On Euler s attempt to compute logarithms by interpolation

Transkript

1 Euler p. 1/1 On Euler s attempt to compute logarithms by interpolation Walter Gautschi wxg@cs.purdue.edu Purdue University

2 Leonhard Euler Euler p. 2/1

3 Euler p. 3/1

4 From Euler s letter to Daniel Bernoulli Euler p. 4/1

5 Euler p. 5/1 Ich vermeinte neulich, daß nachfolgende Series m 1 9 (m 1)(m 10) (m 1)(m 10)(m 100) (m 1)(m 10)(m 100)(m 1000) etc. (alwo die Anzahl der nullen im Numeratore und Denominatore einander gleich sind, im übrigen ist die Lex klar) den Logarithmum communem ipsius m exprimire, dann ist m = 1, so ist die gantze Series = 0, ist m = 10 so kommt 1, ist m = 100, kommt 2, und so fortan. Als ich nun daraus den Log [arithmum] 9 finden wollte, bekam ich eine Zahl welche weit zu klein war, ohngeacht diese Series sehr stark convergirte.

6 Euler p. 6/1 Euler s idea Compute f(x) = log x, 1 x < 10, by interpolating f at x r = 10 r, r = 0, 1, 2,...

7 Euler p. 6/1 Euler s idea Compute f(x) = log x, 1 x < 10, by interpolating f at x r = 10 r, r = 0, 1, 2,... Newton interpolation series S(x) = k=1 a k(x 1)(x 10) (x 10 k 1 ) a k = [x 0, x 1,..., x k ]f, k = 1, 2, 3,... Euler: a 1 = 1 9, a 2 = 1 990, a 3 = , a 4 =

8 Euler s idea Compute f(x) = log x, 1 x < 10, by interpolating f at x r = 10 r, r = 0, 1, 2,... Newton interpolation series S(x) = k=1 a k(x 1)(x 10) (x 10 k 1 ) a k = [x 0, x 1,..., x k ]f, k = 1, 2, 3,... Euler: a 1 = 1 9, a 2 = 1 990, a 3 = , a 4 = in general (proof by induction) a k = ( 1) k 1 10 k(k 1)/2 (10 k 1), k = 1, 2, 3,... Euler p. 6/1

9 Euler p. 7/1 Convergence S(x) = k=1 t k(x), 1 x < 10 t k (x) = x 1 k 1 10 k 1 r=1 (1 x/10r ) t k (x) < 9 10 k 1 ( k 1 ) < 9 10 k, k 2

10 Euler p. 7/1 Convergence remainder S(x) = k=1 t k(x), 1 x < 10 t k (x) = x 1 k 1 10 k 1 r=1 (1 x/10r ) t k (x) < 9 10 k 1 ( k 1 ) < 9 10 k, k 2 S(x) n k=1 t k(x) t n+1 (x) + t n+2 (x) + ( < n ) = 10 n 2

11 Euler p. 7/1 Convergence remainder S(x) = k=1 t k(x), 1 x < 10 t k (x) = x 1 k 1 10 k 1 r=1 (1 x/10r ) t k (x) < 9 10 k 1 ( k 1 ) < 9 10 k, k 2 S(x) n k=1 t k(x) t n+1 (x) + t n+2 (x) + ( < n ) = 10 n 2 fast convergence, but S(9) = , log 9 = ?

12 Euler p. 8/1 Go complex! S(z) = k=1 t k(z), z C t k (z) = z 1 k 1 10 k 1 r=1 (1 z/10r ) Convergence is uniform on z R t k (z) R+1 10 k 1 r=1 (1 + R/10r )

13 Euler p. 8/1 Go complex! S(z) = k=1 t k(z), z C t k (z) = z 1 k 1 10 k 1 r=1 (1 z/10r ) Convergence is uniform on z R t k (z) R+1 10 k 1 r=1 (1 + R/10r ) Weierstrass: S(z) is analytic in z R. Since R is arbitrary, S(z) is an entire function.

14 Euler p. 8/1 Go complex! S(z) = k=1 t k(z), z C t k (z) = z 1 k 1 10 k 1 r=1 (1 z/10r ) Convergence is uniform on z R t k (z) R+1 10 k 1 r=1 (1 + R/10r ) Weierstrass: S(z) is analytic in z R. Since R is arbitrary, S(z) is an entire function. Consequence: Let D = {z C : arg z < π} and log z, z D, denote the principal branch of the logarithm. Then S(z) cannot be equal to log z on any set of z-values having an accumulation point in D\{ }.

15 Euler p. 9/1 The q-analogue of the logarithm W. Van Assche and E. Koelink (2006) S q (x) = n=1 q n 1 q n (x; q) n, 0 < q < 1 where (x; q) n is the q-shifted factorial (x; q) n = n 1 k=0 (1 xqk )

16 Euler p. 9/1 The q-analogue of the logarithm W. Van Assche and E. Koelink (2006) S q (x) = n=1 q n 1 q n (x; q) n, 0 < q < 1 where (x; q) n is the q-shifted factorial (x; q) n = n 1 k=0 (1 xqk ) Why q-analogue? Answer: lim q 1 (1 q)s q (x) = ln x, S q (q n ) = n

17 Euler p. 9/1 The q-analogue of the logarithm W. Van Assche and E. Koelink (2006) S q (x) = n=1 q n 1 q n (x; q) n, 0 < q < 1 where (x; q) n is the q-shifted factorial (x; q) n = n 1 k=0 (1 xqk ) Why q-analogue? Answer: lim q 1 (1 q)s q (x) = ln x, S q (q n ) = n Limit of Euler s interpolation series S(x) = S q (x), q = 1 10

18 Euler p. 10/1 Can Euler s idea be salvaged? 1. Take x r = ω r, r = 0, 1, 2,..., ω > 1. Then S(x; ω) = lim n S n (x; ω) = log ω S q (x), q = 1 ω

19 Euler p. 10/1 Can Euler s idea be salvaged? 1. Take x r = ω r, r = 0, 1, 2,..., ω > 1. Then S(x; ω) = lim n S n (x; ω) = log ω S q (x), q = 1 ω From ln 1 q ln 10 S q(x) = there follows lim q 1 ln 1 q (1 q) ln 10 (1 q)s q(x) ln 1 q ln 10 S q(x) = ln x ln 10 = log x

20 Euler p. 10/1 Can Euler s idea be salvaged? 1. Take x r = ω r, r = 0, 1, 2,..., ω > 1. Then S(x; ω) = lim n S n (x; ω) = log ω S q (x), q = 1 ω From ln 1 q ln 10 S q(x) = there follows ln 1 q (1 q) ln 10 (1 q)s q(x) ln lim 1 q q 1 ln 10 S q(x) = ln x ln 10 = log x As a consequence lim S n(x; ω) = log x ω 1,n Euler (1750) studied S 1/ω (x), but missed the connection with the logarithm!

21 Euler p. 11/1 Convergence behavior S(x; ω) = k=1 t k (x; ω), S n (x; ω) = n k=1 t k (x; ω) If ω < 1 then lim k t k = : interpolation process diverges. Therefore, assume ω > 1.

22 Euler p. 11/1 Convergence behavior S(x; ω) = k=1 t k (x; ω), S n (x; ω) = n k=1 t k (x; ω) If ω < 1 then lim k t k = : interpolation process diverges. Therefore, assume ω > 1. The larger ω, the faster convergence, the worse the limit!

23 Euler p. 11/1 Convergence behavior S(x; ω) = k=1 t k (x; ω), S n (x; ω) = n k=1 t k (x; ω) If ω < 1 then lim k t k = : interpolation process diverges. Therefore, assume ω > 1. The larger ω, the faster convergence, the worse the limit! Theorem The error S n (x; ω) log x, 1 x < 10, can be made arbitrarily small by taking ω > 1 sufficiently close to 1 and n sufficiently large. But...

24 Euler p. 12/1... max k N t k (x; ω) may become extremely large! x\ω

25 Euler p. 12/1... max k N t k (x; ω) may become extremely large! x\ω Errors achievable in d-digit computation with n terms x\ω d n

26 Euler p. 12/1... max k N t k (x; ω) may become extremely large! x\ω Errors achievable in d-digit computation with n terms x\ω d n If 1 x 5 and ω = 1.1, then t k (x; ω) 72.2 and d = 14, n = 100 yields 10-digit accuracy. For 5 < x < 10, use log x = log(x/2) + log 2. If 1 x 2, then t k (x; ω) < 1, and for ω = 1.1 one obtains 10-digit accuracy with n = 20.

27 2. Take x r = 10 ω/(r+1), r = 0, 1, 2,..., ω > 0, hence x r (1, 10 ω ], all r Euler p. 13/1

28 Euler p. 13/1 2. Take x r = 10 ω/(r+1), r = 0, 1, 2,..., ω > 0, hence x r (1, 10 ω ], all r Theorem Let f C [a, b] and x r [a, b], r = 0, 1, 2,.... Then the interpolation series S(x) converges to f(x) for any x [a, b], provided lim k (b a) k k! M k = 0, where M k = max x [a,b] f (k) (x).

29 Euler p. 13/1 2. Take x r = 10 ω/(r+1), r = 0, 1, 2,..., ω > 0, hence x r (1, 10 ω ], all r Theorem Let f C [a, b] and x r [a, b], r = 0, 1, 2,.... Then the interpolation series S(x) converges to f(x) for any x [a, b], provided lim k (b a) k k! M k = 0, where M k = max x [a,b] f (k) (x). Proof Cauchy s formula for the remainder term of interpolation S n (x) f(x) = f (n+1) (ξ n ) n (n+1)! r=0 (x x r)

30 Euler p. 14/1 If f(x) = log x, a x b then (b a) k k! M k = ( b a 1) k /(k ln 10) 0 iff b a 1 1 geometric convergence with ratio q < 1 if b a 1 + q

31 Euler p. 14/1 If f(x) = log x, a x b then (b a) k k! M k = ( b a 1) k /(k ln 10) 0 iff b a 1 1 geometric convergence with ratio q < 1 if b a 1 + q a = 1, b = 10 ω = ω log(1 + q)

32 Euler p. 14/1 If f(x) = log x, a x b then (b a) k k! M k = ( b a 1) k /(k ln 10) 0 iff b a 1 1 geometric convergence with ratio q < 1 if b a 1 + q a = 1, b = 10 ω = ω log(1 + q) Change of variable Given 1 x < 10, determine the integer k 0 0 such that 10 k 0ω x < 10 (k 0+1)ω. Then t := 10 k 0ω x satisfies 1 t < 10 ω, and log x = log t + k 0 ω.