On calculating the limit of the infinite product $prod_{k=3}^n (1-tan^4frac{pi}{2^k})$












5












$begingroup$



Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?




What I attempted:-



$log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.

Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$

Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $



Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.



Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$



Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.






sequences-and-series







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$endgroup$

















    5












    $begingroup$



    Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?




    What I attempted:-



    $log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.

    Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$

    Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $



    Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.



    Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$



    Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.






    sequences-and-series







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$



      Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?




      What I attempted:-



      $log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.

      Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$

      Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $



      Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.



      Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$



      Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.






      sequences-and-series







      share|cite|improve this question











      $endgroup$





      Let $S_n=prod_{k=3}^n (1-tan^4frac{pi}{2^k})$. What is the value of $lim_{n to infty} S_n$ ?




      What I attempted:-



      $log S_n=sum_{k=3}^n log (1-tan^4frac{pi}{2^k})$.

      Since $lim_{x to 0} frac{tan x}{x}=1$, $tan^4frac{pi}{2^k}approx left(frac{pi}{2^k}right)^4$

      Thus, $log S_n=sum_{k=3}^n log (1-frac{pi^4}{2^{4k}})approx sum_{k=3}^nleft( -frac{pi^4}{2^{4k}}right) $



      Taking limit as $n to infty$, $lim_{n to infty} log S_n=frac{-pi^4}{3840}$.



      Finally, $lim_{n to infty}S_n=e^{frac{-pi^4}{3840}}approx 1-frac{pi^4}{3840}$



      Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $frac{pi^3}{4},frac{pi^3}{16}, frac{pi^3}{32},frac{pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.






      sequences-and-series




      sequences-and-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 16:52









      Robert Z

      94.5k1063134




      94.5k1063134










      asked Jan 3 at 15:39









      BhargobBhargob

      663415




      663415






















          2 Answers
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          $begingroup$

          You obtained an approximation of the exact value. In order to find such exact vale, note that
          $$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
          frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
          Hence, as $n$ goes to infinity,
          $$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
          where we used the known fact that
          $$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
          frac{1}{2^{n-1}sin(pi/2^n)}$$
          (see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
            $endgroup$
            – Robert Z
            Jan 3 at 16:50



















          1












          $begingroup$

          What you're describing is how to arrive at an approximation of the actual product.
          But you can compute the exact value for the product using trigonometric identities.



          $$begin{split}
          S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
          &=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
          &=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
          &=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
          &= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
          &=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
          end{split}$$
          Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
          you can verify that, for any $0<theta<pi$,
          $$
          prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
          $$
          Which yields, for $theta=frac pi 4$,
          $$
          prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
          $$
          So finally,
          $$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
          In other words,
          $$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$






          share|cite|improve this answer











          $endgroup$













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            4












            $begingroup$

            You obtained an approximation of the exact value. In order to find such exact vale, note that
            $$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
            frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
            Hence, as $n$ goes to infinity,
            $$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
            where we used the known fact that
            $$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
            frac{1}{2^{n-1}sin(pi/2^n)}$$
            (see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
              $endgroup$
              – Robert Z
              Jan 3 at 16:50
















            4












            $begingroup$

            You obtained an approximation of the exact value. In order to find such exact vale, note that
            $$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
            frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
            Hence, as $n$ goes to infinity,
            $$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
            where we used the known fact that
            $$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
            frac{1}{2^{n-1}sin(pi/2^n)}$$
            (see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
              $endgroup$
              – Robert Z
              Jan 3 at 16:50














            4












            4








            4





            $begingroup$

            You obtained an approximation of the exact value. In order to find such exact vale, note that
            $$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
            frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
            Hence, as $n$ goes to infinity,
            $$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
            where we used the known fact that
            $$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
            frac{1}{2^{n-1}sin(pi/2^n)}$$
            (see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).






            share|cite|improve this answer











            $endgroup$



            You obtained an approximation of the exact value. In order to find such exact vale, note that
            $$(1-tan^4(alpha/2))=(1+tan^2(alpha/2))(1-tan^2(alpha/2))=
            frac{4}{cos(alpha)}left(frac{tan(alpha/2)}{tan(alpha)}right)^2.$$
            Hence, as $n$ goes to infinity,
            $$prod_{k=3}^n left(1-tan^4(pi/2^k)right)=frac{4^{n-2}}{prod_{k=3}^ncos(pi/2^{k-1})}cdot left(prod_{k=3}^nfrac{tan(pi/2^k)}{tan(pi/2^{k-1})}right)^2\=4^{n-2}cdot 2^{n-2}sin(pi/2^{n-1})cdot left(frac{tan(pi/2^n)}{tan(pi/2^{2})}right)^2to frac{pi^3}{32}$$
            where we used the known fact that
            $$prodlimits_{k=2}^{n}cosleft(frac{pi }{2^{k}}right)=
            frac{1}{2^{n-1}sin(pi/2^n)}$$
            (see for example How to evaluate $limlimits_{nto infty}prodlimits_{r=2}^{n}cosleft(frac{pi}{2^{r}}right)$).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 16:51

























            answered Jan 3 at 15:56









            Robert ZRobert Z

            94.5k1063134




            94.5k1063134












            • $begingroup$
              Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
              $endgroup$
              – Robert Z
              Jan 3 at 16:50


















            • $begingroup$
              Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
              $endgroup$
              – Robert Z
              Jan 3 at 16:50
















            $begingroup$
            Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
            $endgroup$
            – Robert Z
            Jan 3 at 16:50




            $begingroup$
            Dear downvoter, do you mind to tell me what's wrong with my answer. Please help me to improve it?
            $endgroup$
            – Robert Z
            Jan 3 at 16:50











            1












            $begingroup$

            What you're describing is how to arrive at an approximation of the actual product.
            But you can compute the exact value for the product using trigonometric identities.



            $$begin{split}
            S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
            &=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
            &=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
            &=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
            &= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
            &=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
            end{split}$$
            Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
            you can verify that, for any $0<theta<pi$,
            $$
            prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
            $$
            Which yields, for $theta=frac pi 4$,
            $$
            prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
            $$
            So finally,
            $$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
            In other words,
            $$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              What you're describing is how to arrive at an approximation of the actual product.
              But you can compute the exact value for the product using trigonometric identities.



              $$begin{split}
              S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
              &=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
              &=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
              &=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
              &= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
              &=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
              end{split}$$
              Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
              you can verify that, for any $0<theta<pi$,
              $$
              prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
              $$
              Which yields, for $theta=frac pi 4$,
              $$
              prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
              $$
              So finally,
              $$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
              In other words,
              $$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                What you're describing is how to arrive at an approximation of the actual product.
                But you can compute the exact value for the product using trigonometric identities.



                $$begin{split}
                S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
                &=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
                &=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
                &=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
                &= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
                &=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
                end{split}$$
                Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
                you can verify that, for any $0<theta<pi$,
                $$
                prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
                $$
                Which yields, for $theta=frac pi 4$,
                $$
                prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
                $$
                So finally,
                $$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
                In other words,
                $$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$






                share|cite|improve this answer











                $endgroup$



                What you're describing is how to arrive at an approximation of the actual product.
                But you can compute the exact value for the product using trigonometric identities.



                $$begin{split}
                S_n&=prod_{k=3}^n (1-tan^4frac{pi}{2^k})\
                &=prod_{k=3}^n (1-tan^2frac{pi}{2^k})(1+tan^2frac{pi}{2^k})\
                &=prod_{k=3}^n (frac {cos^2(frac{pi}{2^k}) - sin^2(frac{pi}{2^k})}{cos^2 frac{pi}{2^k}})(frac 1 {cos^2frac{pi}{2^k}})\
                &=prod_{k=3}^n frac {cos(frac {pi}{2^{k-1}})}{cos^4 frac{pi}{2^k}}\
                &= frac{cosfrac pi 4}{cos^4 frac pi 8}frac{cosfrac pi 8}{cos^4 frac pi {16}}frac{cosfrac pi 16}{cos^4 frac pi {32}}...\
                &=cosfrac pi 4 prod_{k=3}^n frac 1 {cos^3 frac pi {2^k}}
                end{split}$$
                Now because $$frac 1 {cos theta}=frac {2 sin theta}{sin 2theta}$$
                you can verify that, for any $0<theta<pi$,
                $$
                prod_{k=1}^{+infty} frac 1 {cos frac theta {2^k}}=frac theta {sin theta}
                $$
                Which yields, for $theta=frac pi 4$,
                $$
                prod_{k=3}^{+infty} frac 1 {cos frac pi {2^k}}=frac pi {4sin frac pi 4}
                $$
                So finally,
                $$lim_{nrightarrow+infty}S_n=left(cos{frac pi 4}right)frac {pi^3} {4^3sin^3 frac pi 4}$$
                In other words,
                $$lim_{nrightarrow+infty}S_n=frac {pi^3}{32}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 16:18

























                answered Jan 3 at 16:12









                Stefan LafonStefan Lafon

                1,07616




                1,07616






























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                    Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

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                    28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
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                    Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

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                    up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
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                    A Topological Invariant for $pi_3(U(n))$

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                    3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
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