solutions to exercises 1.19

Exercise: do the polynomials \(p_0(x)=1,~~p_1(x)=(x-1),~~p_2(x)=(x-1)^2,~~p_3=(x-1)^3\) form a basis of the vector space of cubic polynomials (with real coefficients)? If so, express \(x^3\) in this basis.


Solution: use the binomial: \(x^n = \big((x-1)+1\big)^n = \sum_{m=0}^n {n\choose m}(x-1)^m\) to infer that \(x^n\) is a linear combination of the polynomials \(\{(x-1)^m\}_{m=0}^{n}\) (having the same linear span). Then \(p_0,p_1,p_2,p_3\) have the same span as \(1,x,x^2,x^3\). Specifically in our case, \( x^3 = (x-1)^3 + 3(x-1)^2+3(x-1)+1 \).


Exercise: let \(V\) be the linear space of quadratic polynomials, and define
the function \(e_k^*:V\to \mathbb{C}\) by \(e_k^*p:=p(k-1),~~k=1,2,3\).
Check that these functions are linear (as functions on polynomials). Express \(p(3)\) as a linear combination of \(e_k\)’s.


Solution: recall that if \(p_1(x) = \sum_{k=0}^m a_k x^k\) and \(p_2(x) = \sum_{k=0}^n b_k x^k\) (assuming that \(m\leq n\) without loss of generality, then the polynomial \(\alpha p_1+\beta p_2\) is defined by the linear combination of the coefficients:

\((\alpha p_1+\beta p_2)(x) = \sum_{k=0}^n (\alpha a_k+\beta b_k) x^k = \alpha p_1(x) + \beta p_2(x)\)

where by convention \(a_k:=0\) for all \(m<k\leq n\).

From that we can infer that valuation functionals are indeed  linear:

\(e_k^*(\alpha p_1+\beta p_2) = (\alpha p_1 + \beta p_2)(k-1)  = \alpha p_1(k-1) + \beta p_2(k-1) = \alpha (e_k^*p_1) + \beta (e_k^* p_2) \)

The valuation functional \(e_x^*\) can be represented by the powers (row) covector acting on the coefficients (column) vector:

\(e_x^* p = p(x) = \begin{bmatrix} 1 & x & x^2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \end{bmatrix} \)

The functionals \(e_0^*,~e_1^*\) and \(e_2^*\) are linearly independent, can be stacked  as rows of a \(3\times 3\) matrix:

\[
\begin{bmatrix} e_0^* \\ e_1^* \\ e_2^* \end{bmatrix}
= \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4\end{bmatrix}}_{A}
\]

The functional \(e_3^* = \begin{bmatrix} 1 & 3 & 9 \end{bmatrix}\) is linearly dependent on \(e_0^*, e_1^*, e_2^*\). We wish to solve the linear system

\(e_3^* = c_0 e_0^* + c_1 e_1^* + c_2 e_2^* = \underbrace{\begin{bmatrix} c_0 & c_1 & c_2 \end{bmatrix}}_{c} A \)

and indeed, by inversion of \(A\) we get \(c =\begin{bmatrix} 1 & 3 & 9\end{bmatrix}A^{-1} =\begin{bmatrix} 1 & -3 & 3\end{bmatrix}\),or \(e_3^* = e_0^* -3e_1^* + 3e_2^*\)


Exercise: find the smallest linear subspace of the space of smooth functions that is invariant with respect to shift, and containing \(\exp(2x)-x^2\).


Solution (option #1): note that whenever \(V\) contains smooth functions and is closed under shifts, then it is closed under differentiations (see Note below).

Let \(f(x) = \exp(2x)-x^2\) and let \(u_k(x)=f^{(k)}(x)\). If \(V\) then \(u_k\in V\) for all \(k\geq 0\); let us compute a few of those derivatives:
\(u_0(x) = \exp(2x)-x^2\)
\(u_1(x) = 2\exp(2x)-2x\)
\(u_2(x) = 4\exp(2x)-2\)
\(u_3(x) = 8\exp(2x)\)
\(u_4(x) = 16\exp(2x) = 2u_3(x)\)
\(u_5(x) = 32\exp(2x) = 2u_4(x)\).

Since \(u_0,\ldots,u_3\) are linearly independent the minimal shift invariant space (or derivative invariant space) must contain all of them, i.e \(V=\text{span}\{u_0,\ldots,u_3\}\), which is shift invariant and contains smooth functions (being spanned by smooth functions). The linear independence streak breaks with \(u_4 = 2u_3\), and in general \(u_{n+1}(x) = 2u_{n}(x)\) for all \(n\geq 3\).

Note: recall that in class we mentioned that whenever \(V\) is a space of smooth functions that is also shift invariant, there exists some differential equation \(\sum_{k=0}^d c_k f^{(k)}(x) =0 \) whose solution space is \(V\). If we differentiate again this very differential equation, we get \(\sum_{k=0}^d c_k (f’)^{(k)}(x)=0 \). So whenever \(f(x)\in V\) is a solution of the differential equation, then so is its first derivative \(f'(x)\in V\), and by induction a derivative \(f^{(n)}\in V\) of any order. In this problem, the differential equation is \(f^{(4)}(x) -2 f^{(3)}(x) = 0 \).


Solution (option #2): we look for a linear function space \(V\)  that has the following properties:

  1. Shift invariant
  2. Contains smooth functions
  3. Contains  \(f(x) = \exp(2x)-x^2\)
  4. Has minimal dimension (i.e does not contain a subspace of lower dimension with all the listed properties).

Since the space should contain \(f\) and all its shifts, we first look at what algebraic expressions occur when we shift \(f(x)\):

\(f(x+T) = e^{2T}\exp(2x) -x^2-2Tx-T^2 \)

We can use the function \(f\) itself as a basis vector to \(V\) (and then add other functions as needed), but since \(f\) (and its arbitrary shift) is a sum of a polynomial of degree \(2\) and and exponential function, we might use those (simpler terms) instead as basis vectors.

Let \(u_0(x) = 1,~ u_1(x) = x,~ u_2(x) = x^2\), and \(u_3(x) = \exp(2x)\), and let \(V=\text{span}\{u_0,u_1,u_2,u_3\}\).  This space has all the desired properties:

  1. \(V\) is shift invariant, as its shifted basis vectors remain in \(V\):
    \(u_0(x+T) = 1 = u_0(x) \in V \)
    \(u_1(x+T) = (x+T) = u_1(x)+Tu_0(x) \in V \)
    \(u_2(x+T) = (x+T)^2 = u_2(x)+2Tu_1(x) + T^2u_0(x) \in V \)
    \(u_3(x+T) = \exp(2(x+T)) = e^{2T} u_3(x) \in V \)
  2. \(V\) is spanned by four smooth functions (three polynomials and one exponential), hence contains only smooth functions.
  3. \(V\) contains \(f=u_3-u_2 \in V\) and all its possible shifts:
    \[
    f(x+T) = e^{2T}u_3(x) – u_2(x) -2T u_1(x) – T^2 u_0(x)  = \underbrace{\begin{bmatrix} e^{2T} & -1 & -2T & -T^2 \end{bmatrix}}_{C_T}\begin{bmatrix} u_3(x) \\ u_2(x) \\ u_1(x) \\ u_0(x) \end{bmatrix}
    \]
  4.  Lastly, we should comment on the minimality of \(V\). The dimension of \(V\) is \(4\), being spanned by four linearly independent functions. The question is now whether a space of lower dimension (\(\leq 3\)) can satisfy all the above properties (1)-(3). We will show that \(f(t),f(t+1),f(t-1)\) and \(f(t+2)\) are linearly independent, and indeed
    \(C_0 = \begin{bmatrix} 1 & -1 & 0 & 0 \end{bmatrix} \)
    \(C_1 = \begin{bmatrix} e^2 & -1& -2& -1 \end{bmatrix}\)
    \(C_2 = \begin{bmatrix} e^{-2} & -1& 2& -1 \end{bmatrix} \)
    \(C_3 = \begin{bmatrix} e^{4} & -1& -4& -4 \end{bmatrix} \)The four coefficient vectors above are linearly independent (easily verifiable by taking their determinant), hence their corresponding functions (all of which reside in \(V\)) are also linearly independent – hence \(V\) must be no less than \(4\) dimensional.
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