# Quadratic splines, decreasing indispensable

Hi I have a trouble with square splines, I am intended to locate $ S_1 $ and $S_2$ that interpolates the following points $S (- 1) = 0$, $S (0) = 1$, $S (1) = 2$, and at the same time we want to find $S$ such that $\int _ - 1 ^ 1 \! (S (x)) ^ 2 \, \mathrm d x$ is marginal. The solution gets on the kind

$S_1(x)=a_1 x^2 +b_1x +c_1$ on $[-1,0]$ and also

$S_2(x)=a_2 x^2 +b_2x +c_2$ on $[0,1]$

my solution :

I make use of the information factors and also locate that $a_1=-a_2$, $b_1=b_2$ and also $c_1=c_2=1$, yet I have no suggestion just how to make use of decrease $\int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x$, can I separate it up?

$$\min \int_{-1}^1 \! (S(x))^2 \, \mathrm{d}x=\min (\int_{-1}^0 \! (S_1(x))^2 \, \mathrm{d}x + \int_0^1 \! (S_2(x))^2 \, \mathrm{d}x )$$

I recognize I need to get an expression and also possibly set the by-product to absolutely no yet I simply do not recognize just how to strike the decreasing indispensable given that the function has 2 components. Aid substantially valued.

Here is a tip : given that you require your piecewise expression to be a *spline *, one point you need to have is the charge of an acquired problem at the factor $x=0$ ; particularly, $S_1^{\prime}(0)=S_2^{\prime}(0)$. (This is in addition to the noticeable formulas you get from replacing ideal values of $x$ and also $y$ right into the expressions for $S_1$ and also $S_2$).

So far, you need to have 5 formulas in 6 unknowns ; you can remove some unknowns, and also have the ability to share several of them in regards to simply among the unknowns, such that both $S_1$ and also $S_2$ rely on a solitary parameter. You had the appropriate suggestion to break up the indispensable right into integrals over $[-1,0]$ and also $[0,1]$ ; if you do that, you need to wind up with a square in a solitary variable, whose minimum is unimportant to locate.

Much shorter variation of the tip : you need to have the ability to share $a_1$ (and also hence $a_2$ too) in regards to either $b_1$ or $b_2$. Selecting $b_1$ as the variable of passion, and also replacing right into your indispensable, you need to have an expression of the kind

$$u(b_1)^2+v\cdot b_1+w$$

where $u$, $v$, and also $w$ are particular constants. If $u$ declares, after that $-\frac{v}{2u}$ is the minimum value of $b_1$ you look for. Replace right into all your various other relationships, and also you need to currently have actually the wanted spline.