Continued Fraction Approximations of the Riemann Zeta Function MATH 336. Shawn Apodaca

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1 Continued Fraction Approximations of the Riemann Zeta Function MATH 336 Shawn Apodaca

2 Introduction Continued fractions serve as a useful tool for approximation and as a field of their own. Here we will concern ourselves with results from Cvijovic and Klinowski from Continued-Fraction Expansions for the Riemann Zeta Function and Polylogarithms [3]. From the results, we will be capable of numerically approximating the Riemann zeta function ζ for integer values n, which are special cases of the polylogarithm. 2 Notation We will denote the positive integers N and N {0} as Z +. We will define the polylogarithm function as follows. z k Li ν (z) = k ν () In particular, Li ν () = ζ(ν) where ζ(ν) is the Riemann zeta function. We will denote the set of all real-valued, bounded, monotone non-decreasing functions ϕ(t) with infinitely many values on a t b as Φ(a, b) where a, b are elements of the extended reals R = R {, }. 3 Preliminary Definitions and Results Here we will give necessary definitions and some preliminary results. 3. Continued Fractions We define a continued fraction as follows. Definition 3.. An (infinite) continued fraction K(a k /b k ) is an expression of the form The nth approximate F n is defined K(a k /b k ) = K a k b k = F n = b + n K a k b k = A n B n a a 2 b 2 + a 3 b 3 + We say K(a k /b k ) converges to F if the sequence of approximates converge F in the extended complex plane C = C { }. We call A n the nth numerator and B n the nth denominator. We say K(a k /b k ) diverges if the limit lim n F n does not exist. We call each a k and b k the kth numerator and denominator, respectively. Note that we will be use the convention that a k 0. We say two continued fractions K(a k /b k ) and K(a k /b k ) are equivalent, written K(a k/b k ) = K(a k /b k ), if each approximate F n = F n.

3 A continued fraction of the form K(a k /b k ) = K a k z Is called a regular C-fraction (regular corresponding fraction) and a continued fraction of the form K(a k /b k ) = K a k Is called a modified regular C-fraction. If each a k > 0, then (2) and (3) are called regular S-fraction and modified regular S-fraction (Stieltjes fractions), respectively. A finite continued fraction K n a k (z) b k (z) is said to correspond to the series c k at z = if the zk following formal power series expansions are valid: Where n =, 2, 3,.... F n (z) λ n p=0 3.2 The Stieltjes-Riemann Integral c p z k = constz (λ n+) + Here we will define the Stieltjes-Riemann integral of a function f(x), denoted by b a f(x) dα(x), and give a few preliminary results. Here, we will use Apostol []. We define α k = α(x k ) α(x k ) such that n α k = α(b) α(a) We will also use the notion of a partition P of an interval [a, b]. This will be the same as that discussed in Foland [4]. We now define the Stieltjes-Riemann integral. Definition 3.2. Let P = {x 0, x,..., x k } be a partition of [a, b] and let t k [x k, x k ]. Then the Stieltjes-Riemann sum of f with respect to α is defined as S(P, f, α) = n f(t k ) α k If there exists a unique number A such that for any ϵ > 0, there exists a partition P ϵ of [a, b] such that for every partition P finer than P ϵ and for every choice of t k [x k, x k ], we have that S(P, f, α) A < ϵ. The number A = b a f(x) dα(x). We state without proof that A is uniquely determined whenever it exists. For our proof the main theorem, we will need the following two theorems. Theorem 3.3. Suppose f is continuous on [a, b] and α is any monotonic, increasing function. Then f is integrable with respect to α over [a, b]. (2) (3) 2

4 For a proof, see [2]. We now give criteria where a Stieltjes-Riemann integral simplifies to a Riemann integral. Theorem 3.4. Suppose f is integrable with respect to α on [a, b]. If α is continuously differentiable on [a, b], then b a f(x)α (x) dx exists. Further For proof, see Apostol []. 3.3 The Markov Theorem b a f(x) dα(x) = b a f(x)α (x) dx We will state the Markov theorem, without proof, since it will be used the proof of the main theorem. For a proof, see Perron [6]. However, we will state it as found in Jones and Thron [5]. Theorem 3.5. Suppose ϕ Φ(0, a). Then there is a modified S-fraction which corresponds to the series ( ) k µ k z k where µ k = a 0 t k dϕ(t) (4) at z =, converges to the function for all z C \ [ a, 0]. a 0 z dϕ(t) (5) z + t 3.4 Hankel Determinants Definition 3.6. Suppose {c k } is a sequence. Then the Hankel determinants H(r) m with {c k }, where r Z + and m N are given by c r c r+ c r+m H (r) 0 =, H m (r) c r+ c r+2 c r+m = c r+m c r+m c r+2m 2 associated 4 The Main Theorem Theorem 4.. Suppose that r Z + is a non-negative integer and m, n N. For any fixed r, m, n, define A m (r) (n) as the determinant of an m m matrix A (r) m (n) = det ( ) i+j+r (r + i + j ) n 3 i,j m

5 Where we define A (r) 0 (n) =. Then With a n, =, Proof. Consider the function Li n ( z) = K a n,k z a n,2m = A() m (n)a (0) m (n) A (0) m (n)a () m (n), ϕ n (t) = a A () m n,2m+ = (n)a() m+ (n) A (0) m (n)a () m (n) 0, t = 0 t ( ( (n )! 0 log )) n x dx, 0 < t, t > (6) (7) For n =, the integrand is just, so it is clearly integrable and ϕ n (t) is continuous. Where n N. Prudnikov [7] gives us t ϵ ( log ( )) n n dx = x We apply L Hôptial s rule to get that ϵ log k ϵ 0 as ϵ 0. So t 0 k (n )! ( ) (t(log t) k ϵ(log ϵ) k) k! ( ( )) n n k (n )! log dx = ( ) t log k t (8) x k! L Hôpital s rule gives that ϕ n (t) 0 as t 0 + and ϕ n (t) as t. For 0 < t, log ( x) 0 and continuous and, thus, integrable, so the integral is monotonically increasing and continuous on [0, ]. Further, ϕ n (t) Φ(0, ). Consider the following integral, called the Stieltjes transform of ϕ n (t). f n (z) = 0 dϕ n (t) z + t (9) Where z / [, 0]. Then by (3.3), the integrand is integrable with respect to ϕ n (t). Further, since ϕ n (t) is continuously differentiable on [0, ), by theorem (3.4), we have that f n (z) = (n )! 0 ( log z + t ( )) n dt t We then substitute x = log ( t ). This gives us that t = e x and dt = e x dx. So f n (z) = 0 (n )! x n z + e x ( e x ) dx = z (n )! 0 x n e x + z dx 4

6 This a form of the Fermi-Dirac integral, which has a known polylogarithm representation. In our case ( f n (z) = Li n ) z Using the series representation of the polylogarithm (), we get, for z >, the following. f n (z) = ( ) k k n z k zf n(z) = ( ) k (k + ) n z k = Where we have let c n,k = ( )k (k+) n. Further, Markov tells us there exists a corresponding modified S-fraction that converges to zf n (z) for all z C \ [0, ] and even tells us that c n,k z k c n,k = ( ) k µ n,k (0) Where µ n,k = ( (n )! 0 tk log ( )) n t dt. Jones and Thron [5] give us that whenever a series S = C-fraction C = K a k at z =, we know that c k z k corresponds to a modified a = c 0, Which is exactly what we have, except a n, =, a 2m = H() m H (0) m H (0) m H () m a n,2m = A() m (n)a (0) m (n) A (0) m (n)a () m (n),, a 2m+ = H() m H(0) m+ With each A (r) m (n) as described in the main theorem. We then have zf n (z) = + z + Dividing both sides by z and simple factoring gives us a n, H m (0) H m () a A () m n,2m+ = (n)a(0) m+ (n) A (0) m (n)a () m (n) a n,2 a n,3 + a n,4 z + f n (z) = + a n, (/z) a n,2 (/z) + a n,3(/z) + a n,4(/z) + = K a n,k (/z) 5

7 Thus, Li n ( /z) = K a n,k (/z). So Li n ( z) = K a n,k z z And we are done. 5 Additional Results We conclude with some calculations. Using our results, we may immediately use our results for Li ( z) = log( + z) and Li n ( ) = ( 2 n )ζ(n), for integers n 2. Cvijović works out the first of these for us. log( + z) = K a,k z Where a, =, a,2m = m 2(2m ), a n,2m+ = m 2(2m + ) Take z =. Then we should have an approximation for log(2). We have { {a,k } =, 2, 6, 3, 5, 3 0, 3 4, 2 7, 2 9, 5 8, 5 } 22 So log(2) The more precise value is log(2) For z = 2, that is, log(3), we multiply each of the a,k by 2. This gives us an approximation log(3) as compared to the more precise log(3) More appropriately, let z = e. Using this value will, of course, give us an exact value for log( + z) = log(e) = to compare to. We get log(e) We may even let z = ei. This gives log(ie) i. The exact value of log(ie) = + i π 2. Using Mathematica v.6, we calculate the first 6 numerators of a n,k for n 0 (attached). n \ k

8 With the above table, we calculate the 6th approximants F 6 for a given n of the continued fraction expansion of the Riemann zeta function ζ(n). Below is a table of values for 2 n 0 accompanied by the values found using Mathematica s internal command. n F 6 Mathematica

9 References. T. M. Apostol, Mathematical Analysis (2nd Edition), Adison-Wesley, M. Barigozzi, 3. D. Cvijović and J. Klinowski, Continued-Fraction Expansions for the Riemann Zeta Function and Polylogarithms, Proceedings of the American Mathematical Society 25 (997), G. Folland, Advanced Calculus, Prentice-Hall, New Jersey, W. B. Jones and W. J. Thron, Continued Fractions: Analytic Theory and Applications, Addison-Wesley, O Perron, Die Lehre von den Kettenbrüchen (3rd edition), Vol. I and II, Teubner, Stuttgart, 954 and A. P. Prudnikov, Yu. A Brychkov and O. I. Marichev, Integrals and Series, Vols. and 3, Gordon and Breach Science Publ., New York, 986 and 990 8