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

The Taylor Expansion of $f(x)=\sin x$ with a Lagrange remainder is:

$\sin x = x-{x{3}\over 3!}+{x^{5}\over5!}+\cdots+{(-1)^{m-1}x^{2m-1}\over(2m-1)!}+{(-1)^{m}x^{2m+1}cos \theta x\over(2m+1)!}, 0<\theta<1, -\infty<x<\infty $

which actually contains $2m$ terms and one $R(x)$ since $f^{(2k)}(x)=sin^{(2k)} x=0$:

$\sin x = x+0-{x{3}\over 3!}+0+{x^{5}\over5!}+0+\cdots+{(-1)^{m-1}x^{2m-1}\over(2m-1)!}+0+{(-1)^{m}x^{2m+1}cos \theta x\over(2m+1)!}, 0<\theta<1, -\infty<x<\infty $

That's what I find in most maths books.

My question is:

Must I always regard the Taylor Expansion of $\sin x$ as containing $2m$ terms and one $R(x)$ ?

If the expansion contains only $2m-1$ terms and the $R(x)$, then $R(x)$ is the $2m$th term. So how can I write the $R(x)$ in Lagrange form (Obviously $R(x)$ is not equal zero)? Or I shouldn't do that ?

Any help will be great appreciated.

share|cite|improve this question
How come there's a term $(-1)^{m}\cos \theta x\over(2m+1)!$ in the series? – ᴊ ᴀ s ᴏ ɴ Sep 24 '12 at 8:07
It should be multiplied by $x^{2m+1}$. – Hans Lundmark Sep 24 '12 at 8:27
Could you clarify your questions, please? A Taylor expansion is, by definition, a polynomial of prescribed degree. So, you should always write "the Taylor expansion of orded ..." I cannot understand what you'd like to know. – Siminore Sep 24 '12 at 8:29
@HansLundmark I don't understand... – Boris Sep 24 '12 at 8:29
The remainder term is incorrect. It should be $\frac{(-1)^m x^{2m+1} \cos(\theta x)}{(2m+1)!}$. – Hans Lundmark Sep 24 '12 at 8:32
up vote 0 down vote accepted

You can of course apply Taylor's formula with a remainder of order $2m$, which in general is $f^{(2m)}(\theta x) x^{2m}/(2m)!$, hence in this case $(-1)^m \sin(\theta x) x^{2m}/(2m)!$. But why settle for this when you can get a remainder of order $2m+1$ "for free"?

share|cite|improve this answer
$f^{(2m)}(x)=sin^{(2m)}(x)=0$, and then the remainder disappers. – Boris Sep 24 '12 at 8:35
No, the sine function is not identically zero... (Note that you're not evaluating it at the origin, but at an unknown point $\theta x$.) – Hans Lundmark Sep 24 '12 at 8:36
Ouch ! Seems that's it! – Boris Sep 24 '12 at 8:41

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.