04 Vector Analysis Intro

Building New Functions from Old

Just as we can combine numbers with addition and multiplication, we can combine functions to create new, more complex ones. Understanding these constructions is fundamental to analyzing functions in higher dimensions.

1. Cartesian Product of Functions

This is the most fundamental way we build vector-valued functions. If we have a set of scalar-valued functions that all share the same domain, we can bundle them together.

Definition: Cartesian Product of Functions

Let $f_1,f_2,\dots ,f_m$ be functions, each mapping from ${\mathbb{R}}^n$ to $\mathbb{R}$. The Cartesian product of these functions is a new, vector-valued function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ defined as: $f(\vec{x})=(f_1(\vec{x}),f_2(\vec{x}),\dots ,f_m(\vec{x}))$ Each $f_i$ is called a component function of $f$.

This idea is powerful because it tells us that to understand a vector-valued function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$, we can often just study its $m$ scalar-valued component functions $f_i:{\mathbb{R}}^n\to \mathbb{R}$ individually.

2. Composition of Functions

Composition is a familiar idea from single-variable calculus, and it extends naturally to higher dimensions. We can “pipe” the output of one function into the input of another, provided the dimensions match up.

Definition: Composition

Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ and $g:{\mathbb{R}}^m\to {\mathbb{R}}^s$. The composition $g\circ f$ is a function from ${\mathbb{R}}^n$ to ${\mathbb{R}}^s$ defined by: $(g\circ f)(\vec{x})=g(f(\vec{x}))$ The output of $f$ (a vector in ${\mathbb{R}}^m$) becomes the input for $g$.

3. Functions with Separated Variables

This is a special, yet common, way to construct a multivariable function from single-variable building blocks.

Definition: Separated Variables

Let $f_1,f_2,\dots ,f_n$ be functions of a single real variable ($f_i:\mathbb{R}\to \mathbb{R}$). A function $f:{\mathbb{R}}^n\to \mathbb{R}$ is said to have separated variables if it is formed by taking the product of these functions, each applied to a different coordinate: $f(x_1,x_2,\dots ,x_n)=f_1(x_1)\cdot f_2(x_2)\cdots f_n(x_n)$

Monomials

The simplest and most important example of functions with separated variables are monomials. A monomial in ${\mathbb{R}}^n$ has the form: $M(x_1,\dots ,x_n)=x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\cdots x_n^{{\alpha }_n}$ Here, each $f_i(x_i)$ is just the power function $x_i^{{\alpha }_i}$. A polynomial is simply a sum of such monomials.

Ways of Visualizing Functions

Our single greatest challenge in multivariable calculus is that we cannot easily “see” the objects we are studying. A function $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ requires an $(n+m)$-dimensional space to graph, which is impossible for us beyond very simple cases. We must therefore rely on different kinds of visualizations to build our intuition.

1. The Graph of a Function ($f:{\mathbb{R}}^2\to \mathbb{R}$)

This is the most direct generalization from single-variable calculus. For a function $f(x,y)$, the graph is the set of all points $(x,y,z)$ in 3D space where $z=f(x,y)$. This creates a surface floating over the $xy$-plane.

Definition: Graph

The graph of a function $f:{\mathbb{R}}^2\to \mathbb{R}$ is the set: $\text{graph}(f):=\{(x,y,f(x,y))\in {\mathbb{R}}^3\}$

Even for simple functions, the resulting surfaces can be complex and beautiful.

• Example 1: $f(x,y)=x^2+y^2$. This function’s graph is a smooth, bowl-shaped surface called a paraboloid.
• Example 2: $f(x,y)=\sin (x^2+y^2)$. This creates a chaotic surface with concentric ripples, like waves in a pond.

2. The Parametric Plot ($f:\mathbb{R}\to {\mathbb{R}}^2$ or ${\mathbb{R}}^3$)

When the domain is one-dimensional but the codomain is higher-dimensional, we can visualize the function’s image as a curve traced out in space. Think of the input variable $t$ as “time.” As $t$ evolves, the point $f(t)$ moves through space, drawing a path.

• Example 1: A Circle ($f:\mathbb{R}\to {\mathbb{R}}^2$) The function $f(t)=(\cos t,\sin t)$ traces out the unit circle in the plane. At $t=0$, the point is $(1,0)$. At $t=\pi /2$, it’s at $(0,1)$, and so on.
• Example 2: A Helix ($f:\mathbb{R}\to {\mathbb{R}}^3$) The function $g(t)=(5\cos t,5\sin t,t)$ traces a helix in 3D space. The $(x,y)$ components move in a circle, while the $z$ component steadily increases, lifting the curve upwards.

3. The Vector Plot (Vector Fields, $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$)

For a vector field, where the domain and codomain have the same dimension, we can’t easily graph the function. Instead, we visualize it by drawing the output vector $f(\vec{x})$ with its tail placed at the input point $\vec{x}$. This gives us a “flow” or “field” diagram, showing the direction and magnitude of the function’s output at various points in space.

• Example 1: A Rotational Field ($f:{\mathbb{R}}^2\to {\mathbb{R}}^2$) The function $f(x,y)=(-y,x)$ creates a vector field that rotates counter-clockwise around the origin.
• Example 2: A 3D Field ($f:{\mathbb{R}}^3\to {\mathbb{R}}^3$) The function $F(x,y,z)=(x,-y,z+1)$ creates a more complex field in 3D space.

Continuity in ${\mathbb{R}}^n$

With a feel for multivariable functions, we can now ask the first fundamental question of analysis: what does it mean for such a function to be continuous? The core idea is the same as in one dimension: “small changes in input lead to small changes in output.”

Recalling Continuity in $\mathbb{R}$

A function $f:\mathbb{R}\to \mathbb{R}$ is continuous at a point $x_0$ if:

1. The $ϵ-\delta$ Definition: For every $ϵ>0$, there exists a $\delta >0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<ϵ$.
2. The Limit Definition: The limit of the function exists at $x_0$ and is equal to the function’s value: ${\lim }_{x\to x_0}f(x)=f(x_0)$.
3. The Sequential Definition: For every sequence $(x_n)$ that converges to $x_0$, the sequence of function values $(f(x_n))$ converges to $f(x_0)$. This can be elegantly written as: ${\lim }_{n\to \infty }f(x_n)=f\left({\lim }_{n\to \infty }{x}_n\right)$ Continuity means we can swap the function and the limit.

Extending to ${\mathbb{R}}^n$

To generalize these ideas, we simply replace the absolute value $|\cdot |$ with the Euclidean norm $\Vert \cdot \Vert$. First, let’s define convergence for sequences of vectors.

Definition: Convergence of a Sequence in ${\mathbb{R}}^n$

A sequence of vectors $({\vec{x}}_k)$ in ${\mathbb{R}}^n$ converges to a vector $\vec{y}\in {\mathbb{R}}^n$, written ${\vec{x}}_k\to \vec{y}$, if for every $ϵ>0$, there exists an integer $N\ge 1$ such that for all $k\ge N$, we have: $\Vert {\vec{x}}_k-\vec{y}\Vert <ϵ$ This means that eventually, all points of the sequence lie within an $ϵ$-ball centered at $\vec{y}$.

Equivalent Conditions for Convergence

The following are equivalent for a sequence $({\vec{x}}_k)$ in ${\mathbb{R}}^n$ and a point $\vec{y}$:

1. ${\lim }_{k\to \infty }{\vec{x}}_k=\vec{y}$.
2. Component-wise Convergence: For each component $i\in \{1,\dots ,n\}$, the sequence of real numbers $(x_{k,i})$ converges to $y_i$.
3. The sequence of real numbers $\Vert {\vec{x}}_k-\vec{y}\Vert$ converges to 0.

Now we can define limits and continuity for multivariable functions.

Definition: Limit of a Function in ${\mathbb{R}}^n$

Let $f:X\subset {\mathbb{R}}^n\to {\mathbb{R}}^m$. We say that $f$ has a limit $\vec{y}$ as $\vec{x}$ approaches ${\vec{x}}_0$, written ${\lim }_{\vec{x}\to {\vec{x}}_0}f(\vec{x})=\vec{y}$, if for every $ϵ>0$, there exists a $\delta >0$ such that for all $\vec{x}\in X$ with $0<\Vert \vec{x}-{\vec{x}}_0\Vert <\delta$, we have: $\Vert f(\vec{x})-\vec{y}\Vert <ϵ$

Definition: Continuity of a Function in ${\mathbb{R}}^n$

A function $f:X\subset {\mathbb{R}}^n\to {\mathbb{R}}^m$ is continuous at a point ${\vec{x}}_0\in X$ if: ${\lim }_{\vec{x}\to {\vec{x}}_0}f(\vec{x})=f({\vec{x}}_0)$ The sequential definition also holds: $f$ is continuous at ${\vec{x}}_0$ if and only if for every sequence $({\vec{x}}_k)\to {\vec{x}}_0$, we have $f({\vec{x}}_k)\to f({\vec{x}}_0)$.

Building Continuous Functions

Thankfully, most of the functions we build from familiar, continuous pieces are also continuous.

Properties of Continuous Functions

1. Cartesian Products: The Cartesian product of continuous functions is continuous. This implies a vector-valued function $f=(f_1,\dots ,f_m)$ is continuous if and only if all its component functions $f_i$ are continuous.
2. Linear Maps: Any linear map $f(\vec{x})=A\vec{x}$ is continuous everywhere.
3. Sums and Products: Sums and products of finitely many continuous functions are continuous.
4. Separated Variables: A function with separated variables, $f(x_1,\dots ,x_n)=f_1(x_1)\cdots f_n(x_n)$, is continuous if each factor $f_i$ is continuous.
5. Polynomials: As a consequence of (3) and (4), all polynomials in several variables are continuous everywhere.
6. Compositions: The composition of continuous functions is continuous.

The Perils of Multidimensional Limits

Here we arrive at one of the most significant, and subtle, departures from single-variable calculus. The very concept of a “limit” becomes richer and more demanding in higher dimensions.

The Core Challenge: Infinite Paths of Approach

In one dimension, when we consider the limit of $f(x)$ as $x\to x_0$, there are only two directions of approach: from the left ($x\to x_0^{-}$) and from the right ($x\to x_0^{+}$). The limit exists if and only if both one-sided limits exist and are equal.

In ${\mathbb{R}}^n$ for $n\ge 2$, the situation is dramatically different. To approach a point ${\vec{x}}_0$, we are no longer confined to a single line. There are infinitely many paths we can take. We can approach along straight lines from any angle, along parabolic curves, spiraling in, and so on.

For a limit to exist in this context, the function must approach the exact same value along every single possible path. This is a much stronger condition.

This leads to a powerful method for proving that a limit does not exist.

The Path-Dependence Test for Discontinuity

If you can find two different paths of approach to a point ${\vec{x}}_0$ along which the function $f(\vec{x})$ has different limits, then the overall limit does not exist. Consequently, the function cannot be continuous at ${\vec{x}}_0$.

Let’s see this principle in action with a classic, illustrative example.

A Classic Discontinuous Function

Consider the function $f:{\mathbb{R}}^2\to \mathbb{R}$ defined by: $f(x,y)=\{\begin{array}{ll}\frac{xy}{x^2+y^2}& \text{if }(x,y)\ne (0,0)\\ 0& \text{if }(x,y)=(0,0)\end{array}$ Is this function continuous at the origin $(0,0)$? For it to be continuous, the limit as $(x,y)\to (0,0)$ must exist and be equal to $f(0,0)=0$. Let’s test this by approaching the origin along different paths.

• Path 1: Along the x-axis (where $y=0$) We approach the origin using points of the form $(x,0)$ as $x\to 0$. ${\lim }_{(x,0)\to (0,0)}f(x,0)={\lim }_{x\to 0}\frac{x\cdot 0}{x^2+0^2}={\lim }_{x\to 0}\frac{0}{x^2}=0$

• Path 2: Along the y-axis (where $x=0$) We approach using points of the form $(0,y)$ as $y\to 0$. ${\lim }_{(0,y)\to (0,0)}f(0,y)={\lim }_{y\to 0}\frac{0\cdot y}{0^2+y^2}={\lim }_{y\to 0}\frac{0}{y^2}=0$

So far, the limit seems to be 0. But we have only checked two paths out of infinitely many. Let’s try a more general path.

• Path 3: Along a straight line $y=mx$ This path represents approaching the origin along a line with slope $m$. ${\lim }_{(x,mx)\to (0,0)}f(x,mx)={\lim }_{x\to 0}\frac{x(mx)}{x^2+(mx{)}^2}={\lim }_{x\to 0}\frac{m{x}^2}{x^2(1+m^2)}={\lim }_{x\to 0}\frac{m}{1+m^2}$ The $x^2$ terms cancel, and we are left with a limit that depends entirely on the slope $m$: ${\lim }_{(x,mx)\to (0,0)}f(x,mx)=\frac{m}{1+m^2}$

The result is path-dependent!

• Along the line $y=x$ (where $m=1$), the limit is $\frac{1}{1+1^2}=\frac{1}{2}$.
• Along the line $y=2x$ (where $m=2$), the limit is $\frac{2}{1+2^2}=\frac{2}{5}$.
• Along the x-axis ($y=0x$, where $m=0$), the limit is $0$.

Since we found different limits along different paths, we can definitively conclude that the limit of $f(x,y)$ as $(x,y)\to (0,0)$ does not exist. Therefore, the function is not continuous at the origin. The 3D plot of this function vividly shows the problem: a tear or rip at the origin.

A More Systematic Tool: Polar Coordinates

Checking paths one by one can be tedious. For limits at the origin in ${\mathbb{R}}^2$, switching to polar coordinates provides a more powerful and comprehensive approach.

The transformation is $x=r\cos \theta$ and $y=r\sin \theta$. The key insight is: $(x,y)\to (0,0)⟺r\to 0$ The limit exists if and only if the value the function approaches as $r\to 0$ is the same for all angles $\theta$. If the resulting limit depends on $\theta$, it means the value depends on the path of approach, and the limit does not exist.

The Previous Example in Polar Coordinates

For $f(x,y)=\frac{xy}{x^2+y^2}$: $f(r\cos \theta ,r\sin \theta )=\frac{(r\cos \theta )(r\sin \theta )}{r^2{\cos }^2\theta +r^2{\sin }^2\theta }=\frac{r^2\cos \theta \sin \theta }{r^2({\cos }^2\theta +{\sin }^2\theta )}=\cos \theta \sin \theta$ The limit as $(x,y)\to (0,0)$ is equivalent to the limit as $r\to 0$: ${\lim }_{r\to 0}(\cos \theta \sin \theta )=\cos \theta \sin \theta$ The result is entirely dependent on $\theta$. This confirms that the limit does not exist.

Now, let’s consider a case where the limit does exist.

A Continuous Function

Consider $f(x,y)=\{\begin{array}{ll}\frac{x^{2}y}{x^2+y^2}& (x,y)\ne (0,0)\\ 0& (x,y)=(0,0)\end{array}$. Let’s switch to polar coordinates: $f(r\cos \theta ,r\sin \theta )=\frac{(r\cos \theta {)}^2(r\sin \theta )}{r^2}=\frac{r^3{\cos }^2\theta \sin \theta }{r^2}=r{\cos }^2\theta \sin \theta$ Now, we take the limit as $r\to 0$: ${\lim }_{r\to 0}r{\cos }^2\theta \sin \theta$ The term ${\cos }^2\theta \sin \theta$ is always bounded between -1 and 1, regardless of the angle $\theta$. We have a classic “zero times bounded” situation. $0\le |r{\cos }^2\theta \sin \theta |\le |r|\cdot 1=r$ As $r\to 0$, the function is squeezed to 0. The limit is 0, independent of $\theta$. Since the limit exists and is equal to $f(0,0)=0$, the function is continuous at the origin.

The Sandwich Lemma (Squeeze Theorem)

Proving a limit exists can be tricky. The Sandwich Lemma, or Squeeze Theorem, is an essential tool for this, just as it was in single-variable calculus.

The Sandwich Lemma

Let $f,g,h$ be functions from ${\mathbb{R}}^n\to \mathbb{R}$. Suppose that for all $\vec{x}$ in a neighborhood of $\vec{a}$ (except possibly at $\vec{a}$ itself), we have: $f(\vec{x})\le g(\vec{x})\le h(\vec{x})$ If ${\lim }_{\vec{x}\to \vec{a}}f(\vec{x})={\lim }_{\vec{x}\to \vec{a}}h(\vec{x})=L$, then the limit of $g$ also exists and is equal to $L$: ${\lim }_{\vec{x}\to \vec{a}}g(\vec{x})=L$

Using the Sandwich Lemma

Let’s re-examine $f(x,y)=\frac{x^{2}y}{x^2+y^2}$. We want to show ${\lim }_{(x,y)\to (0,0)}f(x,y)=0$. We need to bound the absolute value of the function. Notice that $x^2\le x^2+y^2$, which means $\frac{x^2}{x^2+y^2}\le 1$. $|f(x,y)|=\left|\frac{x^2}{x^2+y^2}\cdot y\right|=\left|\frac{x^2}{x^2+y^2}\right||y|\le 1\cdot |y|=|y|$ So we have established the inequality: $0\le |f(x,y)|\le |y|$ Now we can apply the Sandwich Lemma.

• Let our lower bound function be $h_1(x,y)=0$. Clearly, ${\lim }_{(x,y)\to (0,0)}{h}_1(x,y)=0$.
• Let our upper bound function be $h_2(x,y)=|y|$. Clearly, ${\lim }_{(x,y)\to (0,0)}{h}_2(x,y)=0$.

Since $|f(x,y)|$ is squeezed between two functions that both go to 0, we must have ${\lim }_{(x,y)\to (0,0)}|f(x,y)|=0$, which implies ${\lim }_{(x,y)\to (0,0)}f(x,y)=0$.

Another Difficulty: The Domain of Definition

In higher dimensions, the set of points where a function is not defined (its set of discontinuities) can be much more complex than in one dimension. In $\mathbb{R}$, this set is typically a collection of isolated points. In ${\mathbb{R}}^n$, it can be points, curves, surfaces, or entire regions.

Domains of Common Functions

1. $f(x,y)=\cos (x^{2}y)$. Since the cosine function and polynomial functions are continuous everywhere, their composition is also continuous. This function is defined and continuous on all of ${\mathbb{R}}^2$.

2. $f(x,y)=\ln (x^2+y^2)$. The natural logarithm $\ln (u)$ is only defined for $u>0$. The expression $x^2+y^2$ is positive everywhere except at the origin, where it is 0. Therefore, this function is defined and continuous on ${\mathbb{R}}^2\setminus \{(0,0)\}$. The set of discontinuity is a single point.

3. $f(x,y)=\ln (\cos (x^2+y^2))$. This is more complex. We need the argument of the logarithm to be positive: $\cos (x^2+y^2)>0$. The cosine function is positive when its argument is in intervals like $(-\pi /2,\pi /2)$, $(3\pi /2,5\pi /2)$, etc. This means the function is not defined when: $x^2+y^2=\frac{(2k+1)\pi }{2}\text{for any integer }k\ge 0$ The set of discontinuities is an infinite collection of concentric circles centered at the origin.