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19-Tanzanite

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

Hello, Everyone.

How are they different ?

Regards.

1 ACCEPTED SOLUTION

Accepted Solutions
23-Emerald II
(To:lvl107)

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

3 REPLIES 3
21-Topaz I
(To:lvl107)

Hello! you could write it so:

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

23-Emerald II
(To:lvl107)

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

19-Tanzanite
(To:LucMeekes)

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

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

because :

furthermore :