Tricky Summation

Calculus Level 2

For all integers n n , we define ξ n \xi_n as follows: { ξ n = 1 if n 0 ( m o d 4 ) or n 1 ( m o d 4 ) ξ n = 1 if n 2 ( m o d 4 ) or n 3 ( m o d 4 ) \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 Z + n \in \mathbb{Z^+} , let f ( n ) = ξ 0 ( n 0 ) + ξ 1 ( n 1 ) + ξ 2 ( n 2 ) + + ξ n ( n n ) . 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 = 0 f ( 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 4 0 ( m o d 4 ) 4 \equiv 0 \pmod{4} , ξ 4 = 1 \xi_4= 1 , whereas ξ 6 = 1 \xi_6 = -1 since 6 2 ( m o d 4 ) 6 \equiv 2 \pmod{4} . Note that ξ 0 = ξ 1 = 1 \xi_0= \xi_1= 1 .

  • The floor function x \lfloor x \rfloor denotes the largest integer less than or equal to x x . 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.


The answer is 375.

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2 solutions

First, consider the Taylor expansions of sin ( x ) \sin (x) and cos ( x ) \cos (x) : sin ( x ) = x x 3 3 ! + x 5 5 ! x 7 7 ! + cos ( x ) = 1 x 2 2 ! + x 4 4 ! x 6 6 ! + \begin{array}{lcl} \sin (x) & = & x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots \\ \cos (x) & = & 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots\end{array} Adding them, sin ( x ) + cos ( x ) = 1 + x x 2 2 ! x 3 3 ! + x 4 4 ! + x 5 5 ! x 6 6 ! x 7 7 ! + = n = 0 ξ n x n n ! . \begin{array}{lcl} \sin (x) + \cos (x) & = & 1 + x - \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!} + \dfrac{x^5}{5!} - \dfrac{x^6}{6!} - \dfrac{x^7}{7!} + \cdots \\ & = & \sum \limits_{n=0}^{\infty} \xi_n \dfrac{x^n}{n!}. \end{array} Now consider the expansion of e x e^x : e x = 1 + x + x 2 2 ! + x 3 3 ! + = n = 0 x n n ! . \begin{array}{lcl} e^x &=& 1+x+\dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots \\ &=& \sum \limits_{n=0}^{\infty} \dfrac{x^n}{n!} .\end{array}

Multiplying, e x ( sin ( x ) + cos ( x ) ) = ( n = 0 x n n ! ) ( n = 0 ξ n x n n ! ) . e^x (\sin (x) + \cos (x)) = \left( \sum \limits_{n=0}^{\infty} \dfrac{x^n}{n!} \right) \left( \sum \limits_{n=0}^{\infty} \xi_n \dfrac{x^n}{n!} \right) . Note that the coefficient of x n x^n in the RHS is equal to ξ 0 1 0 ! 1 n ! + ξ 1 1 1 ! 1 ( n 1 ) ! + + ξ n 1 n ! 1 0 ! = k = 0 n ξ k 1 k ! ( n k ) ! = k = 0 n ξ k ( n k ) n ! = f ( n ) n ! . \begin{array}{lcl} \xi_0 \dfrac{1}{0!} \cdot \dfrac{1}{n!} + \xi_1 \dfrac{1}{1!} \cdot \dfrac{1}{(n-1)!} + \cdots + \xi_n \dfrac{1}{n!} \cdot \dfrac{1}{0!} & = & \sum \limits_{k=0}^{n} \xi_k \dfrac{1}{k! (n-k)!} \\ & = & \sum \limits_{k=0}^{n} \xi_k \dfrac{ \dbinom{n}{k} }{n!} \\ & = & \dfrac{f(n)}{n!}. \end{array} Thus, e x ( sin ( x ) + cos ( x ) ) = n = 0 f ( n ) n ! x n . e^x (\sin (x) + \cos (x) ) = \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!} x^n . Plugging x = 1 x=1 , n = 0 f ( n ) n ! = e ( sin ( 1 ) + cos ( 1 ) ) . \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!}= \boxed{e(\sin (1) + \cos (1))}. Using a scientific calculator (or WolframAlpha for the lazy), we find out that e ( sin ( 1 ) + cos ( 1 ) ) 3.75604 , e (\sin (1) + \cos (1)) \approx 3.75604, and 100 e ( sin ( 1 ) + cos ( 1 ) ) = 375 . \left \lfloor 100 e (\sin (1) + \cos (1)) \right \rfloor = \boxed{375}.

Rajen Kapur
Mar 9, 2014

(1+i)^n is having binomial expansion which is going to give us the function when we add Re and Im parts. Summation is clearly e^(1+i), i.e. when Re and Im are added, answer is 100e(cos1 + sin1) to integer floor.

I don't agree with you. The summation we want is n = 0 f ( n ) n ! \displaystyle \sum \limits_{n=0}^{\infty} \dfrac{f(n)}{n!} , whereas ( 1 + i ) n + ( 1 + i ) n \Re (1+i)^n + \Im (1+i)^n is equal to f ( n ) f(n) . The sum n = 0 ( 1 + i ) n , \displaystyle \sum \limits_{n=0}^{\infty} (1+i)^n, in fact, doesn't even converge. You need to find n = 0 ( 1 + i ) n + ( 1 + i ) n n ! \displaystyle \sum \limits_{n=0}^{\infty} \dfrac{\Re (1+i)^n + \Im (1+i)^n}{n!} , and it's not obvious at all that the sum equals ( e i + 1 ) + ( e i + 1 ) . \Re \left(e^{i+1} \right) + \Im \left( e^{i+1} \right) .

Sreejato Bhattacharya - 7 years, 3 months ago

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what he means to say is that ( 1 + i ) n = ( 2 × i ) n 2 (1+i)^{n} = ( 2 \times i )^{\frac {n}{2}} . So we can replace x of x n n ! \frac {x^{n}}{n!} with ( 2 × i ) 1 2 ( 2 \times i )^{\frac {1}{2}} which is equal to e ( 2 × i ) 1 2 = e 1 + i e^{( 2 \times i )^{\frac {1}{2}}} = e^{1+i} by De Moivre's formula

dp dp - 7 years, 2 months ago

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