Trapezoidal Rule (just need to check my answer)

[DivGradCurl]

Problem

Use the Trapezoidal Rule with $$n=10$$ to approximate

$$\int _0 ^{20} \cos ( \pi x) \: dx$$

Compare your result to the actual value. Can you explain the discrepancy?

My work

The actual value can be found as follows

$$\int _0 ^{20} \cos ( \pi x) \: dx = \left. \frac{\sin ( \pi x)}{\pi} \right] _{x=20} - \left. \frac{\sin ( \pi x)}{\pi} \right] _{x=0} = 0$$

Next, we have the Trapezoidal Rule with $$n=10$$

$$T_{10} = \left( \frac{20-0}{10\cdot 2} \right) \left[ \cos (0\cdot \pi) + 2\cos (2\cdot \pi) + 2\cos (4\cdot \pi) + \dots + 2\cos (16\cdot \pi) + 2\cos (18\cdot \pi) + \cos (20\cdot \pi) \right] = 20$$

Can I just say that the reason for the discrepancy is the low value of $$n$$ ? According to the error bound, we have

$$\left| E_T \right| \leq \frac{K(b-a)^3}{12n^2} < \left| \mbox{ error } \right| \Longrightarrow \frac{1\cdot (20-0)^3}{12 n^2} < 10^{-1} \Rightarrow n \geq 82$$

which shows that in order to achieve accuracy to within one decimal place, it is required to have $$n \geq 82$$.

Is that it?

Thank you

PS: I should have calculated that for the first digit...

$$\left| E_T \right| \leq \frac{K(b-a)^3}{12n^2} < \left| \mbox{ error } \right| \Longrightarrow \frac{1\cdot (20-0)^3}{12 n^2} < 10^{0} \Rightarrow n \geq 26$$

 

[DivGradCurl]

Can I also say that the need for a higher value of n is due to the oscillatory integrand?

Thanks

 

[ehild]

Can I also say that the need for a higher value of n is due to the oscillatory integrand?

Thanks

I think yes, you can.

I do not remember the name of theorem concerning transfer of information which says that the sampling frequency should be at least twice that of the highest-frequency Fourier component of the signal to be transmitted.

Your sampling frequency here is the same as the frequency of the function. It is the worst choice possible as it gives the same values at each point, like a constant function would do.

ehild