MHB Multivariable Differentiation .... McInerney Definition 3.1.1

[Math Amateur]

I am reading Andrew McInerney's book: First Steps in Diofferential Geometry: Riemannian, Contact, Symplectic ... and I am focused on Chapter 3: Advanced Calculus ... and in particular on Section 3.1: The Derivative and Linear Approximation ...

I am trying to fully understand Definition 3.1.1 and need help with an example based on the definition ...

Definition 3.1.1 reads as follows:
View attachment 8913I constructed the following example ...

Let $$f: \mathbb{R} \to \mathbb{R}^2 $$

such that $$f = ( f^1, f^2 )$$

where $$f^1(x) = 2x$$ and $$f^2(x) = 3x + 1$$

We wish to determine $$T_a(h)$$ ... We have $$f(a + h) = ( f^1(a + h), f^2(a + h) )= (2a + 2h, 3a + 3h +1 )$$

and

$$f(a ) = ( f^1(a ), f^2(a ) ) = (2a , 3a +1 )$$
Now ... consider ... $\displaystyle \lim_{ \mid \mid h \mid \mid \to 0} \frac{ \mid \mid f(a + h) - f(a) - T_a(h) \mid \mid }{ \mid \mid h \mid \mid } $$$\Longrightarrow \displaystyle \lim_{ \mid \mid h \mid \mid \to 0} \frac{ \mid \mid (2a + 2h, 3a + 3h +1) - (2a, 3a + 1) - T_a(h) \mid \mid }{ \mid \mid h \mid \mid }$$$$\Longrightarrow \displaystyle \lim_{ \mid \mid h \mid \mid \to 0} \frac{ \mid \mid ( 2h, 3h ) - T_a(h) \mid \mid }{ \mid \mid h \mid \mid }$$... ... but how do I proceed from here ... ?

... can I take $$T_a (h) = T_a.h$$ ... but how do I justify this?Hope someone can help ...

Peter

 

[MathProfessor]

f(x)=(2x,3x+1) so f(a+h)=(2a+2h,3a+1+3h)=(2a,3a+1)+(2h,3h)=f(a)+T_a(h). Where T_a is a linear transformation which satisfies the limit zero.
If you consider the example f(x)=(x²,3x+1) then f(a+h)=(a²+2ah+h²,3a+1+3h)
so f(a+h)=(a²,3a+1)+(2ah,3h)+(h²,0)
f(a+h)=f(a)+T_a(h)+S(h). Where T_a(h)=(2ah,3h) is a linear transformation and
$\lim_{h \to 0} \frac{\parallel f(a+h)-f(a)-T_a(h) \parallel}{\parallel h \parallel}=\lim_{h \to 0} \frac{\parallel S(h) \parallel}{\parallel h \parallel}=0$

 

[HOI]

I am confused by the notation [math]T_a(h)[/math]. From what was said before, we have a "linear transformation $T_a$ from R^n\to R^n". There is no dependence of "h" Was that just T_a times h?

. If "T_a(h)" is simply T_a times h, we can factor out ||h||: \frac{||(2h, 3h)- T_ah||}{||h||}= \frac{||h||||(2, 3)- T_a||}{||h||}= (2, 3)- T_a= 0 so T_a= (2, 3).