Derivative of a function in a particular point

Hi

I have an external function y= function fa(x)

y=sin(x./pi)

end

i want in the main program to find the derivative of fa in numerical point

thank you

George

Answer 1

[george veropoulos]

format  long 
x = 2 ;
h =  sqrt(eps) ;
f1=(sin(x + h)./(x+h) - sin(x)./x) / h % -0.400000
% -0.435397773981094
 f2=cos(x)./x-sin(x)./x^2 
 
 % -0.435397774979992

[Walter Roberson]

format  long 
x = 2 ;
h =  sqrt(eps(x)) ;
f1=(sin(x + h)./(x+h) - sin(x)./x) / h

f1 = 
  -0.435397778831590

 f2=cos(x)./x-sin(x)./x^2 

f2 = 
  -0.435397774979992

[george veropoulos]

yes sqrt(eps(x) )

Answer 2

[Jan]

Use one of the quotients of differences to get a numerical approximation of the derivative:

x  = 1.2345;
h  = sqrt(eps);
dy_right = (fa(x + h) - fa(x)) / (h)
dy_left  = (fa(x) - fa(x - h)) / (h)
dy_both  = (fa(x + h) - fa(x - h)) / (2 * h)

function y = fa(x)
y = sin(x./pi) 
end

[george veropoulos]

nice solution!! it work

[george veropoulos]

why take the sqrt(eps)? h=eps is wrong

[Walter Roberson]

h should not be less than eps(x), rather than being a fixed value like eps or sqrt(eps). Suppose for example that x were 1e9. You want x+h and x-h to always be distinguishable from x.

If you use eps(x) itself then you run across the possibility that your derivatives are dominated by round-off error.

sqrt(eps(x)) might be feasible.

[george veropoulos]

thank you !

[Jan]

If h is tiny, f(x) is very near to f(x+h). Then the difference is almost 0, such that the 16 valid digits of double values are more or less random. Then you divide by a small h and the random value is magnified. Try it:

x = 2;
h = 10 * eps;
(sin(x + h) - sin(x)) / h   % -0.400000
cos(x)                      % -0.416146
h = 1e6 * eps;
(sin(x + h) - sin(x)) / h   % -0.416147   % Much better

If you choose h too large, e.g. pi/2, the result is random again.

The optimal value of h for the quotient of differences depends on the 2nd derivative of the function, but this must be determined by another quotient of differences at first. The determination is not trivial and sqrt(eps) is a smart guess, which fails for some functions tremendously.

[george veropoulos]

Nice ! Thank you

there a more rigorous reason for that choice i mean from numrerical analysis?

[george veropoulos]

i use an other function sin(x)/x and the choice h=sqrt(eps) seem to work

[Jan]

@george veropoulos: The numerical analysis for the optimal choice of h is still worth to write a PhD thesis. As said before, the 2nd derivative of the function matters. To estimate this, you need a further small variation. Because this is expensive, if the function to be evaluated is huge, some heuristics are useful. This can be an important part of the processing time if you optimize an expensive function. A related field is the optimal choice of temproal and spatial steps sizes in finite element problems, e.g. the simulation of the earth clima.

The rule of thumb: If the function and the argument are about 1 and do not explode nearby, sqrt(eps) is a fair choice between the cancellation and discretization error. The "fairness" can be checkd by testing 10*sqrt(eps) and 0.1*sqrt(eps): If they reply the same derivative, you can assume to be on the right side. For a professional simulation, the explanation must sound more seriously. ;-)

[george veropoulos]

thank you all! very helpfull dicussion

George