What's this function?

Geometry Level 4

r = 0 n n C r sin ( r x ) cos ( ( n r ) x ) = f ( n ) sin ( n x ) \large \displaystyle \sum^{n}_{r=0} {^{n}C_{r} \sin(rx) \cos((n-r)x)=f(n)\sin(nx)}

Let f ( n ) f(n) denote a function such that the equation above is an identity.

Find the value of f ( 15 ) f(15) .


The answer is 16384.

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2 solutions

Tanishq Varshney
May 3, 2015

P = r = 0 n P=\displaystyle \sum^{n}_{r=0} n C r s i n ( r x ) c o s ( n r ) x ^{n}C_{r} sin(rx) cos(n-r)x

P = r = 0 n P=\displaystyle \sum^{n}_{r=0} n C n r s i n ( n r ) x c o s ( r ) x ~^{n}C_{n-r} sin(n-r)x cos(r)x

adding

2 P = ( n C 0 + n C 1 + n C 2 + n C 3 + . . . . + n C n ) s i n ( n x ) \large{2P=(^{n}C_{0}+^{n}C_{1}+^{n}C_{2}+^{n}C_{3}+....+^{n}C_{n})sin(nx)}

2 P = 2 n s i n ( n x ) 2P=2^{n}sin(nx)

f ( n ) = 2 n 1 f(n)=2^{n-1}

f ( 15 ) = 2 14 = 16384 f(15)=2^{14}=16384

Nice solution! +1

User 123 - 6 years, 1 month ago

Did you mean c o s ( ( n r ) x ) cos((n-r)x) in the first summation and c o s ( r x ) cos(rx) in the second summation?

Mark Kong - 6 years, 1 month ago

@Tanishq Varshney Hi Tanishq! Try this please.

User 123 - 6 years, 1 month ago

@Tanishq Varshney I deleted the problem and reposted it again since it had a small error the first time. Please do try it now. Many thanks!

User 123 - 6 years, 1 month ago

I don't believe I actually calculated the LHS. Glad that i got the answer , btw Nice solution . : ) :)

Keshav Tiwari - 6 years, 1 month ago
Stewart S.
May 3, 2015

Let k = 0 n ( n k ) sin ( k x ) cos [ ( n k ) x ] = G ( x ) \sum_{k=0}^n \binom{n}{k} \sin(kx) \cos[(n-k)x] = G(x) . It is obvious that

G ( x ) = 2 ( n 1 ) × sin ( n x ) n G(x) = 2^{(n-1)} \times \sin(nx) \forall n positive odd integer. Therefore f ( 15 ) = 2 14 = 16384 f(15) = 2^{14} = \boxed{16384} .

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