Solving f(x)=e^(0.1x^2) with H5, H3 and Error Bounds

[stunner5000pt]

The data \below gives the list of values for $f(x) = e^{0.1x^2}$
Approximate f(1.25) by using H5 (1.25) and H3 (1.25) wqhere H5 uses nodes x0 =1, x1 =2, x2 = 3. and H3 uses nodes x0=1, x1 = 1.5
Find error bounds for those approximations

this questio nand its data table is given in question on this link
http://people.math.yorku.ca/poliakov/3241f05/3241a4.pdf

So first i have to find the lagrange polynomials
But hwat does it mean $L_{2,0} (x)$ or $L_{2,1) (x)$
im sorry I am asking such an elementary question because my textbook is not specific at all and all web searches yield info that is not useful to solving this

Please help me on this. Also can you have a look at number 4. How would i do tha. Similarly ik now i have to find Sj. But how iw Sj defined? So suppose i were to find S0 then Xj in this case is 0? How do you find aj bj and so on however? Please help! I really need to able to solve this...

Thank you for your help!

 

[lippyka]

The Lagrange polynomials L_{2,0} (x) and L_{2,1} (x) are polynomials of degree 2 that pass through the points (x0, f(x0)), (x1, f(x1)), and (x2, f(x2)). In this case, the nodes x0, x1, and x2 are 1, 1.5, and 2, respectively. To find the Lagrange polynomials, you need to solve a system of linear equations. First, solve for the coefficients of the polynomial:L_{2,0}(x) = a_0 + a_1x + a_2x^2Then, substitute the three given points into the equation and solve the resulting system of equations for the coefficients a0, a1, and a2.For example, for L_{2,0} (x):a_0 + a_1(1) + a_2(1)^2 = e^{0.1(1)^2}a_0 + a_1(1.5) + a_2(1.5)^2 = e^{0.1(1.5)^2}a_0 + a_1(2) + a_2(2)^2 = e^{0.1(2)^2}Solve the equations to get the coefficients:a_0 = 0.3536a_1 = 4.0663a_2 = -3.5908Therefore, L_{2,0} (x) = 0.3536 + 4.0663x - 3.5908x^2Similarly, you can solve for L_{2,1} (x).To approximate f(1.25) using H5 (1.25) and H3 (1.25), substitute x = 1.25 into the respective polynomials5 (1.25) = 0.3536 + 4.0663(1.25) - 3.5908(1.25)^2 = 1.4571H3 (1.25) = 0.7444 + 1.8277(1