Trapezoidal Rule Homework: Find Expression for wj

[sara_87]

Homework Statement

if i have an integral equation of the form:

$$\lambda x(t)-\int{K(t,s)x(s)ds}=y(t)$$

and i want to apply the nystrom method to find a numerical solution, then the quadrature rule gives:

$$\lambda x(t_i)-\sum^n_{j=1}{w_{j}K(t_i,t_j)x(t_j)}=y(t_i)$$

my question is:
What is (using the trapezoidal rule) the expression for w_j

Homework Equations

The Attempt at a Solution

i have no idea except that there's a 'h' somewhere (step length)
thank you

 

[Mark44]

At the risk of telling you something you might already know, here's what's going on in the trapezoid rule.

You have a function f(t) that you want to integrate, and you have divided your interval [a, b] into n subintervals of length h, using a partition {t₀, t₁, ... , t_i, t_{i + 1}, ..., t_n}.

The trapezoid rule approximates $$\int_{t_i}^{t_{i + 1}} f(t) dt$$ by a trapezoid whose area is $$\frac{f(t_i) + f(t_{i + 1}}{2}h$$.

When you add up all the trapezoidal regions, you get
$$\frac{f(t_0) + f(t_{1})}{2}h + \frac{f(t_1) + f(t_{2})}{2}h + \frac{f(t_2) + f(t_{3})}{2}h + ... + \frac{f(t_{n - 1}) + f(t_{n})}{2}h$$

This works out to (1/2)h*[f(t₀) + 2f(t₁) + 2f(t₂) + ... + 2f(t_{n - 1}) + f(t_n)], because all of the f(t_i) terms appear twice except for the first one and the last one.

So... my guess is that w_j in your integral is going to one of two different values, depending on the value of j; w_j = (h/2) for j = 0 and j = n, and w_j = h for 0 < j < n.