How wide is the learning gap between completing first and second/second and third/etc problems?
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This is a question about studying for the Putnam examination (and, secondarily, other high-difficulty proof-based math competitions like the IMO). It is not about the history of the competition, the advisability of participating, the career trajectories of former participants, or other such things. This is a question about how one can most effectively prepare to do well.   Many such questions have been asked before on this site. The answers, while helpful, all seem to boil down to recommendations of the same standard canon of books with the encouragement to do more problems. This is very good advice, but I want to approach the topic from a different angle.   We have a wealth of talent on this website, and in particular many users who have done quite well at these competitions. I would like to hear their personal stories. In particular, how did you prepare, and what kind of time commitment did you put forth? And advice, especially practical study tips, is always welcome. :)  edit: Also, it seems to me that the majority of people who do very well at the Putnam had developed the majority of their skill in high school, and focused mainly on their classes in college (while attending whatever Putnam seminar their college offered). I would appreciate comments on this matter too.   (Motivation: It seems to me like the role of talent is vastly overestimated in mathematics, and in mathematics competitions in particular, to the point where the Putnam exam gets used as a sort of pseudo-IQ test. Of course the people who do well have gifts, but it also seems that, without fail, they all have a history of doing many hours of mathematics a day for years on end. For example, I recall reading an interview with Tao where he admitted his childhood consisted of nothing but math and computer games. I am trying to gather some evidence on this matter [and advice for Putnam preparation!]. Please keep the answers focused on the actual question, though.)
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How should I rewrite the sequence so that we have a form that we can easily calculate the limit?
51309
Problem: James has a pile of n stones for some positive integer n ≥ 2. At each step, he   chooses one pile of stones and splits it into two smaller piles and writes the product   of the new pile sizes on the board. He repeats this process until every pile is exactly   one stone.      For example, if James has n = 12 stones, he could split that pile into a pile of size   4 and another pile of size 8. James would then write the number 4 · 8 = 32 on the   board. He then decides to split the pile of 4 stones into a pile with 1 stone and a   pile with 3 stones and writes 1 · 3 = 3 on the board. He continues this way until he   has 12 piles with one stone each.      Prove that no matter how James splits the piles (starting with a single pile of n   stones), the sum of the numbers on the blackboard at the end of the procedure is   always the same.   Hint: First figure out what the formula for the final sum will be. Then prove it using strong induction.  The formula I came up with is $n(n-1) / $2. It could be totally wrong...  My (Partial) Solution:  1) Base Case: (n=2) James has one pile of 2 stones. Suppose he takes the top-most stone and puts in one pile A, which now has size 1. He takes the remaining stone and places it into another pile B, which now has size 1. Then the product of the sizes is 1. The sum of all the products on his board is 1. Now, if he were to start again, and take the bottom stone first and put it in pile A, he has a pile of size 1, and then takes the top stone and puts it in pile B, he has a pile of size 1, and the product is 1, with sum of all products on the board is 1.   2) Inductive Hypothesis (Strong Induction): Suppose for some $k \geq 2$, $n$ stones can be split in $ 2 \leq n \leq k $ stones and $k-n$ stones.  3) Inductive Step:  Consider $n=k+1$. What do I do now?
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I am working on some of these types of problems for fun, just want to see a couple solved as examples.
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I am trying to solve this problem and looking for any tips. I have 5 D's, 6 E's and 3 F's and I have two conditions.  The first D must be before the first E.  The first E must be before the first F.  I thought I could solve this through determining the ways to make a string of letters with a D before an E before an F, however, that is not enough because it does not cover instances like DEFDEF... or DEFFD... since there can be letters that come after the initial string that meets the conditions.  I also contemplated using the Inclusion Exclusion principle as follows  cases where a D comes before an E + cases where an E comes before an F subtract their intersection (which I'm not sure how I'd calculate) however I still cannot account for cases where there are letters after the initial D, E, F...  Any ideas?
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Is there a benefit to using one over the other? In Python 2, they both seem to return the same results:  &gt;&gt;&gt; 6/3 2 &gt;&gt;&gt; 6//3 2
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1) How do we decide if we split into 2 subnodes, or more? Or is it always 2?  2) How do we decide what threshold is the cutoff? Specifically, you have a continuous variable, do you do a binary search, or how is the cutoff decided?
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I have started solving problems from the American Mathematical Monthly. While I will definitely continue this, I am not sure how much this will help for math graduate school admissions (assuming I submit my solutions and they are recognized by the Monthly). How significant would multiple solutions to Monthly problems be in admissions? What about having the published solution to a problem? One issue is that I don't know how difficult the problems are presumed to be: all I know is that I personally have solved at least one of them!
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Take it for granted that enumerations have secondly, …; thirdly, …; finally, … Is there any reason, except tradition, to prefer the traditional first to firstly?
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Given that Rubik's cube has finitely many positions, one possible "brute force" method to solve it would be to determine once a sequence of moves which eventually reaches every possible position of the cube, and then whenever you want to solve a cube, you just mindlessly follow the sequence until you eventually reach the solved cube. This strategy is absolutely failsafe, provided you have an extraordinary memory and enough time (so, it's not a strategy for humans, but maybe for bored gods waiting for their just created universe to finally develop intelligent life ;-)).  However, the question is: How many moves does the shortest possible sequence which visits every reachable position have? Of course, it's easy to give a lower bound: At least as many moves as there are reachable positions, which according to Wikipedia is $43\,252\,003\,274\,489\,856\,000 \approx 4{.}3 \cdot 10^{19}$. However, I doubt that there's a sequence which gives a not yet visited position at every move, so the actual number of moves is likely even higher.  Is there maybe even a constructive method to create this "bored god's algorithm" (so that even gods with less than stellar memory can apply it)?
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Being a foreign language speaker (or writer and reader to be more exact) I would prefer to see the changes in a side by side diff tool.  The current format with the two above each other doesn't make it easy to recognize the improvements.
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The problem is: Find $C_n$, the number of all ternary strings (length $n≥0$) that don't have consecutive 1's and 2's, using combinatorics.   Any tips or advices?
28836
I have created my own neural net which is using batch gradient descent. In other words, it trains on batches of examples all at once.  My issue is trying to figure out when to stop the training of the batch. I'll try to make things as understandable as possible since there are so many options, but please comment if you need clarification.  Each batch allows me to compute a list of output errors that correspond to the error the computer has on each example. Each error is a vector with elements equal to the number of output neurons.  Let each error vector be notated by e1, e2, e3, ... so the list of output errors is [e1, e2, ...]. Let an error vector in general be notated by e.  Now there are 2 things we can do to a vector or a list:   Average: average all of the elements in a vector or list. This will be notated by the function a(). For example to find the average error of an example we would do a(e) Check: make sure that each element in the vector or list is below a certain value. If any of the elements are too high, we continue to train on the entire batch. (This function can also be applied to a single value.) This will be notated by the function c(). For example, if we wanted to make sure the average error in e was low enough, we would do c(a(e)) Sum: add up all of the elements in the vector or list. This will be notated by the function s(). For example, s(e) would be totaling the error found in error vector e.   Here are the methods I have determined for deciding when to stop gradient descent:   Iterative: just run the training a certain number of times Average of averages: Find the average error of each example, then find the average of those and make sure it's below a certain amount. This is given by the formula c(a([a(e1), a(e2), ...])) Specific error: Check to make sure the error in each output neuron is below a certain amount. This is given by the formula c(e1), c(e2), ... Specific averages: Check to make sure the average error in each example is below a certain amount. This is given by the formula c([a(e1), a(e2), ...]) Sum of sums: Total up all the error in the entire batch and then make sure it's low enough. This is given by the formula c(s([s(e1), s(e2), ...]) Sum of averages: Find the average error of each example, then add up all those and make sure it's below a certain amount. This is given by the formula c(s([a(e1), a(e2), ...])) Average of sums: Compute the average total error in each example, and then make sure it's below a certain amount. This is given by the formula c(a([s(e1), s(e2), ...])) Specific sums: Check to make sure the total of the error in each example is below a certain amount. This is given by the formula c([s(e1), s(e2), ...])   So my question is this:  Which one of these options is most commonly used when training neural nets? Why? (Is there another option that I've missed?)  (Edit: I have found a very general version of this question , where the answers seem to say there is no specifically good way. If that is still true, please say so in your answer. Note that this is not a duplicate, as I explain the specific methods I would like to use.)  Edit: as A.D pointed out, summing is pretty useless because it depends on how many values are in each list/array/batch. And if you know how many examples there are, you can just use averaging. So, the methods involving "summing" have been cut.
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I just came across this question:      This tree (be) _____ planted by the settlers who (found) _____ our city over four hundred years ago.    I think "over four hundred years ago" is an unspecified time. My answer was that the two gaps require present perfect. So I was a little surprised with the answer that both of them require simple past. Why is this the case?
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So my partner and I thought it would be great if we could do some maths problems over Skype while we are apart (long distance). We'd like some recommendations on problem books that are ideal for working on together. We know there are quite a few questions online for best puzzle books in general but we'd like to know of any that have been written for or are suited for couple solving.   We are both fairly proficient in general but are not experts in any one topic. Ideally, the book should cover multiple fields with fairly difficult logic/maths problems.  As a starter list, what do people think of:   Raymond M. Smullyan - The Godelian puzzle book Peter Winkler - Mathematical puzzles a connoisseur's collection Raymond M. Smullyan - What is the name of this book?   I'm aware this is a rather strange request but thought it was worth asking just in case.
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I'm a little confused about the following sentences. Could you please tell me which one(s) is/are correct?        The first and the second chapter are difficult.   The first and the second chapters are difficult.   The first and second chapters are difficult.   The first and second chapter are difficult.      I believe #1 &amp; #3 are correct, while #2 &amp; #4 are incorrect.  But when seeing the answers to the Quora question , I wonder why they say #2 is correct.
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After playing a little with a Rubik cube I thought of the following problem :      Suppose we start with a solved Rubik cube (a general one , with $n^3$ cubes) . Then we choose one of the moves , each having a probability of $\frac{1}{6n}$ of being chosen , and apply it on our cube . We continue doing so ( choosing a new move randomly and then applying it and so on...) , until we reach the solved position again . What is the expected number of moves needed to solve the cube in this way ?   I thought about it and I think the answer should be $\infty$ (at least for $n \geq 3$) because there are a lot of random sequences of moves that will take a long time to solve the cube but I don't have a rigorous way to prove it.  For $n=2$ I'm not that sure if the answer should be finite or infinite (but I still tend to think it's infinite).  Thank you for your time to help me!