M2221-L04 — Symbolic reasoning and honest mathematical visualization

Prerequisites: algebra, differentiation and elementary complex numbers. The symbolic MATLAB examples require Symbolic Math Toolbox.

1. A symbolic expression carries a domain and assumptions

A symbolic system manipulates formal expressions subject to declared mathematical rules. Before accepting a simplification, identify the variable domain, parameter restrictions, branch conventions and excluded denominator zeros. An equality of formulas on a common domain is not automatically equality of two functions with different domains.

For real $x$, $\sqrt{x^2}=|x|$, not $x$ without the assumption $x\ge0$. For complex $z$, the principal square root is branch dependent and $\sqrt{z^2}=z$ fails, for example at $z=-1$. Cancelling

$$(x^2-1)/(x-1)=x+1$$

is valid only for $x\ne1$ in the original expression. A system returning $x+1$ has not proved the original function is defined at one. A continuous extension is a new explicitly defined function.

MathWorks documents both assumption management and simplification, including the distinction between expressions formed numerically first and expressions formed symbolically. Consult the official symbolic simplification guide and assumptions guide. A safe illustrative session uses separate variables for separate assumptions:

syms xr real
syms xp positive
s_real = simplify(sqrt(xr^2));
s_positive = simplify(sqrt(xp^2));
assert(isAlways(s_real == abs(xr)));
assert(isAlways(s_positive == xp));
q_exact = sym(1)/10;

The last line forms the exact rational one tenth, rather than first computing a binary floating-point approximation and then converting it. Existing assumptions may persist independently of a workspace variable; record and reset them through the documented symbolic workflow rather than relying on an unexplained interactive history.

2. Symbolic verification needs a mathematical interpretation

For a proposed polynomial identity, expand the difference and check every coefficient is zero. For a rational identity, record the nonzero-denominator domain, multiply by a common denominator and verify the resulting polynomial identity. For a proposed antiderivative $F$, differentiate and compare $F'$ with $f$ on a stated interval or domain.

Worked example: partial fractions. On $x\ne1,-2$,

$$\frac1{(x-1)(x+2)}=\frac1{3(x-1)}-\frac1{3(x+2)}.$$

Multiplying by $(x-1)(x+2)$ gives $1=(x+2)/3-(x-1)/3$, an exact polynomial identity. Integrating on each real interval separated by the poles gives

$$F(x)=\tfrac13\log|x-1|-\tfrac13\log|x+2|+C.$$

The constant may differ between disconnected intervals. A complex logarithm antiderivative instead needs a selected branch on a permitted domain; silently exchanging real absolute-value logs and complex branch logs is invalid.

A numerical spot check can disprove an identity by finding a valid point of disagreement, but agreement at finitely many arbitrary points does not prove a general functional identity. For a polynomial of known degree at most $d$, agreement at $d+1$ distinct exact points does prove equality by the root bound; floating-point near-agreement is not that exact theorem.

Worked example: extraneous roots. Squaring $\sqrt{x+2}=x$ gives $x^2-x-2=0$, with candidates two and minus one. The original square root is nonnegative, requiring $x\ge0$; substitution accepts two and rejects minus one. A solver's polynomial candidate list is not the final solution set of the original radical equation.

3. Symbolic stationary points are not automatically global extrema

For $f(x)=x^4-2x^2$ on $[-2,2]$, $f'(x)=4x(x^2-1)$ gives interior critical points $-1,0,1$. The function is continuous on a compact interval, so a global maximum and minimum occur at critical points or endpoints. Values are $f(\pm2)=8$, $f(\pm1)=-1$, $f(0)=0$. Hence the global minimum is minus one at $\pm1$ and maximum is eight at the endpoints. Merely solving $f'=0$ would miss the maxima.

When a derivative formula divides by an expression, check its zero set separately. On an open or unbounded interval, a finite stationary-point list alone cannot guarantee a global extremum; endpoint limits or behaviour at infinity may matter. Symbolic software can assist each calculation, but the compactness and completeness argument is mathematical.

4. Plot discontinuities rather than drawing fictitious bridges

A graph is a finite sample plus a rendering rule. Joining sampled points by line segments asserts visual continuity between them. For $f(x)=1/x$, a single line crossing from a negative sample to a positive sample near zero can create a fictitious bridge across a vertical asymptote.

xleft = linspace(-2,-0.05,400);
xright = linspace(0.05,2,400);
plot(xleft,1./xleft,'b-',xright,1./xright,'b-');
xlabel('x'); ylabel('1/x'); grid on;
title('Two domain components; x=0 excluded');

The exclusion width is a visualization choice, not a declaration that the function is undefined on that whole gap. Report it. A second view with a smaller gap checks whether important near-pole behaviour has been clipped. Alternative workflows insert a missing-value separator, but must avoid accidentally rejoining components.

For $\sqrt{1-x^2}$ as a real function, plot only $|x|\le1$; complex outputs outside that interval should not be silently discarded as though they were real values. State whether the graph is of the function, its real part, modulus or another derived quantity.

Aliasing example. Sampling $\sin(20x)$ at $x_k=k\pi/10$ gives zero at every node, although the function is not zero between them. A flat sampled plot can therefore hide oscillations completely. Increase sampling density and vary the grid offset; where a frequency bound is known, use an appropriate sampling theorem. Without such a bound, a finite sample cannot certify absence of narrow spikes or arbitrarily fast oscillations.

5. Two-dimensional visualizations need coordinates, scale and invariants

For a scalar field $f(x,y)$, a surface plot encodes height while contours encode level sets. A faithful plot labels coordinates, units and the function. For a vector field, arrows encode both direction and chosen length scaling; normalizing all arrows removes magnitude information and must be disclosed.

For the linear map $A=\begin{pmatrix}2&1\\0&1\end{pmatrix}$, map a sampled unit circle $(\cos t,\sin t)$ through $A$. The image is an ellipse. The signed area scale is $\det A=2$; an equal-aspect display is essential to compare geometric shapes. Its principal semiaxis lengths are singular values of $A$, not necessarily its eigenvalues. Check geometry with the invariant $A^\mathsf TA$, not a visually guessed angle.

For a complex map $w=z^2$, one possible visualization maps a grid of horizontal and vertical line segments and colours the two families differently. Away from zero the crossing angles remain right angles because the complex derivative $2z$ is nonzero; at zero the derivative vanishes and the local angle-preservation conclusion does not apply. Domain colouring instead displays argument and modulus, which requires an explicit legend and a rule for zero and very large magnitudes. Colour discontinuities at an argument cut can reflect a display convention rather than a discontinuity of the function.

Every figure should have an accessible textual statement of the mathematical conclusion. Colour alone should not distinguish essential groups; combine line style, labels or symbols. If a picture suggests a theorem, write the hypotheses and proof separately.

6. A complete laboratory submission

Submit a clean script and a short report containing: declared domains and assumptions; exact symbolic steps; numerical checks at valid points; plots with sampling ranges, counts and exclusions; and a conclusion distinguishing proved statements from visual or numerical evidence. Keep exact and approximate quantities visibly distinct. If symbolic software is unavailable, perform the exact coefficient, derivative and domain checks by hand; report that limitation rather than inventing software output.

7. Exercises

P01 — Foundation. Simplify $\sqrt{x^2}$ for real $x$ and for positive real $x$. Give a real counterexample to the unconditional simplification $\sqrt{x^2}=x$.

P02 — Core. Solve $\sqrt{x+2}=x$ with the original domain intact.

P03 — Core. Simplify $(x^2-4)/(x-2)$ and solve its equality to four, respecting the original domain.

P04 — Core. Find all global extrema of $x^4-2x^2$ on $[-2,2]$ with a completeness justification.

P05 — Diagnosis. Why can a line plot of $1/x$ containing samples on both sides of zero mislead? Give two repairs.

P06 — Proof. If two polynomials of degree at most four agree at five distinct exact points, prove they are equal. Explain why five approximate floating-point checks do not give the same proof.

P07 — Core. Find the eigenvalues of $A^\mathsf TA$ for the matrix in §5, and hence the ellipse's principal semiaxis lengths.

P08 — Extension. Explain how grid sampling can make $\sin(20x)$ look identically zero. Design a revised experiment that can reveal the hidden oscillation, and state why this remains an experiment rather than a general proof about all functions.

8. Full solutions

P01. For real $x$ the nonnegative square root of $x^2$ is $|x|$. For positive real $x$ it is $x$. At $x=-1$, $\sqrt{x^2}=1\ne-1$, disproving the unconditional rule.

P02. The original equation requires $x+2\ge0$ and its right side must be nonnegative, so $x\ge0$. Squaring gives $(x-2)(x+1)=0$. Only $x=2$ survives the sign condition, and substitution verifies $\sqrt4=2$. Thus the solution set is $\{2\}$.

P03. On $x\ne2$ it simplifies to $x+2$. The equation $x+2=4$ has candidate $x=2$, which is excluded. The original equation has no solution; cancellation did not fill the hole.

P04. Continuity on the compact interval guarantees extrema. All interior extrema must have derivative zero, giving $-1,0,1$; include endpoints $\pm2$. Comparing values yields minima $-1$ at $\pm1$, maxima eight at $\pm2$. This finite candidate list is complete because differentiability holds throughout the interior.

P05. The renderer can join samples across the missing domain point, drawing a bridge that is not part of the graph. Plot the two components as separate curves, or insert an explicit missing-value break between them. Also label the excluded point and state any clipped vertical range.

P06. Their difference has degree at most four and five distinct roots, hence is the zero polynomial. Approximate equalities are not roots: a small nonzero polynomial or rounding error can produce tiny values at every tested point without vanishing identically. Exact coefficients or certified error bounds are needed for a different rigorous conclusion.

P07. $A^{\mathsf T}A=\begin{pmatrix}4&2\\2&2\end{pmatrix}$, with characteristic polynomial $\lambda^2-6\lambda+4$. Its eigenvalues are $3\pm\sqrt5$, both positive. The semiaxis lengths are $\sqrt{3+\sqrt5}$ and $\sqrt{3-\sqrt5}$, whose product is two, agreeing with the area scale $|\det A|$.

P08. At $x_k=k\pi/10$, $20x_k=2\pi k$ and every sample vanishes. A denser grid such as spacing $\pi/100$, combined with an offset grid and direct evaluation of intermediate extrema, reveals nonzero values. The identity $\sin(20x)$ gives an exact known frequency for analysis; without comparable regularity or frequency information, any fixed finite sampling plan can miss another function's narrower or faster features.

9. Summary

Symbolic algebra is trustworthy only with domains, assumptions and exact-input discipline. Visualization is trustworthy only with declared sampling, geometry, exclusions and encoding. Neither a returned expression nor a convincing picture removes the need to prove the mathematical claim it is being used to support.