How to solve this functional equation involving hyperbolic functions?












2












$begingroup$


I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)



$$f(K_1,0)=-frac{1}{2}ln{2sqrt{cosh(2K_1)}}+frac{1}{2}f(lnsqrt{cosh(2K_1)},0).qquad(1)$$



They give the following solution (book eq. 14.2.15)



$$f(K_1,0)=-ln(2cosh K_1).qquad(2)$$



I tried to check the solution was correct simply plugging (2) into (1) but end up with a $ln(cosh(ln(cosh(K_1)))$ thing.



Questions:




  1. How do I obtain the solution (2) from (1)?

  2. Is there an easy way to check the solution is correct?






mathematical-physics functional-equations hyperbolic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
    $endgroup$
    – user458276
    Dec 21 '18 at 21:01
















2












$begingroup$


I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)



$$f(K_1,0)=-frac{1}{2}ln{2sqrt{cosh(2K_1)}}+frac{1}{2}f(lnsqrt{cosh(2K_1)},0).qquad(1)$$



They give the following solution (book eq. 14.2.15)



$$f(K_1,0)=-ln(2cosh K_1).qquad(2)$$



I tried to check the solution was correct simply plugging (2) into (1) but end up with a $ln(cosh(ln(cosh(K_1)))$ thing.



Questions:




  1. How do I obtain the solution (2) from (1)?

  2. Is there an easy way to check the solution is correct?






mathematical-physics functional-equations hyperbolic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
    $endgroup$
    – user458276
    Dec 21 '18 at 21:01














2












2








2





$begingroup$


I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)



$$f(K_1,0)=-frac{1}{2}ln{2sqrt{cosh(2K_1)}}+frac{1}{2}f(lnsqrt{cosh(2K_1)},0).qquad(1)$$



They give the following solution (book eq. 14.2.15)



$$f(K_1,0)=-ln(2cosh K_1).qquad(2)$$



I tried to check the solution was correct simply plugging (2) into (1) but end up with a $ln(cosh(ln(cosh(K_1)))$ thing.



Questions:




  1. How do I obtain the solution (2) from (1)?

  2. Is there an easy way to check the solution is correct?






mathematical-physics functional-equations hyperbolic-functions







share|cite|improve this question











$endgroup$




I'm reading this (physics) book. They have the recurrence relation (book eq. 14.2.14)



$$f(K_1,0)=-frac{1}{2}ln{2sqrt{cosh(2K_1)}}+frac{1}{2}f(lnsqrt{cosh(2K_1)},0).qquad(1)$$



They give the following solution (book eq. 14.2.15)



$$f(K_1,0)=-ln(2cosh K_1).qquad(2)$$



I tried to check the solution was correct simply plugging (2) into (1) but end up with a $ln(cosh(ln(cosh(K_1)))$ thing.



Questions:




  1. How do I obtain the solution (2) from (1)?

  2. Is there an easy way to check the solution is correct?






mathematical-physics functional-equations hyperbolic-functions




mathematical-physics functional-equations hyperbolic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 20:18







Saavestro

















asked Dec 21 '18 at 18:51









SaavestroSaavestro

1345




1345












  • $begingroup$
    $cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
    $endgroup$
    – user458276
    Dec 21 '18 at 21:01


















  • $begingroup$
    $cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
    $endgroup$
    – user458276
    Dec 21 '18 at 21:01
















$begingroup$
$cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
$endgroup$
– user458276
Dec 21 '18 at 21:01




$begingroup$
$cosh(ln(x)) = dfrac{x^{2}+1}{2x}$
$endgroup$
– user458276
Dec 21 '18 at 21:01










1 Answer
1






active

oldest

votes


















3












$begingroup$

Note that
begin{align}
cosh x = frac{e^x+e^{-x}}{2}
end{align}
then you see that
begin{align}
2 cosh( ln( sqrt{cosh(2K_1)}) =& 2frac{expleft(ln sqrt{cosh(2K_1)}right)+expleft(-ln sqrt{cosh(2K_1)} right)}{2}\
=& 2frac{sqrt{cosh 2K_1}+ (sqrt{cosh 2K_1})^{-1}}{2} = frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}}.
end{align}
Then it follows
begin{align}
-frac{1}{2}lnleft(2sqrt{cosh 2K_1} right)-frac{1}{2}lnleft(frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}} right)=& -frac{1}{2}lnleft( 2cosh(2K_1)+2right)\
=& -frac{1}{2}ln(4cosh^2 K_1).
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice check, thank you. Any idea on how to derive the solution from equation (1)?
    $endgroup$
    – Saavestro
    Dec 21 '18 at 20:14











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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3












$begingroup$

Note that
begin{align}
cosh x = frac{e^x+e^{-x}}{2}
end{align}
then you see that
begin{align}
2 cosh( ln( sqrt{cosh(2K_1)}) =& 2frac{expleft(ln sqrt{cosh(2K_1)}right)+expleft(-ln sqrt{cosh(2K_1)} right)}{2}\
=& 2frac{sqrt{cosh 2K_1}+ (sqrt{cosh 2K_1})^{-1}}{2} = frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}}.
end{align}
Then it follows
begin{align}
-frac{1}{2}lnleft(2sqrt{cosh 2K_1} right)-frac{1}{2}lnleft(frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}} right)=& -frac{1}{2}lnleft( 2cosh(2K_1)+2right)\
=& -frac{1}{2}ln(4cosh^2 K_1).
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice check, thank you. Any idea on how to derive the solution from equation (1)?
    $endgroup$
    – Saavestro
    Dec 21 '18 at 20:14
















3












$begingroup$

Note that
begin{align}
cosh x = frac{e^x+e^{-x}}{2}
end{align}
then you see that
begin{align}
2 cosh( ln( sqrt{cosh(2K_1)}) =& 2frac{expleft(ln sqrt{cosh(2K_1)}right)+expleft(-ln sqrt{cosh(2K_1)} right)}{2}\
=& 2frac{sqrt{cosh 2K_1}+ (sqrt{cosh 2K_1})^{-1}}{2} = frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}}.
end{align}
Then it follows
begin{align}
-frac{1}{2}lnleft(2sqrt{cosh 2K_1} right)-frac{1}{2}lnleft(frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}} right)=& -frac{1}{2}lnleft( 2cosh(2K_1)+2right)\
=& -frac{1}{2}ln(4cosh^2 K_1).
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice check, thank you. Any idea on how to derive the solution from equation (1)?
    $endgroup$
    – Saavestro
    Dec 21 '18 at 20:14














3












3








3





$begingroup$

Note that
begin{align}
cosh x = frac{e^x+e^{-x}}{2}
end{align}
then you see that
begin{align}
2 cosh( ln( sqrt{cosh(2K_1)}) =& 2frac{expleft(ln sqrt{cosh(2K_1)}right)+expleft(-ln sqrt{cosh(2K_1)} right)}{2}\
=& 2frac{sqrt{cosh 2K_1}+ (sqrt{cosh 2K_1})^{-1}}{2} = frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}}.
end{align}
Then it follows
begin{align}
-frac{1}{2}lnleft(2sqrt{cosh 2K_1} right)-frac{1}{2}lnleft(frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}} right)=& -frac{1}{2}lnleft( 2cosh(2K_1)+2right)\
=& -frac{1}{2}ln(4cosh^2 K_1).
end{align}






share|cite|improve this answer









$endgroup$



Note that
begin{align}
cosh x = frac{e^x+e^{-x}}{2}
end{align}
then you see that
begin{align}
2 cosh( ln( sqrt{cosh(2K_1)}) =& 2frac{expleft(ln sqrt{cosh(2K_1)}right)+expleft(-ln sqrt{cosh(2K_1)} right)}{2}\
=& 2frac{sqrt{cosh 2K_1}+ (sqrt{cosh 2K_1})^{-1}}{2} = frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}}.
end{align}
Then it follows
begin{align}
-frac{1}{2}lnleft(2sqrt{cosh 2K_1} right)-frac{1}{2}lnleft(frac{cosh 2K_1 +1}{sqrt{cosh 2K_1}} right)=& -frac{1}{2}lnleft( 2cosh(2K_1)+2right)\
=& -frac{1}{2}ln(4cosh^2 K_1).
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 19:54









Jacky ChongJacky Chong

19.3k21129




19.3k21129












  • $begingroup$
    Nice check, thank you. Any idea on how to derive the solution from equation (1)?
    $endgroup$
    – Saavestro
    Dec 21 '18 at 20:14


















  • $begingroup$
    Nice check, thank you. Any idea on how to derive the solution from equation (1)?
    $endgroup$
    – Saavestro
    Dec 21 '18 at 20:14
















$begingroup$
Nice check, thank you. Any idea on how to derive the solution from equation (1)?
$endgroup$
– Saavestro
Dec 21 '18 at 20:14




$begingroup$
Nice check, thank you. Any idea on how to derive the solution from equation (1)?
$endgroup$
– Saavestro
Dec 21 '18 at 20:14


















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