- #1
jegues
- 1,097
- 3
Homework Statement
Please assume that what I have for the remainder is correct, and we are on the domain 2 < x < 4 around 2.
Homework Equations
The Attempt at a Solution
[tex] 0 \leq |R_{n} (2,x)| = \frac{1}{n+1} |\frac{x-2}{z_{n}}|^{n+1}[/tex]
Since,
[tex]2 < x < 4[/tex] then, [tex] 0 < x-2 < 2[/tex] and [tex] 0 < \frac{x-2}{z_{n}} < \frac{2}{z_{n}}[/tex]
But, [tex] 2 < z_{n} < 4[/tex]
So we can see that, [tex]\frac{2}{z_{n}} < 1[/tex]
Therefore, [tex]\frac{x-2}{z_{n}} < 1[/tex].
Up to here makes perfect sense. Then he writes the following,
[tex] 0 \leq |R_{n}(2,x)| < \frac{1}{n+1} \cdot (1)^{n+1}[/tex]
I'm confused about the, [tex](1)^{n+1}[/tex] how does he get that term? We showed that,
[tex]\frac{x-2}{z_{n}} < 1[/tex], and if we take that and raise it to any power, say n+1, we will get 0. How does he get that 1 there?
Everything else makes perfect sense, it's just that one part.
Can someone please explain?