Approximation Theorem Synthesis

Combining Approximation with Exact Constraints

Polynomial approximation can provide more than closeness. In the previous tutorial, polynomial density in $C^m[a,b]$ was obtained by approximating the highest derivative and integrating. That construction matches derivative data at an endpoint. In applications, however, the required data may be prescribed at several points, and may include both function values and derivatives. We now combine smooth-norm approximation with finite Hermite interpolation to meet such constraints exactly.

The strategy has two stages. First choose a polynomial close to the target in the $C^m$ norm. Its values and derivatives at finitely many points are then close to the target data. Next add a correction polynomial that fixes the remaining discrepancies. The correction is small because its finitely many coefficients are small, while the interpolation points and derivative orders are fixed.

Definition: Let $x_1,\ldots,x_s$ be distinct points in $[a,b]$, and let $r_1,\ldots,r_s$ be nonnegative integers. The finite Hermite data of a function $f$ at these nodes are the numbers $f^{(j)}(x_i)$, for $1\leq i\leq s$ and $0\leq j\leq r_i$. A polynomial interpolates these data if its derivatives satisfy $p^{(j)}(x_i)=f^{(j)}(x_i)$ for every indicated pair $i,j$.

The orders $r_i$ must not exceed the smoothness order $m$ when $f\in C^m[a,b]$. The number of prescribed data is $n=\sum_{i=1}^s(r_i+1)$. The crucial finite-dimensional fact is that these $n$ data can be assigned independently by a polynomial of degree less than $n$.

Lemma (Finite Hermite Interpolation): Let $x_1,\ldots,x_s$ be distinct real numbers, and let $r_1,\ldots,r_s$ be nonnegative integers. Put $n=\sum_{i=1}^s(r_i+1)$. For any real numbers $d_{i,j}$, there is a unique polynomial $h$ of degree less than $n$ such that $h^{(j)}(x_i)=d_{i,j}$ for $1\leq i\leq s$ and $0\leq j\leq r_i$.

Proof. Let $\mathcal P_{n-1}$ be the vector space of polynomials of degree less than $n$, and define the linear map $T:\mathcal P_{n-1}\to\mathbb R^n$ by sending a polynomial to the vector of its prescribed derivative values, listed in a fixed order. Both spaces have dimension $n$. It is enough to show that $T$ is injective.

Suppose $T(p)=0$. At each node $x_i$, the equations $p^{(j)}(x_i)=0$ for $0\leq j\leq r_i$ imply that $(x-x_i)^{r_i+1}$ divides $p$. Since the nodes are distinct, the product $\prod_{i=1}^s(x-x_i)^{r_i+1}$ divides $p$. This product has degree $n$. A nonzero polynomial of degree less than $n$ cannot be divisible by a polynomial of degree $n$, so $p=0$. Thus $T$ is injective. Since its domain and codomain have the same finite dimension, $T$ is surjective as well. Every data vector therefore has a unique preimage, proving the lemma. $\square$

Controlling the Size of the Correction

For each data coordinate, apply the lemma to the vector that is $1$ in that coordinate and $0$ in every other coordinate. Denote the resulting polynomial by $H_{i,j}$. These fixed polynomials satisfy $H_{i,j}^{(k)}(x_\ell)=1$ when $(\ell,k)=(i,j)$, and $0$ for all other prescribed coordinates. Thus a correction with data $d_{i,j}$ is $h=\sum_{i,j}d_{i,j}H_{i,j}$.

Proposition (Hermite Correction Bound): Fix the nodes and derivative orders above, with $r_i\leq m$. Define $$ C=\sum_{i=1}^s\sum_{j=0}^{r_i}\|H_{i,j}\|_{C^m}. $$ If a polynomial $q$ has discrepancies $d_{i,j}=f^{(j)}(x_i)-q^{(j)}(x_i)$, then its Hermite correction $h=\sum_{i,j}d_{i,j}H_{i,j}$ satisfies $$ \|h\|_{C^m}\leq C\max_{i,j}|d_{i,j}|. $$

Proof. The $C^m$ norm is a norm and is therefore absolutely homogeneous and satisfies the triangle inequality. Hence $$ \|h\|_{C^m} \leq\sum_{i=1}^s\sum_{j=0}^{r_i}|d_{i,j}|\|H_{i,j}\|_{C^m} \leq \left(\max_{i,j}|d_{i,j}|\right)C. $$ This proves the bound. $\square$

Each discrepancy is bounded by the original approximation error. Indeed, for every prescribed $i,j$, $$ |d_{i,j}|=|f^{(j)}(x_i)-q^{(j)}(x_i)| \leq\|f^{(j)}-q^{(j)}\|_\infty \leq\|f-q\|_{C^m}. $$ The correction bound therefore turns closeness in $C^m$ into a quantitative guarantee that the exact interpolation adjustment remains small.

Constrained Polynomial Approximation

Theorem (Constrained Polynomial Approximation in $C^m[a,b]$): Let $a<b$, let $m\geq0$, and let $f\in C^m[a,b]$. Fix distinct nodes $x_1,\ldots,x_s\in[a,b]$ and integers $0\leq r_i\leq m$. For every $\varepsilon>0$, there is a polynomial $p$ such that $$ \|f-p\|_{C^m}<\varepsilon $$ and $$ p^{(j)}(x_i)=f^{(j)}(x_i) \qquad(1\leq i\leq s,\ 0\leq j\leq r_i). $$

Proof. If $m=0$, the constraints involve only function values. If $m\geq1$, use Polynomial Density in $C^m[a,b]$ from the previous tutorial. In either case, polynomial approximation in the applicable norm gives a polynomial $q$ with $\|f-q\|_{C^m}<\eta$, where $\eta>0$ will be chosen below.

Let $H_{i,j}$ be the Hermite polynomials for the specified nodes and orders, and let $C$ be the constant in the Hermite Correction Bound. Set $$ d_{i,j}=f^{(j)}(x_i)-q^{(j)}(x_i), \qquad h=\sum_{i=1}^s\sum_{j=0}^{r_i}d_{i,j}H_{i,j}, \qquad p=q+h. $$ The defining data of the $H_{i,j}$ give $h^{(j)}(x_i)=d_{i,j}$, so $p^{(j)}(x_i)=f^{(j)}(x_i)$ for every required constraint.

As shown above, $\max_{i,j}|d_{i,j}|\leq\|f-q\|_{C^m}<\eta$. The correction estimate and the triangle inequality now yield $$ \|f-p\|_{C^m} \leq\|f-q\|_{C^m}+\|h\|_{C^m} <(1+C)\eta. $$ Choose $0<\eta<\varepsilon/(1+C)$. Then $\|f-p\|_{C^m}<\varepsilon$, and all the interpolation conditions hold. $\square$

The constant $C$ depends on the fixed nodes, derivative orders, interval, and norm, but not on $f$, $q$, or $\varepsilon$. This is the error-budget point: the initial approximation must be closer than the final target because the correction can increase the error. The factor $1+C$ explicitly accounts for that increase.

Worked Examples: Exact Data and Small Corrections

Worked Example: Matching Two Values of an Exponential Approximation

On $[0,1]$, let $f(x)=e^x$ and take $q_N(x)=\sum_{k=0}^N x^k/k!$, where $N\geq1$. This polynomial already satisfies $q_N(0)=f(0)=1$ and $q_N'(0)=f'(0)=1$. To impose the additional value at $x=1$, put $$ d=e-q_N(1)=e-\sum_{k=0}^N\frac1{k!}, \qquad p_N(x)=q_N(x)+dx^2. $$ Then $p_N(0)=1$, $p_N'(0)=1$, and $p_N(1)=q_N(1)+d=e$. The correction does not alter the data at $0$, because $x^2$ and its first derivative vanish there.

The exponential tail estimate gives $|d|\leq e/(N+1)!$. Also, on $[0,1]$, $$ \|e^x-q_N\|_\infty\leq\frac{e}{(N+1)!}, \qquad \|e^x-q_N'\|_\infty\leq\frac{e}{N!}. $$ Since $\|dx^2\|_\infty=|d|$ and $\|(dx^2)'\|_\infty=2|d|$, we obtain $$ \|f-p_N\|_{C^1} \leq\frac{e}{(N+1)!}+\frac{e}{N!}+\frac{3e}{(N+1)!}. $$ This bound tends to zero, while the three specified data are exact for every $N\geq1$.

Worked Example: Interpolating the Endpoints of a Sine Approximation

Let $z=\pi/2$, $f(x)=\sin x$ on $[0,z]$, and $$ q_N(x)=\sum_{k=0}^N\frac{(-1)^kx^{2k+1}}{(2k+1)!}. $$ At $x=0$, $q_N(0)=f(0)=0$. At $x=z$, define $d=\sin z-q_N(z)=1-q_N(z)$, and set $$ p_N(x)=q_N(x)+d\frac{x}{z}. $$ Then $p_N(0)=0$, while $p_N(z)=q_N(z)+d=1=\sin z$. The correction is the linear Lagrange interpolation correction for the two value constraints.

Taylor's theorem gives $\|f-q_N\|_\infty\leq z^{2N+2}/(2N+2)!$, so $|d|$ is bounded by the same quantity. The derivative of the correction is the constant $d/z$, and its supremum norm is $|d|/z$. Consequently, $$ \|f-p_N\|_{C^0} \leq\frac{z^{2N+2}}{(2N+2)!} +\frac{z^{2N+2}}{(2N+2)!}, $$ which tends to zero. If derivative control is also desired, $q_N'$ is the Taylor polynomial for $\cos x$ through degree $2N$, and Taylor's theorem gives $\|f'-q_N'\|_\infty\leq z^{2N+2}/(2N+2)!$. Adding the derivative correction bound $|d|/z$ shows that $p_N$ also converges to $f$ in $C^1$.

Worked Example: Matching Values and Slopes at Both Ends

For data $d_0,d_1,d_2,d_3$ prescribing a value and slope at $0$ and $1$, the cubic Hermite correction is $$ h(x)=d_0(2x^3-3x^2+1)+d_1(x^3-2x^2+x) +d_2(-2x^3+3x^2)+d_3(x^3-x^2). $$ Substitution gives $h(0)=d_0$, $h'(0)=d_1$, $h(1)=d_2$, and $h'(1)=d_3$. For example, the first basis polynomial has values $1,0$ at $0,1$, and derivative $0$ at both endpoints; the other three have the corresponding single unit datum, as direct differentiation verifies.

Take $f(x)=\sin x$ on $[0,1]$ and $q(x)=x-x^3/6$. At $0$, the value and slope already agree. At $1$, the discrepancies are $$ d_2=\sin 1-\frac56,\qquad d_3=\cos 1-\frac12. $$ The corrected polynomial $p=q+d_2(-2x^3+3x^2)+d_3(x^3-x^2)$ therefore matches both endpoint values and both endpoint slopes exactly. Taylor's theorem gives $|d_2|\leq1/120$ and $|d_3|\leq1/24$. On $[0,1]$, the value and derivative supremum norms of $-2x^3+3x^2$ are at most $1$ and $3/2$, respectively; those of $x^3-x^2$ are at most $1$ and $1$. Thus $$ \|p-q\|_{C^1}\leq\frac52|d_2|+2|d_3| \leq\frac{1}{48}+\frac{1}{12}. $$ The correction is controlled by the endpoint Taylor remainders, and higher Taylor polynomials make those discrepancies, and hence the corrections, tend to zero.

Why the Synthesis Is Useful

The theorem packages two different requirements into one conclusion: uniform control of every derivative through order $m$, and exact satisfaction of finitely many derivative constraints. This is useful when an approximation must respect measured values, boundary values, initial slopes, or other finite data without sacrificing convergence in the function-space norm.

A common pitfall is to assume that exact interpolation automatically preserves a chosen error tolerance. It does not: the correction may increase the norm of the error. The proof avoids that gap by choosing the initial approximation with a smaller error budget. Another limitation is that the correction constant depends on the nodes. If nodes approach one another, the Hermite basis can become poorly controlled, so this theorem makes no uniform claim over changing node configurations.

Takeaway: Approximate first in the desired smooth norm, then correct the finite data with a Hermite polynomial. Fixed finite constraints add only a bounded correction, so arbitrarily accurate approximation remains possible.

Check Your Understanding

Use the Hermite interpolation and error-budget arguments to answer the following questions.

Why does vanishing of the derivatives through order $r_i$ at $x_i$ imply divisibility by $(x-x_i)^{r_i+1}$?
Why does the Hermite data map have a polynomial inverse on the space of degree less than $n$?
How does the constant $C$ in the correction bound depend on the problem data?
Why must the initial approximation error be chosen smaller than the final tolerance?
What feature of the correction $dx^2$ preserves both the value and slope at zero in the exponential example?