On the tribonacci constant $T$ and...











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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







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    up vote
    3
    down vote

    favorite
    3












    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


























      up vote
      3
      down vote

      favorite
      3









      up vote
      3
      down vote

      favorite
      3






      3





      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















      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




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      edited Dec 13 '17 at 8:31

























      asked Dec 12 '17 at 13:38









      Tito Piezas III

      26.7k364168




      26.7k364168






















          1 Answer
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          up vote
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          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





















          • Thanks for the nice initial analysis.
            – Tito Piezas III
            Nov 25 at 1:19











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          1 Answer
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          active

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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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          up vote
          2
          down vote













          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





















          • Thanks for the nice initial analysis.
            – Tito Piezas III
            Nov 25 at 1:19















          up vote
          2
          down vote













          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





















          • Thanks for the nice initial analysis.
            – Tito Piezas III
            Nov 25 at 1:19













          up vote
          2
          down vote










          up vote
          2
          down vote









          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












          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 at 19:46









          Frpzzd

          20.9k638104




          20.9k638104












          • Thanks for the nice initial analysis.
            – Tito Piezas III
            Nov 25 at 1:19


















          • Thanks for the nice initial analysis.
            – Tito Piezas III
            Nov 25 at 1:19
















          Thanks for the nice initial analysis.
          – Tito Piezas III
          Nov 25 at 1:19




          Thanks for the nice initial analysis.
          – Tito Piezas III
          Nov 25 at 1:19


















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