Basic Concepts for nth Order Linear Equations — Question 3

Question 3 – Solution

Strategy. Build the derivative matrix and use its normalization at the initial point.

Step 1: Verify solutions and a nonzero determinant. Every function is cubic or lower, so its fourth derivative is zero. With $h=x-1$, their derivative matrix is $$F(x)=\begin{pmatrix} 1&h&h^2/2&h^3/6\\0&1&h&h^2/2\\0&0&1&h\\0&0&0&1 \end{pmatrix}.$$ It is triangular with unit diagonal, giving $\boxed{W=\det F=1}$ at every $x$.

Step 2: Use the identity matrix at the center. Since $F(1)=I$, the coefficient vector is exactly the prescribed derivative vector. Therefore $$\boxed{y=2-h+\frac 32h^2+\frac 23h^3,\qquad h=x-1.}$$ The factorials in the basis make its initial derivative matrix the identity; without them, derivative values would not equal coefficients.

Step 3: Change coordinates without changing the solution. Expansion gives $$\boxed{y=\frac{23}{6}-2x-\frac 12x^2+\frac 23x^3.}$$ Here $y'=-2-x+2x^2$, $y''=-1+4x$, $y'''=4$. At $1$, these are $-1,3,4$, while $y(1)=2$, verifying all data independently of the shifted expression.

Step 4: Establish the initial-data isomorphism. Differentiation and evaluation are linear. For any vector in $\mathbb R^4$, the corresponding combination of $f_0,f_1,f_2,f_3$ realizes it, proving surjectivity. If all four derivatives at $1$ vanish, the coefficient vector in this basis is zero, proving injectivity. Thus the solution space has dimension four. A change of basis changes coordinates, not the number of free constants, the initial data themselves or the resulting function.