# Message Boards

Posted 2 months ago
1681 Views
|
8 Replies
|
10 Total Likes
|
 There are some daily Youtubers who write out their symbolic solutions to a short straight-forward (non-quiz like type) math problem. Oftentimes Mathematica can solve those problems too. Maybe we could share here the ones which our software cannot solve easily?Problem#1. Let me go first. In the "Putnam Exam 2004 | B5" video the youtuber claims that $L = \frac{2}{e}$ $$L=\lim_{x\to 1^-} \prod _{n=0}^{\infty } \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}=\ldots =\frac{2}{e}$$Imho the Wolfram L code for this expression should be: Limit[Product[((1 + x^(n + 1))/(1 + x^n))^(x^n), {n, 0, Infinity}], x -> 1, Direction -> "FromBelow"] However, Mathematica computes forever without returning a result, i must abort the computation. Maybe my code line is wrong, can you get the desired result?Obviously this (hopefully collective) thread (feel free to memorize the short-URL, if you can't follow/subscribe to the thread for updates) is nothing urgent. It is entertaining to let Mathematica have a crack at such symbolic math problems which get ten thousands of views within a week. Thanks for taking an interest!
8 Replies
Sort By:
Posted 1 month ago
 Table[NProduct[((1 + x^(n + 1))/(1 + x^n))^(x^n), {n, 0, Infinity}, WorkingPrecision -> 15], {x, .99999999, 1, .000000001}] looks like it goes to 2/e.
Posted 1 month ago
 To give you a hint to look for something over e, Limit[Product[((x^(n + 1))/(x^n))^(x^n), {n, 0, Infinity}], x -> 1, Direction -> "FromBelow"] quickly gives 1/e.
Posted 1 month ago
 Problem#2. 2nd problem in this collective thread, see the "What would make this equation true??"-video: Commenting the OddQ[] part outputs all solutions, from which one could pick the odd y solution, namely y=57: In[1]:= FindInstance[{(*OddQ[y] == True,*) 0 <= x <= 100, 0 <= y <= 100, 1/x + 4/y == 1/12}, {x, y}, Integers, 100] Out[1]= {{x -> 24, y -> 96}, {x -> 28, y -> 84}, {x -> 30, y -> 80}, {x -> 36, y -> 72}, {x -> 44, y -> 66}, {x -> 48, y -> 64}, {x -> 60, y -> 60}, {x -> 76, y -> 57}, {x -> 84, y -> 56}} But when I uncomment in the above code, there is no result. Basically I am asking: How do we define in set of equations a variable to assume odd integer values only, for use in mathematical expressions or equations within Solve[], Reduce[], Simplify[], FindInstance[]etc.?
Posted 1 month ago
 Hi Raspi,Seems like FindInstance requires numerical equations, booleans do not work. This works FindInstance[{0 <= x <= 100, 0 <= y <= 100, 1/x + 4/y == 1/12, Mod[y, 2] == 1}, {x, y}, Integers, 100] (* {{x -> 76, y -> 57}} *) 
Posted 1 month ago
 @Mike Thanks for your post and workaround code! I absolutely understand what you mean with generating the full set and then filtering the solution. I used this approach a lot in the stochastics text project which I was working on for 7.0 months (completed in mid April, some 800-1000 pages total). Looks like our legendary Rohit has beaten us to it, I will try to remember the trick of defining an odd number through the Mod function, thank you @Rohit, works like a charm! See you guys in the next problem :-)
Posted 1 month ago
 Raspi:On the 2nd problem.In my day job writing and optimizing SQL, I run into interesting situations like this all the time. The direct and logical approach just doesn't work, or doesn't work in a reasonable amount of time. In those situations, I have found that I can sometimes get what I need by first getting more than I need and then filtering, What amazes me most is that sometimes I can get what I need sub-second using the round-about approach when the direct and logical approach can run for hours.Using this indirect approach, I think this solves this problem:  Select[ FindInstance[{(*OddQ[y]==True,*)0 <= x <= 100, 0 <= y <= 100, 1/x + 4/y == 1/12}, {x, y}, Integers, 100], OddQ[#[[2]][[2]]] & ] Of course, I would agree with anyone who said that this approach is cheating. But, it has helped me overcome challenges that others gave up on.
 Problem#3.In the "An interesting integral with the floor function."-video the youtuber claims that the value of the integral is $I=1-2 \ln (2)$, which is a negative value: $$I=\int_0^1 (-1)^{\left\lfloor \frac{1}{x}\right\rfloor } \ dx=\ldots =1-2 \ln (2)\approx -0.386294$$ Imho the Wolfram L code for this expression should be: \[CapitalIota] = Integrate[(-1)^Floor[1/x], {x, 0, 1}] (* //N out=0.4375 *) However, Mathematica outputs the integral uncomputed and the NIntegrate[] function returns the positive value $0.4375$. WolframAlpha outputs $0.4375$ for the value of the integral, too. Did I make a user error or is our software indeed wrong?