McInerney Example 3.1.5: Multivariable Differentiation Q&A

In summary: So f(x,y)=xy is differentiable for all a andDf(a1,a2)(h1,h2)=a1h2+a2h1In summary, In Section 3.1 of Andrew McInerney's book, "First Steps in Differential Geometry: Riemannian, Contact, Symplectic", the concept of differentiability for a function is introduced. The derivative of a function at a point is also defined. Several examples are given, including Example 3.1.5 which raises two questions. The first question asks for clarification on why a function is differentiable for all points and how the derivative is calculated. The second question asks for an explanation of the limit expression in the example.
  • #1
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I am reading Andrew McInerney's book: First Steps in Differential 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 ...

In Section 3.1 McInerney defines what is meant by a function \(\displaystyle f: \mathbb{R}^n \to \mathbb{R}^m\) being differentiable and also defines the derivative of \(\displaystyle f\) at a point \(\displaystyle a \in \mathbb{R}^n\) ...

... see the scanned text below for McInerney's definitions and notation ...

... McInerney then gives several examples ... I need help with several aspects of Example 3.1.5 which reads as follows:
View attachment 8914I have two questions with respect to Example 3.1.5 ...Question 1

In the above text from McInerney we read the following:

"... ... Then \(\displaystyle \mu\) is differentiable for all \(\displaystyle a = (a_1, a_2) \in \mathbb{R}^2\) and \(\displaystyle D \mu (a_1, a_2) (h_1, h_2) = a_1 h_2 + a_2 h_1\) ... .. Can someone explain exactly how/why ...

(a) \(\displaystyle \mu\) is differentiable for all \(\displaystyle a = (a_1, a_2) \in \mathbb{R}^2\)

and ...

(b)\(\displaystyle D \mu (a_1, a_2) (h_1, h_2) = a_1 h_2 + a_2 h_1\) ... .. that is how/why is this true ...(especially given that McInerney has only just defined differentiable and the derivative!)
Question 2

In the above text from McInerney we read the following:

"... ... \(\displaystyle \displaystyle \lim_{ \mid \mid h \mid \mid \to 0} \frac{ \mid h_1 h_2 \mid }{ \sqrt{ h_1^2 + h_2^2 } } = 0\) ... ... Can someone please show and explain exactly how/why it is that \(\displaystyle \displaystyle \lim_{ \mid \mid h \mid \mid \to 0} \frac{ \mid h_1 h_2 \mid }{ \sqrt{ h_1^2 + h_2^2 } } = 0 \)... ...
Help will be much appreciated ... ...

Peter==========================================================================

It may help members reading the above post to have access to the text at the start of Section 3.1 of McInerney ... if only to give access to McInerney's terminology and notation ... so I am providing access to the text at the start of Section 3.1 ... as follows:
View attachment 8915
Hope that helps ...

Peter
 

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  • #2
When $f(x,y)=xy$ then if $a=({a}_{1},{a}_{2})$ and $h=({h}_{1},{h}_{2})$,
$f(a+h)=f({a}_{1}+{h}_{1},{a}_{2}+{h}_{2})=({a}_{1}+{h}_{1})({a}_{2}+{h}_{2})$
$={a}_{1}{a}_{2}+{a}_{1}{h}_{2}+{a}_{2}{h}_{1}+{h}_{1}{h}_{2}$
$=f({a}_{1},{a}_{2})+{T}_{a}({h}_{1},{h}_{2})+{h}_{1}{h}_{2}$
where ${T}_{a}({h}_{1},{h}_{2})={a}_{1}{h}_{2}+{a}_{2}{h}_{1}$ is a linear transformation in h.
So $f(a+h)=f(a)+{T}_{a}(h)+{h}_{1}{h}_{2}$ you can let $ \phi(h)={h}_{1}{h}_{2}$ .
Now $\frac{\parallel f(a+h)-f(a)-{T}_{a}(h)\parallel}{\parallel h \parallel}$
$=\frac{\parallel \phi(h)\parallel}{\parallel h \parallel}=\frac{\parallel {h}_{1}{h}_{2}\parallel}{\parallel h \parallel}$.
In $R$ you have $\parallel {h}_{1}{h}_{2}\parallel=\left|{{h}_{1}{h}_{2}}\right|$ and in ${R}^{2}$
$\parallel h \parallel=\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$. And
$\left|{{h}_{1}}\right|\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$ and $\left|{{h}_{2}}\right|\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$.
So $0\le \frac{\left|{{h}_{1}{h}_{2}}\right|}{\parallel h \parallel}\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$
And $\lim_{{h}\to{0}} \frac{\left|{{h}_{1}{h}_{2}}\right|}{\parallel h \parallel}=0$
 
  • #3
MathProfessor said:
When $f(x,y)=xy$ then if $a=({a}_{1},{a}_{2})$ and $h=({h}_{1},{h}_{2})$,
$f(a+h)=f({a}_{1}+{h}_{1},{a}_{2}+{h}_{2})=({a}_{1}+{h}_{1})({a}_{2}+{h}_{2})$
$={a}_{1}{a}_{2}+{a}_{1}{h}_{2}+{a}_{2}{h}_{1}+{h}_{1}{h}_{2}$
$=f({a}_{1},{a}_{2})+{T}_{a}({h}_{1},{h}_{2})+{h}_{1}{h}_{2}$
where ${T}_{a}({h}_{1},{h}_{2})={a}_{1}{h}_{2}+{a}_{2}{h}_{1}$ is a linear transformation in h.
So $f(a+h)=f(a)+{T}_{a}(h)+{h}_{1}{h}_{2}$ you can let $ \phi(h)={h}_{1}{h}_{2}$ .
Now $\frac{\parallel f(a+h)-f(a)-{T}_{a}(h)\parallel}{\parallel h \parallel}$
$=\frac{\parallel \phi(h)\parallel}{\parallel h \parallel}=\frac{\parallel {h}_{1}{h}_{2}\parallel}{\parallel h \parallel}$.
In $R$ you have $\parallel {h}_{1}{h}_{2}\parallel=\left|{{h}_{1}{h}_{2}}\right|$ and in ${R}^{2}$
$\parallel h \parallel=\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$. And
$\left|{{h}_{1}}\right|\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$ and $\left|{{h}_{2}}\right|\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$.
So $0\le \frac{\left|{{h}_{1}{h}_{2}}\right|}{\parallel h \parallel}\le\sqrt{{h}_{1}^{2}+{h}_{2}^{2}}$
And $\lim_{{h}\to{0}} \frac{\left|{{h}_{1}{h}_{2}}\right|}{\parallel h \parallel}=0$
Thanks MathProfessor ...

Appreciate your help ... but just another question ...

How would you find/calculate the derivative of $f(x,y)=xy$ ... in the most efficient way ... indeed, what formula would you use ... ?

Thanks again,

Peter
 
  • #4
For every function which is linear with respect to x and y like f(x,y)=xy, the derivative is
Ta(h1,h2)=f(a1,h2)+f(h1,a2)
which satisfies $\lim_{h \to 0}\frac{\parallel f(x)-f(a)-T_a(h)\parallel}{\parallel h \parallel}=0$. this uses the fact that there exists some positive constant C such that $\parallel f(h_1,h_2) \parallel \le C \parallel h_1 \parallel \parallel h_2 \parallel $
If f is not linear with respect to x and y, f must have the first partial derivatives defined at point a. In this case the derivative of f at point a is given by
$T_a(h_1,h_2)=\frac {\partial f}{\partial x}(a_1,a_2)h_1+\frac {\partial f}{\partial y}(a_1,a_2)h_2$ if it satisfies the limit
$\lim_{h \to 0}\frac{\parallel f(x)-f(a)-T_a(h)\parallel}{\parallel h \parallel}=0$.
 

FAQ: McInerney Example 3.1.5: Multivariable Differentiation Q&A

1. What is multivariable differentiation?

Multivariable differentiation is a mathematical concept that involves finding the rate of change of a function with respect to multiple variables. It is an extension of single variable differentiation, which deals with finding the slope of a curve at a single point.

2. What is the purpose of McInerney Example 3.1.5?

McInerney Example 3.1.5 is an example problem that demonstrates the application of multivariable differentiation in a real-world scenario. It helps to illustrate the process of finding the partial derivatives of a function with respect to multiple variables.

3. What are partial derivatives?

Partial derivatives are the derivatives of a multivariable function with respect to one of its variables, while holding the other variables constant. They represent the rate of change of the function in a specific direction.

4. How is multivariable differentiation used in science?

Multivariable differentiation is used in many fields of science, including physics, economics, and engineering. It is used to model and analyze complex systems that involve multiple variables, such as the motion of objects, the behavior of markets, and the flow of fluids.

5. What are some common applications of multivariable differentiation?

Multivariable differentiation has many practical applications, such as optimization problems, curve fitting, and vector calculus. It is also used in fields such as computer graphics, where it is used to create 3D models and animations based on mathematical equations.

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