> Let's say I want to solve this:
>
> a = 10
> 2(SqrRt(a) * Pi) / 2 + 4 = _answer_
>
> Is this possible in Maxima? =) I'm sure it is, but how?
(C1) a:10;
(D1) 10
(C2) 2*(sqrt(a)*%pi)/2+4;
(D2) SQRT(10) %PI + 4
Notice that Maxima normally gives results in symbolic form.
You can now evaluate that result as an ordinary floating-point number
(14 digits):
(C3) float(%);
(D3) 13.9345882657961
or as a high-precision floating-point number (in this case 50 digits):
(C6) bfloat(d5),fpprec:50;
(D6) 1.3934588265796101234433550670320351124771143722743B1
You can even express your approximations as continued fractions:
(C7) cfdisrep(cf(d3));
1
(D7) 13 + ----------------------
1
1 + ------------------
1
14 + -------------
1
3 + ---------
1
2 + -----
1
9 + -
2
Things get more interesting when your equations don't give the answer
directly:
(C8) 1+1/x=x;
1
(D8) - + 1 = x
x
(C9) solve(d8,x);
SQRT(5) - 1 SQRT(5) + 1
(D9) [x = - -----------, x = -----------]
2 2
(C10) float(d9);
(D10) [x = - 0.61803398874989, x = 1.618033988749895]
Have fun!
-s