Forward difference and central difference help - expected rates of convergence

Question

Savannah Phillips 2020-3-17

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https://ww2.mathworks.cn/matlabcentral/answers/511333-forward-difference-and-central-difference-help-expected-rates-of-convergence

回答： Ayush Gupta 2020-6-4

Hi, if anyone could help me with this question on my coursework I'd be really grateful! I've already had help on the first part, now it's just the second part:

Within your file , test your difference quotient formulas with h = 10^−i for i = 1, 2, . . . , 10. Plot the errors versus h in a figure with double-log axes. (This part I've already done)

Can you observe the expected rate of convergence (asymptotic behaviour as h → 0)? What happens for very small h? Hint: It may be helpful to include in your plot the graphs of functions r(h) = h^α with α the expected rates of convergence.

The mark scheme says that I need two error plots. Which is the second one? Is it r(h) = h^α? Do I need to plot the expected error? How would I go about doing that? Thank you in advance for your help!

Here is my code:

f = @(x) sin(x); % function handle

for i = 1:10

h(i) = 10.^(-i);

[fd(i),cd] = FDCD(f,0.9,h(i));

end

% f'(0) = 1 for this question

errors = 1 - fd; % Approximation error for forward difference quotient

% Setting the figure

figure

loglog(errors, h);

grid on;

% Formatting the plot

(just title, axes labels, etc)

I know that all this is right, but i have no idea what the second plot is spoken about in the mark sceme and how to go about making it, if it is r(h) = h^a. Thank you!

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darova 2020-3-18

What is this

Savannah Phillips 2020-3-18

that's just a function used to find the forward and central difference (it's all correct)

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Answer 1

Ayush Gupta 2020-6-4

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/511333-forward-difference-and-central-difference-help-expected-rates-of-convergence#answer_446141

在 MATLAB Online 中打开

Try using with this code for section 2

x=[-10:0.01:20]; 
dt = 0.00001; 
values_y = [-10:0.01:20]; 
y1 = [-10:0.01:20]; 
values_error = [-10:0.01:20]; 
y2 = [-10:0.01:20]; 
for i =1:length(x) 
    values_y(i) = 1/(10^(x(i))); 
    y1(i) = log(values_y(i)); 
    values_error(i) = (1/((10^(x(i)+dt)) - values_y(i)))/dt; 
    y2(i) = log(values_error(i)); 
end 
hold on 
plot(x,y1) 
plot(x,y2) 
hold off 