Find all functions f such that f ( x − f ( y ) ) = f ( f ( x ) ) − f ( y ) − 1So far, I've managed to prove that if f is linear, then either f ( x ) = x + 1 or f ( x ) = − 1 must be true. I did this by plugging in x = 0 to the above equation, which yields f ( − f ( y ) ) = f ( f ( 0 ) ) − f ( y ) − 1and plugging in x instead of y and subtracting, this becomes f ( − f ( y ) ) − f ( − f ( x ) ) = f ( x ) − f ( y )Assuming f ( x ) = a x + b then gives f ( − a y − b ) − f ( − a x − b ) = a x − a y ⇒ − a 2 y − a b + a 2 x + a b = a ( x − y ) ⇒ a 2 ( x − y ) = a ( x − y ) . Thus a = 1 or a = 0. If a = 0, then the original equation becomes b = b − b − 1, thus b = − 1. If a = 1, the original equation becomes x − y − b + b = x + 2 b − y − b − 1 ⟹ b = 1.I briefly tried finding a quadratic function that works but didn't find anything. So my question is: how can I either show that f must be linear or find all other representations?

Question

Find all functions f such that   f  (  x  −  f  (  y  )  )  =  f  (  f  (  x  )  )  −  f  (  y  )  −  1So far, I&#039;ve managed to prove that if f is linear, then either   f  (  x  )  =  x  +  1 or   f  (  x  )  =  −  1 must be true. I did this by plugging in   x  =  0 to the above equation, which yields  f  (  −  f  (  y  )  )  =  f  (  f  (  0  )  )  −  f  (  y  )  −  1and plugging in x instead of y and subtracting, this becomes  f  (  −  f  (  y  )  )  −  f  (  −  f  (  x  )  )  =  f  (  x  )  −  f  (  y  )Assuming   f  (  x  )  =  a  x  +  b then gives                    f        (        −        a        y        −        b        )        −        f        (        −        a        x        −        b        )        =        a        x        −        a        y                            ⇒        −                  a          2                y        −        a        b        +                  a          2                x        +        a        b        =        a        (        x        −        y        )                                          ⇒                  a          2                (        x        −        y        )        =        a        (        x        −        y        )        .            Thus   a  =  1 or   a  =  0. If   a  =  0, then the original equation becomes   b  =  b  −  b  −  1, thus   b  =  −  1. If   a  =  1, the original equation becomes  x  −  y  −  b  +  b  =  x  +  2  b  −  y  −  b  −  1  ⟹  b  =  1.I briefly tried finding a quadratic function that works but didn&#039;t find anything. So my question is: how can I either show that f must be linear or find all other representations?

Travis Sellers · Accepted Answer

Step 1if we can say that f is polynomial, then since   f  (  −  f  (  y  )  )  −  f  (  −  f  (  x  )  )  =  f  (  x  )  −  f  (  y  ), we can say that   f  (  −  f  (  y  )  )  +  f  (  y  )  =  f  (  −  f  (  x  )  )  +  f  (  x  ) for any   x  ,  y  ∈  Z. Hence,   f  (  −  f  (  x  )  )  +  f  (  x  ) is constant for all   x  ∈  Z.If   d  e  g  (  f  )  &amp;gt;  0 then f has infinitely many values, so we can say   f  (  −  x  )  +  x is constant, say   f  (  −  x  )  +  x  =  c.Step 2Then   f  (  −  x  )  =  c  −  xIf   d  e  g  (  f  )  =  0 then f is constant and you know the answer.

Aldo Ashley · Accepted Answer

Step 1Here&#039;s a partial solution.If 0 is a possible value of f, then by taking   f  (  y  )  =  0 we get   f  (  f  (  x  )  )  =  f  (  x  )  +  1. That is,   f  (  t  )  =  t  +  1 for   t  ∈  f  (      Z    ), and in particular   f  (      Z    ) contains all nonnegative integers. If we define   g  (  x  )  =  f  (  x  )  −  x, the equation becomes  g  (  x  −  f  (  y  )  )  =  g  (  x  )  +  g  (  x  +  g  (  x  )  )  −  1Step 2But note that the right side does not depend on y. Thus, since f(y) can be any nonnegative integer, g must be constant, and taking   x  =  0 we see that this constant must be 1. So the only solution where 0 is a possible value of f is   f  (  x  )  =  x  +  1.

Find all functions f such that f(x-f(y))=f(f(x))-f(y)-1.

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