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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

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

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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    @HansLundmark $Y$es, I made a mistake.2012-09-24

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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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    Ouch ! Seems that's it!2012-09-24