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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 $V_1$.



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 $\in V_1$?


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

Thank you in advance!

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up vote 1 down vote accepted

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.

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Can anyone tell me why my "gather" environnement won't make the three lines $((1),(2)$ and $(3))$ as I wish they would be? I always use those environments when I TeX but it doesn't work here right now, don't know why. – Patrick Da Silva Jun 13 '11 at 19:54
The easiest way to achieve it is to write $$\begin{gather*}...\end{gather*}$$ (that's what I did). Another way is to replace the double backslashes \\ by triple backslashes \\\. The problem is that the backslash \ character has a special meaning in the various engines used to produce the displayed text and this leads to some incompatibilities. – t.b. Jun 13 '11 at 20:45
@Patrick-Da-Silva Thank you for your great answer! What I still don't understand is your explanation for b). If I set Q(x) = 1 d is 0 and not infinite, isn't it? – muffel Jun 14 '11 at 19:49
@Patrick-Da-Silva You said that for (3) $\lambda x \in \mathbb{C}$ but in your proof you say $\frac{\lambda P(x)}{Q(x)} \in V_1$. Which one is correct (and my does $\frac{\lambda P(x)}{Q(x)}$ implies that the result is in $V_1$ or $\mathbb{C}$? – muffel Jun 14 '11 at 21:25
@Theo : I have been struggling with this for the last two weeks. I'll check it out and edit this comment later, but thanks. @muffel : $\lambda P(x) / Q(x)$ is in $V_1$ because $\lambda P(x)$ is also a polynomial when $P(x)$ is one, so that the ratio of $\lambda P(x)$ and $Q(x)$ is a ratio of polynomial with complex coefficients, hence I can consider it is an element of $V_1$. This ratio is in $\mathbb C$ only if you substitute a complex value for $x$, but unless you do that, $\lambda P(x) / Q(x)$ is a ratio of complex polynomials. Are you okay? About b) : Can you explain your problem more? – Patrick Da Silva Jun 15 '11 at 4:59

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 $x \in V_1$.

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