
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let $f^{(n)}(x)$ be the $n$th derivative of the function $f(x)$ and $f^{(2)}(x) = [f'(x)]'$, $f^{(n)}(x) = [f^{(n-1)}(x)]'$ ($n \geqslant 3$, $n \in \mathbb{N}^{*}$). If $f^{(n)}(x)$ exists, then $f(x)$ is said to be $n$th order differentiable. The British mathematician Taylor found that if $f(x)$ is $(n+1)$-times differentiable near $x_{0}$, then we can construct ${T_n}(x) = f(x_{0}) + \frac{{f'(x_{0})}}{{1\;!}}(x-x_{0}) + \frac{{f^{(2)}(x_{0})}}{{2\;!}}{(x-x_{0})^2} + \cdots + \frac{{f^{(n)}(x_{0})}}{{n\;!}}{(x-x_{0})^n}$ (called the $n$th order Taylor polynomial) to approximate the value of $f(x)$ near $x_{0}$. Based on this, calculate the 3rd order Taylor polynomial of $f(x) = e^{x}$ at $x_{0} = 0$ as $T_{3}(x)=$____; and the coefficient of $x^{3}$ in the 10th order Taylor polynomial of $f(x) = -\frac{1}{x}$ at $x_{0} = -1$ is ____.

Present the answer in LaTex format: \boxed{Your answer}
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