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What is... $$\lim_{\omega \to \infty} \prod_{N=1}^{\omega} {{1+e^{b \cdot c^{-N}}} \over 2}$$

This is similar to my other question. However, there is a constant factor rather than variable in the product.

My attempt: I have absolutely no clue except for the case of $c=2$ and $b=b$ Create a line integral over the unit line evaluated with a uniform measure... $$\int_L e^{b \cdot x} d \mu=\int_{L/2} e^{b \cdot x} \ d\mu+\int_{L/2} e^{b \cdot (x+1/2)} \ d\mu$$ This identity should be evident by self-similarity. Prepare for recursion... $$\int_L e^{b \cdot x} d \mu=\int_{L/2} e^{b \cdot x}+e^{b \cdot (x+1/2)} \ d\mu=(1+e^{b/2}) \cdot \int_{L/2} e^{b \cdot x} \ d\mu$$ $$\Rightarrow \int_L e^{b \cdot x} d \mu=(1+e^{b/2}) \cdot \left( \int_{L/4} e^{b \cdot x} \ d\mu+\int_{L/4} e^{b \cdot (x+1/4)} \ d\mu \right)$$ $$\Rightarrow \int_L e^x d \mu=(1+e^{b/2}) \cdot (1+e^{b/4}) \cdot \left( \int_{L/4} e^{b \cdot x} \ d\mu \right)$$ It wouldn't be hard to prove by induction then that... $$\Rightarrow \int_L e^{b \cdot x} d \mu=\lim_{\omega \to \infty} \prod_{N=1}^{\omega} (1+e^{b \cdot 2^{-N}}) \cdot \int_{L/{2^{\omega}}} e^{b \cdot x} \ d\mu$$ Yet we know what the left hand side equals, since it can be evaluated as a definite integral, also we know what the integral on the right equals. Since the measure is uniform and the number of values x will be allowed to take on the interval decreases to just the value, namely $0$... $${{e^b-1} \over b}= \lim_{\omega \to \infty} \prod_{N=1}^{\omega} {{{1+e^{2^{-N}}}} \over 2}$$

Numerical analysis For $b=2$, the product for $c=2$ is $3.1045...$, $c=3$ is $1.7532..$, $c=4$ is $1.4424$ and $c=5$ is $1.3108$. Plotting seems to suggest an exponential in the denominator.

Motivation: Getting an answer will allow me to derive methods to integrate a function like $e^x$ over fractals. If you doubt this, look here.

Specifically, $$\int_C e^{b \cdot x} \ d \mu=\lim_{\omega \to \infty} \prod_{N=1}^{\omega} {{1+e^{2 \cdot b \cdot {3}^{-N}}} \over 2}$$ Where the region of integration $C$ is the cantor set

Community
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Zach466920
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1 Answers1

3

The case $c=2$ is equivalent to the classical formula

$$ \sin(x) = x \prod_{k=1}^\infty \cos(x/2^k)$$ and its hyperbolic form $$ \sinh(x) = x \prod_{k=1}^\infty \cosh(x/2^k)$$ I'm not aware of anything similar for $c \ne 2$, however.

If $P(x) = \prod_{k=1}^\infty \cos(x c^{-k})$, we have the functional equation $P(cx) = \cos(x) P(x)$. For $c=2$, $P(x) = \sin(x)/x$ is a solution to this. I don't know if there's a closed-form solution for $c \ne 2$.

EDIT: In the correspondence with your formula, $x = b/2$. Note that

$$ \dfrac{1 + e^{b c^{-k}}}{2} = e^{b c^{-k}/2} \cosh(b c^{-k}/2)$$ Now for $c > 1$,
$$ \prod_{k=1}^\infty e^{b c^{-k}/2} = \exp\left(\sum_{k=1}^\infty b c^{-k}/2\right) = \exp\left(\frac{b}{2c-2}\right)$$ so $$ \prod_{k=1}^\infty \dfrac{1 + e^{b c^{-k}}}{2} = \exp\left(\frac{b}{2c-2}\right) \prod_{k=1}^\infty \cosh(bc^{-k}/2) $$

EDIT: To prove the "classical formula", you can start with

$$ \sin(x/2^n) \prod_{k=1}^n \cos(x/2^k) = \sin(x)/2^n$$ which is easy to prove using induction. Divide by $\sin(x/2^n)$ and take the limit as $n \to \infty$.

Robert Israel
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  • Good start, but some things need to be cleared up. First, does the $x$ in the above correspond to the $b$ in the original post? Second, you'll have to prove, or give reference to a proof, that your formula is equivalent to mine. For now, just (+1) – Zach466920 Aug 07 '15 at 18:29
  • You added the relation, but where's the proof for $$ \sinh(x) = x \prod_{k=1}^\infty \cosh(x/2^k)$$? – Zach466920 Aug 08 '15 at 20:45
  • It comes from $\sinh(x)=2\cosh(x/2)\sinh(x/2)=4\cosh(x/2)\cosh(x/4)\sinh(x/4)=\ldots$ and $\lim_{a\rightarrow\infty}a\sinh(x/a)=x$. – Enredanrestos Aug 11 '15 at 17:37
  • I've decided that your answer adds nothing new that I didn't already know. But, it's nice that it's been expressed in a different form, so you'll be awarded half the bounty. – Zach466920 Aug 14 '15 at 15:14