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

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 further help in fully understanding some remarks by Browder made after Definition 8.9 ...

Definition 8.9 and the following remark read as follows:




Browder - Definition 8.9  ... Differentials ....png




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 lets \(\displaystyle t \to 0\) ... 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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
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 lets \(\displaystyle t \to 0\) ... 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 $Lk-Mk = 0$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 ...

Appreciate your help ...

Peter