Question

Let \(V\) be the real vector space of all continuous functions \( f : [0,2] \to \mathbb{R} \) such that the restriction of \(f\) to the interval \([0,1]\) is a polynomial of degree \(\le 2,\) the restriction of \(f\) to \([1,2]\) is a polynomial of degree \(\le 3,\) and \(f(0) = 0.\) Then the dimension of \(V\) is _________.

Hint:

Always subtract one dimension for each linear constraint like continuity or boundary conditions when combining polynomial segments.

Answer 1

Correct Answer: 5

Step 1: Represent functions on both intervals.
For \(x \in [0,1],\ f_1(x) = a_0 + a_1x + a_2x^2.\) Since \(f(0) = 0,\) we get \(a_0 = 0.\) Thus, \(f_1(x) = a_1x + a_2x^2.\) For \(x \in [1,2],\ f_2(x) = b_0 + b_1x + b_2x^2 + b_3x^3.\)
Step 2: Continuity condition at \(x=1.\)
We must have \(f_1(1) = f_2(1).\) That gives one linear relation between the coefficients: \[ a_1 + a_2 = b_0 + b_1 + b_2 + b_3. \]
Step 3: Count degrees of freedom.
- \(f_1\) has 2 free parameters \((a_1, a_2)\). - \(f_2\) has 4 free parameters \((b_0, b_1, b_2, b_3)\). Total = 6 parameters minus 1 continuity constraint \(= 5.\) Wait, we need to recheck boundary continuity: Actually, \(f_1\) and \(f_2\) must be continuous on \([0,2]\), but not necessarily differentiable, so only one constraint exists (continuity at \(x=1\)). Hence, dimension \(= 6 - 1 = 5.\) But if we also require *continuity at both endpoints* and *\(f(0)=0\)* already included, then the final dimension is \(\boxed{5}.\)