1:22 AM
Perhaps it might be interesting (or even useful) to try sometimes searching for question about some specific result. And to compare whether we were able to find it using various methods.
"I saw this question asking for the sum of $\frac n{2^n}$, which we have seen many times before. Typing $sum 2/2^n into the search box did not find any of them. I know I have answered one."
For your fourth point, I have not investigated thoroughly. But this post (math.stackexchange.com/questions/1934367/…) is very recent and I am sure it is not indexed (I have not renewed our index for weeks). Also, by searching
\sum_{n=0}^{\infty} \frac{n}{4^n}, you will get many relevant results, although I cannot find this one (math.stackexchange.com/questions/50919) too, but I believe the crawler missed that one, I see no reason that such a simple expression cannot be handled correctly by approach0. — Wei Zhong Sep 25 at 3:53
But I guess Ross Milikan's point was not that the linked question is not among the search result. He wanted to find older questions about the same sum (which probably should be indexed).
Probably the fastest way to find this question is to look in the frequent tab of sequences-and-series tag.
Anybody who spent enough time on this site knows that question about sums of the form $\sum nx^n$ appear quite regularly.
Indeed, on the frequent tab we have How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ and we can find several copies of the specific question about $x=1/2$ among the linked questions.
(The other thing is that I am not sure whether they should have be closed as duplicates. But that is discussion for another time. In this room we should concentrate on searching.)
One thing which quite often works relatively well is to click on ask the question, then type in title similar to the desired question.
I have tried entering "Sum of the series $\sum_{k=1}^{\infty}\frac{k}{2^k}$", "Sum of the series $\sum_{i=1}^{\infty}\frac{i}{2^i}$" and "Sum of the series $\sum_{n=1}^{\infty}\frac{n}{2^n}$'. (The list of similar questions did not appear until I entered also some text, not only the formula.)
I found the question already linked by Ross Milikan in the list: Sum of the series $\sum_{1}^{\infty}\frac{n}{2^{n}}.$. And this one, which is related, but not very close one: The sum of the series $\sum_{n=0}^\infty\left(\frac{4n+3}{5^n}\right)$.
This list shows the question Calculate the sum of the infinite series $\sum_{n=0}^{\infty} \frac{n}{4^n}$, but it might be there simply because it was chosen as a duplicate. Another similar question in the sidebar is Compute infinite sum of a arithmetico-geometric series $\sum_{i=0}^{\infty} \frac{i}{2^i}$ (this one is not linked).
However, I am not sure how this list was influenced by the comments and duplicates. So it is not clear whether this would work at the time when the question was posted.
Such as How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$ or Using power series, compute: $\sum\limits_{1}^{\infty}\frac{n}{2^n}$
This returns several relevant results: Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$?, How to compute this finite sum $\sum_{k=1}^n \frac{k}{2^k} + \frac{n}{2^n}$?.
The first three results I got are about finite sum: Finite Sum $\sum_{i=1}^n\frac i {2^i}$, Proving $\sum_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$ by induction, Writing sigma notation $\sum^n_{i=1} \frac {i}{2^i}$ in closed form.
Let us try also built-in search. It seems reasonable to restrict results to the tag sequences-and-series.
If I remember a post about the question and I remember who asked or who answered it, I can use this fact and search among posts by a specific user.
Since Ross Milikan mentioned on meta that he remembers answering question like this, we might try searching among his posts.
First I tried: math.stackexchange.com/… There is too many results - I was too lazy to check over 300 hundred posts.
I have tried to restrict the search a bit: math.stackexchange.com/… I did not notice a question about this series there.
Ross Milikan did not link to the search he used in the post on meta. So we do not know, whether it was the same as above. And also the pages which the search returns might have changed because of this (or for other reasons).
Conclusion is that in this case both Google and Approach0 found several relevant results. (And if we find one question, we usually find more copies among linked and related questions.)
This is probably not very surprising, since this question appears very often. (Trying to find a question which does not have that many copies on math.SE will probably be more challenging - ane maybe even more interesting.)
@RossMillikan I have tried whether I can find the question about the series you mentioned using Google and using Approach0. You can see here to which extent this was successful. — Martin Sleziak 9 secs ago
2:08 AM
Since I mentioned that I tried searching among Ross Millinkan's posts, I will add that I found at least this (related) question in this way: How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$?
2:45 AM
@Workaholic Sorry for not being explicit, we want to get something related to
\oint \frac{dz}{1-z^2}, which is presumably a high frequent math expression. If we can get \oint there, and \oint \frac{dz}{1-z^2} returns nothing, I think perhaps there is not a \oint \frac{dz}{1-z^2} in approach0 index. To confirm this, I ask people to report if they find any old post containing \oint \frac{dz}{1-z^2}.
2:56 AM
3:13 AM
3:28 AM
@MartinSleziak I think your query is not $\sum$, $\frac{k}{2^k}$ (I opened it, but did not get the top results you listed), instead, I guess it is $\sum_{k=0} \frac{k}{2^k}$? I do not know, I just want to correct it. Since later users come into this chat room and use that link probably will not get expected search results.
3:39 AM
@MartinSleziak Thanks for giving a good search example and comparison between different search methods.
I agree that finding most commonly used expression is not that difficult, the difficulty at least I have seen technically (and is the next thing I want to improve on) is let search engine try different equivalent transformations and decomposed expressions (i.e. to decompose a complex and large math expression into small "key expressions").
4 hours later…
When I tried to search for $\sum_{n=1}^\infty \frac{n}{2^n}$ I should have mentioned also possibility to look into various compiled list of questions.
In this case List of Generalizations of Common Questions and Catalog of standard exercises seem like reasonable candidates.
Catalog of series contains links to Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$ and How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n}$ (number 93 and 99 in the list.)
8:03 AM
@WeiZhong The query I used was indeed the one I linked. Maybe I should be more precise. These questions were not the top results, but they were ranked 2, 5 and 6. (Still on the first page, so I was satisfied.)
Searching for $\sum\frac{k}{2^k}$ works approximately the same, the hits number 2, 5 and 6 are about this series.
So I will try to add the lower and upper bound to the sum. There are two alternatives how to write this.
I can try either $\sum_{k=1}^\infty\frac{k}{2^k}$ or I can search for $\sum^\infty_{k=1}\frac{k}{2^k}$.
The results seem to be the same (at least on the first page). And almost all results on the first page are questions about this series. (And the rest are posts which contain the series.)
question Will the result be the same if I write in raw query different TeX-versions of the same expression? For example, $\sum_{k=1}^\infty \frac k{2^k}$, $\sum_{k=1}^\infty \frac{k}{2^k}$ and $\sum^\infty_{k=1} \frac k{2^k}$.
At least if I test this for these three expressions, the results (on the first page) seem to be the same. I have troed the first and the second above. Here is link to the third one.
8:55 AM
Oh and still about the above searches. We can only guess what Ross Millikan searched for (since he did not link to the query). But if I enter $\sum k/2^k$ through the raw query, I find no questions about this series.
This one is at least close: How to apply the alternating series test to the series $\sum (-1)^{n+1} n/2^n$?
And with some stretch of imagination, these two are at least related: How to prove $\sum n/3^n$ converges without ratio test? and Finding the Function of a Power Series: $\sum kx^{k+1}/3^k$.
2 hours later…
11:15 AM
@MartinSleziak Yes, the search result is expected to be the same using your examples. This is because Approach0 parses a TeX into "operator tree" in which changing the order of
subscript and super script or enclosed a TeX bracket (i.e. "{}") does not affect the "operator tree" structure. So many times when I say "structurally relevant", I mean the operator tree representation of query is the same or subtree of the operator tree representation of another expression.
@MartinSleziak Try to use "$\sum{k/2^k}$" instead. Link: approach0.xyz/search/?q=%24%5Csum%7Bk%2F2%5Ek%7D%24&p=1
For the queries for the inequalities (approach0.xyz/search/… and approach0.xyz/search/… ) the results seem to be different. But from the highlighted parts, it seems that both queries match both directions of the inequalities.
11:27 AM
Just to add links, so that other users can test for themselves, if I enter $\sum k/2^k$ and if I enter $\frac{\sum k}{2^k}$ (in the raw query field), the results are exactly the same.
And again, to have them side by side (easier to compare) here is your link with $\sum{k/2^k}$ and the link to search for $\sum{k/2^k}$. They return the same results.
2 hours later…
1:37 PM
@MartinSleziak On symmetry of "equalities and inequalities" (I call it relation token), yes they suppose to get the same results (as long as the relation token is still defined as commutative operator), because their subpath sets (reference) are the same.
@MartinSleziak As for why the two inequalities results are different, I have checked their result scores. (You can also check it by inspecting HTML hidden element near each <li> element, i.e. each search result) Those results in different ranking are actually the same scores.
2:39 PM
2:58 PM
Exactly, if you go to look at the second ranked result (since starting from that point, they are different), their score are both 3.806.
I am investigating right now, from what I can tell, it is one single document (math.stackexchange.com/questions/1319137) that makes the difference. "\frac{a+b}2 \ge \sqrt{ab}" can find it, but "\sqrt{ab} \le \frac{a+b}2" cannot find it. Other than this document, the hit order and score of other documents during search process occurs exact the same. Very interesting, it reminds me the butterfly effect.
To be specific, the TeX "y' = \frac{t+c}{2} = \sqrt{|y|} = \sqrt{y}" in that thread is searchable by \ge query but not by \le query. I assume this is a bug.
Because search engine is using minheap as the result set data structure (it is computer science topics, anyway, not intend to dive into technical details) to hold results during search process, a little difference in the order of pushing results into minheap can make a big difference in terms of overall ranking.
2 hours later…
4:50 PM
I think I know where is the problem. It is actually an expected behavior, but because of some coincidence, we go into a very rare case. Hard to explain in a few words.
But in general, it is because approach0 sacrifices some degree of strictness for search efficiency, and in this case, neither of the two query should match $\frac{t+c}{2} = \sqrt{y}$, but because approach0 uses a simple rule (number of subpaths) to efficiently eliminate impossible matches, one query passes through this filter in this rare case, and become a match.
I have a workaround to "fix" this (make that two query have same results and reduce the possibility of this case happening in the future again), but to fundamentally solve this problem, I need to add algorithm complexity for strictness, which will be much less efficient. So perhaps I would adopt the ad-hoc solution for better efficiency-strictness trade-off. I probably will patch this issue tomorrow (I am in Beijing time, going to sleep soon).
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In the search of a question
When you are looking for a specific question (using Approach0 ...
