←back to thread

Thomas E. Kurtz has died

(computerhistory.org)
618 points 1986 | 1 comments | | HN request time: 0s | source
Show context
peagreen ◴[] No.42144368[source]
Tom Kurtz and John Kemeny and BASIC changed my life, too. I wrote my first BASIC program in 1970 [0] and starting in high school the next year spent hours with the Model 33 Teletype in our school's computer room, programming in BASIC via the school district's HP-2000 time-sharing system. Ultimately I decided to go to Dartmouth because of their undergraduate computer philosophy. Any kind of computer access was a big deal back then, and being able to program really distinguished you compared to the rest of one's age cohort when it came to applying for grad. schools, jobs, etc. So I feel like I've been riding the crest of that early 1970s wave ever since, despite the explosion of skilled people in younger cohorts.

It was a remarkable and fleeting time. If I were 13 years old now, I don't know of a comparable skill that could so effortlessly propel a person forward.

[0] Here it is:

   10 LET N=5^2.5

   20 PRINT N

   30 END
The answer (55 and something) was a revelation. I didn't know about logarithms then, so the meaning of fractional exponents was a complete mystery. I had to ask my math teacher to make sense of the answer.
replies(1): >>42149822 #
fuzztester ◴[] No.42149822[source]
Being rusty at school math, I was going to ask you how to solve that, then decided to google it.

I typed "5 to the power of 2.5", and after a few clicks, got this (with 'math solver' shown in the search box:

https://www.google.com/search?q=5%5E(2.5)

There is a button to show solving steps.

There is also a Quora answer that shows two different ways of solving it, using properties of exponents.

Cool. Now I need to go brush up on my school math. :)

replies(1): >>42152643 #
peagreen ◴[] No.42152643[source]
Simplest way = 5 * 5 * sqrt(5). But logarithms offer a more general approach. The math comes back fast. :-)
replies(1): >>42169194 #
1. fuzztester ◴[] No.42169194{3}[source]
Thanks :)