4 ms·
While the typesetting might not be very portable or compact, this is beautifully useful for teaching. The two extra identities found appear difficult to prove,
by yunruse 6y ago
While the typesetting might not be very portable or compact, this is beautifully useful for teaching. The two extra identities found appear difficult to prove, but in this notation are blindingly obvious.
I wonder if this can be used to apply to other sets of functions, or if the geometry of chaining functions so can be extended to other such geometrically-obvious proofs.
- adamjb 6y agoI don't know how hard they would be to come up with ex nihilo, but they're fairly straightforward to prove if you know that logb(x) = log(x)/log(b) and log(x^y)=y * log(x). For logy√z(z) = y logy√z(z) = log(z)/log(z^1/y) = log(z)/(1/y × log(z)) = y * log(z)/log(z) = y For logx(z)√z = x logx(z)√z = z^(1/(logx(z))) log(LHS) = log(z^(1/logx(z)) = log(z)/logx(z) = log(z)/(log(z)/log(x)) = log(x) = log(RHS) Therefore LHS = RHS
- contravariant 6y agoThe two other identities are only tricky because they're using logarithms with a different base and they're writing their roots with the radical symbol. If you write them using log_b(x) = log(x)/log(b) they become: log(z)/log(z^(1/y)) = y z^(log(x)/log(z)) = x the first is obvious from the fact that log(x^y) = y log(x) and the latter is why log(x)/log(z) is also considered the base z logarithm. The only reason it looks nonobvious is because the notation they chose makes it non-obvious that log_b(y) = 1/log_y(b) (and that the yth root of x is x^(1/y)).