Calculus
# Taylor Series

*Remark:* \(\phi = \frac{1+\sqrt{5}}{2}\)

Find the value of \[S= \dfrac{\cos 0}{e^0 \cdot 0!}+ \dfrac{\cos 1}{e^1 \cdot 1!}+\dfrac{\cos 2}{e^2 \cdot 2!}+ \cdots \]

\(\textbf{Details and Assumptions}\)

- In \(\cos n\), \(n\) is considered to be in \(\textbf{radians}\).

Suppose a particle moves in a right-angled left spiral on an \(xy\)-grid. That is, it moves a distance \(D_{1}(x)\) in a straight line, stops, makes a right-angled turn to it's "left", travels a distance \(D_{2}(x)\) in a straight line, stops, makes a right angled turn to its "left", travels a distance \(D_{3}(x)\) in a straight line and continues in this fashion forever.

If \(D_{n}(x) = \dfrac{x^{n-1}}{(n-1)!}\) for \(n \ge 1,\) and if \(x = 2015,\) then find the magnitude of the straight line distance between the particle's starting and finishing points.

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.

If \(\alpha\) and \(\beta\) are roots of the equation \(x^{2}-px+q=0\) then

\((\alpha+\beta)x-\frac{1}{2}(\alpha^{2}+\beta^{2})x^{2}+\frac{1}{3}(\alpha^{3}+\beta^{3})x^{3}- \ldots \) is equal to?

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