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1 3rd January 15:13
nma124
External User
 
Posts: 1
Default How to convert to hyperpolic function?


hello; this is mma 5.0

I am having hard time figuring how to tell mma to convert the
output to a form I want.

In Maple, I can use the 'convert' command, which is really nice, but
mma has no such command to convert an expression between different
forms. So other than using Simplify, FullSimplify, ExpToTrig, TrigToExp,
I have no idea what to do.

This is an example, the result of this I want to be expressed as
arcsinh(y/c), not in the way it is generated, (which is correct
but different form).

So I guess I am asking one general question: What is the mma command
that is eqivelant to maple 'convert'? and a specific question is how
to make the output of this example expressed in arcsinh instead of in
terms of ln?

--- mma code -----
In[4]:= sol = Integrate[1/Sqrt[c^2+y^2],y]

2 2
Out[4]= Log[y + Sqrt[c + y ]]

In[5]:= Simplify[sol,Elements[c,Reals]]

2 2
Out[5]= Log[y + Sqrt[c + y ]]

---- maple code -----

In maple, in this example I did not have to use the convert
command, but in other examples I used it.

assume(c,real);

/ 1 \1/2
sol := arcsinh(|----| y)
| 2 |
\ c /

Now I can if needed convert the above to ln using convert:
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2 3rd January 15:13
dan
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Posts: 1
Default How to convert to hyperpolic function?


In Mma, if I make the assumption that c>0, I can not show that both are
equal.
Did I type the second equation correctly?

$Assumptions = c > 0;

v1 = FullSimplify[Integrate[1/Sqrt[c^2 + y^2], y]]
Log[y + Sqrt[c^2 + y^2]]

v2 = FullSimplify[TrigToExp[ArcSinh[y*Sqrt[1/c^2]]]]
Log[(y + Sqrt[c^2 + y^2])/c]

--
Dana DeLouis
Using Windows XP & Office XP
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3 3rd January 15:13
ken pledger
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Posts: 1
Default How to convert to hyperpolic function?


I can't help you with mma, but your question may be essentially
mathematical: how can you express a log in terms of an inverse sinh?

If that's it, then in the definition

sinh(u) = (e^u - e^-u)/2

write u = ln(t) so e^u = t.

Then sinh(ln(t)) = (t - (1/t))/2

so ln(t) = arcsinh((t - (1/t))/2).

That should enable you to make the conversions you want, just by
plugging in different values of t.

Ken Pledger.
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