If, for n = 0, 1, 2, … you're given f 0 ( x ) = 1 2 − x and f n + 1 = f 0 ( f n ( x ) ), how do you prove that your formula for f n ( x ) is correct by mathematical induction?I have computed the first few terms: f 1 ( x ) = 2 − x 3 − 2 x f 2 ( x ) = 3 − 2 x 4 − 3 x f 3 ( x ) = 4 − 3 x 5 − 4 x f 4 ( x ) = 5 − 4 x 6 − 5 x Then, f n ( x ) = n + 1 − n x n + 2 − ( n + 1 ) x How to prove this with mathematical induction as it has variable x in it as well as the n.

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If, for   n  =  0,   1,   2, … you&#039;re given       f    0    (  x  )  =      1          2      −      x       and       f          n      +      1        =      f    0    (      f    n    (  x  )  ), how do you prove that your formula for       f    n    (  x  ) is correct by mathematical induction?I have computed the first few terms:      f    1    (  x  )  =            2      −      x              3      −      2      x            f    2    (  x  )  =            3      −      2      x              4      −      3      x            f    3    (  x  )  =            4      −      3      x              5      −      4      x            f    4    (  x  )  =            5      −      4      x              6      −      5      x      Then,      f    n    (  x  )  =            n      +      1      −      n      x              n      +      2      −      (      n      +      1      )      x      How to prove this with mathematical induction as it has variable   x in it as well as the   n.

Mathias Allen · Accepted Answer

You have already shown the base case for your induction. Suppose that your proposition      f    n    (  x  )  =            n      +      1      −      n      x              n      +      2      −      (      n      +      1      )      x        ,holds for   n  =  N. We must show that the result also holds for   n  =  N  +  1. Following the formula you were given for computing       f          N      +      1        (  x  ) we have,      f          N      +      1        (  x  )  =      f    0    (      f    N    (  x  )  )  =      f    0        (                  N        +        1        −        N        x                    N        +        2        −        (        N        +        1        )        x              )    ,where we have used the inductive assumption to substitute       f    N    (  x  ). Now we apply       f    0  ,      f          N      +      1        (  x  )  =      1          2      −                        N          +          1          −          N          x                          N          +          2          −          (          N          +          1          )          x                      .We now simplify this expression,      f          N      +      1        (  x  )  =            N      +      2      −      (      N      +      1      )      x              2      (      N      +      2      −      (      N      +      1      )      x      )      −      (      N      +      1      −      N      x      )      and collect together all of the like terms      f          N      +      1        (  x  )  =            (      N      +      1      )      +      1      −      (      N      +      1      )      x              (      (      N      +      1      )      +      2      )      −      (      (      N      +      1      )      +      1      )      x        .This completes the inductive argument.

atarentspe · Accepted Answer

In fact having       f    n    (  x  ) in hand, we need to prove that      f          n      +      1        (  x  )  =            n      +      2      −      (      n      +      1      )      x              n      +      3      −      (      n      +      2      )      x      while this is easy since      f          n      +      1        (  x  )      =          f      0        (          f      n        (    x    )    )        =          f      0              (                        n          +          1          −          n          x                          n          +          2          −          (          n          +          1          )          x                    )            =                  1                  2          −                                    n              +              1              −              n              x                                      n              +              2              −              (              n              +              1              )              x                                                =                  1                                            2              n              +              4              −              2              n              x              −              2              x              −              n              −              1              +              n              x                                      n              +              2              −              (              n              +              1              )              x                                                =                  1                                            n              +              3              −              n              x              −              2              x                                      n              +              2              −              (              n              +              1              )              x                                                =                  n        +        2        −        (        n        +        1        )        x                    n        +        3        −        (        n        +        2        )        x            and the proof is complete

If, for n=0, 1, 2, … you're given f_0(x)=1/2−x and f_(n+1)=f_(0)(fn_(x)), how do you prove that your formula for f_n(x) is correct by mathematical induction?

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