boolean return value in mathematica

Question

why does the following instruction ok in mathematica, and it returns a boolean value

In[398]:= catenary[x, 1, 0, 0] == catenary[-x, 1, 0, 0]
Out[398]:= True

and not this one?

In[406]:= catenary[-x, 1, 0, 0] == -catenary[x, 1, 0, 0]
Out[406]:= Cosh[x] == -Cosh[x]

I wanted to have a boolean value too.. to prove the symmetry.. Thanks for your help!

Wrap the entire thing in TrueQ[] if need be. Less facetiously: your last equation is true for x an odd multiple of $\dfrac{\pi i}{2}$ and false otherwise. — J. M.'s persistent exhaustion, Jun 19, 2012 at 15:52

yulinlinyu · Accepted Answer · 2012-06-20 15:36:18Z

1

I think you should use three "="s, i.e.,

Cosh[x]===Cosh[-x]

(* True *)

If you need the Boolean value, then

Cosh[x] === Cosh[-x] // Boole

(* 1 *)

answered Jun 19, 2012 at 16:28

yulinlinyu

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$\begingroup$ Cosh[x] == Cosh[-x] works fine; Mathematica is not too dumb to not know that the hyperbolic cosine is even. $\endgroup$
– J. M.'s persistent exhaustion ♦
Jun 19, 2012 at 16:30
$\begingroup$ @J.M. But Sinh[x]==Sinh[-x] return itself, which is not the purpose of the programmer, whereas, Sinh[x]===Sinh[-x] works fine. $\endgroup$
– yulinlinyu
Jun 21, 2012 at 1:28
$\begingroup$ Again: Sinh[] is not the function in the OP. $\endgroup$
– J. M.'s persistent exhaustion ♦
Jun 21, 2012 at 1:29
$\begingroup$ This is an example to illustrate the way to cope with the problem. I don't think the author need Cosh ONLY but no other functions. $\endgroup$
– yulinlinyu
Jun 21, 2012 at 1:31
1

$\begingroup$ You'd think that, until you see OP's other questions... $\endgroup$
– J. M.'s persistent exhaustion ♦
Jun 21, 2012 at 1:32

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