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(e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

SOLVED
lvl107
Peridot

(e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

  Hello, Everyone.

Different.PNG

         How are they different ?

   Thanks in advance for your time and help.

        Regards.

1 ACCEPTED SOLUTION

Accepted Solutions

Re: (e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

I think it's a quirk of the Mupad symbolic engine.

In Mathcad 11 the (Maple) symbolic engine produces the following quirky output:

It appears that e or the exp() function has a special meaning:

The first line gave the error "No symbolic result was found". which is understandable because e^x is always immediatley translated to exp(x). But the second line then surprises with the same error message; unexpected because exp(i)^i indeed IS in there. At least (third line) with any other variable we get expected results, giving clues for the way to deal with e as shown in the fourth line.

Luc

View solution in original post

3 REPLIES 3

Re: (e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

Hello! you could write it so:

aii.jpg

It performs the exponential only if a = e. or to  powers of e, for example a=3:

aii.jpg

Re: (e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

I think it's a quirk of the Mupad symbolic engine.

In Mathcad 11 the (Maple) symbolic engine produces the following quirky output:

It appears that e or the exp() function has a special meaning:

The first line gave the error "No symbolic result was found". which is understandable because e^x is always immediatley translated to exp(x). But the second line then surprises with the same error message; unexpected because exp(i)^i indeed IS in there. At least (third line) with any other variable we get expected results, giving clues for the way to deal with e as shown in the fourth line.

Luc

View solution in original post

Re: (e^1i)^1i simplify ==> e^(-1) and (a^1i)^1i simplify ==> (a^1i)^1i . How are they different ?

  Thanks for your response, P.M. and Luc.

(1).PNG

  It seems my Mathcad 14 don't work (accept) with:

(1).PNG

because :

(4).PNG

furthermore :

(3).PNG

graph.PNG

and I wish learning,wish having your help in Mathcad 14.

       Best Regards

          Loi.

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