On the tribonacci constant $T$ and...












3












$begingroup$


We have,



$$4arctanleft(frac1{sqrt{phi^3}}right)color{red}-arctanleft(frac1{sqrt{phi^6-1}}right)=frac{pi}2$$



$$4arctanleft(frac1{sqrt{T^3}}right)color{blue}+arctanleft(frac1{sqrt{(2T+1)^4-1}}right)=frac{pi}2$$



with golden ratio $phi$ and tribonacci constant $T$.



Note that,



$$begin{aligned}left(sqrt{phi^3}+iright)^4left(sqrt{phi^6-1}color{red}-iright)&=(phi^3+1)^2sqrt{-phi^6}\ left(sqrt{T^3}+iright)^4left(sqrt{(2T+1)^4-1}color{blue}+iright)&=(T^3+1)^2sqrt{-(2T+1)^4}end{aligned}$$



The first one is from this post, while the second is from a session with Mathematica. However, I'm having a little trouble with the tetranacci constant.




Q: What would be analogous formulas using the tetranacci constant (and higher)?







trigonometry pi golden-ratio constants







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    We have,



    $$4arctanleft(frac1{sqrt{phi^3}}right)color{red}-arctanleft(frac1{sqrt{phi^6-1}}right)=frac{pi}2$$



    $$4arctanleft(frac1{sqrt{T^3}}right)color{blue}+arctanleft(frac1{sqrt{(2T+1)^4-1}}right)=frac{pi}2$$



    with golden ratio $phi$ and tribonacci constant $T$.



    Note that,



    $$begin{aligned}left(sqrt{phi^3}+iright)^4left(sqrt{phi^6-1}color{red}-iright)&=(phi^3+1)^2sqrt{-phi^6}\ left(sqrt{T^3}+iright)^4left(sqrt{(2T+1)^4-1}color{blue}+iright)&=(T^3+1)^2sqrt{-(2T+1)^4}end{aligned}$$



    The first one is from this post, while the second is from a session with Mathematica. However, I'm having a little trouble with the tetranacci constant.




    Q: What would be analogous formulas using the tetranacci constant (and higher)?







    trigonometry pi golden-ratio constants







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      3



      $begingroup$


      We have,



      $$4arctanleft(frac1{sqrt{phi^3}}right)color{red}-arctanleft(frac1{sqrt{phi^6-1}}right)=frac{pi}2$$



      $$4arctanleft(frac1{sqrt{T^3}}right)color{blue}+arctanleft(frac1{sqrt{(2T+1)^4-1}}right)=frac{pi}2$$



      with golden ratio $phi$ and tribonacci constant $T$.



      Note that,



      $$begin{aligned}left(sqrt{phi^3}+iright)^4left(sqrt{phi^6-1}color{red}-iright)&=(phi^3+1)^2sqrt{-phi^6}\ left(sqrt{T^3}+iright)^4left(sqrt{(2T+1)^4-1}color{blue}+iright)&=(T^3+1)^2sqrt{-(2T+1)^4}end{aligned}$$



      The first one is from this post, while the second is from a session with Mathematica. However, I'm having a little trouble with the tetranacci constant.




      Q: What would be analogous formulas using the tetranacci constant (and higher)?







      trigonometry pi golden-ratio constants







      share|cite|improve this question











      $endgroup$




      We have,



      $$4arctanleft(frac1{sqrt{phi^3}}right)color{red}-arctanleft(frac1{sqrt{phi^6-1}}right)=frac{pi}2$$



      $$4arctanleft(frac1{sqrt{T^3}}right)color{blue}+arctanleft(frac1{sqrt{(2T+1)^4-1}}right)=frac{pi}2$$



      with golden ratio $phi$ and tribonacci constant $T$.



      Note that,



      $$begin{aligned}left(sqrt{phi^3}+iright)^4left(sqrt{phi^6-1}color{red}-iright)&=(phi^3+1)^2sqrt{-phi^6}\ left(sqrt{T^3}+iright)^4left(sqrt{(2T+1)^4-1}color{blue}+iright)&=(T^3+1)^2sqrt{-(2T+1)^4}end{aligned}$$



      The first one is from this post, while the second is from a session with Mathematica. However, I'm having a little trouble with the tetranacci constant.




      Q: What would be analogous formulas using the tetranacci constant (and higher)?







      trigonometry pi golden-ratio constants




      trigonometry pi golden-ratio constants






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '17 at 8:31







      Tito Piezas III

















      asked Dec 12 '17 at 13:38









      Tito Piezas IIITito Piezas III

      26.9k365169




      26.9k365169






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          This is NOT AN ANSWER, but an identity that might help.



          In general, if $psigt 1.8$, we have that
          $$4arctan(psi^{-3/2})+arctanbigg(frac{1}{sqrt{beta-1}}bigg)=frac{pi}{2} tag{i}$$
          where
          $$beta=frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}=bigg(1+frac{8psi^3}{psi^6-6psi^3+1}bigg)^2 tag{ii}$$



          Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $ngt 24$, since if the expression
          $$frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}$$
          were rewritten as a rational function of $psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $psi$ of degree under $24$. Thus, your problem has no nice solution for $ngt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the nice initial analysis.
            $endgroup$
            – Tito Piezas III
            Nov 25 '18 at 1:19











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          1 Answer
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          1 Answer
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          active

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          This is NOT AN ANSWER, but an identity that might help.



          In general, if $psigt 1.8$, we have that
          $$4arctan(psi^{-3/2})+arctanbigg(frac{1}{sqrt{beta-1}}bigg)=frac{pi}{2} tag{i}$$
          where
          $$beta=frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}=bigg(1+frac{8psi^3}{psi^6-6psi^3+1}bigg)^2 tag{ii}$$



          Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $ngt 24$, since if the expression
          $$frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}$$
          were rewritten as a rational function of $psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $psi$ of degree under $24$. Thus, your problem has no nice solution for $ngt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the nice initial analysis.
            $endgroup$
            – Tito Piezas III
            Nov 25 '18 at 1:19
















          2












          $begingroup$

          This is NOT AN ANSWER, but an identity that might help.



          In general, if $psigt 1.8$, we have that
          $$4arctan(psi^{-3/2})+arctanbigg(frac{1}{sqrt{beta-1}}bigg)=frac{pi}{2} tag{i}$$
          where
          $$beta=frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}=bigg(1+frac{8psi^3}{psi^6-6psi^3+1}bigg)^2 tag{ii}$$



          Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $ngt 24$, since if the expression
          $$frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}$$
          were rewritten as a rational function of $psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $psi$ of degree under $24$. Thus, your problem has no nice solution for $ngt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the nice initial analysis.
            $endgroup$
            – Tito Piezas III
            Nov 25 '18 at 1:19














          2












          2








          2





          $begingroup$

          This is NOT AN ANSWER, but an identity that might help.



          In general, if $psigt 1.8$, we have that
          $$4arctan(psi^{-3/2})+arctanbigg(frac{1}{sqrt{beta-1}}bigg)=frac{pi}{2} tag{i}$$
          where
          $$beta=frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}=bigg(1+frac{8psi^3}{psi^6-6psi^3+1}bigg)^2 tag{ii}$$



          Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $ngt 24$, since if the expression
          $$frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}$$
          were rewritten as a rational function of $psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $psi$ of degree under $24$. Thus, your problem has no nice solution for $ngt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).






          share|cite|improve this answer









          $endgroup$



          This is NOT AN ANSWER, but an identity that might help.



          In general, if $psigt 1.8$, we have that
          $$4arctan(psi^{-3/2})+arctanbigg(frac{1}{sqrt{beta-1}}bigg)=frac{pi}{2} tag{i}$$
          where
          $$beta=frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}=bigg(1+frac{8psi^3}{psi^6-6psi^3+1}bigg)^2 tag{ii}$$



          Notice that (ii) suggests that no "nice simplifications" like in your two examples will be possible for $n$-fibonacci constants with $ngt 24$, since if the expression
          $$frac{(psi^3+1)^4}{(psi^6-6psi^3+1)^2}$$
          were rewritten as a rational function of $psi$ whose numerator and denominator had smaller degree, then that would provide a polynomial equation for $psi$ of degree under $24$. Thus, your problem has no nice solution for $ngt 24$, but I suspect that one for $n=4$ or $n=5$ might be obtainable (if you have the force of will to make yourself go through all of the messy algebra, or ask a computer to do it for you).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '18 at 19:46









          FrpzzdFrpzzd

          22.3k840108




          22.3k840108












          • $begingroup$
            Thanks for the nice initial analysis.
            $endgroup$
            – Tito Piezas III
            Nov 25 '18 at 1:19


















          • $begingroup$
            Thanks for the nice initial analysis.
            $endgroup$
            – Tito Piezas III
            Nov 25 '18 at 1:19
















          $begingroup$
          Thanks for the nice initial analysis.
          $endgroup$
          – Tito Piezas III
          Nov 25 '18 at 1:19




          $begingroup$
          Thanks for the nice initial analysis.
          $endgroup$
          – Tito Piezas III
          Nov 25 '18 at 1:19


















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