Let f ( x ) = <munder> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD



Answered question


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!

Answer & Explanation



Beginner2022-06-27Added 24 answers

Here'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'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.


Beginner2022-06-28Added 8 answers

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

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?