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Given the general term of geometric sequence: $a_n = \dfrac{x}{2^n}$ I would like to solve for the value of n that makes $a_n =1$. My work so far: \begin{align*} a_n &= \frac{x}{2^n}\\ 2^n &= \frac{x}{a_n}\\ n &=\log_2\left(\frac{x}{a_n}\right) \end{align*} Then I let $a_n = 1$ thus: $$n=\log_2(x)$$ However for the specific problem I'm working on $n$ has to be an integer so I applied the ceiling function to both sides of the equation: $$\lceil n \rceil = \lceil \log_2(x) \rceil \,\,$$
At this point the value of $\lceil n \rceil$ should be the value I'm looking for. The important part of this problem is finding the smallest value of the form $2^a$ that satisfies $2^a \ge x$. I believe that $2^{\lceil n \rceil}$ is the smallest number of the form $2^a$ that satisfies the inequality however I am unsure of how to more rigorously prove it. Although I think it might have something to do with the definition of the ceiling function.

Please point out any errors in my work so far. Also any suggestions on how to prove that $2^{\lceil n \rceil}$ is the smallest value that satisfies the inequality would be fantastic!

Thanks for any assistance!

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What you’ve done is fine, and the proof isn’t hard.

Let $k=\lceil\log_2x\rceil$. By definition $k-1<\log_2x\le k$, so $2^{k-1}<x\le 2^k$, and it follows immediately (a) that $2^k\ge x$ and (b) that no smaller integer has that property.

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  • $\begingroup$ @Chris: You’re welcome. $\endgroup$ – Brian M. Scott Jul 8 '13 at 1:30
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All you need to make this proof "rigorous" is to note either that $2^x$ or $\log_2(x)$ are strictly monotonically increasing functions.

So for example, once you show that $2^x$ is increasing you could say that because $2^n=x$, we know that for any integers $m<n$, $2^m<2^n=x$ and for any integers $m>n$, $2^m>2^n=x$. Your statement follows.

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  • $\begingroup$ Thanks for the proof and the LaTeX edits. :) $\endgroup$ – Chris Jul 8 '13 at 1:33

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