Hello,

You've made a slight mistake in C.pm(T) - check the function name - should "C" be in uppercase or lowercase.

Your first "solve" will not work because you used integration, and the unknown is the upper integration limit. This will not work with neither "solve" nor "roots". Those function expect that the function you are trying to find the root is analytically given. Integration is numerically performed in SMath. In those cases you can use some "home made" root finding procedure. For instance if you made this function and insert it into your worksheet (directly or as a snippet).

file secant.sm
you can solve you problem by, say, defining this function first
f(T.2)←int(C.vm(T)/T,T,T.1,T.2)+R*ln(V.2/V.1)
and finding its root
T2←secant(f(_T.2),T.2,10^{-3})
checking
T2=308.1659 f(T2)=-3.2387*10^{-4}

As for the second "solve" I tried to plot the
S(V.1,T.1)-S(V.2,x)
and could not find where the root could be - it was always negative. There might be mistake in defining this function.

Regards,
Radovan

Hello,

The simplest answer to your question would be - just try it. If you are convinced that there are some real roots of the given function try "solve" and "roots" first. If they do not work then try "secant". If this fail as well - then try something else. There is no numerical procedure which will give you the roots for every nonlinear functions. Some functions will fail, some will not.

On the other hand, regarding SMath, one of the situation when "solve" and "roots" will not work is when you use "eval" function.
Here is an example:
f1(x)←2*x^2+5*x-3
solve(f1(x),x,0,2)=0.5 f1(0.5)=0
This will not work:
f2(x)←eval(2*x^2+5*x-3)
solve(f2(x),x,0,2)=#
Although:
f2(0.5)=0

Regards,
Radovan

Edited by user 23 November 2010 19:53:51(UTC)  | Reason: Not specified