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

Forum|Forum|10 years ago
December 14, 2015
3 replies
1656 views

Hello, Everyone.

How are they different ?

Thanks in advance for your time and help.

Regards.

1_Difference-xmcd.zip

Other

Best answer by LucMeekes

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

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-MFra-

21-Topaz II

Hello! you could write it so:

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

L

LucMeekes

Answer

23-Emerald IV

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