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 E. E. Escultura on Turing Crackpottery!
 E. E. Escultura on Turing Crackpottery!
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 E. E. Escultura on E. E. Escultura and the Field Axioms
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Hi Mark,
I'm a guy from China, I love your site, and have learnt a lot from it, that's a great blog, I spread your blog with my colleagues, and they all love it and keep it in the bookmark and visit it at times.
Would you please the "Select Category" function? Seems that it doesn't work on your new homepage, whatever I select from the popup list, either "Haskell" or "Category Theory", they all show me nothing, that's the reason for me, maybe for lots or other guys, still visit your old blog from scienceblogs π
Looking forward to reading more great math/computer science articles from you.
We would support you always, just as we have done in the last few years!
Best Regards,
Harry
Bees and NPHard problem:
http://www.qmul.ac.uk/media/news/items/se/38864.html
My gut says that the PR is overstating the paper. Possible article?
Hi, Mark,
Watching Jeopardy last night, I saw that one of the IBM scientists involved with the Watson project, or perhaps a spokesperson for IBM in general, was your wife!
Niw, I don't know you or her, so this is a bit of an assumpton, and perhaps presumption, but I do remember you discussing on your blog, how you came to select your names. I became so excited seeing your (and my) surname on the TV, that I didn't actually hear what she had to say. I was too busy explaining to my wife my (minimal) connection to you.
It's been a while since I read this blog, but I recall commenting once, a few years ago, using the moniker, "Bob from Yorktown Hts" or some such. Another minimal connection  I don't work for IBM, but have visited the Watson Research Center as a neighbor (and sometime research chemist). My neighbor and friend Vishnu Patel acted as guide.
It was fun seeing the structure of Watson on Jeopardy. I was a bit surprised to hear that the memory of that massive computer was only about 15 terabytes, if I heard correctly. Of course this is a lot of memory, but within reach for a PC user... It's amazing how cheap memory storage has become since the advent of the original IBM Winchester drive.
I'm rooting for Ken Jennings. It was interesting to note that the contest (the first day, anyway,) was not a runaway success by Watson.
After the show, I described the Turing test to my wife, opining that Watson could pass this test, or at least give a good account of itself!
I'll be reading your site more regularly again. Math is a better recreation than just about any other intellectual activity.
Bob Carroll
Code reviews are probably implicitly discouraged in undergraduate classes, in part because grades are meant to represent individual achievement. And I haven't heard much about code reviews in software engineering classes where people are grouped together. Perhaps the closest thing students get to a code review is that from their professor.
Hey Mark,
I saw your site while trying to do some numerology debunking. Any thoughts about karmic debt numbers? It looks like scare tactics to me but I wanted a real mathematical opinion on this.
Thanks,
 Alex
SQUARE β CIRCULAR ARC β CIRCLE
Arc represents the relationship between the circumference of the circle and the
perimeter on the subscribed squares the circle.
Expressed as a formula:
Arc = C : P
Arc β circular arc
C β circumference of the circle
P β perimeter on the subscribed squares the circle
Arc is also a relationship between the value of the approximation and the second
root from eight.
Expressed as a formula:
Arc = Ap : β8
Ap β approximation
From this it follows that the value of the circular arch as constants, in addition to the value of the approximation can be used to calculate the value of the c
METHOD OF CALCULATION THE CIRCUMFERENCE
OF THE CIRCLE WITH THE CIRCULAR ARC
a * arc = a1
P = 4 * a1 = 2r * Ap = C
a β side on the subscribed squares the circle
a1 β side on the squares with same perimeter as circle
P β perimeter on the squares
C β circumference of the circle
D * arc = D1
P = D1 * β8 = 2r * Ap = C
D β diagonal on the subscribed squares the circle
D1 β diagonal on the squares with same perimeter as circle
METHOD OF CALCULATING THE AREA OF A CIRCLE
WITH A CIRCULAR ARC
a * arc = a1 D * arc = D1
Area rectangle = a * D1 = (r * r) * Ap = Area circle
Area rectangle = a1 * D = (r * r) * Ap = Area circle
a β side on the subscribed squares on the circle
a1 β side on the squares with same perimeter as circle
D β diagonal on the subscribed squares on the circle
D1 β diagonal on the squares with same perimeter as circle
This method of calculating the circumference and area of the circle can beapplied in all approximations, but geometrically, we can only draw an approximation of the results, the following value (3,142696805...)
Ap = β8 * 10/9 = 3,142696805.. or Ap = β8 + (β8/9) = 3,142696805
hi, i thought you might be interested in this crock
http://vortexmath.webs.com/
im sorry lily you didn't invent the Rodin coil . Marko Rodin did . and it works .unlike the useless whining from your comment.
Hi Mark, here's another one for the doubtless growing pile of cracked pottery.
http://existics101.com/
Greetings,
I'm not calling this a crock... however, something just doesn't seem correct about this and I can't put my finger on in.
http://phys.org/news/201308cosmologistuniverse.html
There's some pretty good bad math and physics here:
http://www.wildheretic.com/