ...

Peter

 

[Math Amateur]

Thanks fresh_42 ...

Most helpful ...

Peter