Simple formula for the n-ary version of ( x , y ) ↦ x + y 1 − x y Let x ∗ y = x + y 1 − x y . I want a single formula for x 1 ∗ x 2 ∗ … ∗ x n , for all natural n.In order to generate plausible candidates, let's see what happens at small values of n: x 1 ∗ x 2 ∗ x 3 = x 1 + x 2 + x 3 − x 1 x 2 x 3 1 − x 1 x 2 − x 1 x 3 − x 2 x 3 x 1 ∗ x 2 ∗ x 3 ∗ x 4 = x 1 + x 2 + x 3 + x 4 − x 1 x 2 x 3 − x 1 x 2 x 4 − x 1 x 3 x 4 − x 2 x 3 x 4 1 − x 1 x 2 − x 1 x 3 − x 1 x 4 − x 2 x 3 − x 2 x 4 − x 3 x 4 + x 1 x 2 x 3 x 4 Then I conjecture the following: x 1 ∗ x 2 ∗ … ∗ x n = ∑ I ∈ T 1 ∏ i ∈ I x i − ∑ I ∈ T 3 ∏ i ∈ I x i ∑ I ∈ T 0 ∏ i ∈ I x i − ∑ I ∈ T 2 ∏ i ∈ I x i Where T k = { I ∈ ℘ ( { 1 , 2 … n } ) : k ≡ | I | ( mod 4 ) }Is there a nice proof of the above not involving transcendental functions?

Question

Simple formula for the   n-ary version of   (  x  ,  y  )  ↦            x      +      y              1      −      x      y      Let   x  ∗  y  =            x      +      y              1      −      x      y      . I want a single formula for       x    1    ∗      x    2    ∗  …  ∗      x    n  , for all natural   n.In order to generate plausible candidates, let&#039;s see what happens at small values of   n:      x    1    ∗      x    2    ∗      x    3    =                              x          1                +                  x          2                +                  x          3                −                  x          1                          x          2                          x          3                            1        −                  x          1                          x          2                −                  x          1                          x          3                −                  x          2                          x          3                          x    1    ∗      x    2    ∗      x    3    ∗      x    4    =                              x          1                +                  x          2                +                  x          3                +                  x          4                −                  x          1                          x          2                          x          3                −                  x          1                          x          2                          x          4                −                  x          1                          x          3                          x          4                −                  x          2                          x          3                          x          4                            1        −                  x          1                          x          2                −                  x          1                          x          3                −                  x          1                          x          4                −                  x          2                          x          3                −                  x          2                          x          4                −                  x          3                          x          4                +                  x          1                          x          2                          x          3                          x          4                    Then I conjecture the following:      x    1    ∗      x    2    ∗  …  ∗      x    n    =                              ∑                      I            ∈                          T              1                                                ∏                      i            ∈            I                                    x          i                −                  ∑                      I            ∈                          T              3                                                ∏                      i            ∈            I                                    x          i                                      ∑                      I            ∈                          T              0                                                ∏                      i            ∈            I                                    x          i                −                  ∑                      I            ∈                          T              2                                                ∏                      i            ∈            I                                    x          i                    Where       T    k    =  {    I  ∈  ℘  (  {  1  ,  2  …  n  }  )  :  k  ≡      |    I      |      (  mod    4  )    }Is there a nice proof of the above not involving transcendental functions?

Nia Molina · Accepted Answer

Expanding on my comment...Suppose all the       x    j   are real. We proceed by induction. We write       ℑ    ℜ    (  t  ) to represent the ratio of the imaginary to real parts of the complex number   t. Then                              x          1                                    =                  ℑ          ℜ                (        (        1        +                  i                          x          1                )        )        =                  x          1                 and                                       x          1                ∗                  x          2                                    =                  ℑ          ℜ                (        (        1        +                  i                          x          1                )        (        1        +                  i                          x          2                )        )        =                                            x              1                        +                          x              2                                            1            −                          x              1                                      x              2                                      ,            establishing the result for the smallest allowed   n. (Actually, I&#039;ve assumed this is what you want for   n  =  1, since you didn&#039;t specify.) Suppose now   n  &amp;gt;  2 and       ∗          j      =      1              n      −      1            x    j    =      ℑ    ℜ        (          ∏              j        =        1                    n        −        1              (    1    +          i              x      j        )    )  . Then                              ℑ          ℜ                          (                      ∏                          j              =              1                                      n                                (          1          +                      i                                x            j                    )          )                                    =                              ℑ                          (              (              1              +                              i                                            x                n                            )                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                                            ℜ                          (              (              1              +                              i                                            x                n                            )                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                                                                        =                              ℑ                          (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                        +                          x              n                        ℜ                          (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                                            ℜ                          (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                        −                          x              n                        ℑ                          (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                                                                        =                                            ℑ              ℜ                                      (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                        +                          x              n                                            1            −                          x              n                                      ℑ              ℜ                                      (                              ∏                                  j                  =                  1                                                  n                  −                  1                                            (              1              +                              i                                            x                j                            )              )                                                                        =                  ℑ          ℜ                          (                      ∏                          j              =              1                                      n              −              1                                (          1          +                      i                                x            j                    )          )                ∗                  x          n                                                  =                  (                      ∗                          j              =              1                                      n              −              1                                            x            j                    )                ∗                  x          n                                                  =                  ∗                      j            =            1                                n                                    x          j                .            (It&#039;s traditional to write this last display in reverse order, but I think the manipulations are easier to follow in this order.) The step yielding   …      −      x    n    ℑ  … in the denominator may be less than obvious: we want the negative imaginary part here because (in the previous line) when it is multiplied by       i        x    n  , it must yield a positive real part. (Trying to stumble across this manipulation in the traditional order is why I think the order above is easier to follow.) This completes our induction.Having done that, it should be no great challenge to show that       ℑ    ℜ        (          ∏              j        =        1                    n              (    1    +          i              x      j        )    )   splits the even and odd degree terms into the denominator and numerator, respectively, and also has the signs you want for your     (  mod    4  ) representation.

Simple formula for the n -ary version of ( x , y ) ↦

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