Find the tangent of the sum of inverse tangents

Question

Background

It can be shown that for any integer k >= 0, f(k) = tan(atan(0) + atan(1) + atan(2) + ... + atan(k)) is a rational number.

Goal

Write a complete program or function which when given k >= 0, outputs f(k) as a single reduced fraction (the numerator and denominator are coprime).

Test cases

The first few values are

f(0) = (0,1)
f(1) = (1,1)
f(2) = (-3,1)
f(3) = (0,1)
f(4) = (4,1)
f(5) = (-9,19)
f(6) = (105,73)

Rules

• Standard loopholes are forbidden.
• Input and output may be in any convenient format. You may output f(k) as a string numerator/denominator, as a tuple of two integers, a fraction or rational object, etc. If you output a string, give two integers only, that is, output 3/2 instead of 1 1/2.
• This is code-golf, the shortest answer (in bytes) wins.

Answer 1

Mathematica, 28 bytes

Fold[+##/(1-##)&,0,Range@#]&

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A longer, but more interesting approach (32 bytes):

Im@#/Re@#&@Product[1+n I,{n,#}]&

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Answer 2

Python 2, 76 72 bytes

from fractions import*
f=lambda k:Fraction(k and(k+f(k-1))/(1-k*f(k-1)))

Use the identity:

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) * tan(B))

We have:

f(k) = 0                                    if k = 0
     = (k + f(k - 1)) / (1 - k * f(k - 1))  if k > 0

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Thanks to Luis Mendo, save 4 bytes.

Answer 3

M, 11 bytes

×C÷@+
R0;ç/

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Uses the OEIS formula x(n) = (x(n-1)+n)/(1-n*x(n-1)) with x(0) = 0.

Answer 4

APL (Dyalog), 14 bytes

Requires ⎕FR←1287 (128 bit Floating-point Representation) for small input. Takes k as right argument.

1(∧÷,)3○¯3+.○⍳

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⍳ integers one through k (zero is not needed as 0 = arctan 0)

¯3+.○ sum of arcus tangents

3○ tangent

1(…) apply the following tacit function with 1 as left argument and the above as right argument:

 ∧ the lowest common multiple (of 1 and the right argument); gives the numerator

 ÷ divided by

 , the concatenation (of 1 and the right argument); gives the numerator and the denominator

Answer 5

Haskell, 52 bytes

This uses the OEIS series expansion:

import Data.Ratio
f 0=0%1
f n|p<-f$n-1=(p+n)/(1-n*p)

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Or a pointfree version:

(scanl1(\f n->(f+n)/(1-n*f))[0%1..]!!)

Answer 6

JavaScript (ES6), 80 bytes

f=n=>n?([a,b]=f(n-1),g=(a,b)=>a?g(b%a,a):b,c=g(d=b*n+a,e=b-n*a),[d/c,e/c]):[0,1]

Returns a [numerator, denominator] pair. Explanation: Let f(n-1) = a/b then f(n) = atan(tan(n)+tan(a/b)) = (n+a/b)/(1-n*a/b) = (b*n+a)/(b-n*a). It then remains to reduce the fraction to its lowest terms.

Online ES6 Environment

Answer 7

Pari/GP, 36 bytes

n->imag(x=prod(i=0,n,1+i*I))/real(x)

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Or the same length:

n->fold((x,y)->(x+y)/(1-x*y),[0..n])

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Answer 8

05AB1E, 33 26 bytes

0X)Iƒ©`N*+®`sN*-‚D¿D_i\¤}/

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Explanation

0X)                          # initialize stack with [0,1]
   Iƒ                        # for N in range [0 ... input] do:
     ©                       # store a copy of the current pair in the register
      `                      # push the pair separately to the stack
       N*                    # multiply the denominator with N
         +                   # add the numerator
          ®`s                # push the denominator then the numerator to the stack
             N*              # multiply the numerator by N
               -             # subtract it from the denominator
                D¿D          # get 2 copies of the greatest common divisor
                   0Qi  }    # if the gcd equals 0
                      \¤     # replace it with the denominator
                         /   # divide the pair with this number

Answer 9

Octave, 30 bytes

format rat
tan(sum(atan(0:k)))

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Answer 10

Casio-Basic, 35 bytes

tExpand(tan(sum(seq(tan⁻¹(n),n,0,k

tan^-1 should be entered as the one on the Trig keyboard; or the ^-1 can be entered separately from the abc>Math keyboard. According to the fx-CP400's manual, it's a single two-byte character (764).

Function, 34 bytes for the code, +1 byte to add k as an argument.

Explanation

seq(tan-1(n),n,0,k) generates all the values for tan-1(n) from 0 to k.

sum adds them all together, then tan does the tangent function on them.

tExpand will then turn them into a single fraction.

Answer 11

Forth (gforth), 130 bytes

: g dup if tuck mod recurse then ;
: f 0 1 rot for 2dup i * + -rot swap i * - next
2dup g + abs over 0< 2* 1+ * >r i / swap r> / ;

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Uses my own implementation of gcd, and the OEIS formula. Takes k from the stack and returns two integers denom num on the stack. It took a lot of effort (and bytes) to normalize the fraction.

: g ( x y -- gcd 0 )
  dup if tuck mod recurse then
;

: f ( k -- denom num )
  0 1               \ push initial fraction with num=0, denom=1
  rot for           \ ( num denom ) loop from k to 0 inclusive
    2dup i * + -rot \ ( denom*k+num num denom )
    swap i * -      \ ( denom*k+num denom-num*k )
  next
  2dup g + abs      \ ( num denom gcd ) evaluate positive gcd and remove 0
  over 0< 2* 1+ *   \ ( num denom gcd' ) negate gcd if denom < 0
  >r i / swap r> /  \ ( denom/gcd' num/gcd' )
;

Answer 12

Julia 0.6.0 40 bytes

k->rationalize(tan(sum(x->atan(x),1:k)))

It's a straightforward implementation of the question. The precision of rationalize can sometimes be weird but works fine 99% of the time.