Answer by jlewk for Sum of 'the first k' binomial coefficients for fixed $N$
A nice result in probability is Tusnady's inequality. If $B$ has binomial distribution $(n,1/2)$ then there exists $Z\sim N(0,1)$ such that almost surely$$|B - \frac n2 - \frac{\sqrt{n}}{2}Z| \le 1 +...
View ArticleAnswer by Max Alekseyev for Sum of 'the first k' binomial coefficients for...
Two bounds that work uniformly for all $k$:$$f(n,k) \leq \binom{n+1}{k} + \binom{n+k-1}{k-1},$$$$f(n,k)\leq \binom{n+k}{k},$$where the former is tighter for $k\geq 2$.For a proof, see this answer.
View ArticleAnswer by Pietro Majer for Sum of 'the first k' binomial coefficients for...
A relevant fact, which has not been mentioned, not explicitly at least: the integral remainder formula in the Taylor expansion,$$\sum_{i=0}^k{N\choose i}=2^N- N{N-1\choose...
View ArticleAnswer by Vili for Sum of 'the first k' binomial coefficients for fixed $N$
Since you are interested in information theory you might want to consider the following bound.In the limit of large N and setting k/N = p, there is a simple trick (I learned it from "Information,...
View ArticleAnswer by Erel Segal-Halevi for Sum of 'the first k' binomial coefficients...
Let $H(N)$ be a random variable representing the number of heads in $N$ fair coin-tosses.Then, on one hand:$$\Pr[H(N)\leq k] = {1\over 2^N} f(N,k) $$On the other hand, by Hoeffding's...
View ArticleAnswer by Brendan McKay for Sum of 'the first k' binomial coefficients for...
Here's one from an old paper of mine. It has the property of being precise all the way from the middle to the end.Define$$ Y(x) = e^{x^2/2}\int_x^\infty e^{-t^2/2}dt. $$Define $x=(2k-n)/\sqrt{n}$. Then...
View ArticleAnswer by Matt for Sum of 'the first k' binomial coefficients for fixed $N$
In particular, does it have a closed form or notable algorithm for computing it efficiently?Is an $O(k)$ algorithm efficient enough? If so, here is a C++ implementation:unsigned long long sumbincoef(...
View ArticleAnswer by Barry Cipra for Sum of 'the first k' binomial coefficients for...
The sum without the $i=0$ term arises in the "egg drop" problem -- see Michael Boardman's article, "The Egg-Drop Numbers," in Mathematics Magazine, Vol. 77, No. 5 (December, 2004), pp. 368-372, which...
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Answer by Yaroslav Bulatov for Sum of 'the first k' binomial coefficients for...
Jean Gallier gives this bound (Proposition 4.16 in Ch.4 of "Discrete Math"preprint)$$f(n,k) < 2^{n-1} \frac{{n \choose k+1}}{n \choose n/2}$$where $f(N,k)=\sum_{i=0}^k {N\choose i}$, and $k\le...
View ArticleAnswer by Douglas Zare for Sum of 'the first k' binomial coefficients for...
See A008949"Triangle of partial sums of binomial coefficients."$T(n,k) = \sum_{i-0}^k {N\choose i}$ is the maximal number of regions into which $n$ hyperplanes of co-dimension $1$ divide $\mathbb R^k$...
View ArticleAnswer by Douglas Zare for Sum of 'the first k' binomial coefficients for...
One standard estimate when the sum includes about half of the terms is the Chernoff bound, one form of which gives$$\sum_{k=0}^{(N-a)/2} {N\choose k} \le 2^N \exp\bigg(\frac{-a^2}{2N}\bigg)$$This isn't...
View ArticleAnswer by Michael Lugo for Sum of 'the first k' binomial coefficients for...
I'm going to give two families of bounds, one for when $k = N/2 + \alpha \sqrt{N}$ and one for when $k$ is fixed.The sequence of binomial coefficients ${N \choose 0}, {N \choose 1}, \ldots, {N \choose...
View ArticleAnswer by Alex R. for Sum of 'the first k' binomial coefficients for fixed $N$
If you interested in some back-of-the-hand order of magnitude estimates, you might consider looking at how $\binom{n}{k}$ behaves when $k=k(n)$ has a certain size. The idea I have in mind is to break...
View ArticleAnswer by Justin Melvin for Sum of 'the first k' binomial coefficients for...
Each binomial coefficient satisfies $$\left(\frac{N}{i}\right)^i \leq {N \choose i} < \left(\frac{eN}{i}\right)^i,$$ so if $k \leq N/2$, you can upper bound the sum by $k(\frac{eN}{k})^k$
View ArticleAnswer by Jacques Carette for Sum of 'the first k' binomial coefficients for...
There is no useful closed-form for this. You can write it down as$$2^N - \binom{N}{k+1} {}_2F_{1}(1, k+1-N, k+2; -1)$$but that's really just a rewrite of the sum in a different form.
View ArticleSum of 'the first k' binomial coefficients for fixed $N$
I am interested in the function $$f(N,k)=\sum_{i=0}^{k} {N \choose i}$$ for fixed $N$ and $0 \leq k \leq N $. Obviously it equals 1 for $k = 0$ and $2^{N}$ for $k = N$, but are there any other notable...
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