User oren - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:08:20Z http://mathoverflow.net/feeds/user/11998 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial Identity of binomial series with factorial. Oren 2013-04-29T14:25:55Z 2013-05-02T02:54:43Z <p>I'm looking for a simple identity for the formula:</p> <p>$$\sum_{n = 0}^{p} \binom{p}{n} \cdot n! \cdot x^n$$</p> <p>In words, I have $p$ "players" who can choose to play or not (every player is represented by a unique id). Those who chose to play are lined up in all possible orders. Then every playing player picks an element out of $x$ possibilities (with repetition allowed), in addition to his id.</p> <p>How many sequences can we get? Is there a simple solution for this series? If not, what is the closest upper limit you can think of?</p> <p>I haven't touched combinatorics for a long time, so there could be a simple identity that I'm missing...</p> http://mathoverflow.net/questions/85012/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a Algorithm for calculating the sum-of-squares distance of a rolling window from a given line function Oren 2012-01-05T23:53:58Z 2012-01-06T01:11:32Z <p>Given a line function <code>$y = ax + b$</code>, it is easy to calculate the sum-of-squares distance between the line and a window of samples <code>$(1, y_1), (2, y_2), ..., (n, y_n)$</code> (where <code>$y_1$</code> is the oldest sample and <code>$y_n$</code> is the newest):</p> <p><code>$\sum_{x=1}^{n}(y_x - (ax + b))^2$</code></p> <p>I need a fast algorithm for calculating this value for a rolling window (of length <code>n</code>) - I cannot rescan all the samples in the window every time a new sample arrives.<br> Obviously, some state should be saved and updated for every new sample that enters the window and every old sample leaves the window.<br> Notice that when a sample leaves the window, the indecies of the rest of the samples change as well - every <code>$y_x$</code> becomes <code>$y_{x-1}$</code>. Therefore when a sample leaves the window, every other sample in the window contribute a different value to the new sum: <code>$(y_x - (a(x-1) + b))^2$</code> instead of <code>$(y_x - (ax + b))^2$</code>.<br> Is there a known algorithm for calculating this? If not, can you think of one? (It is ok to have some mistakes due first-order linear approximations).</p> <p>Thanks</p> http://mathoverflow.net/questions/85012/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a/85019#85019 Answer by Oren for Algorithm for calculating the sum-of-squares distance of a rolling window from a given line function Oren 2012-01-06T01:04:41Z 2012-01-06T01:04:41Z <p>solved: <a href="http://stackoverflow.com/questions/8751509/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a">http://stackoverflow.com/questions/8751509/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a</a></p> <p>If I will have a more detailed solution (step-by-step algorithm) I'll publish it.</p> http://mathoverflow.net/questions/51078/what-is-the-expected-length-of-the-sum-of-vectors-in-a-multi-dimensional-sphere What is the expected length of the sum of vectors in a multi-dimensional sphere? Oren 2011-01-04T00:50:30Z 2011-01-07T03:02:02Z <p>Suppose we pick $m$ vectors i.i.d from the surface of a $d$-dimensional unit sphere (they all have length 1). What would be the expected length of their sum?</p> <p>Equivalently, we can ask about the expected length of their average.</p> <p>I managed to solve it for 2 vectors in 2 dimensions, but I need a generic solution...</p> <p>If there's any "known solution" for this problem, even a link will be good enough.</p> http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial/129112#129112 Comment by Oren Oren 2013-04-29T16:39:35Z 2013-04-29T16:39:35Z so a good upper bound would be: $p!x^pe^\frac{1}{x}$ http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial/129112#129112 Comment by Oren Oren 2013-04-29T16:28:22Z 2013-04-29T16:28:22Z I guess you meant $e_p$ in the second formula. http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial Comment by Oren Oren 2013-04-29T16:18:58Z 2013-04-29T16:18:58Z This is more of a question to: math.stackexchange.com ( <a href="http://meta.math.stackexchange.com/questions/41/differences-between-mathoverflow-and-math-stackexchange" rel="nofollow">meta.math.stackexchange.com/questions/41/&hellip;</a> ). Should I remove it from here? http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial Comment by Oren Oren 2013-04-29T15:43:23Z 2013-04-29T15:43:23Z I still don't get it. http://mathoverflow.net/questions/129099/identity-of-binomial-series-with-factorial Comment by Oren Oren 2013-04-29T15:03:57Z 2013-04-29T15:03:57Z @Barry Cipra you're right. It's the players who are reordered, not the choices. http://mathoverflow.net/questions/85012/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a/85018#85018 Comment by Oren Oren 2012-01-06T01:11:07Z 2012-01-06T01:11:07Z I think it's a '$(ax + b)$' in the last sum in the first line. http://mathoverflow.net/questions/85012/algorithm-for-calculating-the-sum-of-squares-distance-of-a-rolling-window-from-a Comment by Oren Oren 2012-01-06T00:34:14Z 2012-01-06T00:34:14Z Yes, you can use O(n) storage. See the remark that I added - when samples leave the window the indexes of the other samples change - i.e. their position in the window moved and they became &quot;older&quot;. http://mathoverflow.net/questions/51078/what-is-the-expected-length-of-the-sum-of-vectors-in-a-multi-dimensional-sphere Comment by Oren Oren 2011-01-04T01:09:16Z 2011-01-04T01:09:16Z If I had a formaula, I could probably proove it with induction.. http://mathoverflow.net/questions/51078/what-is-the-expected-length-of-the-sum-of-vectors-in-a-multi-dimensional-sphere Comment by Oren Oren 2011-01-04T00:59:34Z 2011-01-04T00:59:34Z I need an exact answer