Solveeit Logo
Login

Question

Question: Let $f_n(\theta) = \sum_{r=0}^{n} \frac{1}{4^r} \sin^4(2^r \theta)$. Then which of the following alt...

Let fn(θ)=∑r=0n14rsin⁡4(2rθ)f_n(\theta) = \sum_{r=0}^{n} \frac{1}{4^r} \sin^4(2^r \theta). Then which of the following alternative(s) is/are correct?

A

f2(π4)=12f_2(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}

B

f3(π8)=2+24f_3(\frac{\pi}{8}) = \frac{2+\sqrt{2}}{4}

C

f4(3π2)=1f_4(\frac{3\pi}{2}) = 1

D

f5(π)=0f_5(\pi) = 0

Answer

C, D

Explanation

Solution

The function is given by fn(θ)=∑r=0n14rsin⁡4(2rθ)f_n(\theta) = \sum_{r=0}^{n} \frac{1}{4^r} \sin^4(2^r \theta).

We use the identity for sin⁡4x\sin^4 x: We know sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1 - \cos(2x)}{2}. Then sin⁡4x=(1−cos⁡(2x)2)2=1−2cos⁡(2x)+cos⁡2(2x)4\sin^4 x = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4}. Using cos⁡2(2x)=1+cos⁡(4x)2\cos^2(2x) = \frac{1 + \cos(4x)}{2}, we substitute it into the expression for sin⁡4x\sin^4 x: sin⁡4x=1−2cos⁡(2x)+1+cos⁡(4x)24=2−4cos⁡(2x)+1+cos⁡(4x)24=3−4cos⁡(2x)+cos⁡(4x)8\sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4} = \frac{\frac{2 - 4\cos(2x) + 1 + \cos(4x)}{2}}{4} = \frac{3 - 4\cos(2x) + \cos(4x)}{8}.

Now, substitute x=2rθx = 2^r \theta into this identity: sin⁡4(2rθ)=3−4cos⁡(2⋅2rθ)+cos⁡(4⋅2rθ)8=3−4cos⁡(2r+1θ)+cos⁡(2r+2θ)8\sin^4(2^r \theta) = \frac{3 - 4\cos(2 \cdot 2^r \theta) + \cos(4 \cdot 2^r \theta)}{8} = \frac{3 - 4\cos(2^{r+1} \theta) + \cos(2^{r+2} \theta)}{8}.

Multiply by 14r\frac{1}{4^r}: 14rsin⁡4(2rθ)=14r3−4cos⁡(2r+1θ)+cos⁡(2r+2θ)8\frac{1}{4^r} \sin^4(2^r \theta) = \frac{1}{4^r} \frac{3 - 4\cos(2^{r+1} \theta) + \cos(2^{r+2} \theta)}{8}. Let Cr=cos⁡(2r+1θ)4rC_r = \frac{\cos(2^{r+1} \theta)}{4^r}. Then 14rsin⁡4(2rθ)=38⋅4r−48cos⁡(2r+1θ)4r+18cos⁡(2r+2θ)4r\frac{1}{4^r} \sin^4(2^r \theta) = \frac{3}{8 \cdot 4^r} - \frac{4}{8} \frac{\cos(2^{r+1} \theta)}{4^r} + \frac{1}{8} \frac{\cos(2^{r+2} \theta)}{4^r}. Notice that cos⁡(2r+2θ)4r=4⋅cos⁡(2r+2θ)4r+1=4Cr+1\frac{\cos(2^{r+2} \theta)}{4^r} = 4 \cdot \frac{\cos(2^{r+2} \theta)}{4^{r+1}} = 4 C_{r+1}. So, 14rsin⁡4(2rθ)=38⋅4r−12Cr+18(4Cr+1)=38⋅4r−12Cr+12Cr+1\frac{1}{4^r} \sin^4(2^r \theta) = \frac{3}{8 \cdot 4^r} - \frac{1}{2} C_r + \frac{1}{8} (4 C_{r+1}) = \frac{3}{8 \cdot 4^r} - \frac{1}{2} C_r + \frac{1}{2} C_{r+1}.

Now, sum this expression from r=0r=0 to nn: fn(θ)=∑r=0n(38⋅4r−12Cr+12Cr+1)f_n(\theta) = \sum_{r=0}^{n} \left( \frac{3}{8 \cdot 4^r} - \frac{1}{2} C_r + \frac{1}{2} C_{r+1} \right) fn(θ)=38∑r=0n(14)r+12∑r=0n(Cr+1−Cr)f_n(\theta) = \frac{3}{8} \sum_{r=0}^{n} \left(\frac{1}{4}\right)^r + \frac{1}{2} \sum_{r=0}^{n} (C_{r+1} - C_r).

The first sum is a geometric series: ∑r=0n(14)r=1−(1/4)n+11−1/4=1−1/4n+13/4=43(1−14n+1)\sum_{r=0}^{n} \left(\frac{1}{4}\right)^r = \frac{1 - (1/4)^{n+1}}{1 - 1/4} = \frac{1 - 1/4^{n+1}}{3/4} = \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right). So, 38∑r=0n(14)r=38⋅43(1−14n+1)=12(1−14n+1)\frac{3}{8} \sum_{r=0}^{n} \left(\frac{1}{4}\right)^r = \frac{3}{8} \cdot \frac{4}{3} \left(1 - \frac{1}{4^{n+1}}\right) = \frac{1}{2} \left(1 - \frac{1}{4^{n+1}}\right).

The second sum is a telescoping series: ∑r=0n(Cr+1−Cr)=(C1−C0)+(C2−C1)+⋯+(Cn+1−Cn)=Cn+1−C0\sum_{r=0}^{n} (C_{r+1} - C_r) = (C_1 - C_0) + (C_2 - C_1) + \dots + (C_{n+1} - C_n) = C_{n+1} - C_0. Recall Cr=cos⁡(2r+1θ)4rC_r = \frac{\cos(2^{r+1} \theta)}{4^r}. So, Cn+1=cos⁡(2n+2θ)4n+1C_{n+1} = \frac{\cos(2^{n+2} \theta)}{4^{n+1}} and C0=cos⁡(21θ)40=cos⁡(2θ)C_0 = \frac{\cos(2^1 \theta)}{4^0} = \cos(2\theta). Therefore, 12∑r=0n(Cr+1−Cr)=12(cos⁡(2n+2θ)4n+1−cos⁡(2θ))\frac{1}{2} \sum_{r=0}^{n} (C_{r+1} - C_r) = \frac{1}{2} \left( \frac{\cos(2^{n+2} \theta)}{4^{n+1}} - \cos(2\theta) \right).

Combining both parts, the general formula for fn(θ)f_n(\theta) is: fn(θ)=12(1−14n+1)+12(cos⁡(2n+2θ)4n+1−cos⁡(2θ))f_n(\theta) = \frac{1}{2} \left(1 - \frac{1}{4^{n+1}}\right) + \frac{1}{2} \left( \frac{\cos(2^{n+2} \theta)}{4^{n+1}} - \cos(2\theta) \right) fn(θ)=12−12⋅4n+1+cos⁡(2n+2θ)2⋅4n+1−cos⁡(2θ)2f_n(\theta) = \frac{1}{2} - \frac{1}{2 \cdot 4^{n+1}} + \frac{\cos(2^{n+2} \theta)}{2 \cdot 4^{n+1}} - \frac{\cos(2\theta)}{2} fn(θ)=12−cos⁡(2θ)2+cos⁡(2n+2θ)−12⋅4n+1f_n(\theta) = \frac{1}{2} - \frac{\cos(2\theta)}{2} + \frac{\cos(2^{n+2} \theta) - 1}{2 \cdot 4^{n+1}}.

Now let's check each option:

(A) f2(π4)f_2(\frac{\pi}{4}): n=2n=2, θ=π4\theta = \frac{\pi}{4}. f2(π4)=12−cos⁡(2⋅π4)2+cos⁡(22+2⋅π4)−12⋅42+1f_2(\frac{\pi}{4}) = \frac{1}{2} - \frac{\cos(2 \cdot \frac{\pi}{4})}{2} + \frac{\cos(2^{2+2} \cdot \frac{\pi}{4}) - 1}{2 \cdot 4^{2+1}} f2(π4)=12−cos⁡(π2)2+cos⁡(16⋅π4)−12⋅43f_2(\frac{\pi}{4}) = \frac{1}{2} - \frac{\cos(\frac{\pi}{2})}{2} + \frac{\cos(16 \cdot \frac{\pi}{4}) - 1}{2 \cdot 4^3} f2(π4)=12−02+cos⁡(4π)−12⋅64f_2(\frac{\pi}{4}) = \frac{1}{2} - \frac{0}{2} + \frac{\cos(4\pi) - 1}{2 \cdot 64} f2(π4)=12+1−1128=12+0=12f_2(\frac{\pi}{4}) = \frac{1}{2} + \frac{1 - 1}{128} = \frac{1}{2} + 0 = \frac{1}{2}. Option (A) states f2(π4)=12f_2(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}. Since 12≠12\frac{1}{2} \neq \frac{1}{\sqrt{2}}, (A) is incorrect.

(B) f3(π8)f_3(\frac{\pi}{8}): n=3n=3, θ=π8\theta = \frac{\pi}{8}. f3(π8)=12−cos⁡(2⋅π8)2+cos⁡(23+2⋅π8)−12⋅43+1f_3(\frac{\pi}{8}) = \frac{1}{2} - \frac{\cos(2 \cdot \frac{\pi}{8})}{2} + \frac{\cos(2^{3+2} \cdot \frac{\pi}{8}) - 1}{2 \cdot 4^{3+1}} f3(π8)=12−cos⁡(π4)2+cos⁡(32⋅π8)−12⋅44f_3(\frac{\pi}{8}) = \frac{1}{2} - \frac{\cos(\frac{\pi}{4})}{2} + \frac{\cos(32 \cdot \frac{\pi}{8}) - 1}{2 \cdot 4^4} f3(π8)=12−1/22+cos⁡(4π)−12⋅256f_3(\frac{\pi}{8}) = \frac{1}{2} - \frac{1/\sqrt{2}}{2} + \frac{\cos(4\pi) - 1}{2 \cdot 256} f3(π8)=12−24+1−1512f_3(\frac{\pi}{8}) = \frac{1}{2} - \frac{\sqrt{2}}{4} + \frac{1 - 1}{512} f3(π8)=12−24+0=2−24f_3(\frac{\pi}{8}) = \frac{1}{2} - \frac{\sqrt{2}}{4} + 0 = \frac{2 - \sqrt{2}}{4}. Option (B) states f3(π8)=2+24f_3(\frac{\pi}{8}) = \frac{2+\sqrt{2}}{4}. Since 2−24≠2+24\frac{2 - \sqrt{2}}{4} \neq \frac{2+\sqrt{2}}{4}, (B) is incorrect.

(C) f4(3π2)f_4(\frac{3\pi}{2}): n=4n=4, θ=3π2\theta = \frac{3\pi}{2}. f4(3π2)=12−cos⁡(2⋅3π2)2+cos⁡(24+2⋅3π2)−12⋅44+1f_4(\frac{3\pi}{2}) = \frac{1}{2} - \frac{\cos(2 \cdot \frac{3\pi}{2})}{2} + \frac{\cos(2^{4+2} \cdot \frac{3\pi}{2}) - 1}{2 \cdot 4^{4+1}} f4(3π2)=12−cos⁡(3π)2+cos⁡(64⋅3π2)−12⋅45f_4(\frac{3\pi}{2}) = \frac{1}{2} - \frac{\cos(3\pi)}{2} + \frac{\cos(64 \cdot \frac{3\pi}{2}) - 1}{2 \cdot 4^5} f4(3π2)=12−−12+cos⁡(32⋅3π)−12⋅1024f_4(\frac{3\pi}{2}) = \frac{1}{2} - \frac{-1}{2} + \frac{\cos(32 \cdot 3\pi) - 1}{2 \cdot 1024} f4(3π2)=12+12+cos⁡(96π)−12048f_4(\frac{3\pi}{2}) = \frac{1}{2} + \frac{1}{2} + \frac{\cos(96\pi) - 1}{2048} f4(3π2)=1+1−12048=1+0=1f_4(\frac{3\pi}{2}) = 1 + \frac{1 - 1}{2048} = 1 + 0 = 1. Option (C) states f4(3π2)=1f_4(\frac{3\pi}{2}) = 1. This is correct.

(D) f5(π)f_5(\pi): n=5n=5, θ=π\theta = \pi. f5(π)=12−cos⁡(2π)2+cos⁡(25+2⋅π)−12⋅45+1f_5(\pi) = \frac{1}{2} - \frac{\cos(2\pi)}{2} + \frac{\cos(2^{5+2} \cdot \pi) - 1}{2 \cdot 4^{5+1}} f5(π)=12−12+cos⁡(128π)−12⋅46f_5(\pi) = \frac{1}{2} - \frac{1}{2} + \frac{\cos(128\pi) - 1}{2 \cdot 4^6} f5(π)=0+1−12⋅4096=0+0=0f_5(\pi) = 0 + \frac{1 - 1}{2 \cdot 4096} = 0 + 0 = 0. Option (D) states f5(π)=0f_5(\pi) = 0. This is correct.

The correct alternatives are (C) and (D).

Explanation of the solution:

  1. Transform sin⁡4x\sin^4 x: The key step is to express sin⁡4x\sin^4 x in terms of cos⁡(2x)\cos(2x) and cos⁡(4x)\cos(4x) using the half-angle identities: sin⁡2x=1−cos⁡(2x)2\sin^2 x = \frac{1-\cos(2x)}{2} and cos⁡2x=1+cos⁡(2x)2\cos^2 x = \frac{1+\cos(2x)}{2}. This leads to sin⁡4x=3−4cos⁡(2x)+cos⁡(4x)8\sin^4 x = \frac{3 - 4\cos(2x) + \cos(4x)}{8}.
  2. Substitute into the sum term: Replace xx with 2rθ2^r \theta in the transformed identity and divide by 4r4^r. This yields 14rsin⁡4(2rθ)=38⋅4r−12cos⁡(2r+1θ)4r+12cos⁡(2r+2θ)4r+1\frac{1}{4^r} \sin^4(2^r \theta) = \frac{3}{8 \cdot 4^r} - \frac{1}{2} \frac{\cos(2^{r+1} \theta)}{4^r} + \frac{1}{2} \frac{\cos(2^{r+2} \theta)}{4^{r+1}}.
  3. Identify Telescoping Sum: Define Cr=cos⁡(2r+1θ)4rC_r = \frac{\cos(2^{r+1} \theta)}{4^r}. The term becomes 38⋅4r−12Cr+12Cr+1\frac{3}{8 \cdot 4^r} - \frac{1}{2} C_r + \frac{1}{2} C_{r+1}. This structure allows the sum of the CrC_r terms to telescope.
  4. Evaluate the Sums:
    • The first part, ∑r=0n38⋅4r\sum_{r=0}^{n} \frac{3}{8 \cdot 4^r}, is a geometric series sum.
    • The second part, ∑r=0n(12Cr+1−12Cr)\sum_{r=0}^{n} \left(\frac{1}{2} C_{r+1} - \frac{1}{2} C_r\right), is a telescoping sum which simplifies to 12(Cn+1−C0)\frac{1}{2}(C_{n+1} - C_0).
  5. Derive General Formula: Combine the results of the two sums to get a general formula for fn(θ)f_n(\theta): fn(θ)=12−cos⁡(2θ)2+cos⁡(2n+2θ)−12⋅4n+1f_n(\theta) = \frac{1}{2} - \frac{\cos(2\theta)}{2} + \frac{\cos(2^{n+2} \theta) - 1}{2 \cdot 4^{n+1}}.
  6. Verify Options: Substitute the specific values of nn and θ\theta from each option into the derived general formula to check their correctness.

Answer:

The correct options are (C) and (D).