This Note contains some remarks concerning an approximation formula for functions, which is a generalization of some interpolation formulae given in [3] and [5]. In particular, we shall show that only one of the formulae of this type, mentioned in [5], has a maximal degree of exactness. Some particular cases of such formulae were also mentioned in [5, p.163].

Denote by $I_x$ the closed interval determined by two distinct points $x_0,$ $x$ in $ℝ$. For a $\left(2n+1\right)$-times derivable function $f:I_x\to ℝ$ and $n\in \mathbb{N}$, consider the class $G$ of functions given by

Consider the following problem: Find a function $\overline{g}\in G$ such that

In [5] this problem was solved in some particular cases. We shall show that, for $m=2n,$ this problem has a unique solution and we shall give a representation for the remainder.

For $m=2n,$ we are looking for a function $\overline{g}$ in $G$ verifying conditions (2) and having a maximal degree of approximation.

It is easily seen that conditions (2) lead to the following system, having the real numbers $a_i,b_i,i=\overline{1,n},$ as unknowns:

Consider now a continuous function $\phi :[0,1]\to ℝ$ and let

be a quadrature formula, having ${\{b_i\}}_1^n$ as knots and ${\{a_i\}}_1^n$ as coefficients. Asking that $R\left[{\phi }_k\right]=0$ for ${\phi }_k\left(t\right)=t^k,k=\overline{0,2n-1}$, (4) becomes the classical Gauss quadrature formula. On the other hand, the conditions $R\left[{\phi }_k\right]=0,$ for ${\phi }_k\left(t\right)=t^k,k=\overline{0,2n-1}$, lead again to the system (3), implying that $b_i$ must be the roots of the Legendre polynomial $w_n$ of degree $n,$ i.e., the roots of the equation

The coefficients $a_i$ are given by the following formula

(see [2, p.261]).

Now, it is clear that the following theorem holds:

Consider the approximation formula

where $\overline{g}\in G$ is a function verifying (3) and $r\left[f\right]$ is the remainder.

In the conditions of Theorem 2.1, it follows that

where

and ${\theta }_i$ is a number contained in the open interval determined by $x_0$ and $x_0+b_i\left(x-x_0\right),1\le i\le n.$

From (8) we obtain the equalities

Multiplying equalities (10) by $a_i,$ taking into account solution (3) and summing up, we obtain

Now, using (9) and Lagrange from of the remainder in the Taylor formula, we get

where $\eta \in I_x.$

Setting $\phi \left(t\right)=t^{2n}$ in (4) and taking into account the form of the remainder term in the Gauss quadrature formula [[2],p.259], we get

implying

Suppose now that the $\left(2n+1\right)$-order derivative of $f$ is bounded on $I_x$ and let

Taking into account relations (12) and (13), one obtains the following delimitation for $r\left[f\right]$

• a)

  $n=1.$ In this case, $b_1=\frac{1}{2},$ $a_1=1$ and

  $$g\left(x\right)=f\left(x_0\right)+\left(x-x_0\right)f^{\prime }\left(x_0+\frac{1}{2}\left(x-x_0\right)\right).$$

  From (15) we get

  $$\left|f\left(x\right)-g\left(x\right)\right|\le \frac{7M_3}{24}{\left|x-x_0\right|}^3,$$

  where $M_3={sup}_{t\in I_x}\left|f^{\prime \prime \prime }\left(t\right)\right|.$

• b)

  $n=2.$ In this case, $b_1=\frac{3-\sqrt{3}}{6},$ $b_2=\frac{3+\sqrt{3}}{6},a_1=a_2=\frac{1}{2}$and

  $$g\left(x\right)=f\left(x_0\right)+\frac{1}{2}\left(x-x_0\right)\left[f^{\prime }\left(x_0+\frac{3-\sqrt{3}}{6}\left(x-x_0\right)\right)+f^{\prime }\left(x_0+\frac{3+\sqrt{3}}{6}\left(x-x_0\right)\right)\right],$$

  One also obtains the evaluation

  $$\left|f\left(x\right)-g\left(x\right)\right|\le \frac{71M_5}{4320}{\left|x-x_0\right|}^5,$$

  where $M_5={sup}_{t\in I_x}\left|f^{\left(5\right)}\left(t\right)\right|.$