Formulas using sample points as parameters.

Providing some added information to the examples chapter of the book Numerical Methods for Scientists and Engineers (19.2)

The problem is estimating the following integral from two samples:

The parameters here are w_1, w_2, x_1, and x_2. Just as in the previous posting we use four defining equations. Why four and why not three or less? As it turns our for each parameter we need to construct a set of unique linear equations to solve all of the given parameters...this follows from Linear algebra regarding a problem of 4 unknowns, so we'd need for four sets of unique linearly independent equations to solve each such parameter.

Thus we start using exact equations for f(x) in successive powers of xⁿ where n = 0, 1, 2, and 3 in this case.

So the defining equations are :

Keep in mind the book actually evaluates these integrals, and I'll provide some brief explanation on evaluation techniques here.

For equation (1), we'd need to evaluate the definite integral on the left side of the equation. This could otherwise be rewritten as

Remember that while e^x approaches a infinitely large number the inverse of this approaches zero. Also remember e⁰ = 1, from basic natural logarithmic identities (college level algebra).

I'll cover integration techniques given by integration by parts for equation (2), and then we'll rely on a handy integration tables formula which is

this formula can be seen more clearly in our given methods found in evaluating the integral from the left side of equation (2). Namely integration by parts in successive application occur here.

Also I'd mention at this point, we can also use another handy rule known as L'Hopital's rule which goes as follows:

Thus, for instance,

is in indeterminate form, so we can apply L'hopitals rule. Noting e^-x = 1 / e^x.

Thus we take the first derivatives of f(x) = x and g(x) = e^x which yields

We can apply successive nth derivative applications of L'Hopital's Rule in the case of a nth power of x for the indeterminate case successively occurring so noting the factorial in the numerator after successive derivatives is a constant in the evaluation of such limit and thus is fixed 

By the way a factorial if you had forgotten is n! = n*(n-1)*...(2)*1

Thus, 3! = 3*2*1 for instance.

Let's go ahead and evaluate the integral of equation (2) on the left side

Using integration by parts, we have

For the integral in eq.(3) on the left side, we'll use our handy integration formula eq.(5a) to evaluate (which is basically an integration by parts formula) here. We'll use the results of eq.(7) to further evaluate this integral.

Finally we'll evaluate the integral in eq.(4) on the left side, we'll use our handy integration formula eq.(5a) , and we'll use the results of eq.(8) to further evaluate this integral.

The defining equations (1), (2), (3), and (4) can now be rewritten using equations (6), (7), (8), and (9) as

To the right of the defining equations are coefficients c₀, c₁, and 1.

These are to applied to a polynomial that will be used for defining our sample points.

The polynomial we use is a second degree 

multiply eq. (10) by c₀, eq. (11) by c₁, and eq. (12) by 1, and the add these equations together so

Using the second multiplier column to the right of the first column on eq. (10), (11), (12), and (13),

we reapply the same process again as in (14) to get

We then have from eq. (14) and (15) two pairs of linear equations 

The solution to this is c₁ = -4, and c₀ = 2

These are the coefficients of eq. (13a), so we have

The quadratic equation solves this providing samples points 

Applying these sample points to the defining equations (10), and (11) gives

 

 

Again we solve these linear equations for w₁ and w₂

which yields

Our formula above becomes

indicated “exact for cubics using only two samples of the integrand”.

“How did we know that the weights we found from the first two defining equations would satisfy the last two equations? The answer is simple; we choose the x_i so that the last two equations were linear combinations (with coefficients c₀, c₁, and 1) of the first two equations; hence the w_i automatically satisfied the last two equations.”

Numerical Methods for Scientists and Engineers, Dover, pg.319.