Grandiose Gibberish

Calculus Level 4

Suppose we define a piecewise function, $\text{frankensine}(x)$ as described below.

$\text{frankensine}(x) = \begin{cases} \sin(x) \ \quad, \quad 0 < x < 2\pi \\ \sin(2x) \quad, \quad 2\pi < x < 2\pi\left(1 + \frac12 \right) \\ \sin(3x) \quad, \quad 2\pi\left(1 + \frac12 \right) < x < 2\pi\left(1 + \frac12 + \frac13\right) \\ \sin(4x) \quad, \quad 2\pi\left(1 + \frac12 + \frac13\right) < x < 2\pi\left(1 + \frac12 + \frac13+\frac14\right) \\ \vdots \\ \sin(nx) \quad, \quad 2\pi\left(1 + \frac12 + \frac13+\ldots + \frac1{n-1}\right) < x < 2\pi\left(1 + \frac12 + \frac13+\frac14+\ldots + \frac1n \right) \\ \vdots \end{cases}$

For integer $n$ , let $\displaystyle A = \lim_{n\to\infty} \int_0^n \text{frankensine}(x) \, dx.$

What is the value of $\lfloor 1000A \rfloor$ ?

Inspiration .

The answer is 0.

1 solution

Julian Poon
Sep 17, 2015

$\lim _{ n\to \infty } \int _{ 0 }^{ n } { \text{frankensine} }(x)\, dx=\sum _{ n=1 }^{ \infty }{ \int _{ 2\pi { H }_{ n-1 } }^{ 2\pi { H }_{ n } }{ \sin { (nx) } dx } } =\sum _{ n=1 }^{ \infty }{ \frac { \cos \left( { 2\pi nH }_{ n-1 } \right) }{ n } -\frac { \cos \left( { 2\pi nH }_{ n } \right) }{ n } }$

Where $H_{n}$ is the nth harmonic number.

Using the identity $\cos{u}-\cos{v}=-2\sin{\frac{u+v}{2}}\sin{\frac{u-v}{2}}$

$\sum _{ n=1 }^{ \infty }{ \frac { \cos \left( { 2\pi nH }_{ n-1 } \right) }{ n } -\frac { \cos \left( { 2\pi nH }_{ n } \right) }{ n } } =\sum _{ n=1 }^{ \infty }{ \frac { -2\sin \left( n\pi ({ H }_{ n }+{ H }_{ n-1 }) \right) \sin \left( n\pi ({ H }_{ n-1 }-{ H }_{ n }) \right) }{ n } } =\sum _{ n=1 }^{ \infty }{ \frac { -2\sin \left( n\pi ({ H }_{ n }+{ H }_{ n-1 }) \right) \sin \left( -\pi \right) }{ n } }$ $=\sum _{ n=1 }^{ \infty }{ \frac { 0 }{ n } } =\boxed{0}$

By the way, the function $y=\text{frankensine}(x)$ looks something like this:

Response to Challenge Master's note:

To show that the limit of the function exists, I'll split $\int _{ 2\pi { H }_{ n-1 } }^{ 2\pi { H }_{ n } }{ \sin { (nx) } dx }$ into $\int _{ 2\pi { H }_{ n-1 } }^{ \pi ({ H }_{ n-1 }+{ H }_{ n }) }{ \sin { (nx) } dx } +\int _{ \pi ({ H }_{ n-1 }+{ H }_{ n }) }^{ 2\pi { H }_{ n } }{ \sin { (nx) } dx }$ and show that both terms approach $0$ as $n$ approaches $\infty$ .

$\int _{ 2\pi { H }_{ n-1 } }^{ \pi ({ H }_{ n-1 }+{ H }_{ n }) }{ \sin { (nx) } dx } =\frac { 1 }{ n } \left( \cos { (\pi n({ H }_{ n-1 }+{ H }_{ n })) } -\cos { (2\pi { nH }_{ n-1 }) } \right)$

As $n$ approaches $\infty$ , $\cos { (\pi n({ H }_{ n-1 }+{ H }_{ n })) } = \cos { (2\pi { nH }_{ n-1 }) }$ , since ${ H }_{ n-1 }+{ H }_{ n } \text{ ~ } 2{ H }_{ n }$ .

Therefore, $\frac { 1 }{ n } \left( \cos { (\pi n({ H }_{ n-1 }+{ H }_{ n })) } -\cos { (2\pi { nH }_{ n-1 }) } \right) =0$ as $n$ approaches $\infty$ . A similar method can be done for the other term.

Moderator note:

Let $f(n)$ denote the integral to $n$ . You have not shown that $\lim_{n \rightarrow \infty} f(n)= 0$ . All that you have shown, is that there exists values $H_n$ such that $\lim_{n \rightarrow \infty } f(H_n) = 0$ .

For example, if we consider $\int_0 ^n \sin x \, dx$ , then we would say that the limit to inifinity doesn't exist, because it bounces up to $\frac{\pi}{2}$ and back to 0. Even though with $H_n = 2 n \pi$ , we do have $\in_0 ^ {H_n} \sin x \, dx = 0$ , that doesn't imply that the limit of the function exists.

Do you have the Desmos link? I'm trying to make a part 2 for this question but I need to keep referring to the graph.

Pi Han Goh - 5 years, 9 months ago

No... I deleted the graph... I'll remake it.

Here it is: Here

I don't know about your computer but mine lags a lot for this graph.

Julian Poon - 5 years, 8 months ago

Thanks. It lags my computer as well. Sigh. By the way,was planning to ask for number of points of discontinuity in an interval but I haven't worked out a complete proof yet.

Pi Han Goh - 5 years, 8 months ago

Note that you have not shown that $\lim_{n \rightarrow \infty} f(n)= 0$ . All that you have shown, is that there exists values $H_n$ such that $\lim_{n \rightarrow \infty } f(H_n) = 0$ .

Calvin Lin Staff - 5 years, 8 months ago

Is it sufficient to simply state that as $n$ approaches $\infty$ , $\int _{ 0 }^{ k }{ \sin { nx } } dx$ approaches $0$ where $0<k<\frac{\pi}{n}$ ?

Julian Poon - 5 years, 8 months ago

No, because your statement doesn't (yet) make sense. Is $k$ a constant, or a variable dependent on $n$ ? It can't be a constant since we have $n \rightarrow \infty$ and $0 < k < \frac{ \pi }{n}$ .

If $k$ was a variable, and the statement is proven true for all possible choices of $k_n$ , then yes it will be sufficient. Note that in analysis, the proper presentation of the proof is important, and we cannot say "it is obvious that ...".

Calvin Lin Staff - 5 years, 8 months ago

@Calvin Lin – I have split the integral into 2 parts, mind if you re-look at the solution to see if that is acceptable? Thanks.

Julian Poon - 5 years, 8 months ago

@Julian Poon – No, you are still only evaluating the functions at specific points, and not at every point. Thus, you are finding the limit of a sequence, instead of the limit of the function.

For example, consider $f(x) = \sin ( \pi x )$ . If we evaluated it only at the integers, then we would say that $\lim f(n) = 0$ since each of these terms is 0. However, we know that $\lim f(x)$ does not exist.

Calvin Lin Staff - 5 years, 8 months ago

Little shortcut: Observe that $\cos(2\pi{n}H_n)=\cos(2\pi{n}H_{n-1})$

Otto Bretscher - 5 years, 8 months ago

POST! POST! POST! No details should be omitted this time!

Pi Han Goh - 5 years, 8 months ago

I like Julian's solution! Just streamlining it a bit to make it a one-liner

Otto Bretscher - 5 years, 8 months ago

What about the main part of the problem? The integral $n$ as Challenge Master says. I think using an approximation $H_m \sim \ln(m)$ . And then it can be shown quite easily and clearly. I did it this way.

Kartik Sharma - 5 years, 8 months ago

@Calvin Lin Does this work fine?

Kartik Sharma - 5 years, 8 months ago

No. You are still evaluating a function at specific points, and thus only finding the limit of a sequence. It doesn't matter if the sequence has an accumulation point at infinity.

For example, consider the function $\mathbb{1}_{H_n}$ which is 1 at the points of $H_n$ and 0 otherwise. What is the limit of this function to infinity? If you only evaluate it at $H_n$ , then you would claim that the limit (of the sequence) is 1. However, the limit of the function, doesn't exist.

Calvin Lin Staff - 5 years, 8 months ago

Grandiose Gibberish

The answer is 0.

1 solution

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