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Undefined arctanh evaluating when implementing hyperbolic CORDIC



The Next CEO of Stack OverflowTrigonometric hyperbolic function - sinhxWhen I choose arctanh or arccoth?Hyperbolic growth, deriving from hyperbolic functionsHyperbolic Intuition.Geometric interpretation of hyperbolic functions and the hyperbolic angle/argumentEvaluating Hyperbolic Cotangent (coth) IntegralHyperbolic equationWhen does cosine multiplied by hyperbolic cosine equals one?$mathrmarctanh (x+y) = ?$Hyperbolic Function Differentiation Question.










0












$begingroup$


I am trying to implement an exponential function using the CORDIC method. I am able to get the 'basic' version to work fine, but that only works for a very limited input range (i.e., inputs smaller than 1.11). I found this paper titled 'Expanding the range of convergence of the CORDIC algorithm', which seems to be widely used to expand this range. However, when I try to implement their approach, I am running into an issue. Specifically, with equations 34 and 35. In short, I need to calculate values from a value -M to +N. Based on equation 35, when I calculate the value for i=1, it evaluates arctanh(2^-0). Ofcourse arctanh(1) is undefined and the algorithm cannot proceed.



$i leq 0 : Z_i+1 = Z_i = delta_itanh^-1(1-2^i-2) ... eq(34)$



$i > 0 : Z_i+1 = Z_i = delta_itanh^-1(2^i) ... eq(35)$



It seems this work is widely used as I have seen several other papers implement this. So I feel as though I am doing something wrong. If someone can spot what I am doing wrong, I would greatly appreciate it!










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
    $endgroup$
    – Somos
    Mar 19 at 20:43











  • $begingroup$
    @Somos Thank you! I added the equations from the paper as you suggested.
    $endgroup$
    – user3397008
    Mar 19 at 22:07










  • $begingroup$
    From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
    $endgroup$
    – Somos
    Mar 19 at 23:02
















0












$begingroup$


I am trying to implement an exponential function using the CORDIC method. I am able to get the 'basic' version to work fine, but that only works for a very limited input range (i.e., inputs smaller than 1.11). I found this paper titled 'Expanding the range of convergence of the CORDIC algorithm', which seems to be widely used to expand this range. However, when I try to implement their approach, I am running into an issue. Specifically, with equations 34 and 35. In short, I need to calculate values from a value -M to +N. Based on equation 35, when I calculate the value for i=1, it evaluates arctanh(2^-0). Ofcourse arctanh(1) is undefined and the algorithm cannot proceed.



$i leq 0 : Z_i+1 = Z_i = delta_itanh^-1(1-2^i-2) ... eq(34)$



$i > 0 : Z_i+1 = Z_i = delta_itanh^-1(2^i) ... eq(35)$



It seems this work is widely used as I have seen several other papers implement this. So I feel as though I am doing something wrong. If someone can spot what I am doing wrong, I would greatly appreciate it!










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
    $endgroup$
    – Somos
    Mar 19 at 20:43











  • $begingroup$
    @Somos Thank you! I added the equations from the paper as you suggested.
    $endgroup$
    – user3397008
    Mar 19 at 22:07










  • $begingroup$
    From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
    $endgroup$
    – Somos
    Mar 19 at 23:02














0












0








0





$begingroup$


I am trying to implement an exponential function using the CORDIC method. I am able to get the 'basic' version to work fine, but that only works for a very limited input range (i.e., inputs smaller than 1.11). I found this paper titled 'Expanding the range of convergence of the CORDIC algorithm', which seems to be widely used to expand this range. However, when I try to implement their approach, I am running into an issue. Specifically, with equations 34 and 35. In short, I need to calculate values from a value -M to +N. Based on equation 35, when I calculate the value for i=1, it evaluates arctanh(2^-0). Ofcourse arctanh(1) is undefined and the algorithm cannot proceed.



$i leq 0 : Z_i+1 = Z_i = delta_itanh^-1(1-2^i-2) ... eq(34)$



$i > 0 : Z_i+1 = Z_i = delta_itanh^-1(2^i) ... eq(35)$



It seems this work is widely used as I have seen several other papers implement this. So I feel as though I am doing something wrong. If someone can spot what I am doing wrong, I would greatly appreciate it!










share|cite|improve this question











$endgroup$




I am trying to implement an exponential function using the CORDIC method. I am able to get the 'basic' version to work fine, but that only works for a very limited input range (i.e., inputs smaller than 1.11). I found this paper titled 'Expanding the range of convergence of the CORDIC algorithm', which seems to be widely used to expand this range. However, when I try to implement their approach, I am running into an issue. Specifically, with equations 34 and 35. In short, I need to calculate values from a value -M to +N. Based on equation 35, when I calculate the value for i=1, it evaluates arctanh(2^-0). Ofcourse arctanh(1) is undefined and the algorithm cannot proceed.



$i leq 0 : Z_i+1 = Z_i = delta_itanh^-1(1-2^i-2) ... eq(34)$



$i > 0 : Z_i+1 = Z_i = delta_itanh^-1(2^i) ... eq(35)$



It seems this work is widely used as I have seen several other papers implement this. So I feel as though I am doing something wrong. If someone can spot what I am doing wrong, I would greatly appreciate it!







hyperbolic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 19 at 22:07







user3397008

















asked Mar 19 at 20:20









user3397008user3397008

11




11







  • 1




    $begingroup$
    Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
    $endgroup$
    – Somos
    Mar 19 at 20:43











  • $begingroup$
    @Somos Thank you! I added the equations from the paper as you suggested.
    $endgroup$
    – user3397008
    Mar 19 at 22:07










  • $begingroup$
    From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
    $endgroup$
    – Somos
    Mar 19 at 23:02













  • 1




    $begingroup$
    Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
    $endgroup$
    – Somos
    Mar 19 at 20:43











  • $begingroup$
    @Somos Thank you! I added the equations from the paper as you suggested.
    $endgroup$
    – user3397008
    Mar 19 at 22:07










  • $begingroup$
    From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
    $endgroup$
    – Somos
    Mar 19 at 23:02








1




1




$begingroup$
Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
$endgroup$
– Somos
Mar 19 at 20:43





$begingroup$
Int order to make your question self-contained it would be nice to include equations 34 and 35 as well as a few other needed pieces of context.
$endgroup$
– Somos
Mar 19 at 20:43













$begingroup$
@Somos Thank you! I added the equations from the paper as you suggested.
$endgroup$
– user3397008
Mar 19 at 22:07




$begingroup$
@Somos Thank you! I added the equations from the paper as you suggested.
$endgroup$
– user3397008
Mar 19 at 22:07












$begingroup$
From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
$endgroup$
– Somos
Mar 19 at 23:02





$begingroup$
From you4 eq(35) $i>0$ and so I can't figure out how you got arctanh(2^-0) from that. You need to give us more context for this question.
$endgroup$
– Somos
Mar 19 at 23:02











0






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