I would like to calculate, or find a sharp upper bound for the following sum: $$ \sum_{i=0}^m \frac{\sqrt{a_{i+1}}}{2^i}, $$ if $m\leq b=\sum_{j=1}^{m+1}a_j$.
Find a bound on $ \sum_{i=0}^m \frac{\sqrt{a_{i+1}}}{2^i}, $ given that $m\leq b=\sum_{j=1}^{m+1}a_j$.
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calculus
combinatorics
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0Yes, I ment that $m\leq \sum_{j=1}^{m+1}a_j$ – 2012-04-07
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0What can you tell us about $a_1$ throuth $a_{m+1}$? – 2012-04-07
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0There is no upper bound; fix $m$ and make the $a_i$ as large as you want. – 2012-04-07
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0@bgins: Is this last edit correct? I thought the problem was: find an upper bound for $\sum_{i=0}^m \frac{\sqrt{a_{i+1}}}{2^i}$ in terms of $b=\sum_{j=1}^{m+1}a_j$. – 2012-04-07
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0Yes, I wanted an upper bound in terms of $b$. – 2012-04-07
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0@bgins: Oops, I just realized you were not the one who made that edit. Pardon me. – 2012-04-07
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0@whew! but I guess I should have checked the OP. we good now? – 2012-04-07
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0@bgins: I know that sum $\sum_{i=1}^{m+1}a_i=b$. So $b$ is known. – 2012-04-07
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0There's too much uncertainty. You could have $a_1=b$ or $a_{m+1}=b$ with the others all vanishing. Do we know that all $a_j$ are nonnegative? Is $\{a_j\}$ monotonically increasing? As anon points out, the problem is also ill-posed. Do you mean to say that $b\le M$? Also, do you really mean to have the same $m$ as a lower bound on the sum and as one less than the number of terms? – 2012-04-07
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0@bgins: yes, it is the same $m$. All I know is that $a_i$ are non- negative numbers. We can wven assume that they are naturals with zero. – 2012-04-07
1 Answers
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If $a_1=b, a_j=0$ for $j \gt 1$, then the sum is $ \frac{\sqrt b}2$. Dropping $a_1$ and increasing the others will only reduce the sum. If some of the $a$'s can be negative, I don't think there is a limit: Let $a_1$ increase positive and $a_2$ increase negative keeping $b$ the same.