brilliant.com

I've been getting lazy in this July heat!  Even doing one thing takes an effort just now.

Today was Reading Out Loud.  There were nine people, more than usual.  I read some of the character descriptions at the start of Chaucer's Canterbury Tales and Orwell's essay "Decline of the English Murder." Jane read Beatrix Potter's Peter Rabbit and a couple of Hamlet's soliloquies. (I borrowed her book and read Polonius' speech to his son!) Some other girls read Coleridge's "Kublai Khan" and Tennyson's "The Lady of Shallot."

I have a Facebook app for the website brilliant.com , which sends me a challenging puzzle every day.  There was one involving Fibonacci series, where the first two entries are 1 and the later ones the sum of the previous two:  1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597... This puzzle asked what the sum would be for 1+1/2+2/4+3/8+5/16+8/32+... I figured out that it must be 4. That's because this sum equals 1+1/2+(1+1)/4+(2+1)/8+(3+2)/16+... which equals 1+(1/2+1/4)+(1/4+1/8)+(2/8+2/16)+... which equals 1+ (3/4)*(1+1/2+2/4+3/8+...) (I never look up the answer:  either I know it or I don't.)

Just the other night there was another question where we know that a+b=a*b=3, so what does a^3+b^3 equal?  You can figure out the answer through simple algebra: a^3+b^3=(a+b)^3-3*a*b^2-3*b*a^2=(a+b)^3- 3*(a*b)*(a+b)=27-27=0.  

But I like to do things the hard way, so I imagined the complex numbers that a and be could be.  I figured out that a would be 3/2 +i*[3^(1/2)/2] while b would be 3/2 - i*[3^(1/2)/2] (or vice versa!). Alternatively, you could say that a= 3^(1/2)*e^[i*pi/6] and b=3^(1/2)*e^[i*-pi/6].  And I noticed some other interesting things:  a^2+b^2=3, a^4+b^4=-9, a^6+b^6=-54, a^8+b^8=-81, a^10+b^10=243, a^12+b^12=1458!

That's the sort of puzzle that keeps me awake late at night!  Imagine the puzzle where a+b=a*b=1...