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I'm trying to do test corrections for Calculus 2, but I am stuck on this problem.

Find an upper bound for your error estimate in part (b)

The estimate is of $\sqrt2$ using the 4th degree Taylor series of $(1+x)^{1/3}$ (centered at zero). The esimate was that $\sqrt[3]{2} \approx \frac{302}{243}$ $$T_4(x)=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3-\frac{10}{243}x^4$$

Using the lagrange error $$|E_n(x)|\le\frac{M}{(n+1)!}x^{n+1}$$ Where $M$ is the maximum value of the $n+1^{th}$ derivative of $f(x)$ at the value.

The 5th derivative of $(1+x)^{1/3}$ is $\frac{880}{243}(1+x)^{-14/3}$

Therefore, $$\star \quad M=\frac{880}{243}(2)^{-14/3}\approx0.14258$$ $$M\lt1/2$$ $$|E_4(x)|\le\frac{\frac{1}{2}}{(5)!}1^{5}$$ $$|E_4(x)|\le\frac{1}{240}$$ This is the answer I came up with, but I was marked down two points.

My instructor told me I should have evaluated the 5th derivative at $x=0$ instead of $x=1$ (the starred step). Is he right? Could somebody explain why?

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$M$ is supposed to be the maximum absolute value of the ($n+1$)'th derivative on the interval from $a$ (the centre of the expansion) to $x$ (the point where you're evaluating the function), not just the value at $x$. In this case, the $5$'th derivative is decreasing, so the maximum is at $0$, not at $1$.

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  • $\begingroup$ Ah! I knew it was on the interval, but I thought if you weren't given an interval on which to bound it, you wouldn't use any interval. Thank you for the explanation. $\endgroup$ – Noah Harris Mar 4 '16 at 21:23
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The reason why the 5th derivative is supposed to be evaluated at x = 0 and not x = 1 is because your x is always going to be the number that will maximize your value of M. Because the 5th derivative is raised to a negative power, the way to maximize M is to choose as small as possible a value for M. In this case that would b x = 0 and not x = 1.

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