Let f ( x ) = ∑ n ≥ 0 a n x n = P ( x ) ( 1 − x ) d be a rational function.(a) Prove: There is a polynomial P 2 ( x ) so ∑ n ≥ 0 a 2 n x n = P 2 ( x ) ( 1 − x ) d (b) Let r ≥ 1 ∈ N . Show that an polynomial P r exists so that ∑ n ≥ 0 a r n x n = P r ( x ) ( 1 − x ) d Hint: Use the rth roots of unity which are defined by exp ⁡ ( 2 π i k r ) , 0 ≤ k ≤ r − 1(a) I don't know what this d is about (and no one else did). Might be an absolute term.As f ( x ) is a rational function, it can be defined as a fraction of two polynomials P ( x ) Q ( x ) . But that is unfortunately all I know about this.Could you please help me going on?(b) I don't know how the rth roots of unity (and therefore numbers x for which applies: x r = 1) can help me solving this? I don't find any approach.Could you please help me a bit?Thanks in advance!

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Let  f  (  x  )  =      ∑          n      ≥      0            a    n        x    n    =            P      (      x      )              (      1      −      x              )        d            be a rational function.(a) Prove: There is a polynomial       P    2    (  x  ) so      ∑          n      ≥      0            a                  2        n                  x    n    =                    P        2            (      x      )              (      1      −      x              )        d            (b) Let   r  ≥  1  ∈      N  . Show that an polynomial       P    r   exists so that      ∑          n      ≥      0            a          r      n            x    n    =                    P        r            (      x      )              (      1      −      x              )        d            Hint: Use the rth roots of unity which are defined by   exp  ⁡      (                            2          π          i          k                r              )    ,  0  ≤  k  ≤  r  −  1(a) I don&#039;t know what this d is about (and no one else did). Might be an absolute term.As   f  (  x  ) is a rational function, it can be defined as a fraction of two polynomials                     P        (        x        )                    Q        (        x        )            . But that is unfortunately all I know about this.Could you please help me going on?(b) I don&#039;t know how the   rth roots of unity (and therefore numbers   x for which applies:       x    r    =  1) can help me solving this? I don&#039;t find any approach.Could you please help me a bit?Thanks in advance!

candelo6a · Accepted Answer

Here&#039;s what you need to know about roots of unity. Let   ζ  =      e          2      π      i              /            r      . Then the rth roots of unity are the numbers   1  ,  ζ  ,      ζ    2    ,  …  ,      ζ          r      −      1      . Let m be some integer, and raise all these numbers to the power m, and add them:   1  +      ζ    m    +      ζ          2      m        +  ⋯  +      ζ          (      r      −      1      )      m      . That&#039;s the sum of a geometric progression. If   m is a multiple of   r then each term in the sum is 1 so the sum is   r. If   m is not a multiple of   r then you should check that the formula for the sum of a geometric progression tells you that the sum is zero.OK?

glycleWogry · Accepted Answer

For (a), try comparing   f  (  −  x  ) and   f  (  x  ). This should give you some idea how to use the hint in given in (b).

Let f ( x ) = <munder> ∑ <mrow class="MJX-TeXAtom-ORD

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