# Differentials in R^n .... Another Remark by Browder, Section 8.2 .... ....

• MHB
• Math Amateur
In summary, Browder is discussing differentiable maps in Chapter 8 of his book "Mathematical Analysis: An Introduction". In particular, he focuses on Section 8.2 where he defines differentials and provides some remarks in Definition 8.9. These remarks state that for any fixed vector k and a positive value t, we have an equation involving L(tk) and M(tk). In order to understand these remarks better, the reader asks two questions. First, why does Browder put h = tk and let t approach 0? The answer is that by doing this, he is able to make a distinction between fixed and variable vectors and show that the equation holds for all t approaching 0. Second, how does
Math Amateur
Gold Member
MHB
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 some remarks by Browder made after Definition 8.9 ...

Definition 8.9 and the following remark read as follows:

View attachment 9408

In the above Remark by Browder we read the following:

"for any fixed $$\displaystyle k \neq 0$$ and $$\displaystyle t \gt 0$$, we have $$\displaystyle \frac{1}{ |tk| }( L(tk) - M(tk) ) = \frac{1}{|k|},(Lk - Mk )$$ ... ... ... "
My questions are as follows:Question 1

Browder puts $$\displaystyle h = tk$$ and then let's $$\displaystyle t \to 0$$ ... why is Browder doing this ... what is the logic behind this ... what do we gain by putting $$\displaystyle h = tk$$ ... both $$\displaystyle h$$ and $$\displaystyle k \in \mathbb{R}^n$$ and also isn't $$\displaystyle h$$ just as arbitrary as $$\displaystyle k$$ ... ?
Question 2

How exactly (and in detail) does letting $$\displaystyle t \to 0$$ allow us to conclude that $$\displaystyle Lk = Mk$$ ...
Help will be much appreciated ...

Peter

#### Attachments

• Browder - Definition 8.9 ... Differentials ....png
21.4 KB · Views: 86
Peter said:
In the above Remark by Browder we read the following:

"for any fixed $$\displaystyle k \neq 0$$ and $$\displaystyle t \gt 0$$, we have $$\displaystyle \frac{1}{ |tk| }( L(tk) - M(tk) ) = \frac{1}{|k|},(Lk - Mk )$$ ... ... ... "
My questions are as follows:Question 1

Browder puts $$\displaystyle h = tk$$ and then let's $$\displaystyle t \to 0$$ ... why is Browder doing this ... what is the logic behind this ... what do we gain by putting $$\displaystyle h = tk$$ ... both $$\displaystyle h$$ and $$\displaystyle k \in \mathbb{R}^n$$ and also isn't $$\displaystyle h$$ just as arbitrary as $$\displaystyle k$$ ... ?
Question 2

How exactly (and in detail) does letting $$\displaystyle t \to 0$$ allow us to conclude that $$\displaystyle Lk = Mk$$ ...
You need to make a clear distinction between fixed and variable vectors. In the equation $$\displaystyle \lim_{h\to0}\frac1{|h|}(Lh - Mh) = 0$$, $h$ has to be a variable vector that converges to $0$. But $k$ is a fixed (nonzero) vector. if $h=tk$ then $h$ varies as $t$ varies. And as $t$ goes to $0$, $h$ also goes to $0$. So you can conclude that $$\displaystyle \lim_{t\to0}\frac1{|tk|}(L(tk) - M(tk)) = 0$$. But $$\displaystyle \frac1{|tk|}(L(tk) - M(tk)) = \frac1{|k|}(Lk-Mk)$$, which is constant. It follows that$$\displaystyle \frac1{|k|}(Lk-Mk) = 0$$, and since $|k|$ is nonzero the conclusion is that $Lk-Mk = 0$.

Opalg said:
You need to make a clear distinction between fixed and variable vectors. In the equation $$\displaystyle \lim_{h\to0}\frac1{|h|}(Lh - Mh) = 0$$, $h$ has to be a variable vector that converges to $0$. But $k$ is a fixed (nonzero) vector. if $h=tk$ then $h$ varies as $t$ varies. And as $t$ goes to $0$, $h$ also goes to $0$. So you can conclude that $$\displaystyle \lim_{t\to0}\frac1{|tk|}(L(tk) - M(tk)) = 0$$. But $$\displaystyle \frac1{|tk|}(L(tk) - M(tk)) = \frac1{|k|}(Lk-Mk)$$, which is constant. It follows that$$\displaystyle \frac1{|k|}(Lk-Mk) = 0$$, and since $|k|$ is nonzero the conclusion is that $Lk-Mk = 0$.
Thanks Opalg ...

Peter

## 1. What is the meaning of "differentials in R^n" in the context of Browder's work?

"Differentials in R^n" refers to the study of differential equations in n-dimensional space, as discussed in Section 8.2 of Browder's work. This involves analyzing how functions change over time in multiple dimensions.

## 2. How does Browder define differentials in R^n?

Browder defines differentials in R^n as the study of differential equations in n-dimensional space, which involves the use of partial derivatives and vector calculus to understand how functions change over time in multiple dimensions.

## 3. What is the significance of studying differentials in R^n?

Studying differentials in R^n is important because it allows us to understand and predict how systems in multiple dimensions will change over time. This has applications in various fields such as physics, engineering, and economics.

## 4. What is the main focus of Section 8.2 in Browder's work?

The main focus of Section 8.2 in Browder's work is on the application of differentials in R^n to solve differential equations in multiple dimensions. It covers topics such as partial derivatives, gradient vectors, and the chain rule.

## 5. Are there any real-world examples of differentials in R^n?

Yes, there are many real-world examples of differentials in R^n, such as predicting the movement of celestial bodies in space, analyzing the flow of fluids in pipes, and modeling the spread of diseases in a population. These examples all involve understanding how systems change over time in multiple dimensions.

• Topology and Analysis
Replies
7
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
2K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
1
Views
1K
• Topology and Analysis
Replies
2
Views
1K
• Topology and Analysis
Replies
3
Views
2K
• Topology and Analysis
Replies
2
Views
1K