Help with basics and finite difference method

2014 9 14
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更新时间：2014 9 15
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I have to write a program using the finite-difference formula to calculate the approximate value for the derivative of a function. The test will be tan(x) for x=1, determining the error by comparing with sec^2(x). I have no idea where to begin.

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Image Analyst
Image Analyst 2014-9-14

0 个投票

How about a for loop and taking the delta Y over the delta X where the separation is decreasing until it gets really really small, then compare to sec^2(x) and see how the difference gets smaller and smaller as the separation gets smaller and smaller. That's the finite difference method.
http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers

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Will
Will 2014-9-14
I'm supposed to use a log scale for the h (as in f(x+h)) and for the magnitude of error, would that get in the way of the for loop?
Thank you so much. I feel helpless.
Image Analyst
Image Analyst 2014-9-14
编辑：Image Analyst 2014-9-14
So just start with h = 1, then go to h = 0.1, then to h = 0.01, then to h = 0.001, etc. You'll make h get smaller logarithmically.
x = 1;
for k = 1 : 6
h = 10^(1-k);
leftYValue = tan(x-h);
rightYValue = tan(x+h);
slope = ................
and so on.
Will
Will 2014-9-14
Oh my god, I thought I was going to have to figure out how to actually implement f'(x) = (f(x+h)-f(x))/h. Numerical analysis is so foreign... sheesh. Well thank you so much, I appreciate it immensely.
Image Analyst
Image Analyst 2014-9-14
Well that is what you're implementing. In my code, I went h above and h below the target x value (because I wanted to be symmetric), not just to one side like your equation, so my denominator would be 2*h rather than h. And my leftYValue is your f(x) and my rightYValue is your f(x+h). So the slope is (rightYValue - leftYValue)/(2*h), which is essentially your equation.
Will
Will 2014-9-15
I see, that actually seems to be better than what is suggested by the instructor... I'll ask if I can try something like that. So I use that, and then compare by subtracting the slope from sec^2(x), and then plot the magnitude of error from that subtraction on one axis, and h on the other. That is so much better! Thank you!
Image Analyst
Image Analyst 2014-9-15
OK, great, glad I could help. Can you go ahead and mark the answer as Accepted then?
Will
Will 2014-9-15
I will, but I just talked to my instructor and he explained that though your method is valid, I must use the given formula. Could I just do the same things we talked about, only replace slope with
slope = (tan(x+h)-tan(x))/h
and do the error vs. h plot? I'm going to use loglog(h, error, -s), but this gives me a weird window... Please let me know if you can help with that as well, though I understand if you don't want to. Thanks for helping!

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