How do you find the Taylor remainder term $R_{3} (x; 0)$ $R_3(x;0)$ for $f (x) = \frac{1}{2 + x}$ $f(x)=1/(2+x)$ ?

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Answer 1 · 2016-03-01T09:55:20

This exact value of $R_{n} (x)$ $R_n(x)$ is

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for some value of $c$ $c$ between $a$ $a$ and $x$ $x$

However we can determine an upper bound for $R_{3} (x)$ $R_3(x)$

which is

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Hence for $a = 0$ $a=0$ and $n = 3$ $n=3$ we get

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where

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hence

$f^{4} (0) = \frac{24}{2^{5}} = \frac{3}{4}$ $f^4(0)=24/2^5=3/4$

So finally we have that

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How do you find the Taylor remainder term R3(x;0)R_3(x;0) for f(x)=12+xf(x)=1/(2+x)?