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# Taylor Series

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.

# Taylor Series: Level 4 Challenges

The graph of $$\csc(x)\csc(y)=\phi$$ forms a bunch of squares and circles. The area of one of these circles is most nearly which of the following?

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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