Re: TI92 bug, last question :-)

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Re: TI92 bug, last question :-)

God Dag

Yes, for that application it seems that would be good.
However, it's not that easy when computing with approximate
numbers. For example what should

                2^(1+i) be equal to?

Should the value depend upon degree or radian mode.
In some applications perhaps others no. In some applications
one part of the problem we would want it and another part of
the same problem we would not. How can that be programmed?
We must be careful.

Below I did some thinking about it this afternoon. See what you
think.


 :
 :By definition:
 :
 : e^(a+b* )= e^(a)*cos(b)+ e^(a)*sin(b)*
 :
 :In degrees we want to have:
 :
 : e^(0+30*i)=
 :          e^(0)*cos(30)+ e^(0)*sin(30)*i
 :          =sqrt(3)/2+1/2*i
 :
 :          =.86602540378445+.5*
 :
 :However on the TI-92 using DEGREE for
 :angles, REAL for complex format, APPROXIMATE gets:
 :
 :.15425144988752-.98803162409287*i
 :
 :Clearly something is not right.
 :
 :When we use: RADIAN for
 :angles, REAL for complex format, APPROXIMATE gets:
 :
 :.15425144988752-.98803162409287*i
 :
 :Which is correct.
 :
 :So it seems that when using exponents
 :that are complex one must use RADIAN
 :mode only.
 :
 :With TI-85'S and TI-86'S the same
 :thing happens when using DEGREE mode.
 :WQe must use RADIAM mode there too.
 :
 :A related item occurs, again in DEGREE with AUTO or EXACT modes,
  RECTANGULAR mode for Complex Format.
 :
 : e^(a+b* )=e^a*cos(180*b/(¼))
 :               + e^a*sin(180*b/(¼))*i
 :
 :This does not seem correct one would
 :think it should be:
 :
 : e^(a+b* )= e^a*cos(b)+ e^a*sin(b)*i
 :
 :
 :Does this effect other calculations?
 :
 :Is 2^(1+i) different for DEGREE and
 :RADIAN modes? Do we want them
 :different?
 :
 :1.5384778027279+1.2779225526273*i
 :using RADIAN, RECTANGULAR,APPR.
 :
 :1.5384778027279+1.2779225526273*i
 :using DEGREE, RECTANGULAR,APPR.
 :
 :What does c^(a+b* ) really mean?
 :
 :Let   e^(z)=c^(a+b*i)
 :
 :Solving for z in radians we get:
 :
 :z=2*K*¼*i+ln(c^(a+b*i)) (K an integer)
 :
 :c^(a+b*i)=e^(2*K*¼*i+ln(c^(a+b*i)))
 :             (K an integer)
 :
 :If there is to be a differences for
 :exponentials with complex numbers
 :between radian and degree modes the
 :log. function must be sensitive like
 :the trig. functions are to mode
 :changes, but it is not and many applications it
 :should not be.
 :
 :In conclusion when using exponentials
 :with complex numbers stay in radian
 :mode.
 :
 :

Good Luck

Gary Wardall
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