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Question: If $f_n(x) = \prod_{s=0}^{n-1}(2\cos(2^s x) - 1), n \ge 1$. Then which of the following is/are true-...

If fn(x)=∏s=0n−1(2cos⁡(2sx)−1),n≥1f_n(x) = \prod_{s=0}^{n-1}(2\cos(2^s x) - 1), n \ge 1. Then which of the following is/are true-

A

f5(2π33)=−1f_5(\frac{2\pi}{33}) = -1

B

f5(6π255)=1f_5(\frac{6\pi}{255}) = 1

C

f5(2π31)=1f_5(\frac{2\pi}{31}) = 1

D

f5(4π127)=1f_5(\frac{4\pi}{127}) = 1

Answer

C

Explanation

Solution

Let fn(x)=∏s=0n−1(2cos⁡(2sx)−1)f_n(x) = \prod_{s=0}^{n-1}(2\cos(2^s x) - 1). We can simplify the general term 2cos⁡θ−12\cos\theta - 1 using complex exponentials. Let z=eiθz = e^{i\theta}. Then 2cos⁡θ−1=z+z−1−1=z2−z+1z2\cos\theta - 1 = z + z^{-1} - 1 = \frac{z^2 - z + 1}{z}. We know the identity z3+1=(z+1)(z2−z+1)z^3+1 = (z+1)(z^2-z+1), so z2−z+1=z3+1z+1z^2-z+1 = \frac{z^3+1}{z+1}. Substituting this back, 2cos⁡θ−1=1zz3+1z+12\cos\theta - 1 = \frac{1}{z} \frac{z^3+1}{z+1}. Now, let θ=2sx\theta = 2^s x. So z=ei2sxz = e^{i2^s x}. The general term of the product becomes: 2cos⁡(2sx)−1=1ei2sxei3⋅2sx+1ei2sx+12\cos(2^s x) - 1 = \frac{1}{e^{i2^s x}} \frac{e^{i3 \cdot 2^s x}+1}{e^{i2^s x}+1}.

Now, let's write out the product fn(x)f_n(x): fn(x)=∏s=0n−1(1ei2sxei3⋅2sx+1ei2sx+1)f_n(x) = \prod_{s=0}^{n-1} \left( \frac{1}{e^{i2^s x}} \frac{e^{i3 \cdot 2^s x}+1}{e^{i2^s x}+1} \right) fn(x)=∏s=0n−1(ei3⋅2sx+1)∏s=0n−1ei2sx∏s=0n−1(ei2sx+1)f_n(x) = \frac{\prod_{s=0}^{n-1} (e^{i3 \cdot 2^s x}+1)}{\prod_{s=0}^{n-1} e^{i2^s x} \prod_{s=0}^{n-1} (e^{i2^s x}+1)}.

Let's simplify the terms in the denominator:

  1. ∏s=0n−1ei2sx=eix∑s=0n−12s=eix(2n−1)\prod_{s=0}^{n-1} e^{i2^s x} = e^{i x \sum_{s=0}^{n-1} 2^s} = e^{ix (2^n-1)}.
  2. ∏s=0n−1(ei2sx+1)\prod_{s=0}^{n-1} (e^{i2^s x}+1). We use the identity ∏k=0m−1(A2k+1)=A2m−1A−1\prod_{k=0}^{m-1} (A^{2^k}+1) = \frac{A^{2^m}-1}{A-1}. Here, A=eixA = e^{ix} and m=nm=n. So, ∏s=0n−1(ei2sx+1)=(eix)2n−1eix−1=ei2nx−1eix−1\prod_{s=0}^{n-1} (e^{i2^s x}+1) = \frac{(e^{ix})^{2^n}-1}{e^{ix}-1} = \frac{e^{i2^n x}-1}{e^{ix}-1}. Combining these, the denominator of fn(x)f_n(x) is ei(2n−1)xei2nx−1eix−1e^{i(2^n-1)x} \frac{e^{i2^n x}-1}{e^{ix}-1}.

Now for the numerator: ∏s=0n−1(ei3⋅2sx+1)\prod_{s=0}^{n-1} (e^{i3 \cdot 2^s x}+1). Let X=3xX = 3x. This product is ∏s=0n−1(ei2sX+1)=ei2nX−1eiX−1=ei3⋅2nx−1ei3x−1\prod_{s=0}^{n-1} (e^{i2^s X}+1) = \frac{e^{i2^n X}-1}{e^{iX}-1} = \frac{e^{i3 \cdot 2^n x}-1}{e^{i3x}-1}.

So, the general formula for fn(x)f_n(x) is: fn(x)=ei3⋅2nx−1ei3x−1ei(2n−1)xei2nx−1eix−1f_n(x) = \frac{\frac{e^{i3 \cdot 2^n x}-1}{e^{i3x}-1}}{e^{i(2^n-1)x} \frac{e^{i2^n x}-1}{e^{ix}-1}}. This formula is valid as long as the denominators are non-zero. That means eix≠1e^{ix} \ne 1, ei3x≠1e^{i3x} \ne 1, ei2nx≠1e^{i2^n x} \ne 1.

Let's evaluate the options for n=5n=5.

(C) f5(2π31)f_5(\frac{2\pi}{31}) Here n=5n=5, x=2π31x = \frac{2\pi}{31}. Denominator: ei(25−1)xei25x−1eix−1=ei31xei32x−1eix−1e^{i(2^5-1)x} \frac{e^{i2^5 x}-1}{e^{ix}-1} = e^{i31x} \frac{e^{i32x}-1}{e^{ix}-1}. 31x=31⋅2π31=2π31x = 31 \cdot \frac{2\pi}{31} = 2\pi. So ei31x=1e^{i31x} = 1. 32x=32⋅2π31=64π31=2π+2π3132x = 32 \cdot \frac{2\pi}{31} = \frac{64\pi}{31} = 2\pi + \frac{2\pi}{31}. So ei32x=ei2π/31=eixe^{i32x} = e^{i2\pi/31} = e^{ix}. The denominator is 1⋅eix−1eix−1=11 \cdot \frac{e^{ix}-1}{e^{ix}-1} = 1. (Since eix≠1e^{ix} \ne 1).

Numerator: ei3⋅25x−1ei3x−1=ei96x−1ei3x−1\frac{e^{i3 \cdot 2^5 x}-1}{e^{i3x}-1} = \frac{e^{i96x}-1}{e^{i3x}-1}. 96x=96⋅2π31=192π31=6π+6π3196x = 96 \cdot \frac{2\pi}{31} = \frac{192\pi}{31} = 6\pi + \frac{6\pi}{31}. So ei96x=ei6π/31e^{i96x} = e^{i6\pi/31}. 3x=3⋅2π31=6π313x = 3 \cdot \frac{2\pi}{31} = \frac{6\pi}{31}. So ei3x=ei6π/31e^{i3x} = e^{i6\pi/31}. The numerator is ei6π/31−1ei6π/31−1=1\frac{e^{i6\pi/31}-1}{e^{i6\pi/31}-1} = 1. (Since ei6π/31≠1e^{i6\pi/31} \ne 1).

So, f5(2π31)=11=1f_5(\frac{2\pi}{31}) = \frac{1}{1} = 1. Option (C) states f5(2π31)=1f_5(\frac{2\pi}{31}) = 1. So (C) is true.