(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Please assume that what I have for the remainder is correct, and we are on the domain 2 < x < 4 around 2.

2. Relevant equations

3. 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?

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# Homework Help: Small confusion about Taylor's remainder

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