Convergence rate (with unknown exact solution)

[fred_91]

Homework Statement

I want to compute the following integral

I=\int_0^1 f(x) dx

for a function f(x) such that the integral I cannot be evaluated analytically.
f(x) is a known function.
Therefore we want to obtain I numerically. To do this we want to use the Trapezium method with uniform steps for x.
Then after that we need to compute the convergence rate.

Homework Equations

Using the trapezium method, we will split the nodes for x using equidistant steps x_0,x_1,x_2,...x_N
In general, we have
I=\int_a^b f(x) dx = \frac{b-a}{N}\left( f(x_0)+2f(x_1)+2f(x_2)+ 2f(x_3)+...+ 2f(x_{N-1}) +f(x_n) \right)

Aitken's extrapolation formula can be written as:
\overline{I}=I_i+\frac{(I_{i+1}-I_i)^2}{2I_{i+1}-I_i-I_{i+2}}
for 3 consecutive points: I_{i},I_{i+1},I_{i+2}.

The Attempt at a Solution

In our case, a=0, b=1. We will take 3 cases for N: N=25, N=50, N=100.
Therefore, we will have 3 approximations: I_1, I_2, I_3 corresponding to N=25, 50, 100 respectively.

Now we can find a good approximation to the real value using Aitken's approximation:
\overline{I}=I_1+\frac{(I_2-I_1)^2}{2I_2-I_1-I_3}

However, how does this help us to compute the convergence rate?

Any ideas or guidance is very much appreciated.

Thank you.

 

[mfb]

I think there is a 2 missing in the denominator of your first formula.

However, how does this help us to compute the convergence rate?

You could check the difference between the approximation and the estimated real value for N=25, 50, 100 and see if there is some clear trend (like 1/N, 1/N^2 or similar).

 

[fred_91]

Thank you for your reply.

Which missing 2 do you mean? For the formula of the trapezium method?

Oh right, I see, you mean to compute:
|\bar{I}-I_1|, |\bar{I}-I_2|, |\bar{I}-I_3| ?

 

[mfb]

Which missing 2 do you mean? For the formula of the trapezium method?

Right.

Oh right, I see, you mean to compute:
|\bar{I}-I_1|, |\bar{I}-I_2|, |\bar{I}-I_3| ?

Right. It could give a hint about the convergence rate.

 

[fred_91]

Oh right, I see the missing 2 :)

the formula should be:
I=\int_a^b f(x)dx=\frac{b-a}{2N}\left(f(x_0)+f(x_1)+f(x_2) + f(x_3)+...+f(x_{N-1})+f(x_N) \right).

So, I could say that
|\bar{I}-I_1| is an approximation of the error?

Thanks again.

 

[fred_91]

and similarly for I_2, I_3.

 

[mfb]

So, I could say that
|\bar{I}-I_1| is an approximation of the error?

Sure.