98 posts

## Nicely done, Channel 9 (posted by Dr Herbie).

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• brianbec said:
Sven Groot said:
*snip*

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

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.

• 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.

LOL. Awesome

Niners rule!
C

• 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!

• brianbec said:
Sven Groot said:
*snip*

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

How does WolframAlpha get to 9 so fast?
C

• 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)

• brianbec said:
Charles said:
*snip*

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)

Interesting. Man, I'm really liking WAlpha these days. Thanks for the info, sir. BTW,, maybe we need to start a new series on 9: Mathematical Computing  Game?  I can only imagine the brilliant solutions Niners will come up with if we could manage to provide quartely quizzes of mathematical beauty.

And, no, we won't use these for entrance gates to the forums

C

• Charles said:
brianbec said:
*snip*

How does WolframAlpha get to 9 so fast?
C

FWIW, I ran this in MM5 and I get told the product does not converge.

I based my input expression based on this write-up because I've lost the original captcha image Charles posted, so there's probably a mistake in there.

In[5]:=
\!\(∏\+\(k = 1\)\%∞\((\((k + 1)\)\^3\/\(1 + 3\/k\))\)\)

From In[5]:="Product does not converge. More…"

UPDATE:

I found the problem, I had (1+k)^3 rather than (1+k^-1)^3. When I put that in MM tells me it converges to 6, which given +3 results in 9.

Computation takes about 1500ms on this computer, an Athlon64 3200+ from late-2004.

• W3bbo said:
Charles said:
*snip*

FWIW, I ran this in MM5 and I get told the product does not converge.

I based my input expression based on this write-up because I've lost the original captcha image Charles posted, so there's probably a mistake in there.

In[5]:=
\!\(∏\+\(k = 1\)\%∞\((\((k + 1)\)\^3\/\(1 + 3\/k\))\)\)

From In[5]:="Product does not converge. More…"

UPDATE:

I found the problem, I had (1+k)^3 rather than (1+k^-1)^3. When I put that in MM tells me it converges to 6, which given +3 results in 9.

Computation takes about 1500ms on this computer, an Athlon64 3200+ from late-2004.

3+product_(k=1)^infinity(1+1/k)^3/(1+3/k)

C

• W3bbo said:
Charles said:
*snip*

FWIW, I ran this in MM5 and I get told the product does not converge.

I based my input expression based on this write-up because I've lost the original captcha image Charles posted, so there's probably a mistake in there.

In[5]:=
\!\(∏\+\(k = 1\)\%∞\((\((k + 1)\)\^3\/\(1 + 3\/k\))\)\)

From In[5]:="Product does not converge. More…"

UPDATE:

I found the problem, I had (1+k)^3 rather than (1+k^-1)^3. When I put that in MM tells me it converges to 6, which given +3 results in 9.

Computation takes about 1500ms on this computer, an Athlon64 3200+ from late-2004.

But how long does mathematica take if the upperbound on k is not infinite, but another arbitrary high number?

WolframAlpha won't give you an answer if you do that.

• Sven Groot said:
W3bbo said:
*snip*

But how long does mathematica take if the upperbound on k is not infinite, but another arbitrary high number?

WolframAlpha won't give you an answer if you do that.

I was sure it was 42. Dang.

• Sven Groot said:
W3bbo said:
*snip*

But how long does mathematica take if the upperbound on k is not infinite, but another arbitrary high number?

WolframAlpha won't give you an answer if you do that.

It will run until it gets an answer or user cancels, there's no limit on number lengths.

• Red5 said:
Sven Groot said:
*snip*

I was sure it was 42. Dang.

are those scenes from a movie, perhaps?

EDIT: nevermind

• bureX said:
Sven Groot said:
*snip*

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

If I may interject, I suggest we purchase the following gadget (maybe it's listed on ebay, eh?):

...and calculate the product convergence all the way to k=6.1871539834134776012651492465284 x 10^34 (the reciprocal value of Planck's length). Maybe we can halt the Folding@Home or the Seti@Home projects for a temporary period while we use the available CPU time for the calculation of the said problem. After all, this IS channel9, not channel 8.9999999999873293. I, for one, will not stand for this mathematical discrepancy. If anybody has watched the movie Pi, there is a magic number out there that can make all computers become self-aware, predict the functioning of the stock market and reveal god to the Jewish people (maybe it can even aid in the making of lumpless pudding*). If we all work together, maybe we can find this neatly secluded needle in the haystack... or maybe not... after all, I've got some lumpy pudding to tend to (it has just cooled down enough to please my picky taste buds)... Happy april fools day everybody
*Lumpless pudding is not actually sans lumps if there is at least one lump that surpasses Planck's length (all dimensions below Planck's length make "no physical sense", as it seems). Man... that must be some smooth pudding right there.

> this IS channel9, not channel 8.9999999999873293

In a perfect world yes, I'm sorry but this is channel 8.9999999999873293 my friend, and we are all hooked into the matrix

• Sven Groot said:
W3bbo said:
*snip*

But how long does mathematica take if the upperbound on k is not infinite, but another arbitrary high number?

WolframAlpha won't give you an answer if you do that.

Here's a CPU% plot of the time taken to compute the number for a k=1 to n for n=99999 and n=999999

(This is on an Athlon64 3200+)

I note that MM uses GMP for computation of actual numbers; if you do it to Infinity it computes the answer symbolically, which is a lot faster.

From the graph below, you can see that this isn't an O(n) operation: the 999999 takes longer than 10 times the 99999 calculation.

FWIW, 99999 computations gives 5.99982 and 999999 gives 5.99998.

• Charles said:
W3bbo said:
*snip*

3+product_(k=1)^infinity(1+1/k)^3/(1+3/k)

C

Yes, I love WAlpha too, but sometimes I have problems with the notetion (I wouldn't have know how to enter the formula before someone else did).

• Sven Groot said:
W3bbo said:
*snip*

But how long does mathematica take if the upperbound on k is not infinite, but another arbitrary high number?

WolframAlpha won't give you an answer if you do that.

Yes it does. Although it worked only up to 100000.

• giovanni said:
Sven Groot said:
*snip*

Yes it does. Although it worked only up to 100000.

Just to note that I 100% approve of the title change.

Nice work all of you.

Herbie