iStation said:algorith said:*snip*OK!

Another one,

lim x->0 [{sin(9x)}/x] = ?

Nice one. Here's another

Find the first positive real value of t for which the following is zero

256 cos^9(pi t / 162) - 576 cos^7(pi t / 162) + 432 cos^5(pi t / 162) -120 cos^3(pi t/162) + 9 cos(pi t / 162)

]]>bureX said:There's a a^2 - ab ≠ 0 missing somewhere, I think

(by the way, I suck at math)

if a == b, then a^2 == ab and a^2-ab==0. Division by zero is not permitted here You can get any answer you want if you divide by zero, just as you can "prove" any proposition if you accept a contradiction, just as any statement about members of the empty set is logically true!

]]>Charles said:brianbec said:*snip*How does WolframAlpha get to 9 so fast?

C

They analyze the formula symbolically, look it up, and "know" it's factorial in disguise. They have a huge internal database of series and products -- tens of thousands.

(I was at Caltech when Wolfram started up the Mathematica project circa 1979 (it was called SMP, then) -- they've been keyboarding in mathematical knowledge for 30+ years, now; that's how they got such a huge body of info)

]]>Sven Groot said:brianbec said:*snip*My method was meant to be inefficient. I was deliberately going for the most over the top complex way to attack the problem. It was a response to bureX's suggestion of making a steampunk version.

Also I suck at math. Brute force all the way, baby.

I understand! I was just offering a potentially even more inefficient way to go about it!

]]>Sven Groot said:Sven Groot said:*snip*Some results with various maximum values of k:

Max. value of k Formula value Execution time 100,000,000 5.9999998200000059999997770956 15s

10,000,000,000 5.9999999982000000009369275168 208s 50,000,000,000 5.9999999996400000005856836804 1069s

This is done using decimal. If that's precise enough, and my code has no bugs, we can conclude the product converges quite slowly. Again, this was done on a cluster with 56 nodes (most of which aren't very fast, though). Done locally, it wouldn't taken me about 33 hours. Note that the 100,000,000 job (the first one) takes unnecessarily long because the system I'm using isn't optimized for short jobs.

There, I think I can now safely claim the most elaborate solution to this problem.

Take the log of the big product. Each term looks like [ n log(k+1) + (1 - n) log(k) - log(k + n) ], (n is 3, in our example) and the big product is replaced by a big sum. Now investigate the convergence rate of the big sum, taking a look at http://en.wikipedia.org/wiki/Rate_of_convergence

]]>Richard.Hein said:

A+

]]>CreamFilling512 said:I put the second problem into Wolfram alpha:

http://www.wolframalpha.com/input/?i=3+%2B+(product+((1+%2B+1/k)^3)/(1+%2B+3/k)+k%3D1+to+infinity)">http://www.wolframalpha.com/input/?i=3+%2B+(product+((1+%2B+1/k)^3)/(1+%2B+3/k)+k%3D1+to+infinity)

Now, for a little secret: type the same thing into Wolfram Alpha, just replacing "3" with "n" (and, for even more secrets, leave off the first "n + " -- just type in the big infinite product

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