Should 0^{0 }be 0 or 1?

This has always been a strong frustration source for me.

Beary May 8, 2018

#1**+2 **

hi beary!

if you ask the calculator, it says it is zero.

however, the alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)^n using the binomial theorem, i.e., 0n.

But the alternating sum of the entries of every row except the top row is 0, since 0^k=0 for all k greater than 1.

But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)^0=0^0 if it were defined, should be 1.

so therefore 0^0 is zero.

i hope this helped,

gavin

GYanggg May 8, 2018