Integration is essentially multiplication of a function with an infinitesimal quantity, whereas differentiation is division of the function's change with an infinitesimal quantity.
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travis
United States10 Posts
the pattern is as follows n: 1, 3, 7, 15, 31, 63 i: 2, 4, 6, 8, 10, 12 I need a formula for i as a function of n I am not sure how to do it. I understand how n grows and I understand how i grows and I can relate them positionally in a list but I don't know how to write i as a function of n I know we have some clever people at TL... here is a chance to show off  
hypercube
Hungary3 Posts
On June 15 2017 05:13 travis wrote: I am looking for a relation the pattern is as follows n: 1, 3, 7, 15, 31, 63 i: 2, 4, 6, 8, 10, 12 I need a formula for i as a function of n I am not sure how to do it. I understand how n grows and I understand how i grows and I can relate them positionally in a list but I don't know how to write i as a function of n I know we have some clever people at TL... here is a chance to show off Introduce a new variable k and write n(k) and i(k). Now find the inverse, i.e. k(n). Substitute to get i(k(n)), or i(n). edit: for example you could just say k is the position in the list, so you would have n = 2^k  1 and i = 2k  
Shalashaska_123
United States0 Posts
I'm gonna call the first and second sequences of numbers you listed a_n and b_n, respectively. We can represent them as follows. Solve the second equation for n. Substitute it into the first equation and solve for b_n. Therefore, the final answer is this. EDIT: Fixed.  
JimmyJRaynor
Canada5 Posts
On June 15 2017 04:06 Shalashaska_123 wrote: Hi, JimmyJRaynor. Integration is essentially multiplication of a function with an infinitesimal quantity, whereas differentiation is division of the function's change with an infinitesimal quantity. that's pretty good. i like that one. thx for posting.  
fishjie
United States0 Posts
Shannon Entropy  we also used this in Machine Learning when we had to code a decision tree learner. http://homes.cs.washington.edu/~jrl/teaching/cse599swi16/notes/lecture1.pdf Spectral graph https://courses.cs.washington.edu/courses/cse521/16sp/521lecture12.pdf Spectral algorithm for clustering https://courses.cs.washington.edu/courses/cse521/16sp/521lecture11.pdf  
travis
United States10 Posts
Lets say I have a list of n elements. I analyze my elements, reduce the amount of elements by 1, then cut the elements in half. Now I have 2 lists of elements, 1 of size (n1)/2 and the other of size n/2. I do this over and over until i am left with many lists of 2 elements (and lists of 1 elements.. but for this question we don't consider those). So the question is, how many lists did we go through in total? where n was a list at the top, n/2 and n1/2 were lists, ((n/2)1)/2 and (n/2)/2 were lists, etc etc all the way down to our lists of 2. Lists of 1, left over, do not count.  
Acrofales
Spain355 Posts
100 2*50 4* 25 8*12 (and 4*1 which we discard) 16*6 32*3 32*2 (and 32*1 which we discard) So 95 in total? If so, approximately n/2 + n/4 + ... ~= n A precise answer is harder, because you need to know how many times you end up with an odd number of elements in your list. There's almost certainly a numerical way of figuring that out, but I'm lazy right now.  
hypercube
Hungary3 Posts
On June 16 2017 08:34 DanielReeLee wrote: Hello I'm looking for some problem sets for multivariable and vector calculus. Could I have some recommendations Check out the multivariable calculus course on MIT OCW. http://ocw.mit.edu/courses/mathematics/1802scmultivariablecalculusfall2010/ There's a short problem set at the end of each chapter, as well as a longer list of supplemental problems for all 4 major chapters (with solutions).  
travis
United States10 Posts
On June 17 2017 04:38 Acrofales wrote: Not sure I understand the question. Would this be how you do it for n=100: 100 2*50 4* 25 8*12 (and 4*1 which we discard) 16*6 32*3 32*2 (and 32*1 which we discard) So 95 in total? If so, approximately n/2 + n/4 + ... ~= n A precise answer is harder, because you need to know how many times you end up with an odd number of elements in your list. There's almost certainly a numerical way of figuring that out, but I'm lazy right now. well, for n = 100 100 50 , 49 25, 24 24, 24 12, 12, 11, 12, 11, 12, 11, 12 5, 6, 5, 6, 5, 5, 5, 6, 5, 5, 5, 6, 5, 5, 5, 6 2, 2,  oh god there is a lot, u get the idea it would stop at all 1s and 2s but the 1s don't count and what I want is the count of EVERY list of length > 1, including the original list and the lists in every step hmm yeah that's true what I really want is... uh.. the floor of (n1)/2 .. I think. see above, lol I expect the best way to solve this is to represent it with sums and then simplify them but I am not good enough  
Acrofales
Spain355 Posts
On June 17 2017 04:54 travis wrote: well, for n = 100 100 50 , 49 25, 24 24, 24 12, 12, 11, 12, 11, 12, 11, 12 5, 6, 5, 6, 5, 5, 5, 6, 5, 5, 5, 6, 5, 5, 5, 6 2, 2,  oh god there is a lot, u get the idea it would stop at all 1s and 2s but the 1s don't count and what I want is the count of EVERY list of length > 1, including the original list and the lists in every step hmm yeah that's true what I really want is... uh.. the floor of (n1)/2 .. I think. see above, lol I expect the best way to solve this is to represent it with sums and then simplify them but I am not good enough Oh, ok. That seems incomplete. The 6s would expand to 3, 2, right? And what would then happen? It ends there? Seems like a weird algorithm. I thought I had a simple solution, but it breaks if any list in your subdivisions has length equal to a power of 2 (that adds 1, which can occur at different points in the tree). So doesn't work, and given the weirdness of your algorithm, I'm not sure there's an easy way of figuring out how often you'll run into a power of 2.  
Simberto
Germany13 Posts
On June 17 2017 04:53 Amanebak wrote: Hey. I stumbled across a problem like this: What is the longest gondola that can take a rightangle turn of a Venetian canal? The width of the canal is 2, respectively 3 length units before, resp. after the turn. I apologize for my English. Unless i missed something, the result is 5*sqrt(2). The gondola needs to be able fit into the diagonal of the (2+3)*(2+3) square, which has a length of 5 sqrt (2) At that point, 2sqrt(2) of the gondola is in the thinner canal, and 3sqrt(2) is in the thicker canal. It is obvious that no longer gondola could reach as far into the 3m canal if you draw a picture of the situation, and the 5sqrt(2) gondola can continue onwards from this point on.  
 
