This near drop-in replacement for MS Office (Word, Excel, PowerPoint) has always been quite free (except for the Java part) and had it origins more than twenty years ago as Sun Microsystem's StarOffice. I do not recommend using the related other spin-offs like OpenOffice since LibreOffice is better supported these days. It can usually seamlessly read and save ubiquitous MS-formatted files (like Word, Excel, and PowerPoint).
This is an annotation software that you can use to mark up pdf files. It may take some practice to get it to work right for you. The native save format is not useful for anything but xournal, but when you are done marking up a pdf file, you can "export to pdf" to save it as another pdf file.
They have a very simple homepage: xournal.sourceforge.net
This is a download place: Xournal download at sourceforge.net
It is easy to search for variations, like xournalapp, xournal++, and versions that run on Windows or MacOS.
Yet another simple homepage that will point to ways you can Download it: https://sourceforge.net/projects/maxima
This is a software with a fifty-year history, descended from Macsyma originally developed by MIT. It runs on just about any modern computer. It may have a steep learning curve, but I think is worth it.
It resembles the expensive and proprietary CAS known as maple (maplesoft.com), and I think is a useful alternative.
Find out how much time it takes for an object, initially moving with speed of $v_0=1\,\mbox{m/s}$ and accelerating with $a=3\,\mbox{m/s}^2$ through a distance of ten meters. \begin{align} x_1 &= x_0 + v_0 t + \tfrac12 a_{01}t^2 \\[1ex] (10) &= (0) + (1) t + \tfrac12 (3) t^2 \\[1ex] (0) &= -(10) + (1) t + \left(\tfrac32\right) t^2 \end{align} We end up with a quadratic in time $t=t_1-t_0$, the unknown. Using maxima:
(%i1) eq: 0 = -10 + t + t^2 * 3/2$
(%i2) solve(eq,t);
sqrt(61) + 1 sqrt(61) - 1
(%o2) [t = - ------------, t = ------------]
3 3
(%i3) solve(eq,t),float;
(%o3) [t = - 2.936749891968888, t = 2.270083225302221]
From this, we can see there are two possible answers, and we pick the
one after $t_0=0$, so the answer is $t_1=(\sqrt{(61)}-1)/3=2.27\,\mbox{s}$.
(%i1) g(x):=x^5-10*x^4+35*x^3-50*x^2+24*x;
5 4 3 2
(%o1) g(x) := x - 10 x + 35 x + (- 50) x + 24 x
(%i2) f(x):=factor(g(x));
(%o2) f(x) := factor(g(x))
(%i3) f(x);
(%o3) (x - 4) (x - 3) (x - 2) (x - 1) x
(%i4) expand(f(x));
5 4 3 2
(%o4) x - 10 x + 35 x - 50 x + 24 x
(%i5) solve(f(x),x);
(%o5) [x = 0, x = 1, x = 2, x = 3, x = 4]
(%i6) g(x);
5 4 3 2
(%o6) x - 10 x + 35 x - 50 x + 24 x
(%i7) solve(g(x),x);
(%o7) [x = 3, x = 4, x = 1, x = 2, x = 0]
(%i8) solve([Th-m*g=0,m*g-Tf=m*a,Tf=M*a],[a,m,M]);
(Th - Tf) g Th Tf Th
(%o8) [[a = -----------, m = --, M = -----------]]
Th g (Th - Tf) g
where we can pick off the solutions,
\[ a=\frac{(T_H-T_F)\,g}{T_H} \qquad\qquad m=\frac{T_H}{g} \qquad\qquad M=\frac{T_F T_H}{(T_H-T_F)\,g} \]
The physics part of this example is deriving the formulas and identifying the
unknowns. The mathematical part is made easier using maxima so we can double check
our algebra with some confidence.
These are just indications of how it works. Maxima can be used to solve systems of equations with more unknowns and a whole lot more.
GNU Octave is free software and an effective clone of MathWorks matlab, or that is their goal. If you know matlab, and don't want to buy it, I would recommend you investigate octave.
This program octave has a rich, thirty-year history as they successfully implemented the matlab programming language, itself with a forty-year history. Matlab (the language) is quite popular in many fields.
It is better suited to matrix calculations but it also has a symbolic algebra plugin. Solving the previous problem (done in maxima above) but here in octave:
pkg load symbolic syms t solve(-10+t+t^2*3/2)\[ \left[ \begin{matrix} \displaystyle -\frac13 + \frac{\sqrt{61}}{3} \\ \displaystyle -\frac{\sqrt{61}}{3} - \frac13 \end{matrix}\right] \]
solve(-10+t+t^2*3/2,t>0)\[ t = -\frac13 + \frac{\sqrt{61}}{3} \]
You can see the first time I tried, I got the two answers, but in the second, I asked that $t>0$.
If you want a numerical answer to the quadratic, try the somewhat easier:
f(t) = -10 + t + t^2 *3/2; c = [3/2, 1, -10]; roots(c) -2.9367 2.2701from which we identify the second one as the appropriate answer. The f(t) just shows how to construct the array c (for example, 3/2 is the coefficient of t2), and the semicolon seems to suppress output to the screen.
octave:1> m1=[9.3, 7.4, 5.7, 8.2, 6.9, 7.1, 9.0] m1 = 9.3000 7.4000 5.7000 8.2000 6.9000 7.1000 9.0000 octave:2> m2=[2.1, 6.2, 3.6, 2.7, 4.1, 1.3, 2.6] m2 = 2.1000 6.2000 3.6000 2.7000 4.1000 1.3000 2.6000 octave:3> a=9.8*(m1.-m2)./(m1.+m2) a = 6.18947 0.86471 2.21290 4.94495 2.49455 6.76667 5.40690 octave:4> T=m1.*(9.8.-a) T = 33.578 66.121 43.246 39.811 50.408 21.537 39.538 octave:5> T=m2.*(9.8.+a) T = 33.578 66.121 43.246 39.811 50.408 21.537 39.538
This is just an indication of what is possible. GNU Octave can do very many things.
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jones.3 (R Jones)
Last Modified 23:23:51 14-Feb-2022