# Fortran bisection method

• Fortran
Gold Member
I'm rusty with fortran and programming in general. I can't see my "error" in a code that I wrote from scratch.
Basically I wanted to get some fun and solve for a temperature in thermodynamics where I must get "T_f" which appear in a transcendental equation: ##A\ln \left ( \frac{T_f^2}{T_1T_2} \right )+2B[2T_f-(T_1T_2)]=0##.
Where T_1 is worth 100, T_2 is worth 200, A=2 and B=0.005.
Here is my code:
Code:
program rootfinderbisection
implicit none

real(4)::a,b,xn,fa,fb,fn,f,e,x_n,t_1,t_2

write(*,*)"Enter the temperatures t_1 and t_2"
write(*,*)"Enter the precision you want"
fa=f(a)
fb=f(b)

if (abs(fa)<e) then
write(*,*) "The solution is",a
stop
else if (abs(fb)<e) then
write(*,*)"The solution is",b
stop
end if

if (f(a)*f(b)>0) then
write(*,*)"There's an even number of root(s) in that interval"
stop
endif

xn=0.5*(a+b)
fn=f(xn)

do while (abs(fn)>e)
if (fa*fn<0) then
b=xn
fb=fn
else
a=xn
fa=fn
endif

xn=0.5*(a+b)
fn=f(xn)
end do
write(*,*)"The solution is",x_n

end program
that I compile along with
Code:
function f(x)
implicit none

real(4)::f,x, t_1,t_2
f=2.*log((x**2.)/(t_1*t_2))+2.*0.005*(2.*x-(t_1+t_2))

end function
The problem is that for the values I said above, I get the message "There's an even number of root(s) in that interval". Thus apparently the condition f(a)*f(b)>0 is satisfied. However when I enter the numbers in my pocket calculator I get f(a)=-2.3862... and f(b)=2.3862... where I truncated both values. Thus the condition is not satisfied.
I really don't see what's wrong. Any idea?

Edit: I tried with many different values for a and b. I even changed t_1 and t_2, same message.

Edit2: I've just tested the values of fa and fb. It thinks they are worth infinity. Going to check this out.
Edit3: No idea why it thinks they are worth infinity. o_0.
Edit4: ok it does not plug in t_1 and t_2 inside my function.

Last edited:

SteamKing
Staff Emeritus
Homework Helper
Try including t_1 and t_2 as arguments in the function f and see if this resolves the calculation errors.

Your function "f(x)" does not agree with your formula at the top of your post. Is the last term in the bracket term of your equation "T_1 * T_2" or "T_1 + T_2".

jtbell
Mentor
Edit4: ok it does not plug in t_1 and t_2 inside my function.
t_1 and t_2 in the function are completely separate variables from t_1 and t_2 in the main program. If you want t_1 and t_2 in the function to have the same values as in main, you need to make them arguments to the function along with x.

Gold Member
Thanks all for the tips. To TheoMcCloskey, you're right, I made a latex typo here for T_f. The one in my program is ok.
So it got rid of the problem! However I now get "The solution is 3.02481484E-39" which is basically 0. I'll investigate and post if stuck.

Edit: Worked, problem solved. I had to change the line "write(*,*)"The solution is",x_n" to "write(*,*)"The solution is",xn".
Thanks a lot guys!

Last edited: