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

My attempt: I have absolutely no clue except for the case of $c=2$ and $b=1$ Create a line integral over the unit line evaluated with a uniform measure... $$\int_L e^x d \mu=\int_{L/2} e^x \ d\mu+\int_{L/2} e^{x+1/2} \ d\mu$$ This identity should be evident by self-similarity. Prepare for recursion... $$\int_L e^x d \mu=\int_{L/2} e^x+e^{x+1/2} \ d\mu=(1+e^{1/2}) \cdot \int_{L/2} e^x \ d\mu$$ $$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot \left( \int_{L/4} e^x \ d\mu+\int_{L/4} e^{x+1/4} \ d\mu \right)$$ $$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot (1+e^{1/4}) \cdot \left( \int_{L/4} e^x \ d\mu \right)$$ It wouldn't be hard to prove by induction then that... $$\Rightarrow \int_L e^x d \mu=\lim_{\omega \to \infty} \left(\prod_{N=1}^{\omega} (1+e^{2^{-N}}) \cdot \int_{L/{2^{\omega}}} e^x \ d\mu \right)$$ 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 $0$... $$e-1= \lim_{\omega \to \infty} \left( {1 \over {2^{\omega}}} \cdot \prod_{N=1}^{\omega} (1+e^{2^{-N}}) \right)$$

Motivation: Getting an answer will allow me to derive methods to integrate a function like $e^x$ over fractals.

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Define $y_k = \frac{1}{c^k} \prod_{i=1}^k(1 + e^{b/c^i})$.

Claim 1: If $c>2$, then $\lim_{k\rightarrow\infty} y_k=0$. If $1\leq c<2$, then $\lim_{k\rightarrow\infty} y_k = \infty$.

Proof for case $c>2$: Assume $c = 2 + \delta$ for some $\delta>0$. There exists a value $k^*$ such that $e^{b/c^i} \leq 1 + \delta/2$ for all $i \geq k^*$. Then for all $k>k^*$ we have: \begin{align} 0 \leq y_k &= \frac{1}{c^k} \prod_{i=1}^k (1 + e^{b/c^i})\\ &= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{1}{c^{k-k^*}}\prod_{i=k^*+1}^{k} (1+e^{b/c^i})\\ &\leq \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{(2+\delta/2)^{k-k^*}}{c^{k-k^*}}\\ &= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\left(\frac{2+\delta/2}{2 + \delta}\right)^{k-k^*} \rightarrow 0\\ \end{align} So $\lim_{k\rightarrow\infty} y_k = 0$.

Proof for case $1 < c < 2$: Similar.

Proof for case $c=1$: $y_k = (1+e^{b/c})^k \rightarrow \infty$.


Claim 2: If $0 < c < 1$ and $b\geq 0$ then $\lim_{k\rightarrow\infty} y_k = \infty$.

Proof: We know $1+e^{b/c^i} \geq 1$ for all $i \geq 1$, and so $y_k \geq 1/c^k\rightarrow\infty$.


Claim 3: If $c=2$ and $b\geq 0$ then $1 \leq \lim_{k\rightarrow\infty} y_k \leq e^b$.

Proof: Since $b \geq 0$ we know $1 \leq e^{b/2^i}$ for all $i$, and hence:

\begin{align} y_k &= \frac{1}{2^k} \prod_{i=1}^k(1 + e^{b/2^i}) \\ &\leq \frac{1}{2^k} \prod_{i=1}^k (e^{b/2^i} + e^{b/2^i}) \\ &= \prod_{i=1}^k e^{b/2^i} \end{align}

Thus: \begin{align} \log(y_k) &\leq \sum_{i=1}^k \frac{b}{2^i} \rightarrow b \end{align} and so $\lim_{k\rightarrow\infty} y_k \leq e^b$.


For an exact answer for $c=2, b\neq 0$, why not just apply your same method?

\begin{align} \int_0^1 e^{bx} dx &= \int_0^{1/2} e^{bx}dx + \int_0^{1/2} e^{b(x+1/2)}dx \\ &= (1+e^{b/2})\int_0^{1/2} e^{bx}dx \\ &= (1+e^{b/2})\left[ \int_0^{1/4} e^{bx}dx + \int_0^{1/4} e^{b(x+1/4)}dx\right]\\ &= (1+e^{b/2})(1+e^{b/4})\int_0^{1/4}e^{bx}dx \end{align}

and so on, with the left-hand-side giving the answer of $\frac{e^b-1}{b}$? So if $c=2$ and $b\neq 0$ we get: $$ \lim_{k\rightarrow\infty} y_k = \frac{e^b-1}{b} $$

Notice that this is consistent with Claim 3, since $1 < \frac{e^b-1}{b} < e^b$ whenever $b>0$.

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  • $\begingroup$ Well, that is 100% correct, so you get the bounty. Honestly though, I was hoping that the limit wouldn't be so volatile... $\endgroup$ – Zach466920 May 4 '15 at 17:32

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