Functions whose input is the same as the output?












1












$begingroup$


Given the Dedekind eta function $eta(tau)$ and complex number $tau$. I came across these family of functions,



$${f_2(tau)= frac{i}{sqrt{2}}frac{,_2F_1left(tfrac14,tfrac34,1,,1-alpha_2right)}{,_2F_1left(tfrac14,tfrac34,1,,alpha_2right)}=tau}$$





$${f_3(tau)= frac{i}{sqrt{3}}frac{,_2F_1left(tfrac13,tfrac23,1,,1-alpha_3right)}{,_2F_1left(tfrac13,tfrac23,1,,alpha_3right)}=tau}$$





$${f_4(tau)= frac{i}{sqrt{4}}frac{,_2F_1left(tfrac12,tfrac12,1,,1-alpha_4right)}{,_2F_1left(tfrac12,tfrac12,1,,alpha_4right)}=tau}$$





where,



$$alpha_2 =frac{64}{64+Big(frac{eta(tau)}{eta(2tau)}Big)^{24}},quad
alpha_3 =frac{27}{27+Big(frac{eta(tau)}{eta(3tau)}Big)^{12}},quad
alpha_4 =frac{16}{16+Big(frac{eta(tau)}{eta(4tau)}Big)^{8}},$$



So the input variable is $tau$ and the output is also $tau$. Presumably these are identity functions $f(x)=x$?



Q: What are other not-so-trivial examples of identity functions?



P.S. There is a $f_1(tau)$ using $,_2F_1left(tfrac16,tfrac56,1,,alpha_1right)$ but it uses the j-function, instead of the Dedekind eta function.










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$endgroup$












  • $begingroup$
    Essentially the same eta quotients are used in this post.
    $endgroup$
    – Tito Piezas III
    Jan 4 at 12:10
















1












$begingroup$


Given the Dedekind eta function $eta(tau)$ and complex number $tau$. I came across these family of functions,



$${f_2(tau)= frac{i}{sqrt{2}}frac{,_2F_1left(tfrac14,tfrac34,1,,1-alpha_2right)}{,_2F_1left(tfrac14,tfrac34,1,,alpha_2right)}=tau}$$





$${f_3(tau)= frac{i}{sqrt{3}}frac{,_2F_1left(tfrac13,tfrac23,1,,1-alpha_3right)}{,_2F_1left(tfrac13,tfrac23,1,,alpha_3right)}=tau}$$





$${f_4(tau)= frac{i}{sqrt{4}}frac{,_2F_1left(tfrac12,tfrac12,1,,1-alpha_4right)}{,_2F_1left(tfrac12,tfrac12,1,,alpha_4right)}=tau}$$





where,



$$alpha_2 =frac{64}{64+Big(frac{eta(tau)}{eta(2tau)}Big)^{24}},quad
alpha_3 =frac{27}{27+Big(frac{eta(tau)}{eta(3tau)}Big)^{12}},quad
alpha_4 =frac{16}{16+Big(frac{eta(tau)}{eta(4tau)}Big)^{8}},$$



So the input variable is $tau$ and the output is also $tau$. Presumably these are identity functions $f(x)=x$?



Q: What are other not-so-trivial examples of identity functions?



P.S. There is a $f_1(tau)$ using $,_2F_1left(tfrac16,tfrac56,1,,alpha_1right)$ but it uses the j-function, instead of the Dedekind eta function.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Essentially the same eta quotients are used in this post.
    $endgroup$
    – Tito Piezas III
    Jan 4 at 12:10














1












1








1


1



$begingroup$


Given the Dedekind eta function $eta(tau)$ and complex number $tau$. I came across these family of functions,



$${f_2(tau)= frac{i}{sqrt{2}}frac{,_2F_1left(tfrac14,tfrac34,1,,1-alpha_2right)}{,_2F_1left(tfrac14,tfrac34,1,,alpha_2right)}=tau}$$





$${f_3(tau)= frac{i}{sqrt{3}}frac{,_2F_1left(tfrac13,tfrac23,1,,1-alpha_3right)}{,_2F_1left(tfrac13,tfrac23,1,,alpha_3right)}=tau}$$





$${f_4(tau)= frac{i}{sqrt{4}}frac{,_2F_1left(tfrac12,tfrac12,1,,1-alpha_4right)}{,_2F_1left(tfrac12,tfrac12,1,,alpha_4right)}=tau}$$





where,



$$alpha_2 =frac{64}{64+Big(frac{eta(tau)}{eta(2tau)}Big)^{24}},quad
alpha_3 =frac{27}{27+Big(frac{eta(tau)}{eta(3tau)}Big)^{12}},quad
alpha_4 =frac{16}{16+Big(frac{eta(tau)}{eta(4tau)}Big)^{8}},$$



So the input variable is $tau$ and the output is also $tau$. Presumably these are identity functions $f(x)=x$?



Q: What are other not-so-trivial examples of identity functions?



P.S. There is a $f_1(tau)$ using $,_2F_1left(tfrac16,tfrac56,1,,alpha_1right)$ but it uses the j-function, instead of the Dedekind eta function.










share|cite|improve this question











$endgroup$




Given the Dedekind eta function $eta(tau)$ and complex number $tau$. I came across these family of functions,



$${f_2(tau)= frac{i}{sqrt{2}}frac{,_2F_1left(tfrac14,tfrac34,1,,1-alpha_2right)}{,_2F_1left(tfrac14,tfrac34,1,,alpha_2right)}=tau}$$





$${f_3(tau)= frac{i}{sqrt{3}}frac{,_2F_1left(tfrac13,tfrac23,1,,1-alpha_3right)}{,_2F_1left(tfrac13,tfrac23,1,,alpha_3right)}=tau}$$





$${f_4(tau)= frac{i}{sqrt{4}}frac{,_2F_1left(tfrac12,tfrac12,1,,1-alpha_4right)}{,_2F_1left(tfrac12,tfrac12,1,,alpha_4right)}=tau}$$





where,



$$alpha_2 =frac{64}{64+Big(frac{eta(tau)}{eta(2tau)}Big)^{24}},quad
alpha_3 =frac{27}{27+Big(frac{eta(tau)}{eta(3tau)}Big)^{12}},quad
alpha_4 =frac{16}{16+Big(frac{eta(tau)}{eta(4tau)}Big)^{8}},$$



So the input variable is $tau$ and the output is also $tau$. Presumably these are identity functions $f(x)=x$?



Q: What are other not-so-trivial examples of identity functions?



P.S. There is a $f_1(tau)$ using $,_2F_1left(tfrac16,tfrac56,1,,alpha_1right)$ but it uses the j-function, instead of the Dedekind eta function.







complex-analysis functions terminology special-functions hypergeometric-function






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share|cite|improve this question













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edited Jan 4 at 12:05







Tito Piezas III

















asked Jan 3 at 9:55









Tito Piezas IIITito Piezas III

26.9k365169




26.9k365169












  • $begingroup$
    Essentially the same eta quotients are used in this post.
    $endgroup$
    – Tito Piezas III
    Jan 4 at 12:10


















  • $begingroup$
    Essentially the same eta quotients are used in this post.
    $endgroup$
    – Tito Piezas III
    Jan 4 at 12:10
















$begingroup$
Essentially the same eta quotients are used in this post.
$endgroup$
– Tito Piezas III
Jan 4 at 12:10




$begingroup$
Essentially the same eta quotients are used in this post.
$endgroup$
– Tito Piezas III
Jan 4 at 12:10










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