Calculus

Taylor Series

Taylor Series: Level 4 Challenges

         

The graph of csc(x)csc(y)=ϕ\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: ϕ=1+52\phi = \frac{1+\sqrt{5}}{2}

Find the value of S=cos0e00!+cos1e11!+cos2e22!+S= \dfrac{\cos 0}{e^0 \cdot 0!}+ \dfrac{\cos 1}{e^1 \cdot 1!}+\dfrac{\cos 2}{e^2 \cdot 2!}+ \cdots

Details and Assumptions\textbf{Details and Assumptions}

  • In cosn\cos n, nn is considered to be in radians\textbf{radians}.

Suppose a particle moves in a right-angled left spiral on an xyxy-grid. That is, it moves a distance D1(x)D_{1}(x) in a straight line, stops, makes a right-angled turn to it's "left", travels a distance D2(x)D_{2}(x) in a straight line, stops, makes a right angled turn to its "left", travels a distance D3(x)D_{3}(x) in a straight line and continues in this fashion forever.

If Dn(x)=xn1(n1)!D_{n}(x) = \dfrac{x^{n-1}}{(n-1)!} for n1,n \ge 1, and if x=2015,x = 2015, then find the magnitude of the straight line distance between the particle's starting and finishing points.

For all integers nn, we define ξn\xi_n as follows: {ξn=1if n0(mod4) or n1(mod4)ξn=1if n2(mod4) or n3(mod4)\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 nZ+n \in \mathbb{Z^+}, let f(n)=ξ0(n0)+ξ1(n1)+ξ2(n2)++ξn(nn).f(n)= \xi_0 \dbinom{n}{0} + \xi_1 \dbinom{n}{1} + \xi_2 \dbinom{n}{2} + \cdots + \xi_n \dbinom{n}{n}. Find 100(n=0f(n)n!)\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 40(mod4)4 \equiv 0 \pmod{4}, ξ4=1\xi_4= 1, whereas ξ6=1\xi_6 = -1 since 62(mod4)6 \equiv 2 \pmod{4}. Note that ξ0=ξ1=1\xi_0= \xi_1= 1.

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

  • You might use a scientific calculator for this problem.

If α\alpha and β\beta are roots of the equation x2px+q=0x^{2}-px+q=0 then

(α+β)x12(α2+β2)x2+13(α3+β3)x3(\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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