Continued Fraction for Tan(x)Date: 11/03/2000 at 09:56:45 From: Joseph Subject: "Continued Fraction" for the Tangent Function Dear Dr. Math, I was reading about pi on a math site. I saw something that blew my mind! They said that the first argument that pi has to be irrational was based on this formula (I've copied it down so I can type it out exactly): 1 tan x = ---------------------------- 1 1 - + ----------------------- x 3 1 - + ------------------ x 5 1 - + ------------- x 7 1 - + -------- x 9 - + ... x Can you give me a hint as to why this formula is true? They called it an "infinite continued fraction representation." I know about convergent series and infinite products, so I think I understand how you would define this. But how on earth did they get this particular formula? I can follow the rest of the argument from there. Thanks! Date: 11/03/2000 at 11:24:36 From: Doctor Rob Subject: Re: "Continued Fraction" for the Tangent Function Thanks for writing to Ask Dr. Math, Joseph. The above formula is erroneous. You have written the continued fraction for tanh(x), the hyperbolic tangent of x. Correct is: 1 tan x = ---------------------------- 1 1 - - ----------------------- x 3 1 - - ------------------ x 5 1 - - ------------- x 7 1 - - -------- x 9 - - ... x This can be rewritten in the form x tan x = ----------------------------- x^2 1 - ------------------------ x^2 3 - -------------------- x^2 5 - ---------------- x^2 7 - ------------ x^2 9 - -------- 11 - ... Define infinity Psi(c;z) = SUM 1/[c*(c+1)*...*(c+k-1)] * z^k/k! k=0 where c is a complex constant but not a nonpositive integer, and z is a complex variable. Sometimes this is denoted by 0F1(c;z). Observe that sin(x) = x*Psi(3/2;-x^2/4) cos(x) = Psi(1/2;-x^2/4) Let {a(n)} be a sequence of complex numbers defined by a(n) = 1/[(c+n-1)*(c+n)], n = 1, 2, 3, ... Let P(n) = Psi(c+n;z), n = 0, 1, 2, 3, ... Then it is easily shown that the power series P(n) satisfies the 3-term recurrence relation P(n) = P(n+1) + a(n+1)*z*P(n+2), n = 0, 1, 2, ... Divide both sides by P(n+1), and rearrange to get P(n)/P(n+1) = 1 + a(n+1)*z/[P(n+1)/P(n+2)] Thus, starting with n = 0 and doing chain substitution, you'll see that a(1)*z P(0)/P(1) = 1 + ------------------------ a(2)*z 1 + -------------------- a(3)*z 1 + ---------------- a(4)*z 1 + ------------ a(5)*z 1 + -------- 1 + ... I leave the matter of convergence aside, but it does work for all complex numbers z. Now when c = 1/2, P(0)/P(1) is just P(0)/P(1) = Psi(1/2;z)/Psi(3/2;z) and a(n) = 4/[(2*n-1)*(2*n+1)] so you get the equation 4*z/(1*3) Psi(1/2;z)/Psi(3/2;z) = 1 + -------------------------- 4*z/(3*5) 1 + ---------------------- 4*z/(5*7) 1 + ------------------ 4*z/(7*9) 1 + -------------- 4*z/(9*11) 1 + ---------- 1 + ... Now substitute z = -x^2/4, and you get -x^2/(1*3) cos(x)/[sin(x)/x] = 1 + --------------------------- -x^2/(3*5) 1 + ----------------------- -x^2/(5*7) 1 + ------------------- -x^2/(7*9) 1 + --------------- -x^2/(9*11) 1 + ----------- 1 + ... From here it is not too difficult a rearrangement to get the continued fraction first mentioned in my answer above. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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