For all integers n , we define ξ n as follows: { ξ n = 1 ξ n = − 1 if n ≡ 0 ( m o d 4 ) or n ≡ 1 ( m o d 4 ) if n ≡ 2 ( m o d 4 ) or n ≡ 3 ( m o d 4 ) For all n ∈ Z + , let f ( n ) = ξ 0 ( 0 n ) + ξ 1 ( 1 n ) + ξ 2 ( 2 n ) + ⋯ + ξ n ( n n ) . Find ⌊ 1 0 0 ( n = 0 ∑ ∞ n ! f ( n ) ) ⌋ .
Details and assumptions
As an explicit example, since 4 ≡ 0 ( m o d 4 ) , ξ 4 = 1 , whereas ξ 6 = − 1 since 6 ≡ 2 ( m o d 4 ) . Note that ξ 0 = ξ 1 = 1 .
The floor function ⌊ x ⌋ denotes the largest integer less than or equal to x . For example, ⌊ 3 . 2 5 ⌋ = 3 , ⌊ 4 ⌋ = 4 , ⌊ π ⌋ = 3 .
You might use a scientific calculator for this problem.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
(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 ∑ ∞ n ! f ( n ) , whereas ℜ ( 1 + i ) n + ℑ ( 1 + i ) n is equal to f ( n ) . The sum n = 0 ∑ ∞ ( 1 + i ) n , in fact, doesn't even converge. You need to find n = 0 ∑ ∞ n ! ℜ ( 1 + i ) n + ℑ ( 1 + i ) n , and it's not obvious at all that the sum equals ℜ ( e i + 1 ) + ℑ ( e i + 1 ) .
Log in to reply
what he means to say is that ( 1 + i ) n = ( 2 × i ) 2 n . So we can replace x of n ! x n with ( 2 × i ) 2 1 which is equal to e ( 2 × i ) 2 1 = e 1 + i by De Moivre's formula
Problem Loading...
Note Loading...
Set Loading...
First, consider the Taylor expansions of sin ( x ) and cos ( x ) : sin ( x ) cos ( x ) = = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ⋯ 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + ⋯ Adding them, sin ( x ) + cos ( x ) = = 1 + x − 2 ! x 2 − 3 ! x 3 + 4 ! x 4 + 5 ! x 5 − 6 ! x 6 − 7 ! x 7 + ⋯ n = 0 ∑ ∞ ξ n n ! x n . Now consider the expansion of e x : e x = = 1 + x + 2 ! x 2 + 3 ! x 3 + ⋯ n = 0 ∑ ∞ n ! x n .
Multiplying, e x ( sin ( x ) + cos ( x ) ) = ( n = 0 ∑ ∞ n ! x n ) ( n = 0 ∑ ∞ ξ n n ! x n ) . Note that the coefficient of x n in the RHS is equal to ξ 0 0 ! 1 ⋅ n ! 1 + ξ 1 1 ! 1 ⋅ ( n − 1 ) ! 1 + ⋯ + ξ n n ! 1 ⋅ 0 ! 1 = = = k = 0 ∑ n ξ k k ! ( n − k ) ! 1 k = 0 ∑ n ξ k n ! ( k n ) n ! f ( n ) . Thus, e x ( sin ( x ) + cos ( x ) ) = n = 0 ∑ ∞ n ! f ( n ) x n . Plugging x = 1 , n = 0 ∑ ∞ n ! f ( n ) = e ( sin ( 1 ) + cos ( 1 ) ) . Using a scientific calculator (or WolframAlpha for the lazy), we find out that e ( sin ( 1 ) + cos ( 1 ) ) ≈ 3 . 7 5 6 0 4 , and ⌊ 1 0 0 e ( sin ( 1 ) + cos ( 1 ) ) ⌋ = 3 7 5 .