integral with fractional part.

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Vali

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n is a natural number different from 0.
\(\displaystyle \int_{0}^{1}\left \{ nx \right \}^{2}dx\)
Some questions:
So I noted nx=t and I got \(\displaystyle \frac{1}{n}\int_{0}^{n}\left \{ t \right \}^{2}dt\)
From here I have some questions.How to find to periodicity of a fractional function ?Why the periodicity of {t}^2 = 1 ?
If I take f(x)={x}, I can verify if 1 is the period of f(x) like that f(x)=f(x+1) which is true.
But how to find the period for {2x},{3x},...{nx} ?How to know which value to introduce here f(x)=f(x+?)
I found that for {2x} is 1/2, for {3x} is 1/3 ... {nx} is 1/n.But how I see these values?
How about when these fractional parts are squared?Like in my case, the period is the same?
In my case, I also find out that if x is from (0,1) then {x}=x so because the period of {t}^2 is 1 (which I don't know why) the integral becomes \(\displaystyle \frac{1}{n}\cdot n\int_{0}^{1}t^{2}dt=\frac{1}{3}\)
 
Evidently you are defining {x} as the fractional part of x, which is x - [x], or x - floor(x).

The period of {x} is 1, as you say. The period of {nx} is 1/n, since increasing x by 1/n increases nx by 1.

When you square two numbers that are the same, the squares are the same. That is, if f(x+1) = f(x), then f(x+1)^2 = f(x)^2. So clearly f(x)^2 has the same period as f(x).

It isn't quite clear which things that you are asking about you are sure of, and which you are asking for. I imagine you are asking because you are trying to follow someone else's proof, and you don't know why some steps are valid. Please clarify what you still don't understand.
 
Vali said:
n is a natural number different from 0.
\(\displaystyle \int_{0}^{1}\left \{ nx \right \}^{2}dx\)
Some questions:
So I noted nx=t and I got \(\displaystyle \frac{1}{n}\int_{0}^{n}\left \{ t \right \}^{2}dt\)
From here I have some questions.How to find to periodicity of a fractional function ?Why the periodicity of {t}^2 = 1 ?
If I take f(x)={x}, I can verify if 1 is the period of f(x) like that f(x)=f(x+1) which is true.
But how to find the period for {2x},{3x},...{nx} ?How to know which value to introduce here f(x)=f(x+?)
I found that for {2x} is 1/2, for {3x} is 1/3 ... {nx} is 1/n.But how I see these values?
How about when these fractional parts are squared?Like in my case, the period is the same?
In my case, I also find out that if x is from (0,1) then {x}=x so because the period of {t}^2 is 1 (which I don't know why) the integral becomes \(\displaystyle \frac{1}{n}\cdot n\int_{0}^{1}t^{2}dt=\frac{1}{3}\)
Click to expand...
Frankly, I am not sure what to make of this?
The fractional part function \(\displaystyle \left\{ x \right\} = x - \left\lfloor x \right\rfloor\).
Because \(\displaystyle 0\le\{x\}<1\) then \(\displaystyle 0\le\{x\}^2<\{x\}<1\).
That means \(\displaystyle \int_{0}^{1}\left \{ x \right \}^{2}dx\le\int_{0}^{1}\left \{ x \right \}dx\).
Surely you know that if \(\displaystyle x\in(0,1)\) then \(\displaystyle \{x\}=x\) SEE HERE

 
Dr.Peterson said:
Evidently you are defining {x} as the fractional part of x, which is x - [x], or x - floor(x).

The period of {x} is 1, as you say. The period of {nx} is 1/n, since increasing x by 1/n increases nx by 1.

When you square two numbers that are the same, the squares are the same. That is, if f(x+1) = f(x), then f(x+1)^2 = f(x)^2. So clearly f(x)^2 has the same period as f(x).

It isn't quite clear which things that you are asking about you are sure of, and which you are asking for. I imagine you are asking because you are trying to follow someone else's proof, and you don't know why some steps are valid. Please clarify what you still don't understand.
Click to expand...

That thing I still don't understand.There's an easier way to understand this?Like to draw on an axis or something.
Thank you for your answer.
 
Dr.Peterson said:
The period of {nx} is 1/n, since increasing x by 1/n increases nx by 1.
Click to expand...
Vali said:
That thing I still don't understand.There's an easier way to understand this?Like to draw on an axis or something.
Thank you for your answer.
Click to expand...

What does it mean to have a period of p? It means that f(x+p) = f(x) for all x.

Let p = 1/n, and f(x) = {nx}. Then f(x + p) = f(x + 1/n) = {n(x + 1/n)} = {nx + 1} = {nx} = f(x) because {x} has period 1.

If that is not clear, graph the function {3x} and observe that it repeats every 1/3. You may also observe that it is the graph of {x}, compressed by a factor of 3, which explains the period.
 
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