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Calculus Level 3

1 2 × 1 2 + 1 2 1 2 × 1 2 + 1 2 1 2 + 1 2 1 2 × ⋯ = m π m = ? \sqrt{\frac{1}{2}}\times\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\times\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\times\cdots = \frac m \pi \\ m = \, ?


The answer is 2.

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Parth Lohomi by Parth Lohomi , 20, India

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2 solutions

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Parth Lohomi
Dec 29, 2014

cos ⁡ π 4 = 1 2 , π 4 = x \cos{\dfrac{\pi}{4}}=\sqrt{\dfrac{1}{2}},\dfrac{\pi}{4}= x

1 2 + 1 2 1 2 = 1 + cos ⁡ x 2 = cos ⁡ x 2 \sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}}}=\sqrt{\dfrac{1+\cos{x}}{2}}=\cos{\dfrac{x}{2}}

1 2 + 1 2 1 2 + 1 2 1 2 = cos ⁡ x 4 \sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}+\dfrac{1}{2}\sqrt{\dfrac{1}{2}}}}=\cos\dfrac{x}{4}

Thus it becomes

P ( n ) = cos ⁡ x 2 n cos ⁡ x 2 n − 1 ⋯ cos ⁡ x P(n) =\cos{\dfrac{x}{2^n}}\cos{\dfrac{x}{2^{n-1}}}\cdots\cos{x}

P ( n ) = 2 sin ⁡ x 2 n − 1 cos ⁡ x 2 n − 1 cos ⁡ x 2 n − 2 ⋯ cos ⁡ x 2 2 sin ⁡ x 2 n − 1 P(n) =\dfrac{2\sin{\dfrac{x}{2^{n-1}}}\cos{\dfrac{x}{2^{n-1}}}\cos{\dfrac{x}{2^{n-2}}}\cdots\cos{\dfrac{x}{2}}}{2\sin{\dfrac{x}{2^{n-1}}}}

2 sin ⁡ x cos ⁡ x = sin ⁡ 2 x 2\sin{x}\cos{x}=\sin{2x}

lim ⁡ n → ∞ sin ⁡ 2 x 2 n sin ⁡ x 2 n − 1 \lim_{n\rightarrow\infty}\dfrac{\sin{2x}}{2^n\sin{\dfrac{x}{2^{n-1}}}}

lim ⁡ n → ∞ sin ⁡ 2 x 2 x \lim_{n\rightarrow\infty}\dfrac{\sin{2x}}{2x}

2 sin ⁡ π 2 π = 2 π \dfrac{2\sin\dfrac{\pi}{2}}{\pi}=\dfrac{2}{\pi}

So

m = 2 \boxed{m=2}

29 Helpful 0 Interesting 1 Brilliant 0 Confused

can you please elaborate your explanation a bit further ? I don't understand the part with the second P(n)

Radinoiu Damian - 6 years, 5 months ago

P ( n ) = 2 s i n x 2 n − 1 c o s x 2 n − 1 c o s x 2 n − 2 . . . . . c o s x 2 2 s i n x 2 n − 1 P(n) = \dfrac{2sin\dfrac{x}{2^{n-1}}cos\dfrac{x}{2^{n-1}}cos\dfrac{x}{2^{n-2}}..... cos\dfrac{x}{2}}{2sin\dfrac{x}{2^{n-1}}}

2 s i n x c o s x = s i n 2 x 2sinxcosx=sin2x

thus we get(multiplying and divivding by 2 n times and applying the above identity in each step)

l i m n → ∞ P ( n ) = l i m n → ∞ s i n 2 x 2 n s i n x 2 n − 1 lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sin2x}{2^{n}sin\dfrac{x}{2^{n-1}}}

l i m n → ∞ P ( n ) = l i m n → ∞ s i n x 2 n s i n x 2 n − 1 x 2 n − 1 × x 2 n − 1 lim_{n \to \infty}P(n) = lim_{n \to \infty}\dfrac{sinx}{2^{n}\dfrac{sin\dfrac{x}{2^{n-1}}}{\dfrac{x}{2^{n - 1}}}\times\dfrac{x}{2^{n -1}}}

As ( x ∞ → 0 \dfrac{x}{\infty} \to 0 )

l i m n → ∞ P ( n ) = s i n 2 x 2 x lim_{n \to \infty}P(n) = \dfrac{sin2x}{2x}

2 s i n π 2 π = 2 π \dfrac{2sin\dfrac{\pi}{2}}{\pi} = \dfrac{2}{\pi}

U Z - 6 years, 5 months ago

Thanks, I see your solution now. I've written mine also, though I'm not that good with Latex (it is my first attempt)

Radinoiu Damian - 6 years, 5 months ago

@Radinoiu Damian – For posting problems with mathematical expressions refer this

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U Z - 6 years, 5 months ago

Last two steps , kindly explain!

BHinstitute Harendran B - 6 years, 5 months ago

Please referer to @megh choksi 's comment below

Parth Lohomi - 6 years, 4 months ago
Radinoiu Damian
Jan 3, 2015

The first part is the same as Parth's solution, you have to substitute cos ⁡ ( π 4 ) = cos ⁡ ( x ) \cos(\frac{\pi}{4}) = \cos(x) . As you follow the terms you get cos ⁡ ( π 4 ) ⋅ cos ⁡ ( π 8 ) ⋅ cos ⁡ ( π 16 ) . . . \cos(\frac{\pi}{4}) \cdot \cos(\frac{\pi}{8}) \cdot \cos(\frac{\pi}{16}) ... and so on. Now we know that sin ⁡ ( x ) x = cos ⁡ ( x 2 ) ⋅ cos ⁡ ( x 4 ) ⋅ cos ⁡ ( x 8 ) . . . \frac{\sin(x)}{x} = \cos(\frac{x}{2}) \cdot \cos(\frac{x}{4}) \cdot \cos(\frac{x}{8}) ... . Thus, sin ⁡ ( x ) x ⋅ cos ⁡ ( x ) = m π \frac{\sin(x)}{x} \cdot \cos(x) = \frac {m}{\pi} . After this, we can use the well known goniometric formula for sin ⁡ ( x ) ⋅ cos ⁡ ( x ) = 1 2 ⋅ [ sin ⁡ ( x + x ) + sin ⁡ ( x − x ) ] = 1 2 ⋅ [ sin ⁡ ( 2 x ) + 0 ] \sin(x) \cdot \cos(x) = \frac{1}{2} \cdot [\sin(x+x) + \sin(x-x)] = \frac{1}{2} \cdot [\sin(2x) + 0] . Now, remember that we said x = π 4 x = \frac{\pi}{4} That implies, sin ⁡ ( π 2 ) π 2 = m π \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} = \frac{m}{\pi} Finally, m = 2 m = \boxed{2}

18 Helpful 0 Interesting 0 Brilliant 0 Confused

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Nicely done! For further reading, you may read up this page .

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