Let S = { − 1 , 0 , 1 } and T : P 2 ( R ) → Fun ⁡ ( S ) be the transformation T ( p ( x ) ) = p ′ ( x )and consider the ordered bases E = { 1 , x , x 2 } the standard basis of P 2 ( R ) , F = { 1 − x , x + x 2 , 2 + x 2 } a basis of source P 2 ( R ) , E ′ = { χ − 1 , χ 0 , χ 1 } the standard basis of Fun ⁡ ( S ) , G = { χ − 1 , χ 1 − 2 χ − 1 , χ − 1 + χ 0 − 2 χ 1 } a basis of target Fun ⁡ ( S ) .

Question

Let   S  =  {  −  1  ,  0  ,  1  } and   T  :            P        2    (      R    )  →  Fun  ⁡  (  S  ) be the transformation   T  (  p  (  x  )  )  =      p    ′    (  x  )and consider the ordered bases                    E                            =        {        1        ,        x        ,                  x          2                }                           the standard basis of                                                     P                            2                        (                          R                        )                    ,                                    F                            =        {        1        −        x        ,        x        +                  x          2                ,        2        +                  x          2                }                           a basis of source                                                     P                            2                        (                          R                        )                    ,                                              E          ′                                    =        {                  χ                      −            1                          ,                  χ          0                ,                  χ          1                }                           the standard basis of                       Fun            ⁡            (            S            )                    ,                                    G                            =        {                  χ                      −            1                          ,                  χ          1                −        2        χ                  −          1                ,                  χ                      −            1                          +                  χ          0                −        2                  χ          1                }                           a basis of target                       Fun            ⁡            (            S            )                    .

pyphekam · Accepted Answer

I&#039;m assuming   Fun  ⁡  (  S  ) denotes the set of functions   S  →      R  Now what does the map T do? It takes a polynomial   p  (  x  )  ∈            P        2    (      R    ) and sends it to the map   S  →      R   given by   s  ↦      p    ′    (  s  ). So we get                    T        (        1        )        =        (        s        ↦        0        )                            =        0        ⋅                  χ                      −            1                                          +        0        ⋅                  χ          0                +        0        ⋅                  χ          1                ,                            T        (        x        )        =        (        s        ↦        1        )                            =        1        ⋅                  χ                      −            1                                          +        1        ⋅                  χ          0                +        1        ⋅                  χ          1                ,                            T        (                  x          2                )        =        (        s        ↦        2        s        )                            =                  (          s          ↦                      {                                                            −                  2                                                                      if                                           s                      =                      −                      1                                                                                                                    0                                                                      if                                           s                      =                      0                                                                                                                    2                                                                      if                                           s                      =                      1                                                                                                                )                                                  =        (        −        2        )        ⋅                  χ                      −            1                                          +        0        ⋅                  χ          0                +        2        ⋅                  χ          1                .            Note that in general a function   f  :  S  →      R   can always be written as  f  =  f  (  −  1  )  ⋅      χ          −      1        +  f  (  0  )  ⋅      χ    0    +  f  (  1  )  ⋅      χ    1    .

Let S = <mo fence="false" stretchy="false">{ − 1 , 0 , 1 <mo f

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