Taylor formula of $\sin x$

Question

Recall of Taylor's Formula

The Taylor formula for sin $x$ is therefore as follows:

sin $x$=$x-\dfrac {x^{3}} {3!}+\dfrac {x^{5}} {5!}-\ldots +\left( -1\right) ^{n-1}\dfrac {x^{2n-1}} {(2n-1)!}+R_{2n+1}$

My question is: Is there a typo in the formula? I think, It should be ...''$+\left( -1\right) ^{2n-1}\dfrac {x^{2n-1}} {(2n-1)!}$'', shouldn't it?

Answer 1

$(-1)^{2n-1}=-1$ That suggests, according to you the taylor expansion of $\sin x$ should be :

$$\sin x = -\Bigg(x+\dfrac {x^{3}} {3!}+\dfrac {x^{5}} {5!}+\ldots +\dfrac {x^{2n-1}} {(2n-1)!}+R_{2n+1}\Bigg)$$

Try writing a few terms, you'll get the $n$th term as $\left( -1\right) ^{n-1}\dfrac {x^{2n-1}} {(2n-1)!}$

Since every term must have alternating $+$ and $-$ sign, it should change with every consecutive $n$. Since first term has $+$ sign, it suggests, $n$th term must have $(-1)^{n-1}$; or more accurately $(-1)^{n-1 \pmod{2}}$

Answer 2

$(-1)^{2n-1}=\frac{(-1)^{2n}}{-1}=\frac{((-1)^2)^n}{-1}=\frac{1^n}{-1}=-1$ is always negative, yet the signs should be alternating - $(-1)^{n-1}$ is indeed alternating.

Consider the Taylor series at $0$: $$f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$$ All the even derivatives in the sum are either $sin(0)$ or $-sin(0)$, which both equal $0$, so the terms can be omitted. The odd derivatives are $cos(0)=1$ or $-cos(0)=-1$, which confirms that the signs must alternate.