brian abraham

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These are answers submitted by brian abraham

A further question if arctanh(tanh(A))-arctanh(tanh(B)) with |A|>1 and |B|>1 then I can see that at for e.g. I*Pi/2 I may find a solution for both  arctanh(tanh(A)) and arctanh(tanh(B)) and that this will be real valued as the imaginary parts cancel.  My question is I may have found solutions to arctanh(tanh(A)) and arctanh(tanh(B)) where the imaginary parts were not equal so do I say a real solution exists to arctanh(tanh(A))-arctanh(tanh(B)) without implying that non real solutions do not also exist.  Is this just part of dealing with multi-valued functions.

PS I have difficulties uploading today so apologies for double uploads.

Thank you every one.

It may be a limitation of my mathematical ability but I have to see the relationships which you (Robert) have made explicit.

So I plotted: complexplot3d(tanh(z),z=-3-I*0..3+I*Pi,axes=boxed,view=[default,default,0..2.0] bu the real part of tanh(z) is shown as abs(tanh(z)).

Is there a way to show the real part of tanh(z) as tanh(z) rather than abs(tanh(z)): apologies if this explained in the above and I have missed it.

Thank you every one for helping me to think this through.

(It is may be a limitation on my mathematical abilities but I have to explore the graphs of the functions before I begin to feel comfortable with the equations.)

So I plotted: complexplot3d(tanh(z),z=-3-I*0..3+I*Pi,axes=boxed,view=[default,default,0..2.0]);  I could then view the relationships you indicate (Robert).

My one query is that the real part of tanh(z) is shown as abs(tanh(z)): can I get maple to show the real part as tanh(z) rather than abs(tanh(z)): apologies if this has been covered in the above and I have missed it.

 

 

 

Thank you

 

As always thank you for taking the time over this. 
 
I had wrongly assumed that the appearance of imaginary parts of any size was a warning that a real system was not being represented.   Two consequential questions if I might take up some more of your time.
 
 
Firstly, would I be right in thinking that in general that if I set Digits to a higher value and the already small imaginary part gets smaller then that would always be a confirmation that the apperance of complex numbers is a rounding issue?
 
Secondly, if I using your definitions write
 
evalf(eval(J1 - eval(J1, x=0) - 1, x = v)) = evalf(Int(1/(sin(x+1/2)+1/10),x=0..v))
 
am I to understand that mathematically this is correct but that due to limitations in computer solving methods the left hand side produces the wrong answer? 
 
Once again many thanks.

 

I have got it, if I was evaluating by hand for my example I could:
1) Replace s0 with a number
2) Find the number(s) at which RootOf( )=0
3) Evaluate 2.*arctan(u)+ 6.283185307*n substituting u with the number(s) from 2)
 
(Yes, I automated the above in Maple and it is the same as that given directly by plot.)
 
Many thanks.

 

Thank you Scott and Robert, I now understand _Z1~.
 
The _Z in RootOf(_Z^2-2) I understand – I can plot it and see that as the independent variable changes the function evaluates to zero at two values - but in my expression it is still eluding me.  Any insights to help me see this are most appreciated.
 

 

Dear Robert
 
Thank you for your input. 

I now realise I do not understand what _Z and _Z1~ (the tilde appears when I run solve) represent in the above expression.

I have read through a number of posts on RootOf but have not found a similar form of RootOf.

Are _Z and _Z1~  integers which determine the domain over which s0 is taken?

As always, brilliant help and much appreciated.

I could not find a way to use `if` with map but I came up with this which works quickly

> angles1_s:=[angles1_s]:
angles1_dif:=
seq(
    map(x->x,angles1_s[j][1..-2])-map(x->x,angles1_s[j][2..-1])
   ,j=1..nops(angles1_s)):
 
for i from 1 to nops(angles1_s) do
member(sort(angles1_dif[i])[1],angles1_dif[i],'k'):
print([angles1_s[i][k],angles1_s[i][k+1]]);
end do:

One is always learning and not least I had never seen the index with a negative number so thanks again.

Oh yes, that is lightening fast.

What I am actually doing is 

 [seq(
     [
      seq( `if`(abs(angles1_s[j][i]-angles1_s[j][i+1])<.7
                 ,NULL
                 ,op([angles1_s[j][i],angles1_s[j][i+1]])
              )   
       ,i=1..(nops(angles1_s[j])-1)
        )
     ]
,j=1..10
)
];

so if you had time how would that be written in terms of your code;  if you don't have time then thanks anyway, that is a great help

I know I have read that sets do not maintain order where as lists do. 

What I was less sure of was if that held true when used within solve.

Thanks for your comment.

many thanks

I shall need to spend a little time investigating what is going on.

Brian

 

 

Thank you for both responses.

I could not get Joe's to run as I got the error.

Error, `GraphTheory` is not a module or member

Error, cannot split rhs for multiple assignment

I am using Maple 10.

 

I ran alec's and could see from the results that I have not defined my network correctly.  Never the less this is enough to get me started so thank you.

 

 

Thank you once again.

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