Question

rudin theorem 1.21 explanation

Answer 1

where does the inequality b^n - a^n < n (b-a) b^(n-1) come from?

Answer 2

It comes from the identity in the line right above it where they've factored \(b^n-a^n\). To get a feel for why that's true, distribute \((b-a)\) in that identity.

It's the generalization of:

\[b^2-a^2 = (b-a)(b+a)\]\[b^3-a^3 =(b-a)(b^2+ba+a^2)\]\[\cdots\]

Answer 3

well yes but i meant like how does that identity yields b^n - a^n < n (b-a) b^(n-1)

Answer 4

Right after he says that inequality, he says it holds when:

\[0 < a < b\]
So what he's doing there is using that to replace all the \(b^{n-k}a^k < b^{n-1}\). Since there's \(n\) of them in total, that's where it comes from.

Answer 5

But really I'm cheating you out of learning, you're supposed to figure this kind of stuff out for yourself by playing around and actually thinking about it.