The problem is this:

Use the Lagrange error formula to find the bound for the error in approximating $f(x)=\frac{1}{x^2}$ for $0.9 \le x \le 1.1$ with a second degree Taylor polynomial about $x = 1$.

The answers are the following:





What I have done is found the third derivative is -24/x^5, I then tried finding the bounds of x by doing

0.9 <= x <= 1.1 which gave me my z

z = 0.9

for Rn(x) = (f^3(z)(x-a)^3)/3!

my z I got .9 because its value at f^3(z) = -40.644 while the other bound gives -14.902 so z = the greater bound when non negative.

Since x is about 1, I do

(-40.644(.1^3))/3! which gives 0.00677.. I am still unconvinced this is the answer because I should be using -14.902 since it is the greater number

I have been struggling with this question for about 2 days and feel completely lost so if anyone can help me with this problem it would greatly appreciated.


While I was writing up an answer, you fixed some serious problems in your solution attempt. Well done! Now I'll try to answer the remaining part of the question.

You are not supposed to put an upper bound on the value of $R_n(x)$; you are actually supposed to put an upper bound on $\lvert R_n(x) \rvert$. The absolute value makes a big difference, because $$ \lvert R_2(x)\rvert = \frac{\lvert f'''(z)\rvert \cdot \lvert(x-a)^3\rvert}{3!}. $$

So you don't need to be concerned with the facts that $f'''(0.9) \approx -40.644$ and $f'''(1.1) \approx -14.902$; rather you really should consider the facts that $\lvert f'''(0.9)\rvert \approx 40.644$ and $\lvert f'''(1.1)\rvert \approx 14.902$. Clearly $40.644$ is the larger value and is the one you should use.

  • $\begingroup$ Both answers are perfect, this one explains exactly why I was going wrong. Thank you! $\endgroup$ – Pookie Aug 30 '16 at 23:14

$\displaystyle R_3(x)=\frac{f^{(3}(z)}{3!}(x-1)^3=\frac{24/z^5}{3!}(x-1)^3=\frac{4(x-1)^3}{z^5}$ where $.9<z<1.1$, so

$\displaystyle\left|R_3(x)\right|=\frac{4|x-1|^3}{z^5}<\frac{4(.1)^3}{z^5}=\frac{1}{250z^5}<\frac{(10/9)^5}{250}=\frac{400}{9^5}\approx.00677$ since $\frac{1}{z}<\frac{1}{.9}=\frac{10}{9}$

  • $\begingroup$ Excellent thank you! Despite not clearly defining where I went wrong this helps a ton, wish I could choose two answers $\endgroup$ – Pookie Aug 30 '16 at 23:15

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