Approximation of $sum_{n=0}^infty 2^{-n} tanh(3^n x)$ in $x = 0$












4












$begingroup$


I want to (asymptotic function) approximate the series $f(x) = sum_{n=0}^infty 2^{-n} tanh(3^n x)$ around $x = 0$.



I found the equation for $f(x)$:



$$f(x) = dfrac{1}{2} f(3x) + tanh(x)$$



for $f(x)=dfrac{1}{2}f(3x) Rightarrow f(x) = A x^{log_32}$



let $A = A(x) Rightarrow A(x) = Theta(ln(x))$ where $Theta(x) $ any function with period $ln3$






sequences-and-series functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – orlp
    Dec 26 '18 at 16:19










  • $begingroup$
    @orlp how to approximate this sum
    $endgroup$
    – KotSmile
    Dec 26 '18 at 16:22










  • $begingroup$
    For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
    $endgroup$
    – John Barber
    Dec 26 '18 at 17:34












  • $begingroup$
    I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
    $endgroup$
    – John Barber
    Dec 27 '18 at 15:06












  • $begingroup$
    Related?
    $endgroup$
    – Cosmas Zachos
    Jan 1 at 17:55
















4












$begingroup$


I want to (asymptotic function) approximate the series $f(x) = sum_{n=0}^infty 2^{-n} tanh(3^n x)$ around $x = 0$.



I found the equation for $f(x)$:



$$f(x) = dfrac{1}{2} f(3x) + tanh(x)$$



for $f(x)=dfrac{1}{2}f(3x) Rightarrow f(x) = A x^{log_32}$



let $A = A(x) Rightarrow A(x) = Theta(ln(x))$ where $Theta(x) $ any function with period $ln3$






sequences-and-series functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – orlp
    Dec 26 '18 at 16:19










  • $begingroup$
    @orlp how to approximate this sum
    $endgroup$
    – KotSmile
    Dec 26 '18 at 16:22










  • $begingroup$
    For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
    $endgroup$
    – John Barber
    Dec 26 '18 at 17:34












  • $begingroup$
    I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
    $endgroup$
    – John Barber
    Dec 27 '18 at 15:06












  • $begingroup$
    Related?
    $endgroup$
    – Cosmas Zachos
    Jan 1 at 17:55














4












4








4


1



$begingroup$


I want to (asymptotic function) approximate the series $f(x) = sum_{n=0}^infty 2^{-n} tanh(3^n x)$ around $x = 0$.



I found the equation for $f(x)$:



$$f(x) = dfrac{1}{2} f(3x) + tanh(x)$$



for $f(x)=dfrac{1}{2}f(3x) Rightarrow f(x) = A x^{log_32}$



let $A = A(x) Rightarrow A(x) = Theta(ln(x))$ where $Theta(x) $ any function with period $ln3$






sequences-and-series functional-equations







share|cite|improve this question











$endgroup$




I want to (asymptotic function) approximate the series $f(x) = sum_{n=0}^infty 2^{-n} tanh(3^n x)$ around $x = 0$.



I found the equation for $f(x)$:



$$f(x) = dfrac{1}{2} f(3x) + tanh(x)$$



for $f(x)=dfrac{1}{2}f(3x) Rightarrow f(x) = A x^{log_32}$



let $A = A(x) Rightarrow A(x) = Theta(ln(x))$ where $Theta(x) $ any function with period $ln3$






sequences-and-series functional-equations




sequences-and-series functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 17:30







KotSmile

















asked Dec 26 '18 at 16:07









KotSmileKotSmile

316




316












  • $begingroup$
    What is your question?
    $endgroup$
    – orlp
    Dec 26 '18 at 16:19










  • $begingroup$
    @orlp how to approximate this sum
    $endgroup$
    – KotSmile
    Dec 26 '18 at 16:22










  • $begingroup$
    For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
    $endgroup$
    – John Barber
    Dec 26 '18 at 17:34












  • $begingroup$
    I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
    $endgroup$
    – John Barber
    Dec 27 '18 at 15:06












  • $begingroup$
    Related?
    $endgroup$
    – Cosmas Zachos
    Jan 1 at 17:55


















  • $begingroup$
    What is your question?
    $endgroup$
    – orlp
    Dec 26 '18 at 16:19










  • $begingroup$
    @orlp how to approximate this sum
    $endgroup$
    – KotSmile
    Dec 26 '18 at 16:22










  • $begingroup$
    For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
    $endgroup$
    – John Barber
    Dec 26 '18 at 17:34












  • $begingroup$
    I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
    $endgroup$
    – John Barber
    Dec 27 '18 at 15:06












  • $begingroup$
    Related?
    $endgroup$
    – Cosmas Zachos
    Jan 1 at 17:55
















$begingroup$
What is your question?
$endgroup$
– orlp
Dec 26 '18 at 16:19




$begingroup$
What is your question?
$endgroup$
– orlp
Dec 26 '18 at 16:19












$begingroup$
@orlp how to approximate this sum
$endgroup$
– KotSmile
Dec 26 '18 at 16:22




$begingroup$
@orlp how to approximate this sum
$endgroup$
– KotSmile
Dec 26 '18 at 16:22












$begingroup$
For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
$endgroup$
– John Barber
Dec 26 '18 at 17:34






$begingroup$
For whatever it's worth, I've been playing around with this numerically in Mathematica, and it looks like $f(x) approx a x^p$ for small $x>0$, where $aapprox 3.71549$ and $p$ is suspiciously close to $ln(2) / ln(3) approx 0.63093$. (Of course we know that looks can be deceiving.)
$endgroup$
– John Barber
Dec 26 '18 at 17:34














$begingroup$
I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
$endgroup$
– John Barber
Dec 27 '18 at 15:06






$begingroup$
I find analytically (with the aid of numerics) that for small $x$, $f(x) approx a, x^{ln 2 / ln 3} - 2 x$, where $a approx 3.71549$. I can't figure out what that value for $a$ is.
$endgroup$
– John Barber
Dec 27 '18 at 15:06














$begingroup$
Related?
$endgroup$
– Cosmas Zachos
Jan 1 at 17:55




$begingroup$
Related?
$endgroup$
– Cosmas Zachos
Jan 1 at 17:55










1 Answer
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$begingroup$

An alternative justification for the $log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $lim_{xto 0} x^{-a}f(x)=c$ with $0<c<infty$. The functional equation shows
$$c=frac {3^a}2c+begin{cases} 0 & text{if }a<1 \ 1 &text{if } a=1 \ infty & text{if }a>1end{cases}$$
Which forces $a<1$ and $a=log_3 2$.



Given this value of $a$, instead of looking at $lim x^{-a}f(x)$ as $xto 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $xto -infty$. Putting $g(x)=2^{-x}tanh(3^x)$ shows $F(x)=sum_{nge 0}g(x+n)$. Because $g(x)le(frac 32)^x$, $sum_{nle 1} g(x+n)le 2(frac 32)^x$ and thus $bar F(x)=sum_{n=-infty}^{infty}g(x+n)$ tends to $F$ as $xto -infty$. But $bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.



Since we also have $g(x)le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)le 2^{1-n}$ and $g(x-n)le (frac 32)^{1-n}$ so that
$$left|bar F(x)-sum_{-N}^N g(x+n)right|le 2^{1-N}+3left(frac 23right)^N $$
Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $bar F$ and this in turn approximates $F$ in the limit.






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    $begingroup$

    An alternative justification for the $log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $lim_{xto 0} x^{-a}f(x)=c$ with $0<c<infty$. The functional equation shows
    $$c=frac {3^a}2c+begin{cases} 0 & text{if }a<1 \ 1 &text{if } a=1 \ infty & text{if }a>1end{cases}$$
    Which forces $a<1$ and $a=log_3 2$.



    Given this value of $a$, instead of looking at $lim x^{-a}f(x)$ as $xto 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $xto -infty$. Putting $g(x)=2^{-x}tanh(3^x)$ shows $F(x)=sum_{nge 0}g(x+n)$. Because $g(x)le(frac 32)^x$, $sum_{nle 1} g(x+n)le 2(frac 32)^x$ and thus $bar F(x)=sum_{n=-infty}^{infty}g(x+n)$ tends to $F$ as $xto -infty$. But $bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.



    Since we also have $g(x)le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)le 2^{1-n}$ and $g(x-n)le (frac 32)^{1-n}$ so that
    $$left|bar F(x)-sum_{-N}^N g(x+n)right|le 2^{1-N}+3left(frac 23right)^N $$
    Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $bar F$ and this in turn approximates $F$ in the limit.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      An alternative justification for the $log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $lim_{xto 0} x^{-a}f(x)=c$ with $0<c<infty$. The functional equation shows
      $$c=frac {3^a}2c+begin{cases} 0 & text{if }a<1 \ 1 &text{if } a=1 \ infty & text{if }a>1end{cases}$$
      Which forces $a<1$ and $a=log_3 2$.



      Given this value of $a$, instead of looking at $lim x^{-a}f(x)$ as $xto 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $xto -infty$. Putting $g(x)=2^{-x}tanh(3^x)$ shows $F(x)=sum_{nge 0}g(x+n)$. Because $g(x)le(frac 32)^x$, $sum_{nle 1} g(x+n)le 2(frac 32)^x$ and thus $bar F(x)=sum_{n=-infty}^{infty}g(x+n)$ tends to $F$ as $xto -infty$. But $bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.



      Since we also have $g(x)le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)le 2^{1-n}$ and $g(x-n)le (frac 32)^{1-n}$ so that
      $$left|bar F(x)-sum_{-N}^N g(x+n)right|le 2^{1-N}+3left(frac 23right)^N $$
      Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $bar F$ and this in turn approximates $F$ in the limit.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        An alternative justification for the $log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $lim_{xto 0} x^{-a}f(x)=c$ with $0<c<infty$. The functional equation shows
        $$c=frac {3^a}2c+begin{cases} 0 & text{if }a<1 \ 1 &text{if } a=1 \ infty & text{if }a>1end{cases}$$
        Which forces $a<1$ and $a=log_3 2$.



        Given this value of $a$, instead of looking at $lim x^{-a}f(x)$ as $xto 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $xto -infty$. Putting $g(x)=2^{-x}tanh(3^x)$ shows $F(x)=sum_{nge 0}g(x+n)$. Because $g(x)le(frac 32)^x$, $sum_{nle 1} g(x+n)le 2(frac 32)^x$ and thus $bar F(x)=sum_{n=-infty}^{infty}g(x+n)$ tends to $F$ as $xto -infty$. But $bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.



        Since we also have $g(x)le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)le 2^{1-n}$ and $g(x-n)le (frac 32)^{1-n}$ so that
        $$left|bar F(x)-sum_{-N}^N g(x+n)right|le 2^{1-N}+3left(frac 23right)^N $$
        Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $bar F$ and this in turn approximates $F$ in the limit.






        share|cite|improve this answer









        $endgroup$



        An alternative justification for the $log_3 2$ exponent is as follows: suppose we want to approximate $f$ by some $x^a$, where $a>0$ so that $lim_{xto 0} x^{-a}f(x)=c$ with $0<c<infty$. The functional equation shows
        $$c=frac {3^a}2c+begin{cases} 0 & text{if }a<1 \ 1 &text{if } a=1 \ infty & text{if }a>1end{cases}$$
        Which forces $a<1$ and $a=log_3 2$.



        Given this value of $a$, instead of looking at $lim x^{-a}f(x)$ as $xto 0$ we look at $F(x)=3^{-ax}f(3^x)=2^{-x}f(3^x)$ as $xto -infty$. Putting $g(x)=2^{-x}tanh(3^x)$ shows $F(x)=sum_{nge 0}g(x+n)$. Because $g(x)le(frac 32)^x$, $sum_{nle 1} g(x+n)le 2(frac 32)^x$ and thus $bar F(x)=sum_{n=-infty}^{infty}g(x+n)$ tends to $F$ as $xto -infty$. But $bar F$ has period $1$ so all information about it can be gleaned from its behavior on $[0,1]$.



        Since we also have $g(x)le 2^{-x}$, for any $x$ in the unit interval, $g(x+n)le 2^{1-n}$ and $g(x-n)le (frac 32)^{1-n}$ so that
        $$left|bar F(x)-sum_{-N}^N g(x+n)right|le 2^{1-N}+3left(frac 23right)^N $$
        Therefore, the series converges uniformly, and does so at a geometric rate. This means one can obtain a very good approximation for $bar F$ and this in turn approximates $F$ in the limit.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 2:38









        Guacho PerezGuacho Perez

        3,97911134




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