rudin theorem 1.21 explanation

Question

shawn · Answer

attachment

shawn · Answer

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

kainui · Answer

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:

$$b^2-a^2 = (b-a)(b+a)$$$$b^3-a^3 =(b-a)(b^2+ba+a^2)$$$$\cdots$$

shawn · Answer

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

kainui · Answer

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.

kainui · Answer

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.