Maclaurin series for $arctan^{2}(x)$











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I have a question here that requires me to find the Maclaurin series expansion of
$arctan^{2}(x)$. Now I know how to find it for $arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with
$$frac{df}{dx}=2arctan(x)frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $arctan(x)$) and derivate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.










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    up vote
    5
    down vote

    favorite
    3












    I have a question here that requires me to find the Maclaurin series expansion of
    $arctan^{2}(x)$. Now I know how to find it for $arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with
    $$frac{df}{dx}=2arctan(x)frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $arctan(x)$) and derivate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.










    share|cite|improve this question
























      up vote
      5
      down vote

      favorite
      3









      up vote
      5
      down vote

      favorite
      3






      3





      I have a question here that requires me to find the Maclaurin series expansion of
      $arctan^{2}(x)$. Now I know how to find it for $arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with
      $$frac{df}{dx}=2arctan(x)frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $arctan(x)$) and derivate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.










      share|cite|improve this question













      I have a question here that requires me to find the Maclaurin series expansion of
      $arctan^{2}(x)$. Now I know how to find it for $arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with
      $$frac{df}{dx}=2arctan(x)frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $arctan(x)$) and derivate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.







      sequences-and-series power-series taylor-expansion






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      asked Nov 23 at 0:59









      math101

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      497






















          3 Answers
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          Using the Cauchy Product Formula,
          $$
          begin{align}
          frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
          &=2frac{arctan(x)}{1+x^2}\
          &=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
          &=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
          &=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
          end{align}
          $$

          Therefore,
          $$
          arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
          $$






          share|cite|improve this answer























          • Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
            – math101
            Nov 23 at 18:50


















          up vote
          5
          down vote













          We can try to obtain the series in the following way:



          $$f(x)=arctan^2 x=x^2 int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



          It's easier to consider:



          $$g(x)=int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



          Let's use partial fractions:



          $$frac{1}{(1+x^2u^2)(1+x^2v^2)}=frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$



          We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:



          $$g(x)=int_0^1 int_0^1 left(frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)} right) du ~dv=$$



          $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 left(frac{u^{2n+2}}{u^2-v^2}-frac{v^{2n+2}}{u^2-v^2} right) du ~dv$$



          Now:



          $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$



          Obviously, every integral is finite now, and we can write:



          $$int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 int_0^1 int_0^u frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \ = 2 sum_{k=0}^infty int_0^1 int_0^u u^{2n} left(1-frac{v^{2n+2}}{u^{2n+2}} right) frac{v^{2k}}{u^{2k}} dv ~du =2 sum_{k=0}^infty int_0^1 int_0^1 u^{2n+1} left(1-t^{2n+2} right) t^{2k} dt ~du= \ = 2 sum_{k=0}^infty int_0^1 left(frac{1}{2k+1}-frac{1}{2k+2n+3} right) u^{2n+1}~du= 2sum_{k=0}^infty frac{1}{(2k+1)(2k+2n+3)}$$



          So we get:



          $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n x^{2n} sum_{k=0}^infty frac{1}{(k+frac{1}{2})(k+n+frac{3}{2})}$$



          The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.



          I suppose, a kind of closed form for the general term can also be given in terms of digamma function:



          $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



          Which makes:



          $$arctan^2 x=frac{x^2}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



          Which is essentially the same as robjohn's answer.






          share|cite|improve this answer























          • Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
            – math101
            Nov 23 at 18:51










          • @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
            – Yuriy S
            Nov 26 at 13:05




















          up vote
          2
          down vote













          For the sake of an alternative method, here's a double series.



          For $|x|<1$, we have
          $$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
          As you noted,
          $$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
          So, assuming $|x|<1$,
          $$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
          $$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
          Now we focus on
          $$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
          We recall that for $|x|<1$,
          $$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
          Hence
          $$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
          $$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
          Then we have our (pretty inefficient) result:
          $$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
          $$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$






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            3 Answers
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            3 Answers
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            up vote
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            down vote



            accepted










            Using the Cauchy Product Formula,
            $$
            begin{align}
            frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
            &=2frac{arctan(x)}{1+x^2}\
            &=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
            &=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
            &=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
            end{align}
            $$

            Therefore,
            $$
            arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
            $$






            share|cite|improve this answer























            • Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
              – math101
              Nov 23 at 18:50















            up vote
            11
            down vote



            accepted










            Using the Cauchy Product Formula,
            $$
            begin{align}
            frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
            &=2frac{arctan(x)}{1+x^2}\
            &=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
            &=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
            &=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
            end{align}
            $$

            Therefore,
            $$
            arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
            $$






            share|cite|improve this answer























            • Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
              – math101
              Nov 23 at 18:50













            up vote
            11
            down vote



            accepted







            up vote
            11
            down vote



            accepted






            Using the Cauchy Product Formula,
            $$
            begin{align}
            frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
            &=2frac{arctan(x)}{1+x^2}\
            &=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
            &=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
            &=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
            end{align}
            $$

            Therefore,
            $$
            arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
            $$






            share|cite|improve this answer














            Using the Cauchy Product Formula,
            $$
            begin{align}
            frac{mathrm{d}}{mathrm{d}x}arctan^2(x)
            &=2frac{arctan(x)}{1+x^2}\
            &=2left(x-frac{x^3}3+frac{x^5}5-frac{x^7}7+cdotsright)left(1-x^2+x^4-x^6+cdotsright)\
            &=2left((1)x-left(1+tfrac13right)x^3+left(1+tfrac13+tfrac15right)x^5-left(1+tfrac13+tfrac15+tfrac17right)x^7+cdotsright)\
            &=2sum_{k=1}^infty(-1)^{k-1}left(H_{2k}-tfrac12H_kright)x^{2k-1}
            end{align}
            $$

            Therefore,
            $$
            arctan^2(x)=sum_{k=1}^inftyfrac{(-1)^{k-1}}kleft(H_{2k}-tfrac12H_kright)x^{2k}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 23 at 2:07

























            answered Nov 23 at 1:45









            robjohn

            263k27302623




            263k27302623












            • Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
              – math101
              Nov 23 at 18:50


















            • Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
              – math101
              Nov 23 at 18:50
















            Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
            – math101
            Nov 23 at 18:50




            Thank you. I suspected the series had to be multiplied. It looks like you got it in as closed a form as it can get with the harmonic series inside.
            – math101
            Nov 23 at 18:50










            up vote
            5
            down vote













            We can try to obtain the series in the following way:



            $$f(x)=arctan^2 x=x^2 int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            It's easier to consider:



            $$g(x)=int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            Let's use partial fractions:



            $$frac{1}{(1+x^2u^2)(1+x^2v^2)}=frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$



            We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:



            $$g(x)=int_0^1 int_0^1 left(frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)} right) du ~dv=$$



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 left(frac{u^{2n+2}}{u^2-v^2}-frac{v^{2n+2}}{u^2-v^2} right) du ~dv$$



            Now:



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$



            Obviously, every integral is finite now, and we can write:



            $$int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 int_0^1 int_0^u frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \ = 2 sum_{k=0}^infty int_0^1 int_0^u u^{2n} left(1-frac{v^{2n+2}}{u^{2n+2}} right) frac{v^{2k}}{u^{2k}} dv ~du =2 sum_{k=0}^infty int_0^1 int_0^1 u^{2n+1} left(1-t^{2n+2} right) t^{2k} dt ~du= \ = 2 sum_{k=0}^infty int_0^1 left(frac{1}{2k+1}-frac{1}{2k+2n+3} right) u^{2n+1}~du= 2sum_{k=0}^infty frac{1}{(2k+1)(2k+2n+3)}$$



            So we get:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n x^{2n} sum_{k=0}^infty frac{1}{(k+frac{1}{2})(k+n+frac{3}{2})}$$



            The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.



            I suppose, a kind of closed form for the general term can also be given in terms of digamma function:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which makes:



            $$arctan^2 x=frac{x^2}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which is essentially the same as robjohn's answer.






            share|cite|improve this answer























            • Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
              – math101
              Nov 23 at 18:51










            • @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
              – Yuriy S
              Nov 26 at 13:05

















            up vote
            5
            down vote













            We can try to obtain the series in the following way:



            $$f(x)=arctan^2 x=x^2 int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            It's easier to consider:



            $$g(x)=int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            Let's use partial fractions:



            $$frac{1}{(1+x^2u^2)(1+x^2v^2)}=frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$



            We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:



            $$g(x)=int_0^1 int_0^1 left(frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)} right) du ~dv=$$



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 left(frac{u^{2n+2}}{u^2-v^2}-frac{v^{2n+2}}{u^2-v^2} right) du ~dv$$



            Now:



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$



            Obviously, every integral is finite now, and we can write:



            $$int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 int_0^1 int_0^u frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \ = 2 sum_{k=0}^infty int_0^1 int_0^u u^{2n} left(1-frac{v^{2n+2}}{u^{2n+2}} right) frac{v^{2k}}{u^{2k}} dv ~du =2 sum_{k=0}^infty int_0^1 int_0^1 u^{2n+1} left(1-t^{2n+2} right) t^{2k} dt ~du= \ = 2 sum_{k=0}^infty int_0^1 left(frac{1}{2k+1}-frac{1}{2k+2n+3} right) u^{2n+1}~du= 2sum_{k=0}^infty frac{1}{(2k+1)(2k+2n+3)}$$



            So we get:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n x^{2n} sum_{k=0}^infty frac{1}{(k+frac{1}{2})(k+n+frac{3}{2})}$$



            The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.



            I suppose, a kind of closed form for the general term can also be given in terms of digamma function:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which makes:



            $$arctan^2 x=frac{x^2}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which is essentially the same as robjohn's answer.






            share|cite|improve this answer























            • Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
              – math101
              Nov 23 at 18:51










            • @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
              – Yuriy S
              Nov 26 at 13:05















            up vote
            5
            down vote










            up vote
            5
            down vote









            We can try to obtain the series in the following way:



            $$f(x)=arctan^2 x=x^2 int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            It's easier to consider:



            $$g(x)=int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            Let's use partial fractions:



            $$frac{1}{(1+x^2u^2)(1+x^2v^2)}=frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$



            We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:



            $$g(x)=int_0^1 int_0^1 left(frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)} right) du ~dv=$$



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 left(frac{u^{2n+2}}{u^2-v^2}-frac{v^{2n+2}}{u^2-v^2} right) du ~dv$$



            Now:



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$



            Obviously, every integral is finite now, and we can write:



            $$int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 int_0^1 int_0^u frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \ = 2 sum_{k=0}^infty int_0^1 int_0^u u^{2n} left(1-frac{v^{2n+2}}{u^{2n+2}} right) frac{v^{2k}}{u^{2k}} dv ~du =2 sum_{k=0}^infty int_0^1 int_0^1 u^{2n+1} left(1-t^{2n+2} right) t^{2k} dt ~du= \ = 2 sum_{k=0}^infty int_0^1 left(frac{1}{2k+1}-frac{1}{2k+2n+3} right) u^{2n+1}~du= 2sum_{k=0}^infty frac{1}{(2k+1)(2k+2n+3)}$$



            So we get:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n x^{2n} sum_{k=0}^infty frac{1}{(k+frac{1}{2})(k+n+frac{3}{2})}$$



            The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.



            I suppose, a kind of closed form for the general term can also be given in terms of digamma function:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which makes:



            $$arctan^2 x=frac{x^2}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which is essentially the same as robjohn's answer.






            share|cite|improve this answer














            We can try to obtain the series in the following way:



            $$f(x)=arctan^2 x=x^2 int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            It's easier to consider:



            $$g(x)=int_0^1 int_0^1 frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$



            Let's use partial fractions:



            $$frac{1}{(1+x^2u^2)(1+x^2v^2)}=frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$



            We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:



            $$g(x)=int_0^1 int_0^1 left(frac{u^2}{(u^2-v^2)(1+x^2u^2)}-frac{v^2}{(u^2-v^2)(1+x^2v^2)} right) du ~dv=$$



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 left(frac{u^{2n+2}}{u^2-v^2}-frac{v^{2n+2}}{u^2-v^2} right) du ~dv$$



            Now:



            $$g(x)=sum_{n=0}^infty (-1)^n x^{2n} int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$



            Obviously, every integral is finite now, and we can write:



            $$int_0^1 int_0^1 frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 int_0^1 int_0^u frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \ = 2 sum_{k=0}^infty int_0^1 int_0^u u^{2n} left(1-frac{v^{2n+2}}{u^{2n+2}} right) frac{v^{2k}}{u^{2k}} dv ~du =2 sum_{k=0}^infty int_0^1 int_0^1 u^{2n+1} left(1-t^{2n+2} right) t^{2k} dt ~du= \ = 2 sum_{k=0}^infty int_0^1 left(frac{1}{2k+1}-frac{1}{2k+2n+3} right) u^{2n+1}~du= 2sum_{k=0}^infty frac{1}{(2k+1)(2k+2n+3)}$$



            So we get:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n x^{2n} sum_{k=0}^infty frac{1}{(k+frac{1}{2})(k+n+frac{3}{2})}$$



            The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.



            I suppose, a kind of closed form for the general term can also be given in terms of digamma function:



            $$g(x)=frac{1}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which makes:



            $$arctan^2 x=frac{x^2}{2} sum_{n=0}^infty (-1)^n frac{psi left(n+frac32 right)-psi left(frac12 right)}{n+1} x^{2n}$$



            Which is essentially the same as robjohn's answer.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 at 13:06

























            answered Nov 23 at 2:19









            Yuriy S

            15.5k433115




            15.5k433115












            • Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
              – math101
              Nov 23 at 18:51










            • @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
              – Yuriy S
              Nov 26 at 13:05




















            • Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
              – math101
              Nov 23 at 18:51










            • @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
              – Yuriy S
              Nov 26 at 13:05


















            Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
            – math101
            Nov 23 at 18:51




            Thank you. I got it now. You went with a bit more advanced path. e.g. I haven't used digamma functions before. But I see the reasoning.
            – math101
            Nov 23 at 18:51












            @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
            – Yuriy S
            Nov 26 at 13:05






            @math101, the thing is: the expression with harmonic numbers involves subtraction of two divergent series, so it's numerically unstable. But using digamma function (which has a lot of definitions, see Wikipedia for example) or directly the series for $k$ I have, allows us a stable algorithm for computing the coefficients. Even if it doesn't look as pretty
            – Yuriy S
            Nov 26 at 13:05












            up vote
            2
            down vote













            For the sake of an alternative method, here's a double series.



            For $|x|<1$, we have
            $$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
            As you noted,
            $$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
            So, assuming $|x|<1$,
            $$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
            $$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
            Now we focus on
            $$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
            We recall that for $|x|<1$,
            $$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
            Hence
            $$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
            $$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
            Then we have our (pretty inefficient) result:
            $$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
            $$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$






            share|cite|improve this answer

























              up vote
              2
              down vote













              For the sake of an alternative method, here's a double series.



              For $|x|<1$, we have
              $$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
              As you noted,
              $$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
              So, assuming $|x|<1$,
              $$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
              $$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
              Now we focus on
              $$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
              We recall that for $|x|<1$,
              $$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
              Hence
              $$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
              $$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
              Then we have our (pretty inefficient) result:
              $$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
              $$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                For the sake of an alternative method, here's a double series.



                For $|x|<1$, we have
                $$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
                As you noted,
                $$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
                So, assuming $|x|<1$,
                $$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
                $$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
                Now we focus on
                $$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
                We recall that for $|x|<1$,
                $$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
                Hence
                $$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
                $$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
                Then we have our (pretty inefficient) result:
                $$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
                $$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$






                share|cite|improve this answer












                For the sake of an alternative method, here's a double series.



                For $|x|<1$, we have
                $$arctan x=sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}$$
                As you noted,
                $$arctan^2x=2intfrac{arctan x}{x^2+1}mathrm{d}x$$
                So, assuming $|x|<1$,
                $$arctan^2x=2intfrac1{x^2+1}sum_{kgeq0}frac{(-1)^kx^{1+2k}}{1+2k}mathrm{d}x$$
                $$arctan^2x=2sum_{kgeq0}frac{(-1)^k}{1+2k}intfrac{x^{2k}}{1+x^2}xmathrm{d}x$$
                Now we focus on
                $$I=intfrac{x^{2k+1}}{1+x^2}mathrm{d}x$$
                We recall that for $|x|<1$,
                $$frac1{1+x^2}=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)x^n$$
                Hence
                $$I=frac12sum_{ngeq0}i^nbig(1+(-1)^nbig)int x^{2k+n+1}mathrm{d}x$$
                $$I=frac12sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
                Then we have our (pretty inefficient) result:
                $$arctan^2x=sum_{kgeq0}frac{(-1)^k}{1+2k}sum_{ngeq0}frac{i^nbig(1+(-1)^nbig)}{2k+n+2}x^{2k+n+2}$$
                $$arctan^2x=sum_{n,kinBbb N}frac{(-1)^{k+frac{n}2}big(1+(-1)^nbig)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 21:33









                clathratus

                2,683325




                2,683325






























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