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ACOS logarithmic form
04-28-2016, 03:07 PM (This post was last modified: 04-28-2016 03:10 PM by Claudio L..)
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ACOS logarithmic form
Here's my dilemma:

I looked at Wikipedia here for the logarithmic forms of ACOS.
The second row is the one that interests me.

ACOS(Z)=-i*ln(Z+sqrt(Z^2-1))

Now let's make Z=2 (real, but outside range so result will be complex).

2^2-1 = 3
ln(2+sqrt(3))=1.3169... (all real numbers so far)

Now -i*1.3169... = (0 -1.3169)

Now go to your 50g and do 2 ACOS, you'll get (0 1.3169...) (positive imaginary part)

Wolfram Alpha agrees with the 50g, so that imaginary part has to be positive.

Then a quick check:
ACOS(Z)=pi/2-ASIN(Z)

Doing that on the 50g we also get positive imaginary part.

Is the formula wrong in wikipedia? Perhaps it should be i*ln(...) instead of -i ?

I just need some help proving it, so it's not just me against the world.

EDIT: BTW, the ASIN() formula in Wikipedia works in agreement with the 50g.
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Messages In This Thread
ACOS logarithmic form - Claudio L. - 04-28-2016 03:07 PM
RE: ACOS logarithmic form - Claudio L. - 04-28-2016, 03:29 PM
RE: ACOS logarithmic form - Claudio L. - 04-28-2016, 05:03 PM
RE: ACOS logarithmic form - Dieter - 04-28-2016, 06:44 PM
RE: ACOS logarithmic form - Claudio L. - 04-28-2016, 08:23 PM
RE: ACOS logarithmic form - Csaba Tizedes - 04-29-2016, 01:33 PM
RE: ACOS logarithmic form - Ángel Martin - 04-28-2016, 06:42 PM
RE: ACOS logarithmic form - Claudio L. - 04-28-2016, 08:29 PM
RE: ACOS logarithmic form - Ángel Martin - 04-29-2016, 06:06 AM
RE: ACOS logarithmic form - Claudio L. - 04-29-2016, 02:39 PM
RE: ACOS logarithmic form - Claudio L. - 04-29-2016, 02:19 AM
RE: ACOS logarithmic form - Sylvain Cote - 04-29-2016, 02:50 AM
RE: ACOS logarithmic form - Ángel Martin - 04-29-2016, 06:00 AM
RE: ACOS logarithmic form - ljubo - 04-29-2016, 09:15 PM
RE: ACOS logarithmic form - Ángel Martin - 04-30-2016, 07:01 AM
RE: ACOS logarithmic form - ljubo - 04-30-2016, 08:57 AM
RE: ACOS logarithmic form - Claudio L. - 05-01-2016, 03:58 AM



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