Approximation of $sum_{n=0}^infty 2^{-n} tanh(3^n x)$ in $x = 0$
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
asked Dec 26 at 16:07
KotSmile
315
add a comment |
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
asked Dec 26 at 16:07
KotSmile
315
What is your question?
– orlp
Dec 26 at 16:19
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
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.)
– John Barber
Dec 26 at 17:34
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.
– John Barber
Dec 27 at 15:06
add a comment |
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
asked Dec 26 at 16:07
KotSmile
315
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
asked Dec 26 at 16:07
KotSmile
315
asked Dec 26 at 16:07
KotSmile
315
edited Dec 26 at 17:30
asked Dec 26 at 16:07
KotSmile
315
asked Dec 26 at 16:07
KotSmile
315
asked Dec 26 at 16:07
KotSmile
315
315
What is your question?
– orlp
Dec 26 at 16:19
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
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.)
– John Barber
Dec 26 at 17:34
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.
– John Barber
Dec 27 at 15:06
add a comment |
What is your question?
– orlp
Dec 26 at 16:19
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
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.)
– John Barber
Dec 26 at 17:34
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.
– John Barber
Dec 27 at 15:06
What is your question?
– orlp
Dec 26 at 16:19
What is your question?
– orlp
Dec 26 at 16:19
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
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.)
– John Barber
Dec 26 at 17:34
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.)
– John Barber
Dec 26 at 17:34
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.
– John Barber
Dec 27 at 15:06
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.
– John Barber
Dec 27 at 15:06
add a comment |
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What is your question?
– orlp
Dec 26 at 16:19
@orlp how to approximate this sum
– KotSmile
Dec 26 at 16:22
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.)
– John Barber
Dec 26 at 17:34
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.
– John Barber
Dec 27 at 15:06