About the use of tables of special functions

Tiberiu Popoviciu
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Tiberiu Popoviciu
Institutul de Calcul

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T. Popoviciu, Über die Verwendung der Tabellen spezieller Funktionen, Numerische Methoden der Approximationstheorie, Band 2 (Tagung, Math. Forschungsinst., Oberwolfach, 1973), pp. 101-109. Internat. Schriftenreihe Numer. Math., Band 26, Birkhäuser, Basel, 1975 (in German).

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1975 c -Popoviciu- Numer. Meth. Approximationstheorie – Uber die verwendung der tabellen spezieller funktionen

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International series of numerical mathematics

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New York : Academic Press

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0373-3149

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[MR0386220, Zbl 0323.65002].

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1975 c -Popoviciu- Numer. Meth. Approximation theory - On the use of tables of special
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On the use of tables of special functions by T. Popoviciu in Cluj

  1. Given a table containing the values ​​of a real function f f fffthe real variables x x xxxfor a finite number m > 1 m > 1 m > 1m>1m>1of values x 1 , x 2 , … , x m x 1 , x 2 , … , x m x_(1),x_(2),dots,x_(m)x_{1}, x_{2}, \ldots, x_{m}x1,x2,…,xmwhich contains independent variables.
The task is to expand the table by interpolation for other values ​​of x to x to xtox \mathrm{to}xtoTo solve this problem, one usually starts with a linear interpolation in the successive intervals defined by the points x i x i x_(i)x_{i}xi, i = 1 , 2 , … , m i = 1 , 2 , … , m i=1,2,dots,mi=1,2, \ldots, mi=1,2,…,mare intended. It is assumed that x 1 < x 2 < … < x m x 1 < x 2 < … < x m x_(1) < x_(2) < dots < x_(m)x_{1}<x_{2}<\ldots<x_{m}x1<x2<…<xmholds, the linear interpolation consists in using the function f f fffin the completed interval [ x 1 , x m ] x 1 , x m [x_(1),x_(m)]\left[x_{1}, x_{m}\right][x1,xm]the polygonal function P P PPPused, which is in the curve y = f ( x ) y = f ( x ) y=f(x)quady=f(x) \quady=f(x)is inscribed and the corners ( x i , f ( x i ) x i , f x i x_(i),f(x_(i))x_{i}, f\left(x_{i}\right)xi,f(xi)), i = 1 , 2 , … , m i = 1 , 2 , … , m i=1,2,dots,mi=1,2, \ldots, mi=1,2,…,mExpressed analytically, this means that the function f f fffin each interval [ x i , x i + 1 ] x i , x i + 1 [x_(i),x_(i+1)]\left[x_{i}, x_{i+1}\right][xi,xi+1]by the first-degree Lagrange polynomial L ( x i , x i + 1 ; f ) L x i , x i + 1 ; f L(x_(i),x_(i+1);f)L\left(x_{i}, x_{i+1} ; f\right)L(xi,xi+1;f)which is replaced in the endpoints x i , x i + 1 x i , x i + 1 x_(i),x_(i+1)x_{i}, x_{i+1}xi,xi+1the same values ​​as the function f f fffaccepts.
Since we have assumed that the values ​​of the function f f fffoutside the points x i x i x_(i)x_{i}xi, i = 1 , 2 , … , m i = 1 , 2 , … , m i=1,2,dots,mi=1,2, \ldots, mi=1,2,…,mare not known, we cannot generally say anything about the quality of the approximation
(1) f ( x ) ≈ L ( x i , x i + 1 ; f ) ( x ) , (1) f ( x ) ≈ L x i , x i + 1 ; f ( x ) , {:(1)f(x)~~L(x_(i),x_(i+1);f)(x)",":}\begin{equation*} f(x) \approx L\left(x_{i}, x_{i+1} ; f\right)(x), \tag{1} \end{equation*}(1)f(x)≈L(xi,xi+1;f)(x),
where x x xxxis a point that lies between the points x i x i x_(i)x_{i}xiand x i + 1 x i + 1 x_(i+1)x_{i+1}xi+1The task posed at the beginning is treated quite differently if it
is also known that the function f f fffhas a certain behavior; e.g., if the function f f fffis positive, then the approximation functions L ( x i , x i + 1 ; f ) L x i , x i + 1 ; f L(x_(i),x_(i+1);f)L\left(x_{i}, x_{i+1} ; f\right)L(xi,xi+1;f), in the interval
[ x i , x i + 1 ] x i , x i + 1 [x_(i),x_(i+1)]\left[x_{i}, x_{i+1}\right][xi,xi+1]for each i = 1 , 2 , … , m − 1 i = 1 , 2 , … , m − 1 i=1,2,dots,m-1i=1,2, \ldots, m-1i=1,2,…,m−1positive. If the function f f fffincreasing, the above approximation functions (for all x x xxx), and also the polygonal function mentioned above P P PPP, increasing etc.
We will examine another important special case of such behavior below.
It should be noted that the tables commonly used do not contain the exact values ​​of the function represented, but rather approximate values. This is the case, for example, with logarithmic tables. We will continue to bear this fact in mind.
2. We now want to consider the interpolation problem for the case of a function f f fffwhich in one the points x i , i = 1 , 2 , … , m x i , i = 1 , 2 , … , m x_(i),i=1,2,dots,mx_{i}, i=1,2, \ldots, mxi,i=1,2,…,m, is continuous and concave. For such a function, the divided differences of second order are negative. This is the case, for example, for a function whose second derivative is negative. This case is, for example, for the function ln ⁡ x ln ⁡ x ln x\ln xln⁡xbefore, which (for x > 0 x > 0 x > 0x>0x>0) is concave, so that the following can be applied to the logarithm tables.
The problem to be investigated can then be formulated as follows:
Let it be a continuous concave function f f fffin an interval that contains the points a , b ( a < b ) a , b ( a < b ) a,b(a < b)a, b(a<b)a,b(a<b)The approximation of this function by a polynomial of degree 1 in the interval [ a , b a , b a,ba, ba,b] should be examined.
The Lagrange polynomial L = L ( a , b ; f ) L = L ( a , b ; f ) L=L(a,b;f)L=L(a, b ; f)L=L(a,b;f), which represents the values ​​of f f fffin the endpoints a , b a , b a,ba, ba,byields such an approximation. The approximation of f ( x ) f ( x ) f(x)f(x)f(x)through L ( a , b ; f ) ( x ) L ( a , b ; f ) ( x ) L(a,b;f)(x)quadL(a, b ; f)(x) \quadL(a,b;f)(x)is for every x ∈ [ a , b ] x ∈ [ a , b ] x in[a,b]x \in[a, b]x∈[a,b]an approximation from below. To obtain a better approximation, one tries to use the polynomial L L LLLto be replaced by another approximation polynomial.
According to an idea by EV VORONOVSKAIA [3], instead of L L LLLtake the first-degree polynomial that best approximates the function in the sense of Chebyshev. This choice is justified by the fact that this polynomial, due to the special behavior (concavity) of the function, f f ffffrom the polynomial L L LLLonly by a positive constant. More precisely: the Chebyshev polynomial in question is equal to L + ρ L + ρ L+rhoL+\rhoL+ρ, where
(2) 2 ρ = max [ a , b ] ( f − L ) = f ( ξ ) − L ( ξ ) (2) 2 ρ = max [ a , b ]   ( f − L ) = f ( ξ ) − L ( ξ ) {:(2)2rho=max_([a,b])(f-L)=f(xi)-L(xi):}\begin{equation*} 2 \rho=\max _{[a, b]}(f-L)=f(\xi)-L(\xi) \tag{2} \end{equation*}(2)2ρ=max[a,b](f−L)=f(ξ)−L(ξ)
applies. The point ξ ∈ ] a , b [ ( a < ξ < b ) ξ ∈ ] a , b [ ( a < ξ < b ) xi in]a,b[(a < xi < b)\xi \in] a, b[(a<\xi<b)ξ∈]a,b[(a<ξ<b)is uniquely determined. If the function f f fffdifferentiable, it is the only root of the equation
f ′ ( x ) − L ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a = 0 . f ′ ( x ) − L ′ ( x ) = f ′ ( x ) − f ( b ) − f ( a ) b − a = 0 . f^(')(x)-L^(')(x)=f^(')(x)-(f(b)-f(a))/(b-a)=0.f^{\prime}(x)-L^{\prime}(x)=f^{\prime}(x)-\frac{f(b)-f(a)}{b-a}=0 .f′(x)−L′(x)=f′(x)−f(b)−f(a)b−a=0.
However, the calculation of the number p is generally too complicated to be used in numerical calculations. For example, let us assume that we have a table showing the values ​​of the function ln ⁡ x ln ⁡ x ln x\ln xln⁡xfor the natural numbers up to a sufficiently large number, and we set a = n , b = n + 1 a = n , b = n + 1 a=n,b=n+1a=n, b=n+1a=n,b=n+1, where n n nnndenotes a natural number, then
(3) ξ = 1 ln ⁡ ( 1 + 1 n ) . (3) ξ = 1 ln ⁡ 1 + 1 n . {:(3)xi=(1)/(ln(1+(1)/(n))).:}\begin{equation*} \xi=\frac{1}{\ln \left(1+\frac{1}{n}\right)} . \tag{3} \end{equation*}(3)ξ=1ln⁡(1+1n).
The corresponding number 2p:
ln ⁡ ( 1 + 1 n ) n − ln ⁡ ln ⁡ ( 1 + 1 n ) n − 1 ln ⁡ 1 + 1 n n − ln ⁡ ln ⁡ 1 + 1 n n − 1 ln (1+(1)/(n))^(n)-ln ln (1+(1)/(n))^(n)-1\ln \left(1+\frac{1}{n}\right)^{n}-\ln \ln \left(1+\frac{1}{n}\right)^{n}-1ln⁡(1+1n)n−ln⁡ln⁡(1+1n)n−1
cannot be obtained by rational operations from the data in the table.
Therefore, it is more sensible to modify the approximation polynomial L + ρ L + ρ L+rhoL+\rhoL+ρThere are several ways to make such a change. The ones we will discuss below seem to be the simplest.
3. Let us take the polynomial L + λ L + λ L+lambdaL+\lambdaL+λwhere λ λ lambda\lambdaλis a positive constant given by the equation
(4) 2 λ = f ( c ) − L ( c ) (4) 2 λ = f ( c ) − L ( c ) {:(4)2lambda=f(c)-L(c):}\begin{equation*} 2 \lambda=f(c)-L(c) \tag{4} \end{equation*}(4)2λ=f(c)−L(c)
is intended. c c ccca given number with a < c < b a < c < b a < c < ba<c<ba<c<b. It then applies o < λ ≤ ρ o < λ ≤ ρ o < lambda <= rhoo<\lambda \leq \rhoO<λ≤ρ. Equality λ = ρ λ = ρ lambda=rho\lambda=\rhoλ=ρonly occurs in case c = ξ c = ξ c=xic=\xic=ξIf a positive number λ λ lambda\lambdaλspecified which is smaller than ρ ρ rho\rhoρthere are two points c c ccc, for which equation (4) exists. ρ ρ rho\rhoρand ξ ξ xi\xiξhave the meaning given in the preceding section.
The tendon y = L ( x ) y = L ( x ) y=L(x)y=L(x)y=L(x)parallel line y = L ( x ) + λ y = L ( x ) + λ y=L(x)+lambday=L(x)+\lambday=L(x)+λcuts the bow y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x)in two points ( a 1 , f ( a 1 ) ) , ( b 1 , f ( b 1 ) ) a 1 , f a 1 , b 1 , f b 1 (a_(1),f(a_(1))),quad(b_(1),f(b_(1)))\left(a_{1}, f\left(a_{1}\right)\right), \quad\left(b_{1}, f\left(b_{1}\right)\right)(a1,f(a1)),(b1,f(b1)), where a < a 1 < b 1 < b a < a 1 < b 1 < b a < a_(1) < b_(1) < ba<a_{1}<b_{1}<ba<a1<b1<b. The abscissas a 1 a 1 a_(1)a_{1}a1and b 1 b 1 b_(1)b_{1}b1However, you cannot even use simple functions f f fffdetermine exactly.
Example 1. The function x ( 1 − x ) x ( 1 − x ) x(1-x)x(1-x)x(1−x)is in the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1]concave. We choose a = 0 a = 0 a=0a=0a=0,
b = 1 b = 1 b=1b=1b=1. The Chebyshev polynomial of first degree is then the constant 1 8 1 8 (1)/(8)\frac{1}{8}18. The roots 1 2 − 1 2 2 , 1 2 + 1 2 2 1 2 − 1 2 2 , 1 2 + 1 2 2 (1)/(2)-(1)/(2sqrt2),(1)/(2)+(1)/(2sqrt2)\frac{1}{2}-\frac{1}{2 \sqrt{2}}, \frac{1}{2}+\frac{1}{2 \sqrt{2}}12−122,12+122the equation x ( 1 − x ) = 1 8 x ( 1 − x ) = 1 8 x(1-x)=(1)/(8)x(1-x)=\frac{1}{8}x(1−x)=18, which in this case are the abscissas a 1 , b 1 a 1 , b 1 a_(1),b_(1)a_{1}, b_{1}a1,b1cannot be calculated from the data of the problem using rational operations. However, the abscissas a ′ , b ′ a ′ , b ′ a^('),b^(')a^{\prime}, b^{\prime}a′,b′the intersection points of the lines y = L ( x ) + λ y = L ( x ) + λ y=L(x)+lambday=L(x)+\lambday=L(x)+λwith the straight line y = L ( a , c ; f ) ( x ) y = L ( a , c ; f ) ( x ) y=L(a,c;f)(x)y=L(a, c ; f)(x)y=L(a,c;f)(x)or y = L ( c , b ; f ) ( x ) y = L ( c , b ; f ) ( x ) y=L(c,b;f)(x)y=L(c, b ; f)(x)y=L(c,b;f)(x)can be easily determined.
A simple calculation gives a ′ = a + c 2 , b ′ = c + b 2 a ′ = a + c 2 , b ′ = c + b 2 a^(')=(a+c)/(2),quadb^(')=(c+b)/(2)a^{\prime}=\frac{a+c}{2}, \quad b^{\prime}=\frac{c+b}{2}a′=a+c2,b′=c+b2and because of the concavity of the function f f fffapplies a < a 1 < a ′ < c < b ′ < b 1 < b a < a 1 < a ′ < c < b ′ < b 1 < b a < a_(1) < a^(') < c < b^(') < b_(1) < ba<a_{1}<a^{\prime}<c<b^{\prime}<b_{1}<ba<a1<a′<c<b′<b1<b.
From this follows:
I. In the interval [ a ′ , b ′ ] a ′ , b ′ [a^('),b^(')]\left[a^{\prime}, b^{\prime}\right][a′,b′], which is the length b − a 2 b − a 2 (b-a)/(2)\frac{b-a}{2}b−a2, the approximation of f ( x ) f ( x ) f(x)f(x)f(x)through L ( x ) + λ L ( x ) + λ L(x)+lambdaL(x)+\lambdaL(x)+λalso an approximation from below, but better than the approximation of f ( x ) f ( x ) f(x)f(x)f(x)through L ( x ) L ( x ) L(x)L(x)L(x).
This property also applies to x ∈ ] a 1 , b 1 [ x ∈ ] a 1 , b 1 [ x in]a_(1),b_(1)[x \in] a_{1}, b_{1}[x∈]a1,b1[, but, as noted earlier, the endpoints of this interval are difficult to calculate.
4. Outside the interval ] a 1 , b 1 [ ] a 1 , b 1 [ ]a_(1),b_(1)[] a_{1}, b_{1}[]a1,b1[, so for x ∈ [ a , b ] ∖ ] a 1 , b 1 [ x ∈ [ a , b ] ∖ ] a 1 , b 1 [ x in[a,b]\\]a_(1),b_(1)[x \in[a, b] \backslash] a_{1}, b_{1}[x∈[a,b]∖]a1,b1[results L ( x ) + λ L ( x ) + λ L(x)+lambdaL(x)+\lambdaL(x)+λan approximation from above for f ( x ) f ( x ) f(x)f(x)f(x).
There are now ( a 2 , f ( a 2 ) ) , ( b 2 , f ( b 2 ) ) a 2 , f a 2 , b 2 , f b 2 {:a_(2),f(a_(2))),(b_(2),f(b_(2)))\left.a_{2}, f\left(a_{2}\right)\right),\left(b_{2}, f\left(b_{2}\right)\right)a2,f(a2)),(b2,f(b2)), where a < a 2 < b 2 < b a < a 2 < b 2 < b a < a_(2) < b_(2) < ba<a_{2}<b_{2}<ba<a2<b2<bis the intersection points of the lines y = L ( x ) + 1 2 λ y = L ( x ) + 1 2 λ y=L(x)+(1)/(2)lambday=L(x)+\frac{1}{2} \lambday=L(x)+12λwith the bow y = f ( x ) y = f ( x ) y=f(x)y=f(x)y=f(x), and a ′ ′ a ′ ′ a^('')a^{\prime \prime}a′′or b ′ ′ b ′ ′ b^('')b^{\prime \prime}b′′the abscissa of the intersection point of this line and the line y = L ( a , c ; f ) ( x ) y = L ( a , c ; f ) ( x ) y=L(a,c;f)(x)y=L(a, c ; f)(x)y=L(a,c;f)(x)or y = L ( c , b ; f ) ( x ) y = L ( c , b ; f ) ( x ) y=L(c,b;f)(x)y=L(c, b ; f)(x)y=L(c,b;f)(x)As in the case of a 1 , b 1 a 1 , b 1 a_(1),b_(1)a_{1}, b_{1}a1,b1the abscissas can a 2 , b 2 a 2 , b 2 a_(2),b_(2)a_{2}, b_{2}a2,b2in general cannot be explicitly calculated by rational operations from the data of the problem.
Example 2, For the function considered in Example 1 x ( 1 − x ) x ( 1 − x ) x(1-x)x(1-x)x(1−x)is a 2 = 1 2 − 3 4 , b 2 = 1 2 + 3 4 a 2 = 1 2 − 3 4 , b 2 = 1 2 + 3 4 a_(2)=(1)/(2)-(sqrt3)/(4),quadb_(2)=(1)/(2)+(sqrt3)/(4)a_{2}=\frac{1}{2}-\frac{\sqrt{3}}{4}, \quad b_{2}=\frac{1}{2}+\frac{\sqrt{3}}{4}a2=12−34,b2=12+34.
It applies a n = 3 a + c 4 , b n = c + 3 b 4 a n = 3 a + c 4 , b n = c + 3 b 4 a^(n)=(3a+c)/(4),b^(n)=(c+3b)/(4)a^{n}=\frac{3 a+c}{4}, b^{n}=\frac{c+3 b}{4}an=3a+c4,bn=c+3b4. Furthermore, a < a 2 < a 1 < b 1 < b 2 < b a < a 2 < a 1 < b 1 < b 2 < b a < a_(2) < a_(1) < b_(1) < b_(2) < ba<a_{2}<a_{1}<b_{1}<b_{2}<ba<a2<a1<b1<b2<band a < a ′ ′ < a ′ < c < b ′ < b ′ ′ < b a < a ′ ′ < a ′ < c < b ′ < b ′ ′ < b a < a^('') < a^(') < c < b^(') < b^('') < ba<a^{\prime \prime}<a^{\prime}<c<b^{\prime}<b^{\prime \prime}<ba<a′′<a′<c<b′<b′′<b. On the one hand, if one considers that a 1 < ξ < b 1 a 1 < ξ < b 1 a_(1) < xi < b_(1)a_{1}<\xi<b_{1}a1<ξ<b1is, where ξ ξ xi\xiξis the point determined by (2), and on the other hand known properties of the concave functions, it follows that the function f f fffon the interval [ a , ξ a , ξ a,xia, \xia,ξ] rising and on the interval [ ξ , b ] [ ξ , b ] [xi,b][\xi, b][ξ,b]This function also has the same monotonicity properties on the interval [ a , a 1 ] a , a 1 [a,a_(1)]\left[a, a_{1}\right][a,a1]or [ b 1 , b b 1 , b b_(1),bb_{1}, bb1,b]. This means:
II. In the interval [ a n , b n a n , b n a^(n),b^(n)a^{n}, b^{n}an,bn], which is the length 3 4 3 4 (3)/(4)\frac{3}{4}34(ba), the approximation of f ( x ) f ( x ) f(x)f(x)f(x)through L ( x ) + λ L ( x ) + λ L(x)+lambdaL(x)+\lambdaL(x)+λan absolute error that is smaller than that of the approximation of f ( x ) f ( x ) f(x)f(x)f(x)through L ( x ) L ( x ) L(x)L(x)L(x).
This property also applies to x ∈ ] a 2 , b 2 [ x ∈ ] a 2 , b 2 x in]a_(2),b_(2)[:}x \in] a_{2}, b_{2}\left[\right.x∈]a2,b2[, but regarding the endpoints a 2 , b 2 a 2 , b 2 a_(2),b_(2)a_{2}, b_{2}a2,b2the same remark as the one about a 1 a 1 a_(1)a_{1}a1and b 1 b 1 b_(1)b_{1}b15. To illustrate
an application, we return to the natural logarithm table, which contains values ​​for a sufficiently large number of consecutive natural numbers.
To determine the value of ln ⁡ ( n + r ) ln ⁡ ( n + r ) ln(n+r)\ln (n+r)ln⁡(n+r)(with o < r < 1 o < r < 1 o < r < 1o<r<1O<r<1), one usually interpolates linearly between n n nnnand n + 1 n + 1 n+1n+1n+1using the Lagrange polynomial. One then obtains
(5) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) . (5) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) . {:(5)ln(n+r)~~(1-r)ln n+r ln(n+1).:}\begin{equation*} \ln (n+r) \approx(1-r) \ln n+r \ln (n+1) . \tag{5} \end{equation*}(5)ln⁡(n+r)≈(1−r)ln⁡n+rln⁡(n+1).
If we apply the theory presented in the previous sections and c = 2 n + 1 2 c = 2 n + 1 2 c=(2n+1)/(2)c=\frac{2 n+1}{2}c=2n+12set (i.e. c c cccas the midpoint of the interval [ n , n + 1 n , n + 1 n,n+1n, n+1n,n+1]), we get the approximation
(6) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) + λ 1 , (6) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) + λ 1 , {:(6)ln(n+r)~~(1-r)ln n+r ln(n+1)+lambda_(1)",":}\begin{equation*} \ln (n+r) \approx(1-r) \ln n+r \ln (n+1)+\lambda_{1}, \tag{6} \end{equation*}(6)ln⁡(n+r)≈(1−r)ln⁡n+rln⁡(n+1)+λ1,
where
λ 1 = 1 2 ln ⁡ 2 n + 1 2 − 1 4 ( ln ⁡ n + ln ⁡ ( n + 1 ) ) = 1 4 ln ⁡ ( 2 n + 1 ) 2 4 n ( n + 1 ) = λ 1 = 1 2 ln ⁡ 2 n + 1 2 − 1 4 ( ln ⁡ n + ln ⁡ ( n + 1 ) ) = 1 4 ln ⁡ ( 2 n + 1 ) 2 4 n ( n + 1 ) = lambda_(1)=(1)/(2)ln((2n+1)/(2))-(1)/(4)(ln n+ln(n+1))=(1)/(4)ln(((2n+1)^(2))/(4n(n+1)))=\lambda_{1}=\frac{1}{2} \ln \frac{2 n+1}{2}-\frac{1}{4}(\ln n+\ln (n+1))=\frac{1}{4} \ln \frac{(2 n+1)^{2}}{4 n(n+1)}=λ1=12ln⁡2n+12−14(ln⁡n+ln⁡(n+1))=14ln⁡(2n+1)24n(n+1)=
(7) = 1 4 ln ⁡ ( 1 + 1 4 n ( n + 1 ) ) = 1 16 n ( n + 1 ) ln ⁡ ( 1 + 1 4 n ( n + 1 ) ) 4 n ( n + 1 ) . (7) = 1 4 ln ⁡ 1 + 1 4 n ( n + 1 ) = 1 16 n ( n + 1 ) ln ⁡ 1 + 1 4 n ( n + 1 ) 4 n ( n + 1 ) . {:(7)=(1)/(4)ln(1+(1)/(4n(n+1)))=(1)/(16 n(n+1))ln (1+(1)/(4n(n+1)))^(4n(n+1)).:}\begin{equation*} =\frac{1}{4} \ln \left(1+\frac{1}{4 n(n+1)}\right)=\frac{1}{16 n(n+1)} \ln \left(1+\frac{1}{4 n(n+1)}\right)^{4 n(n+1)} . \tag{7} \end{equation*}(7)=14ln⁡(1+14n(n+1))=116n(n+1)ln⁡(1+14n(n+1))4n(n+1).
In this case, a ′ = n + 1 4 , b ′ = n + 3 4 , a ′ ′ = n + 1 8 , b ′ ′ = n + 7 8 a ′ = n + 1 4 , b ′ = n + 3 4 , a ′ ′ = n + 1 8 , b ′ ′ = n + 7 8 a^(')=n+(1)/(4),b^(')=n+(3)/(4),a^('')=n+(1)/(8),b^('')=n+(7)/(8)a^{\prime}=n+\frac{1}{4}, b^{\prime}=n+\frac{3}{4}, a^{\prime \prime}=n+\frac{1}{8}, b^{\prime \prime}=n+\frac{7}{8}a′=n+14,b′=n+34,a′′=n+18,b′′=n+78
Approximation (6) is a bottom-up approximation that is 1 4 ≤ r ≤ 3 4 1 4 ≤ r ≤ 3 4 (1)/(4) <= r <= (3)/(4)\frac{1}{4} \leq r \leq \frac{3}{4}14≤r≤34is better than (5) and for 1 8 ≤ r ≤ 7 8 1 8 ≤ r ≤ 7 8 (1)/(8) <= r <= (7)/(8)\frac{1}{8} \leq r \leq \frac{7}{8}18≤r≤78an approximation with a smaller absolute error than (5).
Note that for a rational number r r rrrthe approximate value given in (6) can be calculated using elementary operations, namely ln ⁡ n , ln ⁡ ( n + 1 ) ln ⁡ n , ln ⁡ ( n + 1 ) ln n,ln(n+1)\ln n, \ln (n+1)ln⁡n,ln⁡(n+1)and ln ⁡ 2 n + 1 2 = ln ⁡ ( 2 n + 1 ) − ln ⁡ 2 ln ⁡ 2 n + 1 2 = ln ⁡ ( 2 n + 1 ) − ln ⁡ 2 ln((2n+1)/(2))=ln(2n+1)-ln 2\ln \frac{2 n+1}{2}=\ln (2 n+1)-\ln 2ln⁡2n+12=ln⁡(2n+1)−ln⁡2, values ​​that are either given in the table or can be determined immediately from the data in the table.
From (7) it follows that the value of λ 1 λ 1 lambda_(1)\lambda_{1}λ1with increasing n n nnnfalls and for n → + ∞ n → + ∞ n rarr+oon \rightarrow+\inftyn→+∞tends towards zero. Furthermore,
(8) λ 1 < 1 16 n ( n + 1 ) (8) λ 1 < 1 16 n ( n + 1 ) {:(8)lambda_(1) < (1)/(16 n(n+1)):}\begin{equation*} \lambda_{1}<\frac{1}{16 n(n+1)} \tag{8} \end{equation*}(8)λ1<116n(n+1)
and n ( n + 1 ) λ 1 → 1 16 n ( n + 1 ) λ 1 → 1 16 n(n+1)lambda_(1)rarr(1)/(16)n(n+1) \lambda_{1} \rightarrow \frac{1}{16}n(n+1)λ1→116for n → + ∞ n → + ∞ n rarr+oon \rightarrow+\inftyn→+∞, so λ 1 λ 1 lambda_(1)\lambda_{1}λ1asymptotically equal 1 16 n ( n + 1 ) 1 16 n ( n + 1 ) (1)/(16 n(n+1))\frac{1}{16 n(n+1)}116n(n+1).
Example 3. Let n = 2 , r = 3 8 = o , 375 n = 2 , r = 3 8 = o , 375 n=2,r=(3)/(8)=o,375n=2, r=\frac{3}{8}=o, 375n=2,r=38=O,375. From a logarithmic table with 10 decimal places we take the values
ln ⁡ 2 = 0 , 6931471806 ; ln ⁡ 3 = 1 , 0986122887 ; ln ⁡ 5 = 1 , 6094379124 . ln ⁡ 2 = 0 , 6931471806 ; ln ⁡ 3 = 1 , 0986122887 ; ln ⁡ 5 = 1 , 6094379124 . ln 2=0,6931471806;ln 3=1,0986122887;ln 5=1,6094379124.\ln 2=0,6931471806 ; \ln 3=1,0986122887 ; \ln 5=1,6094379124 .ln⁡2=0,6931471806;ln⁡3=1,0986122887;ln⁡5=1,6094379124.
You then receive
λ 1 = 1 2 ln ⁡ 5 − 3 4 ln ⁡ 2 − 1 4 ln ⁡ 3 = 0 , 0102054986 λ 1 = 1 2 ln ⁡ 5 − 3 4 ln ⁡ 2 − 1 4 ln ⁡ 3 = 0 , 0102054986 lambda_(1)=(1)/(2)ln 5-(3)/(4)ln 2-(1)/(4)ln 3=0,0102054986\lambda_{1}=\frac{1}{2} \ln 5-\frac{3}{4} \ln 2-\frac{1}{4} \ln 3=0,0102054986λ1=12ln⁡5−34ln⁡2−14ln⁡3=0,0102054986
and
( 1 − r ) ln ⁡ 2 + r ln ⁡ 3 = 5 8 ln ⁡ 2 + 3 8 ln ⁡ 3 = 0 , 8376965961 . ( 1 − r ) ln ⁡ 2 + r ln ⁡ 3 = 5 8 ln ⁡ 2 + 3 8 ln ⁡ 3 = 0 , 8376965961 . (1-r)ln 2+r ln 3=(5)/(8)ln 2+(3)/(8)ln 3=0,8376965961.(1-r) \ln 2+r \ln 3=\frac{5}{8} \ln 2+\frac{3}{8} \ln 3=0,8376965961 .(1−r)ln⁡2+rln⁡3=58ln⁡2+38ln⁡3=0,8376965961.
Accordingly,
ln ⁡ 2 , 375 ≈ 0 , 8376965961 + 0 , 0103054986 = 0 , 847902 o 947 . ln ⁡ 2 , 375 ≈ 0 , 8376965961 + 0 , 0103054986 = 0 , 847902 o 947 . ln 2,375~~0,8376965961+0,0103054986=0,847902 o 947.\ln 2,375 \approx 0,8376965961+0,0103054986=0,847902 o 947 .ln⁡2,375≈0,8376965961+0,0103054986=0,847902O947.
In the same table we find the value ln ⁡ 2 , 375 = 0 , 8649975374 ln ⁡ 2 , 375 = 0 , 8649975374 ln 2,375=0,8649975374\ln 2,375=0,8649975374ln⁡2,375=0,8649975374; so we have an exact decimal place for ln ⁡ 2 , 375 ln ⁡ 2 , 375 ln 2,375\ln 2,375ln⁡2,3756.
A formula similar to formula (6) can be obtained as follows. We start from the interpolation formula
(9) f ( x ) − L ( x ) = ( x − a ) ( x − b ) [ a , b , x ; f ] ( a < x < b ) (9) f ( x ) − L ( x ) = ( x − a ) ( x − b ) [ a , b , x ; f ] ( a < x < b ) {:(9)f(x)-L(x)=(x-a)(x-b)[a","b","x;f]quad(a < x < b):}\begin{equation*} f(x)-L(x)=(x-a)(x-b)[a, b, x ; f] \quad(a<x<b) \tag{9} \end{equation*}(9)f(x)−L(x)=(x−a)(x−b)[a,b,x;f](a<x<b)
where [ a , b , x ; f ] [ a , b , x ; f ] [a,b,x;f][a, b, x ; f][a,b,x;f]the second-order divided difference of the interval [ a , b ] [ a , b ] [a,b][a, b][a,b]concave function f f fffregarding the points a , b , x a , b , x a,b,xa, b, xa,b,xreferred to. Is f f ffftwice differentiable and the second derivative (negative) ≤ − M ≤ − M <= -M\leq-M≤−Mon the interval [ a , b ] [ a , b ] [a,b][a, b][a,b], where M M MMMis a positive constant, we get from (2)
(10) 2 ρ > ( ξ − a ) ( b − ξ ) M 2 (10) 2 ρ > ( ξ − a ) ( b − ξ ) M 2 {:(10)2rho > (xi-a)(b-xi)(M)/(2):}\begin{equation*} 2 \rho>(\xi-a)(b-\xi) \frac{M}{2} \tag{10} \end{equation*}(10)2ρ>(ξ−a)(b−ξ)M2
In the case of the function ln ⁡ x ln ⁡ x ln x\ln xln⁡xis M = 1 ( n + 1 ) 2 M = 1 ( n + 1 ) 2 M=(1)/((n+1)^(2))M=\frac{1}{(n+1)^{2}}M=1(n+1)2, if you a = n , b = n + 1 a = n , b = n + 1 a=n,b=n+1quada=n, b=n+1 \quada=n,b=n+1selects, while the value of ξ ξ xi\xiξis given by (3). Considering exercise I-168 from the well-known collection of exercises by G. PÓLYA and G. SZEGÖ [1], it turns out that the sequence ( ( 1 + 1 n ) n + 1 2 ) n = 1 + ∞ 1 + 1 n n + 1 2 n = 1 + ∞ ((1+(1)/(n))^(n+(1)/(2)))_(n=1)^(+oo)\left(\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\right)_{n=1}^{+\infty}((1+1n)n+12)n=1+∞is falling 1 ) 1 ) ^(1)){ }^{1)}1)To prove this statement, one can use the function y = ( x + 1 2 ) ln ⁡ ( 1 + 1 x ) y = x + 1 2 ln ⁡ 1 + 1 x y=(x+(1)/(2))ln(1+(1)/(x))y=\left(x+\frac{1}{2}\right) \ln \left(1+\frac{1}{x}\right)y=(x+12)ln⁡(1+1x)use for the y ′ = ln ⁡ ( 1 + 1 x ) − 2 x + 1 x ( x + 1 ) y ′ = ln ⁡ 1 + 1 x − 2 x + 1 x ( x + 1 ) y^(')=ln(1+(1)/(x))-(2x+1)/(x(x+1))y^{\prime}=\ln \left(1+\frac{1}{x}\right)-\frac{2 x+1}{x(x+1)}y′=ln⁡(1+1x)−2x+1x(x+1)and y ′ ′ = 1 2 x 2 ( x + 1 ) 2 > 0 y ′ ′ = 1 2 x 2 ( x + 1 ) 2 > 0 y^('')=(1)/(2x^(2)(x+1)^(2)) > 0y^{\prime \prime}=\frac{1}{2 x^{2}(x+1)^{2}}>0y′′=12x2(x+1)2>0
This results in lim x → + ∞ y ′ = 0 ; lim x → + ∞   y ′ = 0 ; lim_(x rarr+oo)y^(')=0;quad\lim _{x \rightarrow+\infty} y^{\prime}=0 ; \quadlimx→+∞y′=0;so y ′ < 0 y ′ < 0 y^(') < 0quady^{\prime}<0 \quady′<0for x > 0 x > 0 x > 0x>0x>0.
Using the function y = ( x + 1 3 ) ln ⁡ ( 1 + 1 x ) y = x + 1 3 ln ⁡ 1 + 1 x y=(x+(1)/(3))ln(1+(1)/(x))y=\left(x+\frac{1}{3}\right) \ln \left(1+\frac{1}{x}\right)y=(x+13)ln⁡(1+1x)one can show that the sequence ( ( 1 + 1 n ) n + 1 3 ) n = 1 + ∞ 1 + 1 n n + 1 3 n = 1 + ∞ ((1+(1)/(n))^(n+(1)/(3)))_(n=1)^(+oo)\left(\left(1+\frac{1}{n}\right)^{n+\frac{1}{3}}\right)_{n=1}^{+\infty}((1+1n)n+13)n=1+∞Both consequences have increased the number e e eeeas the limit. We therefore obtain
n + 1 3 < ξ < n + 1 2 < n + 2 3 . n + 1 3 < ξ < n + 1 2 < n + 2 3 . n+(1)/(3) < xi < n+(1)/(2) < n+(2)/(3).n+\frac{1}{3}<\xi<n+\frac{1}{2}<n+\frac{2}{3} .n+13<ξ<n+12<n+23.
Accordingly,
( ξ − n ) ( n + 1 − ξ ) = ( ξ − n − 1 3 ) ( n + 2 3 − ξ ) + 2 9 ≥ 2 9 ( ξ − n ) ( n + 1 − ξ ) = ξ − n − 1 3 n + 2 3 − ξ + 2 9 ≥ 2 9 (xi-n)(n+1-xi)=(xi-n-(1)/(3))(n+(2)/(3)-xi)+(2)/(9) >= (2)/(9)(\xi-n)(n+1-\xi)=\left(\xi-n-\frac{1}{3}\right)\left(n+\frac{2}{3}-\xi\right)+\frac{2}{9} \geq \frac{2}{9}(ξ−n)(n+1−ξ)=(ξ−n−13)(n+23−ξ)+29≥29
and (10) gives
(11) 2 ρ > 1 9 ( n + 1 ) 2 (11) 2 ρ > 1 9 ( n + 1 ) 2 {:(11)2rho > (1)/(9(n+1)^(2)):}\begin{equation*} 2 \rho>\frac{1}{9(n+1)^{2}} \tag{11} \end{equation*}(11)2ρ>19(n+1)2
Instead of formula (6) you can also use the formula
(12) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) + λ 2 ( 0 < r < 1 ) (12) ln ⁡ ( n + r ) ≈ ( 1 − r ) ln ⁡ n + r ln ⁡ ( n + 1 ) + λ 2 ( 0 < r < 1 ) {:(12)ln(n+r)~~(1-r)ln n+r ln(n+1)+lambda_(2)quad(0 < r < 1):}\begin{equation*} \ln (n+r) \approx(1-r) \ln n+r \ln (n+1)+\lambda_{2} \quad(0<r<1) \tag{12} \end{equation*}(12)ln⁡(n+r)≈(1−r)ln⁡n+rln⁡(n+1)+λ2(0<r<1)
use, whereby
(13) λ 2 = 1 18 ( n + 1 ) 2 (13) λ 2 = 1 18 ( n + 1 ) 2 {:(13)lambda_(2)=(1)/(18(n+1)^(2)):}\begin{equation*} \lambda_{2}=\frac{1}{18(n+1)^{2}} \tag{13} \end{equation*}(13)λ2=118(n+1)2
is.
Example 4. Returning to Example 3, we get
λ 2 = 1 162 = 0 , 00617283950 … λ 2 = 1 162 = 0 , 00617283950 … lambda_(2)=(1)/(162)=0,00617283950 dots\lambda_{2}=\frac{1}{162}=0,00617283950 \ldotsλ2=1162=0,00617283950…
and
ln ⁡ 2 , 375 ≈ 0 , 8438694356 . ln ⁡ 2 , 375 ≈ 0 , 8438694356 . ln 2,375~~0,8438694356.\ln 2,375 \approx 0,8438694356 .ln⁡2,375≈0,8438694356.
This value differs only by 0 , 0041 < 1 200 0 , 0041 < 1 200 0,0041 < (1)/(200)0,0041<\frac{1}{200}0,0041<1200from the value calculated using formula (6).
Finally, it should be noted that it can be shown that for a given number k > 16 k > 16 k > 16k>16k>16in (12) instead of the value of λ 2 λ 2 lambda_(2)\lambda_{2}λ2the value 1 k ( n + 1 ) 2 1 k ( n + 1 ) 2 (1)/(k(n+1)^(2))\frac{1}{k(n+1)^{2}}1k(n+1)2can be used if n n nnnis sufficiently large.
7. In a previous paper [2], we presented a method for calculating natural logarithms by quadratic interpolation. The results obtained in this way can be better than those obtained by the linear interpolation discussed in this paper.
Example 5. Let us calculate ln ⁡ 2 , 5 ln ⁡ 2 , 5 ln 2,5\ln 2,5ln⁡2,5using formula (6), we get
ln ⁡ 2 , 5 ≈ 1 2 ln ⁡ 5 + 1 4 ( ln ⁡ 3 − ln ⁡ 2 ) . ln ⁡ 2 , 5 ≈ 1 2 ln ⁡ 5 + 1 4 ( ln ⁡ 3 − ln ⁡ 2 ) . ln 2,5~~(1)/(2)ln 5+(1)/(4)(ln 3-ln 2).\ln 2,5 \approx \frac{1}{2} \ln 5+\frac{1}{4}(\ln 3-\ln 2) .ln⁡2,5≈12ln⁡5+14(ln⁡3−ln⁡2).
A table with 10 decimal places then results in
ln ⁡ 2 , 5 ≈ 0 , 9060852332 . ln ⁡ 2 , 5 ≈ 0 , 9060852332 . ln 2,5~~0,9060852332.\ln 2,5 \approx 0,9060852332 .ln⁡2,5≈0,9060852332.
However, the procedure described in [2] yields ln ⁡ 2 , 5 ≈ 0 , 9167130679 ln ⁡ 2 , 5 ≈ 0 , 9167130679 ln 2,5~~0,9167130679\ln 2,5 \approx 0,9167130679ln⁡2,5≈0,9167130679; a value that much more closely approximates the value 0.9162907319 given in the same table.

FOOTNOTE

1 ) 1 ) ^(1)){ }^{1)}1)In our further remarks we only need to note that the consequence ( ( 1 + 1 n ) n + 2 3 ) n = 1 + ∞ 1 + 1 n n + 2 3 n = 1 + ∞ ((1+(1)/(n))^(n+(2)/(3)))_(n=1)^(+oo)\left(\left(1+\frac{1}{n}\right)^{n+\frac{2}{3}}\right)_{n=1}^{+\infty}((1+1n)n+23)n=1+∞However, if one considers the monotonicity of the sequence ( ( 1 + 1 n ) n + 1 2 ) n = 1 + ∞ 1 + 1 n n + 1 2 n = 1 + ∞ ((1+(1)/(n))^(n+(1)/(2)))_(n=1)^(+oo)\left(\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\right)_{n=1}^{+\infty}((1+1n)n+12)n=1+∞. this gives a more accurate upper estimate for ξ ξ xi\xiξ.

LITERATURE

  1. PÓLYA, G. and G. SZEGÖ: Problems and Theorems from Analysis I. Berlin 1925.
  2. POPOVICIU, T.: On the approximation of functions and solutions of an equation by quadratic interpolation. Methods of Approximation Theory, Vol. 1 (1972), 155-163.
  3. VORONOVSKAIA, EV: O vidoizmenenii metoda Ciaplighina dlia differentialnih uravnenii pervovo poriadka.Prikladnaia matematika i mehanika, XIX, (1955), 121-126.
1975

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