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Math Help - Clamped Cubic Spline

  1. #1
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    Clamped Cubic Spline

    A clamped cubic spline s for a function f is defined by

    s(x)=\begin{cases} s_{0}(x) = 1 + Bx + 2x^2 - 2x^3 & \text{if } 0 \le x < 1 \\ s_{1}(x) = 1 + b(x - 1) - 4(x - 1)^2 + 7(x - 1)^3  & \text{if } 1 \le x \le 2 \end{cases}

    Find f'(0) and f'(2)

    So s'_{0}(1) = s'_{1}(1) and I get b = -2. Substitute it into the second derivative and I get f'(2) = 11. s'_{0} = 4x - 6x^2 so f'(0) = 0. But how do I find what B is?
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  2. #2
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    Quote Originally Posted by Makall View Post
    A clamped cubic spline s for a function f is defined by

    s(x)=\begin{cases} s_{0}(x) = 1 + Bx + 2x^2 - 2x^3 & \text{if } 0 \le x < 1 \\ s_{1}(x) = 1 + b(x - 1) - 4(x - 1)^2 + 7(x - 1)^3  & \text{if } 1 \le x \le 2 \end{cases}

    Find f'(0) and f'(2)

    So s'_{0}(1) = s'_{1}(1) and I get b = -2.
    How? s'_0(1)= s'_1(1) is B- 2= b.

    Of course, the function must also be continuous at x= 1 so you must have 1+ B= 1 so that B= 0 and then b= -2.

    Substitute it into the second derivative and I get f'(2) = 11. s'_{0} = 4x - 6x^2 so f'(0) = 0. But how do I find what B is?
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