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Math one-liners from YouTube

Posted 4 months ago
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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
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.

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.

Problem#2.

2nd problem in this collective thread, see the "What would make this equation true??"-video: How do we express y to be odd?

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 4 months 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}} *)

@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 4 months 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?

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