Quasipolynomial vector spaces 
Functions of the form
  $$f(x)=e^{ax}(c_nx^n+...+c_0)$$
  where
  $$n\ge0$$
  is an integer and
  $$a,c_0,...,c_n$$
  are real numbers are called quasipolynomials. If
  $$c_n\ne0$$
  we say the order of the polynomial $f$ is $n$.
  Firstly show that the set of quasipolynomials with fixed exponent $a$ and of order less or equal to $n$ form a vector space. I know that for an ordinary polynomial
  $$f(t)=b_kt^k+...+b_0$$
  $$k\ge0$$
  $$b_0,...,b_k$$
  $$b_k\ne0$$
  $$g(t)=h_kt^n+...+h_0$$
  any linear combination
  $$\mu f(t)+\lambda g(t)=(\mu b_k+\lambda h_k)t^k+...+(\mu b_0+\lambda h_0)$$
  is again a polynomial of order $\le k$. How do I do the same as the above for the quasipolynomial? How do I show the set forms a vector space? Secondly, how do I find a basis of this vector space - the following equations describe the case for the polynomials - how do I achieve this for the set of quasipolynomials? The set of polynomials has the monomials 
  $$1=t^0,t,...,t^k$$
  form a basis set- on the other hand the monomials are linearly dependent. Indeed
  $$\phi (t)=\lambda _kt^k+...+\lambda _0.1=0$$
  $$\frac{d^r\phi}{dx^r}(0)=r!\lambda _r=0 $$
  $$0\le r\le k$$
  $$\textrm{dimension} =k+1$$
  Thirdly is the subset of quasipolynomials of exact order $n$ a vector space? If it does not contain the zero vector then it is not a vector space. Finally is the subset of quasipolynomials of order $\le n$ with $f''(3)=0$ a vector space? I know the following applies for ordinary polynomials of order $\le k$ with $f'(1)=0$ is a vector space:
  $$f'(1)=0$$
  $$f'(1)=g'(1)=0$$
  $$\mu f(t)+\lambda g(t)$$
  $$(\mu f+\lambda g)'(1)=\mu f'(1)+\lambda g'(1)=0$$
  $$f''(3)=0$$

 A: I'll call $f(x) = e^{ax}(\sum_i^n c_i x^i)$ ($c_n \neq 0$) an $a$-polynomial of order $n$. To show that the set of $a$-polynomials of order $\leq n$ form a vector space, first note that it is non-empty, and then consider two such quasipolynomials $f(x) = e^{ax}(\sum_i^n c_i x^i)$, $g(x) = e^{ax}(\sum_i^n d_i x^i)$ (where some or all of the $c_i,d_i$ may be zero), and an arbitrary scalar $\lambda \in \mathbb{R}$. Then
$$ (f+\lambda g)(x) = f(x) + \lambda g(x) = e^{ax}(\sum_i^n c_i x^i) + \lambda e^{ax}(\sum_i^n d_i x^i) = e^{ax}(\sum_i^n(c_i+\lambda d_i)x^i)$$
is evidently also an $a$-polynomial of order $\leq n$.
As for a basis, how about $(e^{ax},e^{ax}x,e^{ax}x^2,\ldots,e^{ax}x^n)$?
For your last question, the answer is evidently no if we are talking about general quasipolynomials. (For instance, $e^{x} + e^{2x}$ is not a quasipolynomial.) So I assume that $a$ is again fixed, i.e., we want to know whether the set of $a$-polynomials of order $\leq n$ with $f''(3) = 0$ is a subspace of the space above. But this is clear, as the derivative is linear:
$$ (f+\lambda g)''(3) = f''(3) + \lambda g''(3) = 0.$$
