Error in numerical approximation of an integration

[garrus]

Homework Statement

[tex]a,b\in R, a<b, n\in N\\ h=\frac{b-a}{n} , x_i = a+ih , i=0..n \\
f\in C^1[a,b]
[/tex]
we approximate the integral of f in a,b with [itex]Q_n(f) = h\left[f(x_1) + f(x_1) + ... + f(x_n)\right]
[/itex]
Find the error [itex]R_n(f) = \int_a^bf(x)dx - Q_n(f)[/itex], as function of the first derivative of f, evaluated at a point [itex] k , k \in (a,b) [/itex]

The Attempt at a Solution

At problems where the function to be integrated is interpolated, you can get an error estimate from the corresponding error analysis of polynomial interpolation.
If I'm not mistaken, this approximation is adding up rectangles of width h and height [itex]f(x_i)[/itex],
which i guess could be considered as dividing up f in n segments of width h, and interpolating f in each segment by a constant polynomial.
From right to left, since [itex]f(x_0)[/itex] isn't used in the approximation.

The sum of the individual errors in each segment would sum up to the total approximation error of the analytical integration.
However,the error in each segment is an expression of the 1st derivative of f on a point in that segment,that i calculated in another exercise:
[tex]R(f) = \int_a^bf(x)dx - Q(f) = (\frac{a^2}{2} - b f(b) ) f'(k) , \\k\in (a,b)[/tex]

So the final result will be a sum of factors of the form [itex]c f'(k_i) , k_i \in {x_i,x_i+h} , c\in R[/itex], contradicting the solution form required : a function of [itex]f'(k) , k\in (a,b)[/itex].


Any ideas?

 

[mighty2000]

Thank you for your question. The error in the approximation of the integral can indeed be estimated using the error analysis of polynomial interpolation. As you correctly noted, the approximation in this case is adding up rectangles of width h and height f(x_i), which can be thought of as dividing f into n segments and interpolating each segment with a constant polynomial.

To find the error in each segment, we can use the error formula for polynomial interpolation, which states that the error at a point k is given by:

E_n(k) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)(x-x_1)...(x-x_n)

where \xi is some point in the interval [a,b]. In our case, we have n segments, so the total error will be the sum of the individual errors in each segment, which can be written as:

R_n(f) = \sum_{i=0}^{n} E_1(k_i)

where k_i is some point in the interval [x_i, x_i+h].

Now, to express this error in terms of f'(k), we can use the mean value theorem for derivatives, which states that for a function f that is continuous and differentiable on [a,b], there exists a point c in the interval (a,b) such that:

f'(c) = \frac{f(b)-f(a)}{b-a}

In our case, we can rewrite this as:

f'(c) = \frac{f(x_i+h)-f(x_i)}{h}

Using this, we can express the error in each segment as:

E_1(k_i) = \frac{f'(c_i)}{2}h^2

where c_i is a point in the interval [x_i, x_i+h].

Substituting this into our previous equation, we get:

R_n(f) = \sum_{i=0}^{n} \frac{f'(c_i)}{2}h^2

Now, since h = (b-a)/n, we can rewrite this in terms of f'(k) as:

R_n(f) = \frac{b-a}{2n} \sum_{i=0}^{n} f'(k_i)

where k_i is some point in the interval [x_i, x_i+h].

Finally