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The Taylor series is a polynomial of infinite degree used to represent functions like sine, cube roots, and the exponential function. They're how some calculators (and Physicists) make approximations.

\[f\left( x \right) =\sum _{ n=0 }^{ \infty }{ \frac { \sin { \left( xn \right) } }{ { 7 }^{ n } } }\] For all real numbers \(x\), let \(f(x) \) be a function with fundamental period \(P \).

Let \( \displaystyle a = \int_0^{P/2} f(x) \, dx \). If \(f \left( \pi e^{|a|} \right) \) can be expressed as \( -\dfrac{\alpha\sqrt \beta}{\gamma}\), where \(\alpha, \beta\) and \(\gamma\) are positive integers with \(\alpha, \gamma\) coprime and \(\beta\) square-free, find \(\alpha + \beta + \gamma\).

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\[ \frac {1^7}{1!} + \frac {1^7 + 2^7}{2!} + \frac {1^7 + 2^7 + 3^7}{3!} + \frac {1^7 + 2^7 + 3^7 + 4^7}{4!} + \ldots \]
###### Image Credit: Flickr Gingertwist

If the series above equals to \(W\), what is the value of \(\frac {24}{e} \times W \)?

Note: \(e = \displaystyle \lim_{n \to \infty} \left (1 + \frac 1 n \right )^n \)

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For all integers \(n\), we define \(\xi_n\) as follows: \[\begin{cases} \xi_n = 1 & \text{if } n \equiv 0 \pmod{4} \text{ or } n \equiv 1 \pmod{4} \\ \xi_n= -1 & \text{if } n \equiv 2 \pmod{4} \text{ or } n \equiv 3 \pmod{4} \end{cases} \] For all \(n \in \mathbb{Z^+}\), let \[f(n)= \xi_0 \dbinom{n}{0} + \xi_1 \dbinom{n}{1} + \xi_2 \dbinom{n}{2} + \cdots + \xi_n \dbinom{n}{n}.\] Find \(\left \lfloor 100 \left( \displaystyle \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!} \right) \right \rfloor\).

**Details and assumptions**

As an explicit example, since \(4 \equiv 0 \pmod{4}\), \(\xi_4= 1\), whereas \(\xi_6 = -1\) since \(6 \equiv 2 \pmod{4}\). Note that \(\xi_0= \xi_1= 1\).

The floor function \(\lfloor x \rfloor \) denotes the largest integer less than or equal to \(x\). For example, \(\lfloor 3.25 \rfloor = 3, \lfloor 4 \rfloor= 4, \lfloor \pi \rfloor = 3\).

You might use a scientific calculator for this problem.

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\[ \large\displaystyle \lim_{x \to 0} \frac {(\cos x)^{\sin x} - \sqrt{1-x^3}}{x^6} =\ ?\]

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Find \(\dfrac{f^{(2016)}(0)}{2016!}\) for \(f(x)=\dfrac{x^3}{(1-x)^3(1+x+x^2)}\).

**Clarifications**:

\(f^{(k)}(x)\) denotes the \(k^\text{th}\) derivative of \(f(x) \).

\(!\) denotes the factorial notation. For example, \(8! = 1\times2\times3\times\cdots\times8 \).

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