For $ B^2_1(t) $ with knot values $ t_1 = 1, ..., t_4 = 4 $ Using the determinant method $ B^d_i(t) = (-1)^{d+1} (t_{i+d+1}-t_i) \frac1D A $
where D is the determinant of $\begin{bmatrix} 1 \ t_i \ ... \ t_1^d+1 \\ ... \\ 1 \ t_{i+d+1} \ ... \ t_{i+d+1} ^{d+1} \end{bmatrix}$ = 12
and A is the determinant of $\begin{bmatrix} 1 \ t_i \ ... \ t_i^d \ g(t_i) \\ ... \\ 1 \ t_{i+d+1} \ ... \ t_{i+d+1}^d \ g(t_{i+d+1}) \end{bmatrix}$ s.t. g(x) is the spline fn $ (t-x)_+^d $
thus by expanding along the last column (ie the column with g(t)) $ A = 2(t-1)_+^2 - 6(t-2)_+^2 + 6(t-3)_+^2 - 2(t-4)_+^2 $
then $ B^2_1(t) = \frac{-1}{4}[2(t-1)_+^2 - 6(t-2)_+^2 + 6(t-3)_+^2 - 2(t-4)_+^2] \\ = \frac12[(t-4)_+^2 - (t-1)_+^2] + \frac32[(t-2)_+^2 - (t-3)_+^2]$
Then the with the recursion formula $ B^d_i(t) = \frac{t-t_i}{t_{i+d} - t_i} B_{i}^{d-1}(t) + \frac{t_{i+d+1}-t}{t_{i+d+1} - t_{i+1}} B_{i+1}^{d-1}(t) $
I get: $ B^2_1(t) = \frac12[(t-1)[(t-3)_+^1 - 2(t-2)_+^1 + (t-1)_+^1] + (t-4)[(t-4)_+^1 - 2(t-3)_+^1 + (t-2)_+^1]] \\ = \frac12[ - (t-4)_+^2 + (t-1)_+^2] + \frac32[ - (t-2)_+^2 + (t-3)_+^2]$
However my two $ B^2_1(t) $ aren't the same. (note: the signs are different) Where did I go wrong?
The $B$ splines are usually taken to be nonnegative. The easiest way to check nonnegativity is to look at the sign of the coefficient of $(t-t_i)_+^d$ with the smallest $t_i$. This coefficient should be positive.
So, the first formula has the wrong sign. It seems that when expanding the denominator $A$ along the last column, you did not account for it being the last column (remember $(-1)^{i+j}$ factors in the expansion).