A Capstone in Approximation and Error Control
Approximation arguments often end with a polynomial that must do more than be close to a target function. It may need to preserve positivity, support a calculation involving products or reciprocals, or match exact values and derivatives. The central skill in this examination is to keep track of how each operation changes the error. A bound that looks plausible is not enough: its derivation must account for every difference between the target and its approximant.
The problems below bring together tools developed throughout Approximation Theory, including polynomial approximation on compact intervals, the \(C^m\) norm, and constrained approximation by Hermite correction. Earlier results such as the Real Stone-Weierstrass Theorem and Constrained Polynomial Approximation in \(C^m[a,b]\) will be used by name. The new emphasis is on quantitative stability: how an approximation error behaves when functions are multiplied or inverted.
Problem 1: Does a Product Preserve Approximation?
Suppose \(f,g,p,q\in C(K)\), where \(K\) is nonempty and compact, and suppose \(\|f-p\|_\infty\leq\delta\) and \(\|g-q\|_\infty\leq\eta\). Find a bound for \(\|fg-pq\|_\infty\). A useful first step is to insert an intermediate product, but the choice of that product must leave no error term unaccounted for.
Proof. At each \(x\in K\), add and subtract \(p(x)g(x)\). Then $$ f(x)g(x)-p(x)q(x) =(f(x)-p(x))g(x)+p(x)(g(x)-q(x)). $$ Taking absolute values and using the triangle inequality gives $$ |f(x)g(x)-p(x)q(x)| \leq |f(x)-p(x)|\,|g(x)| +|p(x)|\,|g(x)-q(x)|. $$ The assumptions imply \(|f(x)-p(x)|\leq\delta\) and \(|g(x)-q(x)|\leq\eta\). Also, $$ |p(x)|\leq |f(x)|+|p(x)-f(x)| \leq \|f\|_\infty+\delta. $$ Consequently, $$ |f(x)g(x)-p(x)q(x)| \leq \delta\|g\|_\infty+ \bigl(\|f\|_\infty+\delta\bigr)\eta. $$ The right-hand side is independent of \(x\), so taking the supremum proves the result. \(\square\)
The first term measures the error in approximating \(f\), multiplied by the size of the unapproximated factor \(g\). The second measures the error in approximating \(g\), multiplied by a bound for \(p\). That bound must include the error \(\delta\): from \(\|f-p\|_\infty\leq\delta\), one can conclude \(\|p\|_\infty\leq\|f\|_\infty+\delta\), not merely \(\|p\|_\infty\leq\|f\|_\infty\).
Worked Example: Multiplying by an Approximation to Cosine
On \([0,1]\), take \(f(x)=1+x\), \(g(x)=\cos x\), \(p(x)=1+x\), and \(q(x)=1-x^2/2\). Here \(\delta=0\), since \(p=f\), and \(\|f\|_\infty=2\). Taylor's theorem applied to \(\cos x\) at \(0\), through degree \(3\), gives $$ |\cos x-(1-x^2/2)|\leq \frac{x^4}{4!}\leq\frac1{24} \qquad (0\leq x\leq1). $$ Thus \(\eta=1/24\). Since \(\|g\|_\infty=1\), the product bound yields $$ \|fg-pq\|_\infty \leq 0\cdot1+(2+0)\frac1{24} =\frac1{12}. $$ In this example the product difference is exactly \((1+x)(\cos x-q(x))\). Its absolute value is at most \(2/24\), which verifies the bound directly.
Problem 2: Can an Approximation Preserve Positivity?
A strictly positive continuous function on a compact set has a positive minimum, by the Extreme-Value Theorem. That minimum creates a margin: an approximation closer than the margin cannot cross zero. The lower bound matters. Mere pointwise positivity, without a uniform lower bound, would not provide the same quantitative guarantee on a general domain.
Proof. For every \(x\in K\), the definition of the supremum norm gives $$ p(x)\geq f(x)-|f(x)-p(x)| \geq c-\|f-p\|_\infty. $$ If \(\|f-p\|_\infty<c\), the final expression is positive, so \(p(x)>0\). If the stronger bound \(\|f-p\|_\infty<c/2\) holds, then \(p(x)>c/2\). \(\square\)
Worked Example: A Positive Polynomial Approximation to the Exponential
Let \(f(x)=e^x\) on \([-1,1]\), and let $$ p_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}. $$ Since \(e^x\geq e^{-1}\) on this interval, \(f\) has lower bound \(c=e^{-1}\). Taylor's theorem, using that the fourth derivative of \(e^x\) is \(e^x\leq e\) on \([-1,1]\), gives $$ \|f-p_3\|_\infty\leq\frac{e}{4!}=\frac{e}{24}. $$ This is less than \(e^{-1}/2\), because \(e/24<1/(2e)\) is equivalent to \(e^2<12\), and \(e<3\) implies \(e^2<9<12\). Therefore $$ p_3(x)\geq e^{-1}-\frac{e}{24}>\frac{e^{-1}}2>0 \qquad (-1\leq x\leq1). $$ The approximation is thus positive everywhere, with a quantitative margin. Checking positivity from the error bound avoids having to locate the minimum of the cubic directly.
Problem 3: How Stable Is Taking a Reciprocal?
Positivity alone is not enough to control reciprocal errors. If positive functions can approach zero, their reciprocals can become large. A uniform lower bound prevents that instability and gives a direct estimate.
Proof. For every \(x\in K\), both denominators are positive and at least \(c\), so $$ \left|\frac1{f(x)}-\frac1{g(x)}\right| =\frac{|f(x)-g(x)|}{f(x)g(x)} \leq\frac{|f(x)-g(x)|}{c^2} \leq\frac{\|f-g\|_\infty}{c^2}. $$ Taking the supremum over \(x\) proves the estimate. \(\square\)
Combining this result with Stability of Strict Positivity gives a practical approximation procedure. If \(f\geq c>0\) and \(\|f-p\|_\infty<c/2\), then \(p\geq c/2\) and both reciprocals are defined. The reciprocal estimate, with lower bound \(c/2\) for both functions, gives $$ \left\|\frac1f-\frac1p\right\|_\infty \leq\frac{4}{c^2}\|f-p\|_\infty. $$ Thus approximation of a function bounded away from zero also controls the reciprocal of its approximant. This estimate does not say that \(1/p\) is a polynomial; it quantifies the error between the two reciprocal functions.
Worked Example: Controlling Reciprocal Error
On \([0,1]\), let \(f(x)=2+x\) and \(p(x)=2+x+x^2/10\). Both are at least \(2\), and $$ \|f-p\|_\infty=\sup_{0\leq x\leq1}\frac{x^2}{10}=\frac1{10}. $$ The reciprocal stability estimate with \(c=2\) yields $$ \left\|\frac1f-\frac1p\right\|_\infty \leq\frac{1/10}{2^2} =\frac1{40}. $$ For a direct check of the algebra, at every \(x\in[0,1]\), $$ \left|\frac1{2+x}-\frac1{2+x+x^2/10}\right| =\frac{x^2/10}{(2+x)(2+x+x^2/10)} \leq\frac{1/10}{4} =\frac1{40}. $$ The denominator estimate uses \(2+x\geq2\) and \(2+x+x^2/10\geq2\).
Problem 4: Approximate While Matching Exact Data
A final synthesis problem combines uniform or smooth approximation with exact constraints. Suppose a function and its derivative are prescribed at selected points. A polynomial chosen only for closeness will not usually meet those data exactly. Constrained Polynomial Approximation in \(C^m[a,b]\), proved earlier in this course, resolves both requirements at once: for \(f\in C^m[a,b]\), fixed nodes \(x_i\), and derivative orders \(r_i\leq m\), it provides a polynomial arbitrarily close in \(C^m\) while matching every prescribed derivative value. The example below displays the correction explicitly in a simple case.
Worked Example: Matching an Endpoint Value for Cosine
On \([0,1]\), let \(f(x)=\cos x\). For each integer \(N\geq1\), set $$ q_N(x)=\sum_{k=0}^{N}\frac{(-1)^k x^{2k}}{(2k)!}, \qquad d_N=\cos 1-q_N(1), \qquad p_N(x)=q_N(x)+d_Nx^2. $$ At \(x=0\), \(q_N(0)=1\), \(q_N'(0)=0\), and the correction \(d_Nx^2\) has value and derivative zero. Therefore \(p_N(0)=\cos0\) and \(p_N'(0)=-\sin0\). At \(x=1\), the definition of \(d_N\) gives \(p_N(1)=\cos1\). These three conditions hold exactly for every \(N\).
Taylor's theorem on \([0,1]\) gives $$ \|f-q_N\|_\infty\leq\frac1{(2N+2)!}, \qquad \|f'-q_N'\|_\infty\leq\frac1{(2N+1)!}. $$ The first estimate uses the Taylor polynomial for cosine through degree \(2N+1\), which equals \(q_N\); the second uses the Taylor polynomial for \(-\sin x\) through degree \(2N\), which equals \(q_N'\). In particular, $$ |d_N|=|f(1)-q_N(1)|\leq\frac1{(2N+2)!}. $$ Using the \(C^1\) norm \(\|u\|_{C^1}=\|u\|_\infty+\|u'\|_\infty\), the correction satisfies $$ \|d_Nx^2\|_{C^1} =|d_N|+2|d_N| =3|d_N|, $$ because \(0\leq x^2\leq1\) and \(0\leq2x\leq2\). Consequently, $$ \|f-p_N\|_{C^1} \leq \frac1{(2N+2)!}+\frac1{(2N+1)!} +\frac3{(2N+2)!}. $$ The right-hand side tends to zero as \(N\) tends to infinity, so \(p_N\) converges to \(f\) in \(C^1[0,1]\) while preserving the stated data exactly.
How to Audit an Approximation Argument
The worked problems illustrate several checks that apply well beyond these examples:
- Expand the difference exactly. For products, write an identity that accounts for both factors’ errors before estimating anything.
- Bound the approximant when it appears. If an estimate contains \(\|p\|_\infty\), derive a bound for \(p\) from the hypotheses instead of replacing it by \(\|f\|_\infty\).
- Check denominators before taking reciprocals. A positive lower bound is what makes a uniform reciprocal estimate possible.
- Separate closeness from exact constraints. An approximation theorem may give small error, but exact data require either a constrained theorem or an explicit correction.
- Verify the final norm. In \(C^1\), for example, control both the function error and the derivative error; a uniform bound on the function alone is insufficient.
The product estimate also illustrates why different valid decompositions can lead to different, but compatible, bounds. The chosen decomposition controls the error in \(f\) using \(g\), and the error in \(g\) using \(p\). Interchanging which intermediate product is added and subtracted gives another valid estimate, provided its terms are derived from that identity and all factors are bounded. An estimate should never be accepted merely because it resembles a familiar inequality: test the algebra, verify each norm bound, and check extreme cases such as a zero factor or zero approximation error.
Check Your Understanding
Use the estimates and proof strategies in this examination to answer the following questions.
- In the Product Error Bound, where does the term \(\|f\|_\infty+\delta\) come from?
- Why does a uniform lower bound \(f\geq c>0\) allow a sufficiently close approximant to remain positive?
- What fails in the reciprocal estimate if no positive lower bound is available?
- Why does the cosine correction \(d_Nx^2\) preserve the value and derivative at zero?
- In the product example, what bounds justify \(\|f\|_\infty=2\) and \(\|g\|_\infty=1\) on \([0,1]\)?