1
$\begingroup$

Let $\mathbb{C}[[x]] := \{\sum_{n\geq 0} a_n x^n | a_n \in \mathbb{C}\}$ be the set of formal power series of x and $V$ be the vector space of all series over $\mathbb{C}$.

Let in addition $V_1$ be the set of series for which applies

$f_a(x) = \sum_{n\geq 0} a_n x^n = \frac{P(x)}{Q(x)}$ with $Q(x) = 1 + \alpha_1 t +\cdots + \alpha_d t^d$ and a polynomial $P(x)$ having a degree < $d$.

a) Prove that $V_1$ is a vector subspace of $V$.

b) Show that the dimension of $V_1$ is $d$ and identify a base of $V_1$.


Hi!

a)

As I know I need to show three things:

  1. $V_1 \neq \emptyset$
  2. $\forall x,y \in V_1 : x + y \in V_1$
  3. $\forall x \in V_1, \forall y \in V : x \cdot y \in V_1$

For (1) I need to show that at least one element exists and therefore $V_1$ is not empty.

If I pick $P(x) = x^2, \; Q(x) = 1 + x + x^2 + x^3$ I get $P(x) /Q(x) = 0$. This by definition is $\in V$, isn't it?

2) How do I add two series?

3) According to Wikipedia the product of two series is given by $\left( \sum_{n=0}^\infty a_n x^n \right) \left( \sum_{n=0}^\infty b_n x^n \right) = \sum_{n=0}^\infty c_n x^n, \; c_n = \sum_{k=0}^n a_k b_{n-k}$

I tried to play around with some values, but this all didn't make any sense. How do I show that multiplying an element of $V_1$ with any other series ($\in \mathbb{C}$) is again $\in V_1$?

b)

What is the "dimension" of a formal power series?


Thank you in advance!

2 Answers 2

1

Oh god, ok some clarification here.

V is a vector space over $\mathbb C$, so that $V_1$ being a sub-space of $V$ means you have to prove that $\begin{gather*} \begin{aligned} (1) & 0_V \in V_1, \text{ i.e the } 0 \text{ serie is in } V_1 \\ (2) & \forall x,y \in V_1, x+y \in V_1 \\ (3) & \forall x \in V_1, \lambda \in \mathbb C, \lambda x \in \mathbb C. \\ \end{aligned} \end{gather*}$

This is way more simple : for $(1)$, you can write $0 = 0/1 \in V_1$. For $(2)$, you have $ \frac{P(x)}{Q(x)} + \frac{R(x)}{S(x)} = \frac{P(x) S(x) + Q(x) R(x)}{Q(x)S(x)} \in V_1 $

and for $(3)$, you have that $ \lambda \frac{P(x)}{Q(x)} = \frac{\lambda P(x)}{Q(x)} \in V_1. $

As for $b)$, you are not trying to think of the dimension of the serie, but of the dimension of the vector space. You can think of the dimension as the number of elements that are in the basis of the vector space. For instance, you can use the fact that there is a subspace of $V_1$, call it $V_2$, which consists of those elements in $V_1$ for which $Q(x) = 1$. If you show that this is a subspace of $V_1$ (in the same manner that I did up there), you will notice that the set of vectors $ \mathcal B = \{ 1,x,x^2, ..., x^n \} $ is a basis of $V_2$, which means that the dimension of $V_2$ is infinite, but since it is a subspace of $V_1$, the dimension of $V_1$ is also infinite. Read your definitions more carefully is my best tip, though.

  • 0
    I think you're just messing up somewhere : The $d$ is not related at all to the dimension of $V$ nor $V_1$ ; it is simply a way to state that if an element of $V_1$ can be written as a quotient of polynomials, the denominator must have finite degree, it must not necessarily have "degree $d$" for some fixed $d$. For instance, $(x^2+1)/(x-3)$ is in $V_1$, and so is $(x-7)/(x^4+22)$ or $(x^43-1)/(x^27+x^18 + 42)$. A "non-trivial" example of an element of $V_1$ would be $ \sum_{n \ge 0} x^n = \frac 1{1-x} \in V_1. $ Is it more clear now?2011-06-21
3

Your requirement "3." is wrong. It would actually say that $V_1$ is an ideal of $V$. What you actually need is $c x \in V_1$ for any $c \in {\mathbb C}$ and $x \in V_1$.