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.


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 $b>0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.