So in estimating $(1.6)^{4/5}$ using the first 3 terms of the Taylor Series for function $f(x) = (1+x)^{4/5}$ we want to use the remainder theorem to obtain the error estimate.

$|R_2(0.6)| ≤ $Max $0 ≤ z ≤ 0.6$ $|\frac{f'''(z)}{3!}(0.6)^3|$

Now $f'''(x) = \frac{24}{125(1+x)^{11/5}}$ I assume you would sub in the value for $z$ in place of $x$ before applying to the formula, but how do you calculate it if $z$ can take on multiple values?


1 Answer 1


I think you were trying to write

$$\lvert R_2(0.6)\rvert \leq \max_{0 \leq z \leq 0.6} \left\lvert\frac{f'''(z)}{3!}(0.6)^3\right\rvert.$$

The point of the "max" is to choose the $z$ that makes the expression the largest.

If $f'''(z) = \frac{24}{125(1+z)^{11/5}},$ which value in the range $0 \leq z \leq 0.6$ produces the largest value of $f'''(z)$? That's the value to use.


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