MHB Evaluate Integral: \(\cos x \cdot \cdot \cdot \cos 2^{2018}x\)

[lfdahl]

Evaluate

\[\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx\]

 

[Prove It]

Evaluate

\[\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx\]

Is that -1 supposed to be there?

 

[lfdahl]

Is that -1 supposed to be there?

Yes!

 

[I like Serena]

Evaluate

\[\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx\]

Spoiler

We can prove with induction and the product-to-sum-identity that:
$$\prod_{k=0}^n \cos 2^k x = \frac 1{2^n}\sum_{k=0}^{2^n-1}\cos (2k+1)x
$$
Thus:
$$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\
=\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\
=\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi} 1 + \sum_{k=1}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\
=\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1}\frac{\sin(2^{k+1}-2)x}{2^{k+1}-2} \right]_{0}^{2\pi} \\
=\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}\\
$$

 

[Opalg]

Evaluate

\[\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \: dx\]

[sp]Using the identity $\cos\theta \cos\phi = \frac12\bigl(\cos(\theta+\phi) + \cos(\theta-\phi)\bigr)$, $$\begin{aligned} \cos x\cos 2x &= 2^{-1}(\cos x + \cos 3x) \\ \cos x\cos 2x\cos 4x &= 2^{-1}(\cos x + \cos 3x)\cos 4x = 2^{-2}(\cos x + \cos 3x + \cos 5x + \cos 7x) \\ \cos x\cos 2x\cos 4x \cos8x &= 2^{-3}(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos 15x) \\ &\vdots \\ \cos x\cos 2x\cos 4x \cdots \cos(2^nx) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr). \end{aligned}$$ Then $$\begin{aligned}\cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr)\cos((2^{n+1}-1)x) \\ &= 2^{-n-1}\bigl(1 + \cos 2x + \cos 4x + \cos 6x + \cos 8x + \ldots + \cos ((2^{n+2}-2)x)\bigr). \end{aligned}$$ But $$\int_0^{2\pi}1\,dx = 2\pi,$$ and $$\int_0^{2\pi}\cos2k\pi\,dx = 0$$ for all nonzero integers $k$. Therefore $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x)\,dx = \frac{2\pi}{2^{n+1}} = \frac\pi{2^n}.$$ Finally, put $n=2017$ to get $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^{2017}x) \cos((2^{2018}-1)x)\,dx = \frac\pi{2^{2017}}.$$

[/sp]

Edit: Beaten to it by ILS!

 

[lfdahl]

Spoiler

We can prove with induction and the product-to-sum-identity that:
$$\prod_{k=0}^n \cos 2^k x = \frac 1{2^n}\sum_{k=0}^{2^n-1}\cos (2k+1)x
$$
Thus:
$$\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\
=\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\
=\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi} 1 + \sum_{k=1}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \\
=\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1}\frac{\sin(2^{k+1}-2)x}{2^{k+1}-2} \right]_{0}^{2\pi} \\
=\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}\\
$$

Great job, I like Serena! Thankyou for your participation and a correct answer. Would you please help me understand the following step? I´m sure, you are right. It´s just my inert understanding of a detail in your nice solution:

Spoiler

It is the following equality:

\[\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \]

Thankyou for being patient with me!(Blush)

 

[lfdahl]

[sp]Using the identity $\cos\theta \cos\phi = \frac12\bigl(\cos(\theta+\phi) + \cos(\theta-\phi)\bigr)$, $$\begin{aligned} \cos x\cos 2x &= 2^{-1}(\cos x + \cos 3x) \\ \cos x\cos 2x\cos 4x &= 2^{-1}(\cos x + \cos 3x)\cos 4x = 2^{-2}(\cos x + \cos 3x + \cos 5x + \cos 7x) \\ \cos x\cos 2x\cos 4x \cos8x &= 2^{-3}(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos 15x) \\ &\vdots \\ \cos x\cos 2x\cos 4x \cdots \cos(2^nx) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr). \end{aligned}$$ Then $$\begin{aligned}\cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x) &= 2^{-n}\bigl(\cos x + \cos 3x + \cos 5x + \cos 7x + \ldots + \cos ((2^{n+1}-1)x)\bigr)\cos((2^{n+1}-1)x) \\ &= 2^{-n-1}\bigl(1 + \cos 2x + \cos 4x + \cos 6x + \cos 8x + \ldots + \cos ((2^{n+2}-2)x)\bigr). \end{aligned}$$ But $$\int_0^{2\pi}1\,dx = 2\pi,$$ and $$\int_0^{2\pi}\cos2k\pi\,dx = 0$$ for all nonzero integers $k$. Therefore $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^nx) \cos((2^{n+1}-1)x)\,dx = \frac{2\pi}{2^{n+1}} = \frac\pi{2^n}.$$ Finally, put $n=2017$ to get $$\int_0^{2\pi} \cos x\cos 2x\cos 4x \cdots \cos(2^{2017}x) \cos((2^{2018}-1)x)\,dx = \frac\pi{2^{2017}}.$$

[/sp]

Edit: Beaten to it by ILS!

Thankyou, Opalg for a very clear and lucid solution!

Just a small typo:

Spoiler

I´d expect the integrand $\cos 2k\pi$ to be $\cos 2kx$.

 

[I like Serena]

Great job, I like Serena! Thankyou for your participation and a correct answer. Would you please help me understand the following step? I´m sure, you are right. It´s just my inert understanding of a detail in your nice solution:

Spoiler

It is the following equality:

\[\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x \, dx \]

Thankyou for being patient with me!(Blush)

Sure!

Spoiler

It's a reordering of terms:
\[\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \\
= \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos \big(2^{2018}+2\cdot(2^{2017} -1)\big)x\Big) \\
\phantom{==} + \Big(\cos (2^{2018}-2)x + ... + \cos \big(2^{2018}-2\cdot(2^{2017}-1)-2\big)x\Big) \\
= \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\Big) \\
\phantom{==}+ \Big(\cos (2^{2018}-2)x + ... + \cos(2^{0+1}-2)x\Big)\\
= \cos(2^{0+1}-2)x + ... + \cos (2^{2018}-2)x + \cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\\
=\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x
\]

(Angel)

 

[lfdahl]

Sure!

Spoiler

It's a reordering of terms:
\[\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \\
= \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos \big(2^{2018}+2\cdot(2^{2017} -1)\big)x\Big) \\
\phantom{==} + \Big(\cos (2^{2018}-2)x + ... + \cos \big(2^{2018}-2\cdot(2^{2017}-1)-2\big)x\Big) \\
= \Big(\cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\Big) \\
\phantom{==}+ \Big(\cos (2^{2018}-2)x + ... + \cos(2^{0+1}-2)x\Big)\\
= \cos(2^{0+1}-2)x + ... + \cos (2^{2018}-2)x + \cos 2^{2018}x + \cos (2^{2018}+2)x + ... + \cos (2^{2018+1} - 2)x\\
=\sum_{k=0}^{2^{2018}-1}\cos (2^{k+1}-2)x
\]

(Angel)

Hello again, I like Serena!

I´m still getting a different result, when I reorder the sum ... (Wondering) Please take a look:

Spoiler

To ease the algebra, I put $2^{2018}=N$.

This is the sum to be reordered:

\[S = \sum_{k=0}^{\frac{N}{2}-1}\left ( \cos (N+2k)x + \cos (N-2k-2)x \right )\]

The factors in the cosine arguments (for $k \in \left \{ 0,1,2,...,\frac{N}{2}-2, \frac{N}{2}-1\right \}$):

\[(N+2k) \in \left \{ N,\, N+2, \, N+4, ... ,\, 2N-6,\, 2N-4, \, 2N-2 \right \} \]
\[ (N-2k-2) \in \left \{ 0,\, 2, \, 4, ... ,\, N-6,\, N-4, \, N-2 \right \}\]

This arithmetic progression is also obtained, when I look at the factor $2k$:
\[2k \in \left \{ 0,2,4, ..., \, N-2,\, N,\, N+2, \, N+4, ...,2N-4, 2N-2\right \}\]
for $k \in \left \{ 0,1,2, ... ,\, N-2, \, N-1 \right \}$

So our sum, $S$, reduces to:

\[S = \sum_{k=0}^{N-1} \cos (2kx) = 1 + \sum_{k=1}^{2^{2018}-1} \cos (2kx)\]

 

[lfdahl]

A small correction to I like Serena´s nice solution:

Spoiler

\[\int_{0}^{2\pi}\cos x \cos 2x \cos 4x \cdot \cdot \cos 2^{2017}x \cos (2^{2018}-1)x \, dx \\ =\int_{0}^{2\pi}\left(\prod_{k=0}^{2017}\cos 2^kx\right) \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2017}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\cos (2k+1)x \cos (2^{2018}-1)x \, dx \\ =\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2017}-1}\left(\cos (2^{2018}+2k)x + \cos (2^{2018}-2k-2)x\right) \, dx \\
=\frac 1{2^{2018}}\int_{0}^{2\pi}\sum_{k=0}^{2^{2018}-1}\cos 2kx \, dx \\
=\frac 1{2^{2018}} \left ( \int_{0}^{2\pi} 1 dx + \sum_{k=1}^{2^{2018}-1}\int_{0}^{2\pi}\cos 2kx \, dx \right ) \\
=\frac 1{2^{2018}}\left[ x + \sum_{k=1}^{2^{2018}-1} \frac{\sin 2kx}{2k} \right]_{0}^{2\pi} \\ =\frac {2\pi}{2^{2018}} = \frac\pi{2^{2017}}\]

I have a suggested/alternative solution to the problem, but Opalg´s exemplary solution simply beats it.
Thankyou very much both of you, I like Serena and Opalg for your valuable contributions and participation!(Yes)