Taylor's Theorem with Remainder fails when sequence has zero terms?

Question

If we try to calculate the remainder of a series such as $0+\frac{x}{3}+0-\frac{x^3}{5}+0+\frac{x^5}{7}+0...$

Does the theorem fail?

I was trying to find the remainder centered at $x=\frac{1}{2}$ after the first two nonzero terms. Since the first two nonzero terms are at $n=1$ and $n=3$, I tried to find $R_3(\frac{1}{2})$ and I ended up with $0$ (incorrect).

However if I move on to the next nonzero term and calculate $R_4(\frac{1}{2})$ then I get the correct answer of $\frac{1}{224}$.

Is there supposed to be a stipulation in the theorem? I'm failing to see how $R_3(\frac{1}{2})=0$ doesn't break the theorem. Thanks!

EDIT

Well I'm still a bit confused so let me post the actual question and my procedure:

So the actual function $g$ is not given. However, by comparing the terms of the sequence to the definition of McLaurin/Taylor series, namely:

$f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+...$

I am able to determine that $a=0,f(a)=0,f'(a)=\frac{1!}{3},f''(a)=0,f^{(3)}(a)=\frac{-3!}{5},f^{(4)}(a)=0,f^{(5)}(a)=\frac{5!}{7}$

Then I apply Taylor's Theory with Remainder, which says:

$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$

It says $c$ can be anything between $a$ and $x$, so I let $c=0$ and start plugging in to get

$R_3(\frac{1}{2})=\frac{0}{24}(\frac{1}{2})^4=0$

$R_4(\frac{1}{2})=\frac{\frac{120}{7}}{120}(\frac{1}{2})^5=\frac{1}{224}$

Answer 1

My initial suspicion was correct. The theorem should have a stipulation: The Taylor Remainder can only be calculated using the next non-zero term.

Answer 2

You have misundestood "after the first two non-zero terms." The function is $$f(x)=0+\frac{x}{3}+0-\frac{x^3}{5}+0+\frac{x^5}{7}+0...$$ The Taylor series is $$f(1/2)+(x-1/2)f'(1/2)+\frac{(x-1/2)^2}{2!}f''(1/2)+\frac{(x-1/2)^3}{3!}f'''(1/2)+...$$ You are asked to find the remainder after the first two non-zero terms of this series, not after the first two terms of the series defining f(x).