Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Replace the following function by its taylor polynomial of the given grade, and approximate the error in the given interval:

$$f(x) = \sin(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$

My solution and thoughts

We only need the first three derivatives:

$$ f'(x) = \cos(x) \\ f''(x) = -\sin(x) \\ f'''(x) = -\cos(x) $$

And by definition we know that

$$ T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k $$

we get

$$ = 0 + \frac{1}{1!}(x-0)^1 + 0 - \frac{1}{3!}(x-0)^3 $$

Is this right so far? It looks suspiciously simple, which merely confuses me.

$$ = x - \frac{x^3}{6} $$

I am clueless when it comes to the error. In which points do I have to calculate $T_3(x)$ in order to get the error as

$$ R_3(x) = sin(x) - T_3(x) $$

in the given interval $[-\frac{1}{10};\frac{1}{10}]$?


Oh, I'm reading in a book about the Lagrange representation of the error. Can I use it? $f(x)$ looks endlessly differentiable in $0$.

share|cite|improve this question
up vote 1 down vote accepted

Yes, you computed $T_3$ correctly.

Towards estimating the error, you may (might) use the fact that for $-1/10<x<1/10$, the error is given by $$\tag{1} R_3(x) ={f^{(iv)}(c)\over 4!}x^4, $$ for some $c$ between $0$ and $x$.

Note that $c$ depends on $x$. In general, the value of $c$ cannot be found easily (without knowing the exact value of $\sin x$). But that's ok; you were asked to estimate the error. And here it suffices to find an upper bound of the absolute value of $(1)$ that is valid over $(-1/10,1/10)$.

Thus, you need to find a number $M$ so that$$ \Bigl|{f^{(iv)}(c)\over 4!}x^4\Bigr|\le M $$ for all $x\in(-1/10,1/10)$. Towards this end, it's useful to note that $|f^{(iv)}(x)|\le 1$ for all $x$.

Can you take it from here?

Once you've found $M$, this will be the desired estimate; you'll know that $|f(x)-P_3(x)|\le M$ for all $x\in(-1/10,1/10)$.

share|cite|improve this answer
"Can you take it from here?" - Hm, but your relation $|R_3|\le M$ does look like the definition of $\lim$, so I just have to look for the $\lim R_3$ over c from 0 to 1/10. Did I misunderstand something (I certainly had)? – Flavius Dec 6 '12 at 13:26
@Flavius No, it's not a limit. Just do an estimate: Since $|f^{(iv)}(x)|=|\sin(x)|\le1$, we have $$|R_3(x)|\le {1\over 4!}|x^4|;$$ and since $|x|\le 1/10$, we have $$|R_3(x)|\le {1\over 4!}\cdot {1\over 10^4}.$$ Note the first of these is the better estimate for a particular $x$, but the second is an estimate that is valid over the entire interval $(-1/10,1/10)$. – David Mitra Dec 6 '12 at 13:31
So to wrap it up, if I would start to play with these relations in sagemath: 1. If I plot $T_3$, I get a function similar to sin, but with the given error $R_3$, and 2. if I would calculate $\sin(x) - T_3(x)$ I would get something close to $R_3$? Wow, this sounds really powerful - I've got to test it. Especially if I think about throwing those coefficients of $T_n$ into a matrix and do it like in linear algebra to calculate sin based on that. Now I also start to see where those matrices representing differentials in computer graphics were coming from. – Flavius Dec 6 '12 at 13:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.