MaplePrimes - answers and comments on Question, Finding the roots isn't so easy

avoid

pagan — Sat, 04 Jul 2009 00:28:03 Z

> r1 := fsolve(x^2/20-10*x-15*cos(x+15),x);
                               r1 := 1.274092075

> fsolve(x^2/20-10*x-15*cos(x+15),x=0..infinity,avoid={x=r1});
                                  200.1193789

Also, RootFinding:-Analytic might be a possible approach

gulliet — Sat, 04 Jul 2009 00:37:28 Z


> restart;
> RootFinding:-Analytic((1/20)*x^2-10*x-15*cos(x+15), x, re = -1000 .. 1000, im = -1 .. 1);
         1.27409207502708, -1.80895918513793 - 0.959196521549030 I, 

           -1.80895918513793 + 0.959196521549025 I, 200.119378915794

HTH, -- Jean-Marc

Student[Calculus1]

gkokovidis — Sat, 04 Jul 2009 00:55:42 Z

Use the numeric option when calling Roots, from the Student[Calculus1] package, over the range of interest.

>restart:
>with(Student[Calculus1]):
>Roots(x^2/20-10*x-15*cos(x+15),x=0..210,numeric);

[1.274092075, 200.1193789]

Regards,
Georgios Kokovidis

Dräger Medical

Student[Calculus1]

Christopher2222 — Sat, 04 Jul 2009 03:38:47 Z

Thanks for all the options. It seems like using Student[Calculus1] Roots seems to be the best option.

Using fsolve with avoid will still miss some roots if there are more than one root within the range.

In my opinion I think fsolve should return all roots within a specified range. Even if I use solve, I get a RootOf answer that I can't seem to coax out the real numbers for the roots.

solve(x^2/20-10*x-15*cox(15+x));

-15 + RootOf( -300 cos(_Z) + 3225 - 230 _Z + Z^2)

Intersection of y=x^2-10-10*sin(x) and y=2

Christopher2222 — Sat, 04 Jul 2009 03:56:05 Z

This is sort of along the same line of problem.

a:= y=x^2-10-10*sin(x);

b:=y=2;

solve({a,b},{x,y});

Again only returns one solution, x as a RootOf and y=2 answer. Using allvalues(%) returns the same answer. Can't get the other 3 points of intersection with a single command.

and if b:=cos(x)

gkokovidis — Sun, 05 Jul 2009 07:13:07 Z

Using Student[Calculus1]:-Roots(expr, x = -10 .. 10, numeric); where expr is your new expression with b:=cos(x), you will get a diiferent set of answers. So, to answer your question, yes it does. Did you try it, and did it not work for you? Below are the values that I got.

[-4.367042332, -3.390886872, -1.144307303, 3.088083981]

Regards,
Georgios Kokovidis

Dräger Medical