UDC 518.5

MATHEMATICS

A. S. KRONROD, V. A. KRONROD, I. A. FARADZHEV

ON THE CHOICE OF STEP IN COMPUTING DERIVATIVES

(Presented by Academician I. G. Petrovskii on 24 III 1970)

1°. Let (f(x)) be a function differentiable up to order (k+2) inclusive. Let (x_0, x_1, \ldots, x_k) be points forming a grid with equal step (\Delta): (x_{p+1}=x_p+\Delta). Suppose, further, that the operator (L_{\Delta,x_0}^k[f]) is defined by the equality

[
L_{\Delta,x_0}^k[f]=\frac{(-1)^k}{\Delta^k}\sum_{s=0}^k(-1)^s C_k^s f(x_k).
\tag{1}
]

Finally, let (x=\frac12(x_0+x_k)). Then the following holds.

Theorem.

[
L_{\Delta,x_0}^k[f]=f^{(k)}(x)+\frac{k\Delta^2}{24}f^{(k+2)}(x)+o(\Delta^2).
\tag{2}
]

Proof. It is verified directly that

[
L_{\Delta,x_0}^k[f]=\frac1\Delta{L_{\Delta,x_1}^{k-1}[f]-L_{\Delta,x_0}^{k-1}[f]}.
\tag{3}
]

Observe that the operator (L_{\Delta,x_0}^k) is linear and invariant with respect to the change of variable (z=x-\frac12(x_0+x_k)). Therefore in what follows we shall denote the operator (L_{\Delta,x_0}^k), applied to (f(x)) on the grid (x_0,x_1,\ldots,x_k), where (x_0=-x_k), by (L_\Delta^k[f(x)]).

In this notation (3) is written as follows:

[
L_\Delta^k[f(x)]=\frac1\Delta\left{L_\Delta^{k-1}\left[f\left(x+\frac\Delta2\right)\right]-L_\Delta^{k-1}\left[f\left(x-\frac\Delta2\right)\right]\right}.
\tag{4}
]

We shall prove by induction that:

A. (L_\Delta^s(x^m)=0) for (m<s).

B. (L_\Delta^s(x^s)=s!) and (L_\Delta^s(x^{s+1})=0).

C. (L_\Delta^s(x^{s+2})=\dfrac{s\Delta^2}{24}(s+2)!)

For (k=1) we have:

A(^0). (L_\Delta^1(x^0)=0).

B(^0). (L_\Delta^1(x)=\dfrac1\Delta\left(\dfrac\Delta2+\dfrac\Delta2\right)=1!) and (L_\Delta^1(x^2)=\left[\dfrac1\Delta\left(\dfrac{\Delta^2}{2^2}-\dfrac{\Delta^2}{2^2}\right)\right]=0).

C(^0). (L_\Delta^1(x^3)=\dfrac1\Delta\left(\dfrac{\Delta^3}{2^3}+\dfrac{\Delta^3}{2^3}\right)=\dfrac{\Delta^2}{4}=\dfrac{1\cdot\Delta^2}{24}\cdot3!)

Let (A), (B), and (C) be true for all (s\le k). Then

[
L_\Delta^{k+1}(x^m)=\frac1\Delta\left{L_\Delta^k\left[\left(x+\frac\Delta2\right)^m\right]-L_\Delta^k\left[\left(x-\frac\Delta2\right)^m\right]\right}=
]

[
=\frac1\Delta\left{L_\Delta^k(x^m)+C_m^1\frac\Delta2L_\Delta^k(x^{m-1})+C_m^2\left(\frac\Delta2\right)^2L_\Delta^k(x^{m-2})+C_m^3\left(\frac\Delta2\right)^3L_\Delta^k(x^{m-3})+\right.
]

[
\begin{gathered}
+\,L_\Delta^k[P_{m-4}(x)]-L_\Delta^k(x^m)+C_m^1\frac{\Delta}{2}L_\Delta^k(x^{m-1})
-C_m^2\left(\frac{\Delta}{2}\right)^2L_\Delta^k(x^{m-2})+\
+\,C_m^3\left(\frac{\Delta}{2}\right)^3L_\Delta^k(x^{m-3})-L_\Delta^k[Q_{m-4}(x)]
=\frac{2}{\Delta}\left{C_m^1\frac{\Delta}{2}L_\Delta^k(x^{m-1})
+C_m^3\left(\frac{\Delta}{2}\right)^3L_\Delta^k(x^{m-3})+\right.\
\left.
+\frac{1}{2}L_\Delta^k[P_{m-4}(x)]-\frac{1}{2}L_\Delta^k[Q_{m-4}(x)]\right}.
\end{gathered}
\tag{5}
]

Here (P_{m-4}(x)) and (Q_{m-4}(x)) are polynomials of degree (m-4).

From (5) we obtain:

[
\mathrm{A}^1.\quad \text{For } m<k+1\quad L_\Delta^{k+1}(x^m)=0.
]

[
\mathrm{B}^1.\quad \text{For } m=k+1\quad
L_\Delta^{k+1}(x^{k+1})=\frac{2}{\Delta}C_{k+1}^1\frac{\Delta}{2}L_\Delta^k(x^k)=
]

[
=(k+1)k!=(k+1)!\quad \text{and for } m=k+2\quad
L_\Delta^{k+1}(x^{k+2})=\frac{2}{\Delta}C_{k+2}^1\frac{\Delta}{2}L_\Delta^k(x^{k+1})=0.
]

[
\mathrm{C}^1.\quad \text{For } m=k+3\quad
L_\Delta^{k+1}(x^{k+3})=
C_{k+3}^1L_\Delta^k(x^{k+2})+
C_{k+3}^3\left(\frac{\Delta}{2}\right)^2L_\Delta^k(x^k)=
]

[
=(k+3)\frac{k\Delta^2}{24}(k+2)!
+\frac{(k+3)(k+2)(k+1)}{6}\frac{\Delta^2}{4}k!
=\frac{k\Delta^2}{24}(k+3)!+
]

[
+\frac{\Delta^2}{24}(k+3)!
=\frac{(k+1)\Delta^2}{24}(k+3)!
]

From (\mathrm{A}^0,\mathrm{B}^0,\mathrm{C}^0) and (\mathrm{A}^1,\mathrm{B}^1,\mathrm{C}^1) there follow A, B, and C.

Since the function (f(x)) is assumed differentiable up to order ((k+2)), the assertion of the theorem is easily obtained from A, B, and C.

(2^\circ). Let the computations be performed on a machine with an (n)-digit binary mantissa. Suppose that we compute the (k)-th derivative of (f) at the point (x) by the formula

[
f^{(k)}(x)=L_{\Delta,x_0}^k[f],
\tag{6}
]

using the scheme of successive differences, i.e., applying formula (3) (k(k+1)/2) times. Then the absolute error in the right-hand side of (6) as a result of the inaccuracy of machine computation is, on average,

[
\operatorname{err}_M \simeq
\frac{k\cdot 2^{-n}|f(x)|}{\sqrt{2}\,\Delta^k}.
\tag{7}
]

On the other hand, it follows from (2) that, when computing by formula (6), we incur an absolute error

[
\operatorname{err}_B \simeq
\frac{k\Delta^2}{24}\,|f^{(k+2)}(x)|.
\tag{8}
]

Minimizing with respect to (\Delta) the sum of (7) and (8), we obtain an expression for the optimal step (\Delta_k) in the machine computation of (f^{(k)}(x)):

[
\Delta_k=
\sqrt[k+2]{\frac{2^{-n}\cdot 12\,k\,|f(x)|}{\sqrt{2}\,|f^{(k+2)}(x)|}}.
\tag{9}
]

(3^\circ). When computing a one-sided derivative (at the points (x_0,x_1,\ldots,\ldots,x_k)), we obtain an absolute error in (f^{(k)}(x_0)) of order

[
f^{(k)}\left(\frac{x_0+x_k}{2}\right)-f^{(k)}(x_0)
\simeq \frac{k\Delta}{2}f^{(k+1)}(x).
\tag{8′}
]

Hence the expression for the optimal step in computing a one-sided derivative is

[
\Delta_k^{\mathrm{one\text{-}sided}}=
\sqrt[k+1]{\frac{2^{-n}\cdot 2k\,|f(x)|}{\sqrt{2}\,|f^{(k+1)}(x)|}}.
\tag{9′}
]

4°. Numerical example. For (n=40), for (f(x)=\exp(x)) at (x=1), the quantities obtained for (\Delta_k) are

[
\Delta_1 = 0.000198,\qquad
\Delta_2 = 0.00198,\qquad
\Delta_3 = 0.00746.
]

Under the same assumptions, for the step in the case of one-sided derivatives the quantities (\Delta_k^{\mathrm{one}}) are substantially smaller, namely:

[
\Delta_1^{\mathrm{one}} = 0.00000113,\qquad
\Delta_2^{\mathrm{one}} = 0.000137,\qquad
\Delta_3^{\mathrm{one}} = 0.00140.
]

To compute the ((k+2))-nd derivative one may use, for example, the same (L_{\Delta,x_0}^{k+2}[f]). It is only necessary to ensure that, for the given step, not all correct digits disappear. If this nevertheless happens, one should increase (\Delta) when computing (L_{\Delta,x_0}^{k+2}[f]).

Central Research Institute
of Patent Information and Techno-Economic
Studies

Received
4 III 1970