Converging Infinite Sum of Square Roots of Polynomials












0












$begingroup$


I was wondering if anyone knew a way to express a formula for the infinite sum of the following converging series when $d$ is between 0 and 1:



$$
sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}
$$



Thank you!






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    It doesn't converge.
    $endgroup$
    – T. Bongers
    Jan 21 at 23:01










  • $begingroup$
    My bad I left out a term
    $endgroup$
    – Evan Dastin-van Rijn
    Jan 21 at 23:04










  • $begingroup$
    In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
    $endgroup$
    – Claude Leibovici
    Jan 22 at 5:56
















0












$begingroup$


I was wondering if anyone knew a way to express a formula for the infinite sum of the following converging series when $d$ is between 0 and 1:



$$
sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}
$$



Thank you!






sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    It doesn't converge.
    $endgroup$
    – T. Bongers
    Jan 21 at 23:01










  • $begingroup$
    My bad I left out a term
    $endgroup$
    – Evan Dastin-van Rijn
    Jan 21 at 23:04










  • $begingroup$
    In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
    $endgroup$
    – Claude Leibovici
    Jan 22 at 5:56














0












0








0


1



$begingroup$


I was wondering if anyone knew a way to express a formula for the infinite sum of the following converging series when $d$ is between 0 and 1:



$$
sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}
$$



Thank you!






sequences-and-series convergence







share|cite|improve this question











$endgroup$




I was wondering if anyone knew a way to express a formula for the infinite sum of the following converging series when $d$ is between 0 and 1:



$$
sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}
$$



Thank you!






sequences-and-series convergence




sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 23:04







Evan Dastin-van Rijn

















asked Jan 21 at 22:58









Evan Dastin-van RijnEvan Dastin-van Rijn

184




184












  • $begingroup$
    It doesn't converge.
    $endgroup$
    – T. Bongers
    Jan 21 at 23:01










  • $begingroup$
    My bad I left out a term
    $endgroup$
    – Evan Dastin-van Rijn
    Jan 21 at 23:04










  • $begingroup$
    In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
    $endgroup$
    – Claude Leibovici
    Jan 22 at 5:56


















  • $begingroup$
    It doesn't converge.
    $endgroup$
    – T. Bongers
    Jan 21 at 23:01










  • $begingroup$
    My bad I left out a term
    $endgroup$
    – Evan Dastin-van Rijn
    Jan 21 at 23:04










  • $begingroup$
    In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
    $endgroup$
    – Claude Leibovici
    Jan 22 at 5:56
















$begingroup$
It doesn't converge.
$endgroup$
– T. Bongers
Jan 21 at 23:01




$begingroup$
It doesn't converge.
$endgroup$
– T. Bongers
Jan 21 at 23:01












$begingroup$
My bad I left out a term
$endgroup$
– Evan Dastin-van Rijn
Jan 21 at 23:04




$begingroup$
My bad I left out a term
$endgroup$
– Evan Dastin-van Rijn
Jan 21 at 23:04












$begingroup$
In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
$endgroup$
– Claude Leibovici
Jan 22 at 5:56




$begingroup$
In the real domain, there is probably for $d$ an upper bound $<1$ above which the result would be a complex number.
$endgroup$
– Claude Leibovici
Jan 22 at 5:56










1 Answer
1






active

oldest

votes


















0












$begingroup$

Not a real answer since based on numerical computations.



As I wrote in comments, considering
$$S(d)=sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.



I do not think that there is any hope for a closed form.



For the range $0 leq d leq 0.6$, a quick and dirty non linear regression of a power law model
$$S(x)=alpha, d^beta $$gives
$$begin{array}{clclclclc}
text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
alpha & 1.590951 & 0.007556 & {1.575821,1.606081} \
beta & 1.732464 & 0.006246 & {1.719956,1.744972} \
end{array}$$ $(R^2=0.99989)$ with a surprising $beta simeq sqrt 3$.



However, as shown below, the fit is not fantastic at all for small values of $d$
$$left(
begin{array}{ccc}
0.05 & 0.011471 & 0.008873 \
0.10 & 0.033329 & 0.029477 \
0.15 & 0.062983 & 0.059494 \
0.20 & 0.099883 & 0.097921 \
0.25 & 0.143984 & 0.144121 \
0.30 & 0.195496 & 0.197639 \
0.35 & 0.254772 & 0.258124 \
0.40 & 0.322209 & 0.325292 \
0.45 & 0.398040 & 0.398908 \
0.50 & 0.481771 & 0.478770 \
0.55 & 0.570129 & 0.564705 \
0.60 & 0.643821 & 0.656558
end{array}
right)$$






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Not a real answer since based on numerical computations.



    As I wrote in comments, considering
    $$S(d)=sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.



    I do not think that there is any hope for a closed form.



    For the range $0 leq d leq 0.6$, a quick and dirty non linear regression of a power law model
    $$S(x)=alpha, d^beta $$gives
    $$begin{array}{clclclclc}
    text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
    alpha & 1.590951 & 0.007556 & {1.575821,1.606081} \
    beta & 1.732464 & 0.006246 & {1.719956,1.744972} \
    end{array}$$     $(R^2=0.99989)$ with a surprising $beta simeq sqrt 3$.



    However, as shown below, the fit is not fantastic at all for small values of $d$
    $$left(
    begin{array}{ccc}
    0.05 & 0.011471 & 0.008873 \
    0.10 & 0.033329 & 0.029477 \
    0.15 & 0.062983 & 0.059494 \
    0.20 & 0.099883 & 0.097921 \
    0.25 & 0.143984 & 0.144121 \
    0.30 & 0.195496 & 0.197639 \
    0.35 & 0.254772 & 0.258124 \
    0.40 & 0.322209 & 0.325292 \
    0.45 & 0.398040 & 0.398908 \
    0.50 & 0.481771 & 0.478770 \
    0.55 & 0.570129 & 0.564705 \
    0.60 & 0.643821 & 0.656558
    end{array}
    right)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Not a real answer since based on numerical computations.



      As I wrote in comments, considering
      $$S(d)=sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.



      I do not think that there is any hope for a closed form.



      For the range $0 leq d leq 0.6$, a quick and dirty non linear regression of a power law model
      $$S(x)=alpha, d^beta $$gives
      $$begin{array}{clclclclc}
      text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
      alpha & 1.590951 & 0.007556 & {1.575821,1.606081} \
      beta & 1.732464 & 0.006246 & {1.719956,1.744972} \
      end{array}$$       $(R^2=0.99989)$ with a surprising $beta simeq sqrt 3$.



      However, as shown below, the fit is not fantastic at all for small values of $d$
      $$left(
      begin{array}{ccc}
      0.05 & 0.011471 & 0.008873 \
      0.10 & 0.033329 & 0.029477 \
      0.15 & 0.062983 & 0.059494 \
      0.20 & 0.099883 & 0.097921 \
      0.25 & 0.143984 & 0.144121 \
      0.30 & 0.195496 & 0.197639 \
      0.35 & 0.254772 & 0.258124 \
      0.40 & 0.322209 & 0.325292 \
      0.45 & 0.398040 & 0.398908 \
      0.50 & 0.481771 & 0.478770 \
      0.55 & 0.570129 & 0.564705 \
      0.60 & 0.643821 & 0.656558
      end{array}
      right)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Not a real answer since based on numerical computations.



        As I wrote in comments, considering
        $$S(d)=sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.



        I do not think that there is any hope for a closed form.



        For the range $0 leq d leq 0.6$, a quick and dirty non linear regression of a power law model
        $$S(x)=alpha, d^beta $$gives
        $$begin{array}{clclclclc}
        text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
        alpha & 1.590951 & 0.007556 & {1.575821,1.606081} \
        beta & 1.732464 & 0.006246 & {1.719956,1.744972} \
        end{array}$$         $(R^2=0.99989)$ with a surprising $beta simeq sqrt 3$.



        However, as shown below, the fit is not fantastic at all for small values of $d$
        $$left(
        begin{array}{ccc}
        0.05 & 0.011471 & 0.008873 \
        0.10 & 0.033329 & 0.029477 \
        0.15 & 0.062983 & 0.059494 \
        0.20 & 0.099883 & 0.097921 \
        0.25 & 0.143984 & 0.144121 \
        0.30 & 0.195496 & 0.197639 \
        0.35 & 0.254772 & 0.258124 \
        0.40 & 0.322209 & 0.325292 \
        0.45 & 0.398040 & 0.398908 \
        0.50 & 0.481771 & 0.478770 \
        0.55 & 0.570129 & 0.564705 \
        0.60 & 0.643821 & 0.656558
        end{array}
        right)$$






        share|cite|improve this answer









        $endgroup$



        Not a real answer since based on numerical computations.



        As I wrote in comments, considering
        $$S(d)=sum_{n=2}^{infty}sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.



        I do not think that there is any hope for a closed form.



        For the range $0 leq d leq 0.6$, a quick and dirty non linear regression of a power law model
        $$S(x)=alpha, d^beta $$gives
        $$begin{array}{clclclclc}
        text{} & text{Estimate} & text{Standard Error} & text{Confidence Interval} \
        alpha & 1.590951 & 0.007556 & {1.575821,1.606081} \
        beta & 1.732464 & 0.006246 & {1.719956,1.744972} \
        end{array}$$         $(R^2=0.99989)$ with a surprising $beta simeq sqrt 3$.



        However, as shown below, the fit is not fantastic at all for small values of $d$
        $$left(
        begin{array}{ccc}
        0.05 & 0.011471 & 0.008873 \
        0.10 & 0.033329 & 0.029477 \
        0.15 & 0.062983 & 0.059494 \
        0.20 & 0.099883 & 0.097921 \
        0.25 & 0.143984 & 0.144121 \
        0.30 & 0.195496 & 0.197639 \
        0.35 & 0.254772 & 0.258124 \
        0.40 & 0.322209 & 0.325292 \
        0.45 & 0.398040 & 0.398908 \
        0.50 & 0.481771 & 0.478770 \
        0.55 & 0.570129 & 0.564705 \
        0.60 & 0.643821 & 0.656558
        end{array}
        right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 7:20









        Claude LeiboviciClaude Leibovici

        122k1157134




        122k1157134






























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