My question is simple:
Give the table:
| x |0|2|4|6| |f(x)|1|3|5|7|
Why when calculating Lagrange Polynomial Interpolation for:
| x |0|2| |f(x)|1|3|
P1(x) = x+1
And when calculating Lagrange Polynomial Interpolation for:
| x |0|2|4| |f(x)|1|3|5|
P2(x) = 1,5x^2 +1x + 1 ??
And when calculating Lagrange Polynomial Interpolation for:
| x |0|2|4|6| |f(x)|1|3|5|7|
P3(x) = -0,229166666...x^3 + 0,75x^2 + 0,2033333...x + 1 ??
My calculations for P2(x) and P3(x) are wrong??
If you calculate in "Lagrange Calculators" at Internet, if you give the points for some values of the table, or all (P3(X)), the result is always "x + 1".
Please help me...
You second result is wrong, you must obtain $P_2(X)=X+1$.
$$l_0(X)=\frac{X-2}{0-2} \frac{X-4}{0-4}=\frac{1}{8}(X-2)(X-4)$$ $$l_1(X)=\frac{X-0}{2-0} \frac{X-4}{2-4}=\frac{-1}{4}X(X-4)$$ $$l_2(X)=\frac{X-0}{4-0} \frac{X-2}{4-2}=\frac{1}{8}X(X-2)$$ $$P_2(X)=1l_0(X)+3l_1(X)+5l_2(X)=X^2(\frac{1}{8}-\frac{3}{4}+\frac{5}{8})+X(-\frac{2}{8}-\frac{4}{8}+\frac{12}{4}-\frac{10}{8})+\frac{8}{8}=X+1$$
For $P_3$ you also have $P_3(X)=X+1$.