"Identity" of volume of a solid of revolution (Limited Weierstrass function)

Question

The question is: Is this correct?

This is pretty much the first thing I've tried to come up with by myself. I wanted to see what would happen if I tried to calculate volume of a solid of revolution of the Weierstrass-like function. I'm pretty bad at calculus so I'm pretty certain there are lot's errors but I wanted to share it anyway and maybe get some feedback.

Please, let the bashing begin!

Let $w(x) := \displaystyle\sum_{k= 0}^{\infty} a^k\cos(b^k\pi x) $ where $0 < a < 1$ and $ab > 1+ \frac{3\pi}{2}, b$ is an odd integer greater than $1$.

Let us limit $w(x)$ to $n$ so that, $w(x)_n := \displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) $. Consider the volume of the solid formed by rotating the area of $w(x)_n$ on the x-axis in the interval $[\alpha,\beta]$ given by,

$V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$

$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))^2dx =$

$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))dx =$

$\pi\displaystyle\int_{\alpha}^{\beta} [ \cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) +$

$\cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $

$a\cos(b \pi x) (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $ $a^2\cos(b^2 \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $$\ldots +$

$a^n\cos(b^n \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) ] dx =$

$\pi\displaystyle\int_{\alpha}^{\beta}[ (\cos^2(\pi x) + a\cos(\pi x)\cos(b \pi x) + a^2\cos(\pi x)\cos(b^2 \pi x) + \ldots + a^n\cos(\pi x)\cos(b^n \pi x)) +$

$( a^2\cos^2(b \pi x)+a\cos(b \pi x) \cos(\pi x) + a^3\cos(b \pi x)\cos(b^2 \pi x) + \ldots + a^{n+1}\cos(b \pi x)\cos(b^n \pi x)) +$

$(a^4\cos^2(b^2 \pi x) + a^2\cos(b^2 \pi x) \cos(\pi x) + a^3\cos(b^2 \pi x) \cos(b \pi x) + \ldots + a^{n+2}\cos(b^2 \pi x) \cos(b^n \pi x)) + \ldots + $

$( a^{2n}\cos^2(b^n \pi x)) + a^n\cos(b^n \pi x)) \cos(\pi x) + a^{n+1}\cos(b^n \pi x)) \cos(b \pi x) + a^{n+2}\cos(b^n \pi x)) \cos(b^2 \pi x) + \ldots + a^{2n-1}\cos(b^n \pi x)) \cos(b^{n-1} \pi x)) ]dx = $

$ \pi\displaystyle\int_{\alpha}^{\beta} ([ \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x) ] + [\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)])dx =$

$ \pi[\displaystyle\int_{\alpha}^{\beta} \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

[ A miracle happens here When can a sum and integral be interchanged? If this is correct, how can I motivate this? ]

$ \pi[ \displaystyle\sum_{k= 1}^{n+1}\displaystyle\int_{\alpha}^{\beta} a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

Note that

$\int \cos^2(\gamma_1 x)dx = \frac{2\gamma_1 x + \sin(2\gamma_1x )}{4\gamma_1} + c$ and

$\int \cos(\gamma_1 x)cos(\gamma_2 x)dx = \frac{\gamma_1 \sin(\gamma_1 x)\cos(\gamma_2 x) - \gamma_2 \cos(\gamma_1 x)\sin(\gamma_2 x) }{\gamma_1^2 -\gamma_2^2} + c, \gamma_1, \gamma_2 \in \mathbb{R} $

Thus,

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] $

Now restrict $\alpha, \beta \in \mathbb{R}: \alpha, \beta \in \mathbb{Z}$

Now, note that, $ \sin(2\pi b^{k-1}x) = \sin(\pi y) = 0, \forall y \in \mathbb{Z}$ $ \sin(\pi x b^i - \pi x b^j) = \sin(\pi (x b^i - x b^j)) = \sin(\pi y) = 0 \forall y \in \mathbb{Z}$ and $\sin(\pi x b^j + \pi x b^i) = \sin(\pi (x b^j + x b^i)) = \sin(\pi y) = o \forall y \in \mathbb{Z}$

Which implies that,

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$ $ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + 0}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)0 + (b^j -b^i)0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}0 ] = \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + 0] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta] = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta $

Thus $V_{w(x)_n} = \pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta, \alpha,\beta \in \mathbb{Z} $

Can this be "extended" to $w(x)$?

Answer 1

please do not up- or downvote this answer. It's just impossible to do this by pure commenting. I copy pasted your calculus and added some remarks.

$V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$

$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))^2dx =$

$\pi\displaystyle\int_{\alpha}^{\beta} (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x))dx =$

here you wrote the first line doulbe, must be a typo:

$\pi\displaystyle\int_{\alpha}^{\beta} [ \cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) +$

$\cos(\pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $

$a\cos(b \pi x) (\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $ $a^2\cos(b^2 \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) + $$\ldots +$

$a^n\cos(b^n \pi x)(\cos(\pi x) + a\cos(b \pi x) + a^2\cos(b^2 \pi x) + \ldots + a^n\cos(b^n \pi x)) ] dx =$

then you sorted some special summands to the beginning which is fine due to the fact that you have a finite sum

$\pi\displaystyle\int_{\alpha}^{\beta}[ (\cos^2(\pi x) + a\cos(\pi x)\cos(b \pi x) + a^2\cos(\pi x)\cos(b^2 \pi x) + \ldots + a^n\cos(\pi x)\cos(b^n \pi x)) +$

$( a^2\cos^2(b \pi x)+a\cos(b \pi x) \cos(\pi x) + a^3\cos(b \pi x)\cos(b^2 \pi x) + \ldots + a^{n+1}\cos(b \pi x)\cos(b^n \pi x)) +$

$(a^4\cos^2(b^2 \pi x) + a^2\cos(b^2 \pi x) \cos(\pi x) + a^3\cos(b^2 \pi x) \cos(b \pi x) + \ldots + a^{n+2}\cos(b^2 \pi x) \cos(b^n \pi x)) + \ldots + $

$( a^{2n}\cos^2(b^n \pi x)) + a^n\cos(b^n \pi x)) \cos(\pi x) + a^{n+1}\cos(b^n \pi x)) \cos(b \pi x) + a^{n+2}\cos(b^n \pi x)) \cos(b^2 \pi x) + \ldots + a^{2n-1}\cos(b^n \pi x)) \cos(b^{n-1} \pi x)) ]dx = $

from here on I assume by $\cup$ you mean $\Sigma$

$ \pi\displaystyle\int_{\alpha}^{\beta} ([ \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x) ] + [\bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)])dx =$

$ \pi[\displaystyle\int_{\alpha}^{\beta} \displaystyle\sum_{k= 1}^{n+1}a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

[ A miracle happens here When can a sum and integral be interchanged? If this is correct, how can I motivate this? ]

finite sum and integral are always commuting - since you are not considering any limits this is fine.

$ \pi[ \displaystyle\sum_{k= 1}^{n+1}\displaystyle\int_{\alpha}^{\beta} a^{2(k-1)}\cos^2(b^{k-1}\pi x)dx + \displaystyle\int_{\alpha}^{\beta} \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

the following claim I just believe you. (maybe I or someone else will check it later)

Note that

$\int \cos^2(\gamma_1 x)dx = \frac{2\gamma_1 x + \sin(2\gamma_1x )}{4\gamma_1} + c$ and

$\int \cos(\gamma_1 x)cos(\gamma_2 x)dx = \frac{\gamma_1 \sin(\gamma_1 x)\cos(\gamma_2 x) - \gamma_2 \cos(\gamma_1 x)\sin(\gamma_2 x) }{\gamma_1^2 -\gamma_2^2} + c, \gamma_1, \gamma_2 \in \mathbb{R} $

ok up to this point

Thus,

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)}\displaystyle\int_{\alpha}^{\beta} \cos^2(b^{k-1}\pi x)dx + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \displaystyle\int_{\alpha}^{\beta}\cos(b^i\pi x)\cos(b^j\pi x)dx] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] $

Now restrict $\alpha, \beta \in \mathbb{R}: \alpha, \beta \in \mathbb{Z}$

you are restricting to the case $x\in\mathbb{R}:\forall i\in\mathbb{Z}:x\cdot b^{i}\in\mathbb{Z}$. since $b$ is odd (and hence not zero) this is means $x\in\mathbb{Z}$ (take $i=0$).

Now, note that, $ \sin(2\pi b^{k-1}x) = \sin(\pi y) = 0, \forall y \in \mathbb{Z}$ $ \sin(\pi x b^i - \pi x b^j) = \sin(\pi (x b^i - x b^j)) = \sin(\pi y) = 0 \forall y \in \mathbb{Z}$ and $\sin(\pi x b^j + \pi x b^i) = \sin(\pi (x b^j + x b^i)) = \sin(\pi y) = o \forall y \in \mathbb{Z}$

Which implies that,

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + \sin(2\pi b^{k-1}x)}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)\sin(\pi x b^i - \pi x b^j) + (b^j -b^i)\sin(\pi x b^j + \pi x b^i)} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$ $ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \Big| \frac{2\pi b^{k-1}x + 0}{4\pi b^{k-1}} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{(-b^j -b^i)0 + (b^j -b^i)0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j} \Big| \frac{0} {2\pi (b^j -b^i) (b^j + b^i)} \Big|_\alpha^\beta ] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + \bigcup _{i,j=0,i\neq j}^{n}a^{i+j}0 ] = \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta + 0] =$

$ \pi[ \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta] = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta $

Thus $V_{w(x)_n} = \pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx = \pi \displaystyle\sum_{k= 1}^{n+1} a^{2(k-1)} \frac{x}{2} \Big|_\alpha^\beta, \alpha,\beta \in \mathbb{Z} $

Can this be ""extended" to $w(x)$?

since you made a very strong restriction on $x$ generalization will be hard. I have a suggenstion but that might also be a dead end: You might try to write $x=\frac{p}{q}$ and then try to apply integral substitution to get rid of $q$ and then be able to apply your formula. after this (if this really should work) you could try to find a density argument since $\mathbb{Q}$ is dense in $\mathbb{R}$ (not even talking about the limit $n\rightarrow\infty$; therefore you could use that the sum on the right hand side is geometric and then you would probably have to apply dominated convergence on the integral at the left hand side of this expression)