Another formula and other stuff
The latest formula discovered by our friend is
N^k - (N-1)^k = Sum_r=0^k-1 N^r (N-1)^{k-1-r}
I asked him how he found it, he said that he had observed this for small k when he was young (well he is 8 now!), and generalized it to all k recently.
He has also started learning about logarithms. I asked him for the value of log 2.5. His first question was -- to what base? I said 10. After a few seconds he said 0.4. I assumed that he made a guess, he probably knew that log 2 = 0.3, and log 3 is close to 0.5. After a few minutes I asked him how he got it --- he said that he knew that 4^5 = 1024, so 5 log 4 = 3, so log 4 = 0.6. Now log 2.5 = log 10 - log 4 = 0.4. Pretty cool! I asked him for logs of some more numbers, each time he came up with a different way to evaluate it, so he not only knows what a log is, but understands it at a deep conceptual level.
N^k - (N-1)^k = Sum_r=0^k-1 N^r (N-1)^{k-1-r}
I asked him how he found it, he said that he had observed this for small k when he was young (well he is 8 now!), and generalized it to all k recently.
He has also started learning about logarithms. I asked him for the value of log 2.5. His first question was -- to what base? I said 10. After a few seconds he said 0.4. I assumed that he made a guess, he probably knew that log 2 = 0.3, and log 3 is close to 0.5. After a few minutes I asked him how he got it --- he said that he knew that 4^5 = 1024, so 5 log 4 = 3, so log 4 = 0.6. Now log 2.5 = log 10 - log 4 = 0.4. Pretty cool! I asked him for logs of some more numbers, each time he came up with a different way to evaluate it, so he not only knows what a log is, but understands it at a deep conceptual level.
posted by Vineet Gupta at 6:48 PM
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