MaplePrimes - answers and comments on Question, verifying solutions of tan(x)=x

verify or testfloat or fnormal

pagan — Sat, 30 Jan 2010 18:44:31 Z

You could use testfloat (or verify with float option), or apply fnormal first.

I prefer verify,float usually, for floating-point scalar number comparison.

thanks for the reply. i

latentcorpse — Sat, 30 Jan 2010 19:34:27 Z

thanks for the reply. i tried changing

evalb(simplify(tan(this_soln)-this_soln)=0);

verify(tan(this_soln),this_soln,float);

and it came up with an error

i also tried changing it to

evalb(fnormal(simplify(tan(this_soln)-this_soln))=0);

and this gave the first couple as true but the rest remained false.

thanks. that works for me as

latentcorpse — Sat, 30 Jan 2010 23:49:57 Z

thanks. that works for me as well.

i am confused as to what a ulps is. i looked it up, it units in last place. so this means that if the tan(this_soln) and this_soln are within a given tolerance (in this case 100 ulps) verify will come back true.

but what is the value of ulps? what is the last place?

take for example the 1st soln 4.493409458.

the last place is 8x0.000000001

so would 100 ulps = 100 x 0.000000001 = 0.0000001

and then if the two numbers are within 0.0000001 of each other verify will come back true?

what would be the reason for

latentcorpse — Sun, 31 Jan 2010 01:35:04 Z

what would be the reason for Maple giving us very small answers for tan(x)-x and not exactly zero? is it just because tan has to round the answer off?

i'm a bit confused. do you

latentcorpse — Sun, 31 Jan 2010 03:06:13 Z

i'm a bit confused.

do you mean that because we are working in floating point, maple rounds these answers off?

played with it

Axel Vogt — Mon, 01 Feb 2010 02:40:50 Z

I played with it (no, not with the mathematical part, just with the
Numerics, even if the Math would be more interesting - but too hard
for me (have not found Schanuel's conjecture for trigonometrics, I
guess that might be a way).

Since sin(x)/cos(x) = tan(x) is better to solve sin(x) = cos(x)*x and
using asympotics a choice is x = n*Pi + 1/2*Pi - epsilon, n an integer.

  sin(x) - cos(x)*x:
  'eval(%, x=n*Pi + Pi/2 - epsilon)';
  expand(%): 
  simplify(%) assuming n::posint:
  collect(%, sin):
  EQ:=simplify(%, size);

    EQ := (-1)^(1+n)*(((n+1/2)*Pi-epsilon)*sin(epsilon)-cos(epsilon))

The asymptotic series for solve(EQ, epsilon) is 1/(Pi*n) +- ... and
this way even for high values one finds a solution:

  nTst:=1+'`^`(10,10)'; # even or odd does not matter

  eval(EQ, n=nTst);
  epsilon0:=fsolve(%, epsilon=1/Pi/nTst, fulldigits);
  #epsilon0:= convert(%, rational);

Check it (I work with Digits = 14):

  eval(sin(x) - cos(x)*x, x=nTst*Pi + Pi/2 - epsilon0); evalf(%);

                                  0.


Doing the same with 'RootOf' instead of 'fsolve' gives a formal solution,
which (of course) Maple agrees to be correct:

  eval(EQ, n=nTst);
  epsilon0:=RootOf(%, epsilon);
  eval(sin(x) - cos(x)*x, x=nTst*Pi + Pi/2 - epsilon0); simplify(%);

                                  0

and if one wants to know it, then

  'evalf[100](epsilon0)': '%'=%;
                                             -18
     evalf[100](epsilon0) = 0.3183 ... 983 10


To compute or verify it directly for tan(x)=x is not my thing ...