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Research Article

An Approach to Conformal Transformation Using Symbolic Language Facilities: Application in Electrical Engineering

Algorithm 6

Program for calculating the fluxes in terms of the geometrical dimensions of the considered domain in a special case avoiding the occurring indeterminate form.
#   Below, a special case will be examined and an indeterminate form will be removed.
#   The symbols which are supplementary to those of Algorithm 1 are given below and represent expressions occurring in the calculation and have been put in a form for facilitating their usage.
#   They are 𝑦 1 , 𝑦 2 , 𝑦 𝑎 , 𝑦 𝐿 . Another symbol to be added is 𝑘 and serves to establish the sign of the flux in terms of the used abscissa.
𝑏 0 ∶ = 1 . 5 ;
𝑐 ∶ = − 0 .;
𝑥 𝑤 ∶ = − 1 ;
𝑘 ∶ = s i g n u m ( c + 1 ) ;
𝑦 1 ∶ = ( 1 / 2 . ) l n ( 1 . / ( 𝑐 2 − 𝑐 𝑏 2 0 − 2 . 𝑐 + 1 . − 0 . 5 ( − 2 . + 𝑐 𝑏 2 0 + 2 . 𝑐 ) ) ) ;
𝑦 2 ∶ = − ( 1 / 2 . ) l n ( 𝑐 2 / ( 𝑐 2 − 𝑐 𝑏 2 0 − 2 . 𝑐 + 1 . + 0 . 5 ( − 2 . + 𝑐 𝑏 2 0 + 2 . 𝑐 ) ) ) ;
# 𝑦 𝑎 ∶ = a r c t a n h ( ( 1 . / 2 ) ( − 2 . + 𝑐 𝑏 2 0 + 2 . 𝑐 ) / 𝑐 2 − 𝑐 𝑏 2 0 − 2 . 𝑐 + 1 . ) ;
#   The command concerning the variable 𝑦 𝑎 may be enabled only if c is different from zero.
𝑦 𝐿 ∶ = − ( 1 / 2 . ) l n ( − 8 . / ( 𝑏 2 0 + 4 . ) 𝑏 2 0 ) ;
#   For 𝑐 = 0 , in expression below, 𝑦 2 should be replaced by its limit 𝑦 𝐿 ;
𝜎 ∶ = − 𝑘 ( 0 . 3 1 8 3 0 9 8 8 6 1 ( ( 𝑏 2 0 / − 𝑏 2 0 ) l n ( ( − 𝑏 2 0 − 𝑐 𝑏 2 0 + 2 . − 𝑏 2 0 𝑐 2 − 𝑐 𝑏 2 0 − 2 . 𝑐 + 1 . ) / ( 𝑐 − 1 . ) )
  + l n ( − ( 1 . / 2 ) 𝑏 2 0 + 𝑐 − 1 . + 𝑐 2 − 𝑐 𝑏 2 0 − 2 . 𝑐 + 1 . ) − ( 𝑦 1 + 𝑦 2 )
  − ( 𝑏 2 0 / − 𝑏 2 0 ) l n ( ( − 𝑏 2 0 − 1 . 𝑥 𝑤 𝑏 2 0 + 2 . − 𝑏 2 0 𝑥 𝑤 2 − 𝑥 𝑤 𝑏 2 0 − 2 . 𝑥 𝑤 + 1 . ) / ( 𝑥 𝑤 − 1 . ) )
  − l n ( ( − 1 . / 2 ) 𝑏 2 0 + 𝑥 𝑤 − 1 . + 𝑥 𝑤 2 − 𝑥 𝑤 𝑏 2 0 − 2 . 𝑥 𝑤 + 1 . )
+ a r c t a n h ( 0 . 5 0 0 0 0 0 0 0 0 0 ( − 2 . + 𝑥 𝑤 𝑏 2 0 + 2 . 𝑥 𝑤 ) / 𝑥 𝑤 2 − 𝑥 𝑤 𝑏 2 0 − 2 . 𝑥 𝑤 + 1 . ) − l n ( 𝑥 𝑤 ) ) ) ;
𝜎 ∶ = s i m p l i f y ( 𝜎 ) ∶
#   The total value will be 𝜎 𝑇 ∶ = 2 𝜎 .
#   Results after running the program:
                   𝑏 0 ∶ = 1 . 5
                    𝑐 ∶ = − 0 .
                   𝑥 𝑤 ∶ = − 1
                    𝑘 ∶ = 1
             𝑦 2 ∶ = F l o a t ( u n d e fi n e d ) + F l o a t ( u n d e fi n e d ) 𝐼
             𝑦 𝐿 ∶ = 0 . 2 8 2 0 3 5 0 6 9 2 − 1 . 5 7 0 7 9 6 3 2 7 𝐼
              𝜎 ∶ = − 0 . 1 6 5 1 9 1 5 5 0 4 − 1 . 1 0 − 1 0 𝐼             ( 1)