Homework Help: Small confusion about Taylor's remainder

1. Sep 23, 2010

jegues

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

$$0 \leq |R_{n} (2,x)| = \frac{1}{n+1} |\frac{x-2}{z_{n}}|^{n+1}$$

Since,

$$2 < x < 4$$ then, $$0 < x-2 < 2$$ and $$0 < \frac{x-2}{z_{n}} < \frac{2}{z_{n}}$$

But, $$2 < z_{n} < 4$$

So we can see that, $$\frac{2}{z_{n}} < 1$$

Therefore, $$\frac{x-2}{z_{n}} < 1$$.

Up to here makes perfect sense. Then he writes the following,

$$0 \leq |R_{n}(2,x)| < \frac{1}{n+1} \cdot (1)^{n+1}$$

I'm confused about the, $$(1)^{n+1}$$ how does he get that term? We showed that,

$$\frac{x-2}{z_{n}} < 1$$, 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.

2. Sep 23, 2010

Office_Shredder

Staff Emeritus
You can't raise anything to any power and get zero. When you say we take "that", what are you referring to? 1n+1=1 and $$\frac{x-2}{z_n}^{n+1}$$ is going to be some small positive number

if a<b then an<bn for any value of n (you can see this by noticing that the derivative of xn is positive so is an increasing function)

3. Sep 23, 2010

jegues

The part in bold is what I was missing,

Since,

$$\frac{x-2}{z_{n}} < 1$$ then,

$$\left( \frac{x-2}{z_{n}} \right)^{n+1} < 1^{n+1}$$

So we can see that,

$$0 \leq |R_{n}(2,x)| < \frac{1}{n+1} \cdot (1)^{n+1}$$

Applying squeeze theorem,

$$lim_{n \rightarrow \infty} 0 = lim_{n \rightarrow \infty} \frac{1}{n+1} = 0$$

Therefore,

$$lim_{n \rightarrow \infty} |R_{n}(2,x)| = 0$$

So it follows that,

$$lim_{n \rightarrow \infty} R_{n}(2,x) = 0$$ for $$2 < x < 4$$