Hi all,
I'm a final year high school student in Australia and have my final
exams coming up in about 2 weeks. I use maxima since I'm blind and can't
use the standard CAS calculator systems.
I've got a couple of last minute problems that I was hoping to get some
advice about.
1. I use the %solve command from package to_poly_solve. It works well
providing me with the general solutions. However, my class mates can
specify a domain where the solutions could be provided for and they get
these listed.
It seems a bit tedious with maxima as I have to set up the two general
solutions as separate functions and sub the various integers in that
domain, check all the solutions are in the domain and keep track of
everything. Is there any possible way to automate this process somehow
with lisp perhaps in the maxima-init.mac file? If anyone has any
pointers where I can start that would be really helpful.
2. I've had a problem with simplification. The function f(x)=log(x) has
been defined.
Then I must set up the equation:
2*f(u)=f(2*v)+f(3*v)
Then solve in terms of u:
solve(2*f(u)=f(2*v)+f(3*v),u);
[u = %e^(log(3*v)/2+log(2*v)/2)]
I ratsimp (%) this, but it remains the same.
Shouldn't it simplify down to sqrt(6)v?
If so is there any way to avoid future problems like this, or are there
just particular cases to be aware of?
3. I have two integrals, which must equal 3 and 2 respectively. So I
assign the integrated equation to f(x) in maxima and attempt to solve
for the two integrals like this simultaneously:
f(x):=7*log(x);
(%o3) f(x):=7*log(x)
(%i4) solve([f(m*n)-f(1)=3, f(m/n)-f(1)=2],[m,n]);
(%o4) []
How do I find the solution to this? (I need an exact value)
Here is the question exactly:
Find the exact value of m and n such that
\int _1^{mn} (f(x)) dx = 3 and
\int _1^{\frac{m}{n}} f(x) dx = 2
f(x)= \frac{7}{x}
4. My last question! I have a function h(x) = sin(x/2)+tan(x/2)+2.
I must solve h'(x)=0.
So I differentiate with diff() and attempt to solve the resultant
equation with %solve:
%solve(sec(x/2)^2/2+cos(x/2)/2=0,x);
Nonalgebraic argument given to 'to_poly'
unable to solve
(%o12) %solve([sec(x/2)^2/2+cos(x/2)/2 = 0],[x])
It seems solve () works a bit better:
solve(h(x)=0,x);
solve: using arc-trig functions to get a solution.
Some solutions will be lost.
(%o13) [x = 2*%pi-2*acos(sqrt(3)*%i/2-1/2),x = 2*acos(sqrt(3)*%i/2+1/2),x
= 2
*%pi]
But how can I then find the rest of the solutions? the domain is (-%pi,
%pi) union (%pi, 3*%pi)
If anyone can help with any of this stuff I would be very appreciative.
Thank you very much.
Daniel