# Convergence rate (with unknown exact solution)

1. Jul 13, 2013

### fred_91

1. The problem statement, all variables and given/known data

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.

2. Relevant 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: $I_{i},I_{i+1},I_{i+2}$.

3. 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.

2. Jul 13, 2013

### Staff: Mentor

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

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).

3. Jul 13, 2013

### fred_91

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

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

4. Jul 13, 2013

### Staff: Mentor

Right.

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

5. Jul 13, 2013

### 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.

6. Jul 13, 2013

### fred_91

and similarly for $I_2, I_3$.

7. Jul 13, 2013

Sure.