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

Find the value of n so that the error obtained by approximating sin(x) by the nth degree Maclaurin polynomial Tn(x) on the interval [-0.5,0.5] will be less than 5 x 10^-6.

I'm really stuck and don't even know where to start. Any help would be appreciated.

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

The error is given by the following: $$f(x) - T_{n}(x) = \frac{1}{(n+1)!} x^{n+1} f^{(n+1)} (c_x)$$

where $c_x \in [0,x]$.

share|cite|improve this answer
And how large can the derivatives of sin(x) be? – Ross Millikan Feb 3 '11 at 4:08
Well sin(x) and cos(x) never gets bigger than 1. But I'm not sure how to use this information. – Bill M. Feb 3 '11 at 4:12
@Bill M.: That should let you put an upper bound on the error term. If you know $x$ can't be larger than something and $f^n$ can't be larger than something else... – Ross Millikan Feb 3 '11 at 5:23

There are many ways to find the power series that corresponds to a function. As is often the case in mathematics, you try to reduce the problem at hand to a problem you already know how to solve. So if you know the geometric power series you try to reduce new problems to that, which the examples do. An example would be if you want to do $\sum_{n=0}^{\infty} n x^n$. Since you know $\sum_0^{\infty}x^n=\frac{1}{1-x}, \sum_{n=0}^{\infty} nx^n=\sum_0^{\infty}x\frac{d}{dx}x^n$ Now you need to justify taking the derivative outside the sum. There are beautiful theorems that tell you when this is justified. We physicists ignore them, as all of our functions satisfy them.

Another example would be the series for $\sin(x)$ near 0 (the Maclaurin series). Taylor's theorem says $$\sin(x)=\sum_{n=0}^{\infty}\frac{\sin^{(n)}(0)}{n!}x^n$$ where $\sin^{(n)}(0)$ is the $n^{th}$ derivative of $\sin(x)$ evaluated at 0. If you know that the $n^{th}$ derivative of $\sin (x)$ at 0 is 0 if $n$ is even and $-1^{(n-1)/2}$ if $n$ is odd this becomes the familiar $$\sin(x)=\sum_{n=1,n \text{ odd}}^{\infty}\frac{-1^{\left(\frac{n-1}{2}\right)}x^n}{n!}$$

Once you have this (and the corresponding ones for other functions like e^x, cos(x), etc) you can reduce problems to these as well. Then, as Arjang commented, if you try to operate with differential and integral operators, you need to worry about whether this is justified.

share|cite|improve this answer

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.