Dr Herbie

EDIT: I (Charles) changed the title of this post.

Please tell me that your little game didn't keep anybody from non-stop slaving away on Rev9?

Ion Todirel said:

Can't believe it was a joke... I'm going to find you Charles

I really did not get that I found the answer in the JS file

mastermine said:

Ion Todirel said:

*snip*

I really did not get that I found the answer in the JS file

I didn't bother to look, I was like "knowing the C9 devs they probably did server side validation"

Ion Todirel said:

mastermine said:

*snip*

I didn't bother to look, I was like "knowing the C9 devs they probably did server side validation"

I have been up so long i forgot what day it was

I had to make an Excell sheet an having k going from 1 to 50, I finally got it (though I had problems with decimals)!

I still have no clue what the first equation means though....

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)

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)

That was easy! Your solution is much faster than mine! Excel is not the best tool for this kind of problems, but I didn't have anything handy at the office.

Still not sure what the first equation meant...

giovanni said:

I try a couple of educated guesses, but not 9!

 

I had to make an Excell sheet an having k going from 1 to 50, I finally got it (though I had problems with decimals)!

 

I still have no clue what the first equation means though....

It's the standard Church encoding of the natural numbers into pure lambda calculus.  If you've been exposed to it it's definitely easier than the other problem, since you just need to be familiar with the encoding, not actually solve anything.   =)

La Bomba said:

Nicely done.

My first answer way 5 then 8 then I hit 9... took me 10mins....  very funny!      Was beginning to feel very stupid

La Bomba said:

Nicely done.

double post

Kryptos said:

La Bomba said:

*snip*

My first answer way 5 then 8 then I hit 9... took me 10mins....  very funny!      Was beginning to feel very stupid

Yes, my first guesses were actually lower than 9 because I thought that k at the denominator would drive the multyplying factor to 1 much faster.

contextfree said:

giovanni said:

*snip*

It's the standard Church encoding of the natural numbers into pure lambda calculus.  If you've been exposed to it it's definitely easier than the other problem, since you just need to be familiar with the encoding, not actually solve anything.   =)

If I have ever been exposed to it, my memory definitely removed it very well.

Thanks for the link, very instructive!

Edit: That was brilliant Charles

giovanni said:

contextfree said:

*snip*

If I have ever been exposed to it, my memory definitely removed it very well.

 

Thanks for the link, very instructive!

 

 

Edit: That was brilliant Charles

It broke the CoffeeHouse for me

I can only navigate to the last three threads via the menu on the right side

Maddus Mattus said:

giovanni said:

*snip*

It broke the CoffeeHouse for me

 

I can only navigate to the last three threads via the menu on the right side

Works on my machine

vesuvius said:

Maddus Mattus said:

*snip*

Works on my machine

Thanks!

Where is that machine located?

I'll come over and post from there

Nice.

http://en.wikipedia.org/wiki/Church_numerals

Sven Groot said:

I don't know how to solve those formulas...

I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it

W3bbo said:

Sven Groot said:

*snip*

I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

 

I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

 

Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it

The next captcha should be "Within the context of computational complexity theory, prove or disprove P=NP."

Bass said:

W3bbo said:

*snip*

The next captcha should be "Within the context of computational complexity theory, prove or disprove P=NP."

Or you could bypass the filter and go directly into the topic via bing and find out the answer.  

W3bbo said:

Sven Groot said:

*snip*

I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

 

I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

 

Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it

Yeah, I can't reach any of the forum pages due to this...

The prompt comes up, then IE complains that it cant reach the website

I was able to go to the homepage and get into this thread from there.

W3bbo said:

Sven Groot said:

*snip*

I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

 

I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

 

Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it

Component factory have something similar, you are asked

Answer the question to prove you are not a spam bot, but a human.

Binary number 1001 is what in decimal?

or

how many fingers on a human hand?

Obviously Phil used to lurk around the channel 9 forums a lot so I wonder why the 1001 binary? The long and the short of it is spam in his fora is now almost non-existent, whereas it is a growing problem in the Sandquist fora. Look at the mix comments and here on channel 9, a lot of the posts are begining to get targeted more and more.

kettch said:

Fun stuff.

 

Please tell me that your little game didn't keep anybody from non-stop slaving away on Rev9?

 

This is all I've been doing for months.

Rev9 was an elaborate cover for this new feature.

ManipUni said:

I just solved it.

I got it on my first guess, like a real scientist  

(Second guess was going to be 42).

Herbie

Dr Herbie said:

ManipUni said:

*snip*

I got it on my first guess, like a real scientist  

 

(Second guess was going to be 42).

 

Herbie

Ditto

ScanIAm said:

W3bbo said:

*snip*

Yeah, I can't reach any of the forum pages due to this...

 

The prompt comes up, then IE complains that it cant reach the website

 

I was able to go to the homepage and get into this thread from there.

Bash the stop button, fill in 9, then no more worries

To be clear, I had little to do with this clever prank. Dan and Duncan came up with the basic idea, I recruited the best two people in the world IMO, Erik and Brian, to come up with the complex expressions used in the gate (Erik provided the Church numeral encoding(nice job, W3bbo!), Brian the beautiful mathematical expression, of course). Geoff, a recent stellar addition to our dev team, implemented this in record time. Rock star.

So, nicely done Dan, Duncan, Geoff, Erik and Brian.

Happy April Fools,

C

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

giovanni said:

CreamFilling512 said:

*snip*

That was easy! Your solution is much faster than mine! Excel is not the best tool for this kind of problems, but I didn't have anything handy at the office.

 

Still not sure what the first equation meant...

This is my naive solution to the second problem, it approaches 9 as k approaches infinity.

static void Main(string[] args) { 
    Func<double, double> f = (k) => (Math.Pow((1 + (1 / k)), 3)) / (1 + (3 / k)); 
    double z = f(1); 
    for (double k = 2; k < double.MaxValue; k++) { 
        z *= f(k);
        Console.WriteLine(3 + z);
    } 
}

giovanni said:

CreamFilling512 said:

*snip*

That was easy! Your solution is much faster than mine! Excel is not the best tool for this kind of problems, but I didn't have anything handy at the office.

 

Still not sure what the first equation meant...

@giovanni -- the first formula was the easiest: just count the f's! 

brianbec said:

CreamFilling512 said:

*snip*

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

Awesome! I have to say, Wolfram Alpha is rocking and rolling, man.

C

brianbec said:

CreamFilling512 said:

*snip*

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

The second problem is n + the gamma function, Γ(n+1), with n = 3.

Here's a graph of the absolute value of the Gamma function on the complex plane, from Wikipedia:

JoshuaAllen said:

giovanni said:

*snip*

@giovanni -- the first formula was the easiest: just count the f's! 

Nice job, Joshua.
C

daSmirnov said:

I viewed the page source, got the URL for this thread.... Nice and simple.

Also a clever solution.
C

W3bbo said:

Sven Groot said:

*snip*

I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

 

I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

 

Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it

Actually that was 3+product()

JoshuaAllen said:

giovanni said:

*snip*

@giovanni -- the first formula was the easiest: just count the f's! 

Yes, it is easy once you know what the notation mean

brianbec said:

CreamFilling512 said:

*snip*

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

Try WolframAlpha on the Church numeral encoding  λf.λx.f(f(f(f(f(f(f(f(f(x))))))))))

The computed result isn't correct.

C

Before that, I tried just entering random things a couple of times to see if it would eventually let you in. After 9 times of entering bad answers you get a subtle hint.

Richard.Hein said:

brianbec said:

*snip*

The second problem is n + the gamma function, Γ(n+1), with n = 3.

 

Here's a graph of the absolute value of the Gamma function on the complex plane, from Wikipedia:

 

 

A+

Richard.Hein said:

giovanni said:

*snip*

This is my naive solution to the second problem, it approaches 9 as k approaches infinity.

 

static void Main(string[] args) { 
    Func<double, double> f = (k) => (Math.Pow((1 + (1 / k)), 3)) / (1 + (3 / k)); 
    double z = f(1); 
    for (double k = 2; k < double.MaxValue; k++) { 
        z *= f(k);
        Console.WriteLine(3 + z);
    } 
}

I did it with PowerShell.  Anyway, way to go Channel9!

Ion Todirel said:

mastermine said:

*snip*

I didn't bother to look, I was like "knowing the C9 devs they probably did server side validation"

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

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

Although I solved it by counting the number of f's in the first formula, then realizing that the number '9' was special to C9.

sushovande said:

Ion Todirel said:

*snip*

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

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

Although I solved it by counting the number of f's in the first formula, then realizing that the number '9' was special to C9.

Great job!

C

Charles said:

sushovande said:

*snip*

Great job!

C

Nice job, Charles

Charles said:

Nicely done, Niners!

 

To be clear, I had little to do with this clever prank. Dan and Duncan came up with the basic idea, I recruited the best two people in the world IMO, Erik and Brian, to come up with the complex expressions used in the gate (Erik provided the Church numeral encoding(nice job, W3bbo!), Brian the beautiful mathematical expression, of course). Geoff, a recent stellar addition to our dev team, implemented this in record time. Rock star.

 

So, nicely done Dan, Duncan, Geoff, Erik and Brian.

 

Happy April Fools,

 

C

Thanks Charles

This was a fun one. I like to make sure that the 'truth' is always somewhere in the source

Geoffreyk said:

Charles said:

*snip*

Thanks Charles

 

This was a fun one. I like to make sure that the 'truth' is always somewhere in the source

My blood started to boil and I was writing a draft email full of abuse in my head!

"Ive been coming to the coffeehouse for six years and......!!!!"

Got me!!!

Charles said:

sushovande said:

*snip*

Great job!

C

Kudos to everyone who figured it out, figure out a way to find out, and even brute forced through the js.

LarryLarsen said:

Charles said:

*snip*

Kudos to everyone who figured it out, figure out a way to find out, and even brute forced through the js.

,,i couldnt get in! ... i tried a few things... then i went - wait... it cant be hard...  ahhh    * 9! *

f^7 f(x) (fλ).(xλ).f

?

SlackmasterK said:

Am I the only one who got:

 

f^7 f(x) (fλ).(xλ).f

 

?

I'm all for april fools, but this makes it so you can't enter the forums or use the site. Don't see the point of it.

I had to use the equation editor in word to figure out what the sign meant.  Once I found out it stands for product, it was just a matter of putting it in excel.  Thanks for the entertainment and challenge.

jamie said:

LarryLarsen said:

*snip*

,,i couldnt get in! ... i tried a few things... then i went - wait... it cant be hard...  ahhh    * 9! *

Nice work, Jamie!!
C

And then I...

But then...

And suddenly...

And then I ended up here... Now what!?

All in all, my algebra teacher would probably scalp me and go with "Y cannot be 9, Y can only be formulated as Y->9"...

bureX said:

First I...

 

And then I...

 

But then...

 

 

And suddenly...

 

 

And then I ended up here... Now what!?

All in all, my algebra teacher would probably scalp me and go with "Y cannot be 9, Y can only be formulated as Y->9"...

Wow. Did you also program your FPGA to try to compute it for you?

Bass said:

bureX said:

*snip*

Wow. Did you also program your FPGA to try to compute it for you?

We'll make a steampunk version eventually 

bureX said:

Bass said:

*snip*

We'll make a steampunk version eventually 

I tried computing the value of the product using the highest value of k possible to see how fast it converges, using a 56 node cluster (I know there are better ways to compute something like this; doing it in a hopelessly convoluted way was kind of the point here). The problem is, doubles aren't precise enough, and Math.Pow doesn't support decimals. Does anyone know a good, fast, high precision floating point library for .Net?

EDIT: Worked around the lack of Math.Pow by simply multiplying the value three times. Definitely more precise than double, but also a lot slower. Even on these 56 nodes it takes me about 3.5 minutes to calculate for k = 1 through 10,000,000,000. The same scenario with double takes only 21 seconds, but yields a clearly different result.

Mind you, doing the same thing on my local machine would've taken roughly 7 hours.

Sven Groot said:

bureX said:

*snip*

I tried computing the value of the product using the highest value of k possible to see how fast it converges, using a 56 node cluster (I know there are better ways to compute something like this; doing it in a hopelessly convoluted way was kind of the point here). The problem is, doubles aren't precise enough, and Math.Pow doesn't support decimals. Does anyone know a good, fast, high precision floating point library for .Net?

 

EDIT: Worked around the lack of Math.Pow by simply multiplying the value three times. Definitely more precise than double, but also a lot slower. Even on these 56 nodes it takes me about 3.5 minutes to calculate for k = 1 through 10,000,000,000. The same scenario with double takes only 21 seconds, but yields a clearly different result.

 

Mind you, doing the same thing on my local machine would've taken roughly 7 hours.

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.

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.

Awesome, in our office we did joke about creating a 1000 node azure cluster to do it but it was luch time and there was curry calling

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.

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

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 

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

Dr Herbie said:

giovanni said:

*snip*

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

 

Nice work all of you.

 

 

Herbie

 

Thanks

C

Dr Herbie said:

giovanni said:

*snip*

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

 

Nice work all of you.

 

 

Herbie

 

I'm glad the math geeks liked it...but the rest of us couldn't get in

Harlequin said:

Dr Herbie said:

*snip*

I'm glad the math geeks liked it...but the rest of us couldn't get in

I am certainly not a maths geek (tried to learn calculus 3 times so far, still don't get it), I thought it was an obvious guess.

Herbie

Dr Herbie said:

Harlequin said:

*snip*

I am certainly not a maths geek (tried to learn calculus 3 times so far, still don't get it), I thought it was an obvious guess.

 

Herbie

 

Pretty sure I put 9 in at one point...could have been  a bug

Point is April Fools or not, it had the ability to shut down the site for some for an entire day....not a good thing.

Dr Herbie said:

Harlequin said:

*snip*

I am certainly not a maths geek (tried to learn calculus 3 times so far, still don't get it), I thought it was an obvious guess.

 

Herbie

 

It's really worth learning. Not so much how to solve integrals by hand (Mathematica is better then you'll ever be), but at least understanding the fundamental theorem at a philosophical level. Are there specific parts of Calculus you are having difficulty understanding?

Dovella said:

Watching this movie right now (for the nth time...)
C

Harlequin said:

Dr Herbie said:

*snip*

I'm glad the math geeks liked it...but the rest of us couldn't get in

Jamie did!

When all else fails, just guess 9. It's a very special number..

C

Dr Herbie said:

Harlequin said:

*snip*

I am certainly not a maths geek (tried to learn calculus 3 times so far, still don't get it), I thought it was an obvious guess.

 

Herbie

 

And you were right! The beautiful thing in all of this is that you didn't even need math to "solve" the complex mathematical expression. Well done!
C