In Taylor's Inequality,

$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$

What exactly goes inside the $|x-a|$ portion of the equation?

The problem I'm given is asking me what would give a bound on the remainder if $T_3(x)$ was used to approximate $f(x) = ln(x)$ on the interval $1\le x\le2.5$. (Where a = 2).

I know that $f^{4}(x)$ is $-6x^{-4}$ and since it's a decreasing function,

$|f^4(x)| = |f^4(1)| = 6\le M$

But I have been having so much trouble understanding what goes in the $|x-a|$ part of Taylor's inequality.


If you're looking to create an upper bound, you should make $|x-a|$ as big as possible.

Since $a=2$ and $1 \leq x \leq 2.5$, the absolute value is maximized when $x=1$, so $|x-a| \leq |1-2|=|-1|=1$.

  • $\begingroup$ Thank you. Are there ever any cases where I should look to create a lower bound? $\endgroup$ – user11892 Nov 29 '14 at 20:01
  • $\begingroup$ I can't really think of one. Of course the lower bound at x=a is 0. :) In fact, since the error is zero at the base point x=a, it's tricky to get a lower bound. $\endgroup$ – Bill Cook Nov 29 '14 at 20:37

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