Representing Scalar Fields with Tables and Formulas

Introduction

A scalar field assigns a number to each point in its domain. A table can record some of those point-value assignments, while a formula can specify how to obtain a value at every point in a stated domain. These are two ways to represent scalar field values, but they do not supply the same amount of information.

A table tells us what has been recorded at its listed coordinates. If a point is absent, the table alone does not tell us the field’s value there. A formula may let us evaluate the field at that point, provided the point is in the formula’s domain. Keeping these distinctions clear is important when reading data and when interpreting a mathematical model.

Scalar Fields, Tables, and Formulas

A scalar field on a set \(D\) assigns one real number to each point in \(D\). In this tutorial, points will usually be in the plane or in three-dimensional space. A table represents a finite list of assignments; a formula specifies a rule for assignments on a domain.

Definition: A scalar field on \(D\) is a function \(f:D\to\mathbb{R}\). A coordinate-value table for \(f\) lists selected points \(\mathbf{p}\in D\) together with their corresponding values \(f(\mathbf{p})\).

For a field on the plane, a table row might list coordinates \(x\) and \(y\), followed by the value \(f(x,y)\). The two coordinates together specify one input point. The output is one number, not a vector. For a field on three-dimensional space, a row lists three coordinates and one scalar value.

A formula-based field might be written \(f(x,y)=x^2+2y\). To use it, substitute the coordinates of the point into the formula, following the usual order of operations. Before doing so, check that the point belongs to the domain on which the formula is defined.

Reading a Coordinate-Value Table

Each table entry reports the field value at the coordinates in that row. The table does not automatically report values between rows, at nearby coordinates, or outside the range of coordinates displayed. Even if the listed values appear to follow a pattern, using that pattern at an unlisted point requires an additional assumption or a specified formula.

Worked Example: Reading a Listed Value and Identifying an Unlisted One

Suppose a table records the scalar field \(T\) at selected points:

\(x\)	\(y\)	\(T(x,y)\)
0	0	12
1	0	15
0	2	10

At the point \((1,0)\), the row lists \(T(1,0)=15\). The point \((0,1)\) does not appear in the table. Therefore, the table supplies no value for \(T(0,1)\). In particular, it does not justify setting that value equal to \(12\), \(15\), or \(10\).

Check: The coordinates \((1,0)\) match the second row, whose output is \(15\). The coordinates \((0,1)\) match no row, so no table entry can be read for that point.

When table entries come from measurements, a recorded number may also reflect the precision of the measuring device or a rounding convention. The table still tells us what was recorded; it need not, by itself, tell us the exact underlying value or how the field behaves between observations.

Evaluating a Formula at Listed Points

If a formula is provided, it gives a way to calculate field values at points in its domain, including points not listed in a table. Keep the coordinates in the correct order, substitute every coordinate, and calculate the resulting scalar. If the expression contains a restriction, such as a denominator that must be nonzero, apply that restriction before evaluating.

Worked Example: Evaluating a Formula at Two Points

Let \(f(x,y)=3x-2y^2\), with domain \(\mathbb{R}^2\). Evaluate \(f(2,1)\) and \(f(-1,2)\).

For \((2,1)\), substitute \(x=2\) and \(y=1\):

\begin{aligned} f(2,1) &=3(2)-2(1)^2\\ &=6-2\\ &=4. \end{aligned}

For \((-1,2)\), substitute \(x=-1\) and \(y=2\):

\begin{aligned} f(-1,2) &=3(-1)-2(2)^2\\ &=-3-2(4)\\ &=-3-8\\ &=-11. \end{aligned}

Check: Substituting the first point into the original expression gives \(6-2=4\), and substituting the second gives \(-3-8=-11\). The formula is defined for all real \(x\) and \(y\), so both evaluations are permitted.

A table and formula can be used together. If a table lists a point and the formula is said to describe the same field, the formula’s value at that point should agree with the listed value, subject to any stated measurement or rounding precision. If a table is described only as measurements and no exact formula is supplied, do not assume that a simple formula fits the entries exactly.

What a Finite Table Does Not Determine

Knowing a field at finitely many points does not, on its own, determine a unique formula for the field. In fact, for any finite table with distinct planar sample points and real recorded values, there is a polynomial formula that matches all the entries. There are also other formulas that match those same entries but differ elsewhere.

Theorem: Let \(p_i=(a_i,b_i)\), for \(i=1,\ldots,n\), be distinct points in \(\mathbb{R}^2\), and let \(c_i\in\mathbb{R}\) be arbitrary recorded values. There is a polynomial \(f\) satisfying \(f(p_i)=c_i\) for every \(i\). Moreover, there is another polynomial agreeing with \(f\) at every sample point and differing from \(f\) at every point outside the sample set.

The result has two parts. First, it establishes that every such finite table is compatible with at least one polynomial formula. Second, it shows that fitting the table does not determine the value away from the samples. The construction and its verification are given below.

Build a factor for each sample point.
For each \(i\), define \(L_i(x,y)\) by multiplying squared distances to all the other sample points and dividing by the corresponding squared distances from \(p_i\). Distinctness ensures that every denominator is positive.

Combine the factors with the recorded values.
Set \(f(x,y)=\sum_{i=1}^{n}c_iL_i(x,y)\). At sample \(p_k\), the factor \(L_k\) equals \(1\), and every \(L_i\) with \(i\ne k\) equals \(0\). Thus \(f(p_k)=c_k\).

Make a second formula with the same samples.
Add the product of squared distances to all sample points. That product is zero at every sample and strictly positive at every point outside the sample set, so the new polynomial agrees at the samples and differs elsewhere.

For completeness, the factors and the verification are:

L_i(x,y)= \prod_{\substack{1\le j\le n\\j\ne i}} \frac{(x-a_j)^2+(y-b_j)^2} {(a_i-a_j)^2+(b_i-b_j)^2}, \qquad f(x,y)=\sum_{i=1}^{n}c_iL_i(x,y).

If \(j\ne i\), distinctness of \(p_i\) and \(p_j\) gives \((a_i-a_j)^2+(b_i-b_j)^2>0\). Thus the denominators are nonzero. At \(p_k\), each factor in \(L_k\) has identical numerator and denominator, so \(L_k(p_k)=1\). If \(i\ne k\), the product defining \(L_i(p_k)\) includes the factor indexed by \(j=k\); its numerator is \((a_k-a_k)^2+(b_k-b_k)^2=0\). Therefore \(L_i(p_k)=0\), and the sum gives \(f(p_k)=c_k\). For \(n=1\), the empty product is defined to be \(1\), and the same conclusion holds.

Now define \(H(x,y)=\prod_{j=1}^{n}((x-a_j)^2+(y-b_j)^2)\) and \(g=f+H\). At each \(p_k\), one factor in \(H(p_k)\) is zero, so \(g(p_k)=f(p_k)=c_k\). At a point outside the sample set, every squared distance in the product is positive. Hence \(H(x,y)>0\) there, so \(g(x,y)\ne f(x,y)\). This proves both claims.

Worked Example: Two Formulas Fitting the Same Table

Consider the two recorded values \(f(0,0)=2\) and \(f(1,0)=5\). We will construct two polynomials agreeing with both entries but not at \((0,1)\).

Using the construction above, the factors for \(p_1=(0,0)\) and \(p_2=(1,0)\) are

L_1(x,y)=(x-1)^2+y^2, \qquad L_2(x,y)=x^2+y^2.

The denominators in both factors equal \(1\), because the squared distance between the two sample points is \((1-0)^2+(0-0)^2=1\). Thus one fitting polynomial is

\begin{aligned} f(x,y) &=2L_1(x,y)+5L_2(x,y)\\ &=2\big((x-1)^2+y^2\big)+5(x^2+y^2). \end{aligned}

Check the first entry: \(L_1(0,0)=1\) and \(L_2(0,0)=0\), so \(f(0,0)=2(1)+5(0)=2\). Check the second: \(L_1(1,0)=0\) and \(L_2(1,0)=1\), so \(f(1,0)=2(0)+5(1)=5\).

For a second polynomial, add \(H(x,y)=(x^2+y^2)((x-1)^2+y^2)\). This product vanishes at both sample points. At the unlisted point \((0,1)\), the first formula gives \(f(0,1)=2(1+1)+5(1)=4+5=9\). The added term is \(H(0,1)=(0+1)(1+1)=2\), so \(g(0,1)=f(0,1)+H(0,1)=9+2=11\).

Check: At \((0,0)\), \(H=0\), so \(g=2\); at \((1,0)\), \(H=0\), so \(g=5\). The two formulas both match the table, but at \((0,1)\) they give \(9\) and \(11\). Therefore the table does not specify the value at that unlisted point.

Common Mistake: Treating Missing Data as a Value

A common error is to read a table entry at the nearest listed point and use it as if it were the field value at a different point. For example, if a table has \(T(0,0)=12\), it is incorrect to conclude \(T(0,1)=12\) merely because \((0,1)\) is nearby. That conclusion would require an additional rule, such as a formula or an explicit assumption that the field is constant on a region.

The correction is to state exactly what is known: the table reports \(T(0,0)=12\), while it does not report \(T(0,1)\). If a formula for \(T\) is supplied, evaluate that formula at \((0,1)\), after checking the domain. If no such information is supplied, the value is undetermined from the table alone. It is not automatically zero, equal to a nearby entry, or undefined; the data simply do not provide it.

Why the Distinction Matters

Tables and formulas serve different purposes. A table is useful for preserving observations at selected coordinates. A formula offers a rule that can be evaluated at any point in its domain. A formula may fit a table, but finitely many entries generally cannot tell us which fitting formula is the intended field. This matters whenever values are interpreted, compared, or used in further calculations: conclusions should come from the supplied data and assumptions, not from an unstated interpolation or guessed pattern.

When given a table, first locate the exact coordinate row and report only its stated value. When given a formula, substitute the specified coordinates and check that evaluation is permitted. When given both, distinguish the recorded entries from values calculated by the formula, and note any stated precision or measurement qualification.

Check Your Understanding

Answer each question using only the information stated.

A table lists \(q(2,1)=7\). What value does it give at \((2,1)\), and what can the table alone tell you about \(q(2,2)\)?
For \(h(x,y)=x^2-y\), evaluate \(h(3,2)\). Show the substitution and arithmetic.
Why must a point be checked against a formula’s domain before evaluating the formula there?
Can two different polynomial formulas agree at every point in a finite table but disagree at an unlisted point? Explain briefly.
If a table does not list a point, does that mean the field value there is zero? State what can be concluded instead.