MHB Find Order of Accuracy: Solve Mistake Worried

[evinda]

Hello! (Wave)

Suppose that we are given $h_1=0.1 \to \delta_1=0.01$ and $h_2=0.05 \to \delta_2=0.025$ and we want to find the order of accuracy of the method.
I have tried the following:

$\delta_1^n= c h_1^{p+1}, \delta_2^n=c h_2^{p+1} \Rightarrow p+1= \frac{\log \left( \frac{\delta_1^n}{\delta_2^n}\right)}{\log \left( \frac{h_1}{h_2} \right)}= \frac{\log 2- \log 5}{\log 2} \Rightarrow p= \frac{- \log 2}{\log 5}$.

But the result should be 1. Where is my mistake? (Worried)

 

[I like Serena]

Hello! (Wave)

Suppose that we are given $h_1=0.1 \to \delta_1=0.01$ and $h_2=0.05 \to \delta_2=0.025$ and we want to find the order of accuracy of the method.
I have tried the following:

$\delta_1^n= c h_1^{p+1}, \delta_2^n=c h_2^{p+1} \Rightarrow p+1= \frac{\log \left( \frac{\delta_1^n}{\delta_2^n}\right)}{\log \left( \frac{h_1}{h_2} \right)}= \frac{\log 2- \log 5}{\log 2} \Rightarrow p= \frac{- \log 2}{\log 5}$.

But the result should be 1. Where is my mistake? (Worried)

Hey! (Smile)

Since $h_2$ is smaller than $h_1$, I'd expect $\delta_2$ to be smaller than $\delta_1$.

Perhaps it should be:
$$p+1= \frac{\log \left( \frac{\delta_1}{\delta_2}\right)}{\log \left( \frac{h_1}{h_2} \right)}
=\frac{\log \left( \frac{0.01}{0.0025}\right)}{\log \left( \frac{0.1}{0.05} \right)}
= \frac{\log 4}{\log 2} = 2 \qquad\Rightarrow\qquad p = 1$$
(Wondering)