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!

  • 1
    $\begingroup$ 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. $\endgroup$ – J. M.'s discontentment Jun 19 '12 at 15:52

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


(* True *)

If you need the Boolean value, then

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

(* 1 *)

| improve this answer | |
  • 1
    $\begingroup$ Cosh[x] == Cosh[-x] works fine; Mathematica is not too dumb to not know that the hyperbolic cosine is even. $\endgroup$ – J. M.'s discontentment Jun 19 '12 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 '12 at 1:28
  • $\begingroup$ Again: Sinh[] is not the function in the OP. $\endgroup$ – J. M.'s discontentment Jun 21 '12 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 '12 at 1:31
  • 1
    $\begingroup$ You'd think that, until you see OP's other questions... $\endgroup$ – J. M.'s discontentment Jun 21 '12 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.