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Small confusion about Taylor's remainder

  1. Sep 23, 2010 #1
    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]


    [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?
  2. jcsd
  3. Sep 23, 2010 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    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 [tex] \frac{x-2}{z_n}^{n+1}[/tex] 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)
  4. Sep 23, 2010 #3
    The part in bold is what I was missing,


    [tex]\frac{x-2}{z_{n}} < 1[/tex] then,

    [tex]\left( \frac{x-2}{z_{n}} \right)^{n+1} < 1^{n+1}[/tex]

    So we can see that,

    [tex] 0 \leq |R_{n}(2,x)| < \frac{1}{n+1} \cdot (1)^{n+1}[/tex]

    Applying squeeze theorem,

    [tex]lim_{n \rightarrow \infty} 0 = lim_{n \rightarrow \infty} \frac{1}{n+1} = 0 [/tex]


    [tex]lim_{n \rightarrow \infty} |R_{n}(2,x)| = 0[/tex]

    So it follows that,

    [tex]lim_{n \rightarrow \infty} R_{n}(2,x) = 0[/tex] for [tex]2 < x < 4[/tex]
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