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first forward and backward central diference

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Adam
Adam 2012 年 4 月 23 日
Hi, I have a task to do by diference of those formulas someone help me? input values are specified they get when I have done.
dy1=((f(x+h)-f(x-h))/(h))-1/2*h;
dy2=(((-3*f(x))+4*f(x+h)-f(x+2*h))/(2*h))+1/3*h^2;
dy3=((f(x+h)-f(x-f))/2*h)-1/6*h^2;
  3 件のコメント
Titus Edelhofer
Titus Edelhofer 2012 年 4 月 23 日
I understand the task. But I have not found a question?
Adam
Adam 2012 年 4 月 23 日
I do not know what I should do next, the result is some kind of strange
t=0.8;
h = 0:t:10^-5;
x=5;
f(x)=(x^2)+3;
dy=((f(x+h)-f(x-h))/(h))-1/2*h

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回答 (2 件)

Titus Edelhofer
Titus Edelhofer 2012 年 4 月 23 日
Hi,
o.k., let's start with the h:
0:t:10^-5
will be the zero. Start with e.g.
h = 0.01;
Take a look at the chapter about defining functions (or search the doc for "function handles" on how to define f.
Titus

Walter Roberson
Walter Roberson 2012 年 4 月 23 日
Algebra. What is f(x)=(x^2)+3 evaluated at (x+h) ? And at (x-h) ? And so on. Simplify and you get down to a nice formula.

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