Partial sum of $sec nx$

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Does there exist a closed form of $$sum_n=1^n sec nx$$
that is expressed in elementary terms?
$$$$




summation




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  • wolframalpha.com/input/?i=summation+sec+nx
    – Piyush Divyanakar
    Jul 27 at 5:54










  • @PiyushDivyanakar I was wondering whether an elementary form exists.
    – John Glenn
    Jul 27 at 5:55










  • Usually if it does it will appear in wolfram alpha.
    – Piyush Divyanakar
    Jul 27 at 5:56










  • That's not very conclusive though
    – John Glenn
    Jul 27 at 5:57






  • 2




    It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
    – Lord Shark the Unknown
    Jul 27 at 5:59















up vote
0
down vote

favorite












Does there exist a closed form of $$sum_n=1^n sec nx$$
that is expressed in elementary terms?
$$$$




summation




share|cite|improve this question



















  • wolframalpha.com/input/?i=summation+sec+nx
    – Piyush Divyanakar
    Jul 27 at 5:54










  • @PiyushDivyanakar I was wondering whether an elementary form exists.
    – John Glenn
    Jul 27 at 5:55










  • Usually if it does it will appear in wolfram alpha.
    – Piyush Divyanakar
    Jul 27 at 5:56










  • That's not very conclusive though
    – John Glenn
    Jul 27 at 5:57






  • 2




    It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
    – Lord Shark the Unknown
    Jul 27 at 5:59













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Does there exist a closed form of $$sum_n=1^n sec nx$$
that is expressed in elementary terms?
$$$$




summation




share|cite|improve this question











Does there exist a closed form of $$sum_n=1^n sec nx$$
that is expressed in elementary terms?
$$$$





summation





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 27 at 5:51









John Glenn

1,618123




1,618123











  • wolframalpha.com/input/?i=summation+sec+nx
    – Piyush Divyanakar
    Jul 27 at 5:54










  • @PiyushDivyanakar I was wondering whether an elementary form exists.
    – John Glenn
    Jul 27 at 5:55










  • Usually if it does it will appear in wolfram alpha.
    – Piyush Divyanakar
    Jul 27 at 5:56










  • That's not very conclusive though
    – John Glenn
    Jul 27 at 5:57






  • 2




    It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
    – Lord Shark the Unknown
    Jul 27 at 5:59

















  • wolframalpha.com/input/?i=summation+sec+nx
    – Piyush Divyanakar
    Jul 27 at 5:54










  • @PiyushDivyanakar I was wondering whether an elementary form exists.
    – John Glenn
    Jul 27 at 5:55










  • Usually if it does it will appear in wolfram alpha.
    – Piyush Divyanakar
    Jul 27 at 5:56










  • That's not very conclusive though
    – John Glenn
    Jul 27 at 5:57






  • 2




    It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
    – Lord Shark the Unknown
    Jul 27 at 5:59
















wolframalpha.com/input/?i=summation+sec+nx
– Piyush Divyanakar
Jul 27 at 5:54




wolframalpha.com/input/?i=summation+sec+nx
– Piyush Divyanakar
Jul 27 at 5:54












@PiyushDivyanakar I was wondering whether an elementary form exists.
– John Glenn
Jul 27 at 5:55




@PiyushDivyanakar I was wondering whether an elementary form exists.
– John Glenn
Jul 27 at 5:55












Usually if it does it will appear in wolfram alpha.
– Piyush Divyanakar
Jul 27 at 5:56




Usually if it does it will appear in wolfram alpha.
– Piyush Divyanakar
Jul 27 at 5:56












That's not very conclusive though
– John Glenn
Jul 27 at 5:57




That's not very conclusive though
– John Glenn
Jul 27 at 5:57




2




2




It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
– Lord Shark the Unknown
Jul 27 at 5:59





It's $sum_1^nfrac2z^n+z^-n$ for $z=exp(ix)$. It looks unlikely that this has a simpler form.
– Lord Shark the Unknown
Jul 27 at 5:59











1 Answer
1






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up vote
3
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Admitting that you look for
$$S_n=sum_k=1^n sec(k x)$$ if $nx$ is small, you could approximate using
$$sec(x)=1+fracx^22+frac5 x^424+frac61 x^6720+frac277
x^88064+Oleft(x^10right)$$ and get
$$S_n=sum_k=1^n 1+fracx^22sum_k=1^n k^2+frac5 x^424sum_k=1^n k^4+frac61 x^6720sum_k=1^n k^6+frac277
x^88064sum_k=1^n k^8+cdots$$ and use Faulhaber's formulae and get
$$S_n=n+fracn12 left(2 n^2+3 n+1right) x^2+fracn144 left(6 n^4+15 n^3+10
n^2-1right) x^4+frac61n left(6 n^6+21 n^5+21 n^4-7 n^2+1right)
30240x^6+frac277n left(10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3right)
725760 x^8+cdots$$



Using for example $n=20$ and $x=frac pi180$, the above would give
$$20+frac287 pi ^26480+frac361333 pi ^42519424000+frac1320380441 pi
^62448880128000000+frac1395698742463 pi ^8634749729177600000000$$ which is $approx 20.45163552$ while the exact summation would give $approx 20.45163645$.



Edit



If, instead of Taylor series, we use a $[4,2]$ Padé approximant built at $x=0$
$$sec(x)=frac1+frac7 75x^2+frac1200x^4 1-frac61 150x^2 $$ which is equivalent to a Taylor series up to $Oleft(x^8right)$ we have
$$S_n=sum_k=1^n frac1+frac7 75i ^2x^2+frac1200i^4x^4 1-frac61 150i^2x^2 $$
and we can get the "surprising" expression
$$1815848, x, S_n=93750 sqrt366 left(H_n+frac5
sqrtfrac661x-H_n-frac5 sqrtfrac661x+pi cot left(frac5 sqrtfrac661 pi
xright)right)-122 (3866 n+9375), x-3721, n (n+1) (2 n+1), x^3$$ where appear generalized harmonic numbers.



For the worked example, this would give $approx 20.45163652$ which is better.



For sure, this could be improved using $[2k,2]$ Padé approximants $(k>2)$ which would lead to similar expressions






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    1 Answer
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    active

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    1 Answer
    1






    active

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    active

    oldest

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    up vote
    3
    down vote













    Admitting that you look for
    $$S_n=sum_k=1^n sec(k x)$$ if $nx$ is small, you could approximate using
    $$sec(x)=1+fracx^22+frac5 x^424+frac61 x^6720+frac277
    x^88064+Oleft(x^10right)$$ and get
    $$S_n=sum_k=1^n 1+fracx^22sum_k=1^n k^2+frac5 x^424sum_k=1^n k^4+frac61 x^6720sum_k=1^n k^6+frac277
    x^88064sum_k=1^n k^8+cdots$$ and use Faulhaber's formulae and get
    $$S_n=n+fracn12 left(2 n^2+3 n+1right) x^2+fracn144 left(6 n^4+15 n^3+10
    n^2-1right) x^4+frac61n left(6 n^6+21 n^5+21 n^4-7 n^2+1right)
    30240x^6+frac277n left(10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3right)
    725760 x^8+cdots$$



    Using for example $n=20$ and $x=frac pi180$, the above would give
    $$20+frac287 pi ^26480+frac361333 pi ^42519424000+frac1320380441 pi
    ^62448880128000000+frac1395698742463 pi ^8634749729177600000000$$ which is $approx 20.45163552$ while the exact summation would give $approx 20.45163645$.



    Edit



    If, instead of Taylor series, we use a $[4,2]$ Padé approximant built at $x=0$
    $$sec(x)=frac1+frac7 75x^2+frac1200x^4 1-frac61 150x^2 $$ which is equivalent to a Taylor series up to $Oleft(x^8right)$ we have
    $$S_n=sum_k=1^n frac1+frac7 75i ^2x^2+frac1200i^4x^4 1-frac61 150i^2x^2 $$
    and we can get the "surprising" expression
    $$1815848, x, S_n=93750 sqrt366 left(H_n+frac5
    sqrtfrac661x-H_n-frac5 sqrtfrac661x+pi cot left(frac5 sqrtfrac661 pi
    xright)right)-122 (3866 n+9375), x-3721, n (n+1) (2 n+1), x^3$$ where appear generalized harmonic numbers.



    For the worked example, this would give $approx 20.45163652$ which is better.



    For sure, this could be improved using $[2k,2]$ Padé approximants $(k>2)$ which would lead to similar expressions






    share|cite|improve this answer



























      up vote
      3
      down vote













      Admitting that you look for
      $$S_n=sum_k=1^n sec(k x)$$ if $nx$ is small, you could approximate using
      $$sec(x)=1+fracx^22+frac5 x^424+frac61 x^6720+frac277
      x^88064+Oleft(x^10right)$$ and get
      $$S_n=sum_k=1^n 1+fracx^22sum_k=1^n k^2+frac5 x^424sum_k=1^n k^4+frac61 x^6720sum_k=1^n k^6+frac277
      x^88064sum_k=1^n k^8+cdots$$ and use Faulhaber's formulae and get
      $$S_n=n+fracn12 left(2 n^2+3 n+1right) x^2+fracn144 left(6 n^4+15 n^3+10
      n^2-1right) x^4+frac61n left(6 n^6+21 n^5+21 n^4-7 n^2+1right)
      30240x^6+frac277n left(10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3right)
      725760 x^8+cdots$$



      Using for example $n=20$ and $x=frac pi180$, the above would give
      $$20+frac287 pi ^26480+frac361333 pi ^42519424000+frac1320380441 pi
      ^62448880128000000+frac1395698742463 pi ^8634749729177600000000$$ which is $approx 20.45163552$ while the exact summation would give $approx 20.45163645$.



      Edit



      If, instead of Taylor series, we use a $[4,2]$ Padé approximant built at $x=0$
      $$sec(x)=frac1+frac7 75x^2+frac1200x^4 1-frac61 150x^2 $$ which is equivalent to a Taylor series up to $Oleft(x^8right)$ we have
      $$S_n=sum_k=1^n frac1+frac7 75i ^2x^2+frac1200i^4x^4 1-frac61 150i^2x^2 $$
      and we can get the "surprising" expression
      $$1815848, x, S_n=93750 sqrt366 left(H_n+frac5
      sqrtfrac661x-H_n-frac5 sqrtfrac661x+pi cot left(frac5 sqrtfrac661 pi
      xright)right)-122 (3866 n+9375), x-3721, n (n+1) (2 n+1), x^3$$ where appear generalized harmonic numbers.



      For the worked example, this would give $approx 20.45163652$ which is better.



      For sure, this could be improved using $[2k,2]$ Padé approximants $(k>2)$ which would lead to similar expressions






      share|cite|improve this answer

























        up vote
        3
        down vote










        up vote
        3
        down vote









        Admitting that you look for
        $$S_n=sum_k=1^n sec(k x)$$ if $nx$ is small, you could approximate using
        $$sec(x)=1+fracx^22+frac5 x^424+frac61 x^6720+frac277
        x^88064+Oleft(x^10right)$$ and get
        $$S_n=sum_k=1^n 1+fracx^22sum_k=1^n k^2+frac5 x^424sum_k=1^n k^4+frac61 x^6720sum_k=1^n k^6+frac277
        x^88064sum_k=1^n k^8+cdots$$ and use Faulhaber's formulae and get
        $$S_n=n+fracn12 left(2 n^2+3 n+1right) x^2+fracn144 left(6 n^4+15 n^3+10
        n^2-1right) x^4+frac61n left(6 n^6+21 n^5+21 n^4-7 n^2+1right)
        30240x^6+frac277n left(10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3right)
        725760 x^8+cdots$$



        Using for example $n=20$ and $x=frac pi180$, the above would give
        $$20+frac287 pi ^26480+frac361333 pi ^42519424000+frac1320380441 pi
        ^62448880128000000+frac1395698742463 pi ^8634749729177600000000$$ which is $approx 20.45163552$ while the exact summation would give $approx 20.45163645$.



        Edit



        If, instead of Taylor series, we use a $[4,2]$ Padé approximant built at $x=0$
        $$sec(x)=frac1+frac7 75x^2+frac1200x^4 1-frac61 150x^2 $$ which is equivalent to a Taylor series up to $Oleft(x^8right)$ we have
        $$S_n=sum_k=1^n frac1+frac7 75i ^2x^2+frac1200i^4x^4 1-frac61 150i^2x^2 $$
        and we can get the "surprising" expression
        $$1815848, x, S_n=93750 sqrt366 left(H_n+frac5
        sqrtfrac661x-H_n-frac5 sqrtfrac661x+pi cot left(frac5 sqrtfrac661 pi
        xright)right)-122 (3866 n+9375), x-3721, n (n+1) (2 n+1), x^3$$ where appear generalized harmonic numbers.



        For the worked example, this would give $approx 20.45163652$ which is better.



        For sure, this could be improved using $[2k,2]$ Padé approximants $(k>2)$ which would lead to similar expressions






        share|cite|improve this answer















        Admitting that you look for
        $$S_n=sum_k=1^n sec(k x)$$ if $nx$ is small, you could approximate using
        $$sec(x)=1+fracx^22+frac5 x^424+frac61 x^6720+frac277
        x^88064+Oleft(x^10right)$$ and get
        $$S_n=sum_k=1^n 1+fracx^22sum_k=1^n k^2+frac5 x^424sum_k=1^n k^4+frac61 x^6720sum_k=1^n k^6+frac277
        x^88064sum_k=1^n k^8+cdots$$ and use Faulhaber's formulae and get
        $$S_n=n+fracn12 left(2 n^2+3 n+1right) x^2+fracn144 left(6 n^4+15 n^3+10
        n^2-1right) x^4+frac61n left(6 n^6+21 n^5+21 n^4-7 n^2+1right)
        30240x^6+frac277n left(10 n^8+45 n^7+60 n^6-42 n^4+20 n^2-3right)
        725760 x^8+cdots$$



        Using for example $n=20$ and $x=frac pi180$, the above would give
        $$20+frac287 pi ^26480+frac361333 pi ^42519424000+frac1320380441 pi
        ^62448880128000000+frac1395698742463 pi ^8634749729177600000000$$ which is $approx 20.45163552$ while the exact summation would give $approx 20.45163645$.



        Edit



        If, instead of Taylor series, we use a $[4,2]$ Padé approximant built at $x=0$
        $$sec(x)=frac1+frac7 75x^2+frac1200x^4 1-frac61 150x^2 $$ which is equivalent to a Taylor series up to $Oleft(x^8right)$ we have
        $$S_n=sum_k=1^n frac1+frac7 75i ^2x^2+frac1200i^4x^4 1-frac61 150i^2x^2 $$
        and we can get the "surprising" expression
        $$1815848, x, S_n=93750 sqrt366 left(H_n+frac5
        sqrtfrac661x-H_n-frac5 sqrtfrac661x+pi cot left(frac5 sqrtfrac661 pi
        xright)right)-122 (3866 n+9375), x-3721, n (n+1) (2 n+1), x^3$$ where appear generalized harmonic numbers.



        For the worked example, this would give $approx 20.45163652$ which is better.



        For sure, this could be improved using $[2k,2]$ Padé approximants $(k>2)$ which would lead to similar expressions







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 28 at 4:43


























        answered Jul 27 at 6:26









        Claude Leibovici

        111k1055126




        111k1055126






















             

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