Told you!

Well, a solution of sorts is up to Derb's math problem.

There's no closed-form solution for the radius of the inner circle in the limit. Its numerical value is approximately 0.1149420448532, but using the product of cosines formula converges horribly slowly. However, using some evil jiggery-pokery and the product of terms -> sum of logs technique adumbrated above, you can get a fast converging series using, mirabile dictu, the Riemann zeta function.

It is a hair advanced for me. In retrospect, I took it as far as was reasonable for me. Just being a dropout, country boy, the likelihood of my taking it as far as the Zeta function was slim to none, with Slim having just left town.

For those of you who don't know John Derbyshire, I will recommend his book Prime Obsession. It is excellent if you like Math at all.

I am the proud owner of a signed copy. I met Mr. Derbyshire (DARB-uh-sure) at the Boulder Bookstore. Had any of them encountered his political writings, I am sure there would have been protests on Pearl Street. But his Math stuff is non-partisan.

Posted by jk at September 7, 2004 09:27 AMComments

I agree, I can say I have heard of d'Alembert and Riemann, but that is about it. Very cool problem though even if it is still a bit non-intuitive for my little brain. My favorite has always been Gauss' ability as a teenager in a math class to compute the sum of the numbers 1-100 in about 10 seconds.

Posted by: Silence Dogood at September 7, 2004 10:52 AMI should have given props to Silence, who emailed me the URL today. Thanks! You would LOVE "Prime Obsession," Silence. I think it would be right down your street.

Derb will post a solution on his website before the end of the month that will work it all the way through.

Eye of the hurricane, Johngalt, just taking the smallest breather from election news. We haven't turned into a math blog or anything. What's that about square concrete windows?

Posted by: jk at September 7, 2004 11:01 AMSumming numbers 1-100 is not that hard once you think about it.

1 + 99 = 100

2 + 98 = 100

3 + 97 = 100

etc etc...

dont forget to include 50 + 100 at the end..

I leave it to you an excercize.

Very clever Alex!

My "guess" at the answer is (49 x 100) plus 5000.

Posted by: johngalt at September 7, 2004 12:08 PMOOPS! plus 150.

Posted by: johngalt at September 7, 2004 12:09 PMUse 1+100, 2+99...down to 50+51 gives you 50 sets of 101 and thus 50x101=5050.

Posted by: Silence Dogood at September 7, 2004 03:01 PMAnother method worthy of praise. Thanks Silence.

Posted by: johngalt at September 7, 2004 03:30 PMOne of the great untaught arithmetic methods is breaking up a problem into smaller ones.

Sure we're taught it (easy example... think how you would work out 46*34 on paper, then "look" at what you're doing)... but we're not actively encouraged to do so.

it wasnt' until i stopped learning math, that I really learned how to do it... unfortunately it was too late!

damned public school.