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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.

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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
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1 Answer 1

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"?

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$f^{(2m)}(x)=sin^{(2m)}(x)=0$, 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
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