Re: [Help-gsl] Can you do this in GSL?

[Yanlong Yang]

assume

F(x)=0.067x3 - int_0^x t2/(exp(t)+1) dx
the soultion of the equation F(x)=0 is what Anders wants
the solution of the equation  F'(x)=0 is where F(x) 's max/min/inflexion,
e.g ,the following steps are not right to find the solution of the first one
x^2=exp(x)  -->
2*x=exp(x) -->
2=exp(x)-->
0=exp(x)
Wrong

So we cannot find a solution of a equation from its derivative form.

Using the following Matlab script,we can find the solution is near
1.8927,
==============
clear all;
clc;

b=linspace(0,20,100);         % b=[0,0.2,0.4,...,20]
f1=0.067*b.^3;

f2=[];
f=@(x)x.^2./(exp(x)+1);
for n=1:length(b)
    int_f=quad(f,0,b(n));      % integral of f from 0 to b(n)
    f2=[f2 int_f];
end

plot(b,f1,b,f2);      % from the plot,we can find the solution is near
2. and reduce the b span step by step to
                             % [1.8926,1.8928]

==================

but the solution of 0.2=1/(exp(x)+1) is 1.3826.

Hope this helps.

Yanlong Yang
address@hidden




zkoza wrote:
> > >   0.067*b^3 = \int_0^b x^2/(exp(x)+1) dx 
>
>
> > Assume that F(b) = \int_0^b x^2/(exp(x)+1) dx. 
> > Your equation is then 0.067*b^3 = F(b) - F(0) = F(b). 
> > Now differentiate with b, and you have 0.2*b^2 = F'(b) = f(b).
>
>
> > You can calculate f(b) as a finite difference, f(b) = (F(b+h)-F(b-h))/2h.
>
>
> You can do even better: since b is used only as the upper limit in the 
> integral, the derivative is simply the integrand:
> f(b) = b^2/(exp(b)+1).
> After substituting it to 0.2*b^2 = f(b), you are left with
> something as simple as exp(b) \approx 4, or 
> b \approx ln(4).
>
> regards,
> Z. Koza
>
>
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