# Proving E[x]=0 for x=cos(2n*pi/n)

• nikki92
In summary, x is cosine function of 2pi divided by n. Each value is equally likely. The expected value is 0.
nikki92

x=cos(2n*pi/n)

E[x]=0;

## The Attempt at a Solution

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In general: $$E[x]=\sum_{n=0}^N p(x_n)x_n$$

You have: $$x\in \{ x_n:x_n=2\cos \frac{2\pi n}{N}\}$$
I'm guessing from your working that ##p(x_n)=1/N## - i.e. each value is equally likely?
... is that correct? (It helps to write the problem statement out in full.)

You can check your answers by sketching the distribution of x out and looking at the symmetry. The expetation should be sort-of a middle-ish value.

i.e. Looking at the first expectation:$$E[x]=\frac{2}{N}\sum_{n=0}^N \cos \frac{2\pi n}{N}$$ ... the cosine function of n has equal weighting to values on either side of zero - for every value of x that is, say x=a:a>0, there is another value that is x=-a:a>0, so the sum comes to zero.

* You may want to check that this is the case: i.e. what if N is an odd number? What if it is not a multiple of 4?
However, you may be doing this in the limit that N is large.

Thus: ##(E[x])^2=0## also.

$$E[x^2]=\frac{4}{N}\sum_{n=0}^N \cos^2 \frac{2\pi n}{N}$$ ... your answer says that ##E[x^2]=2## ... is that correct?

You can check this out the same way ... ##A\cos^2(\theta)## is evenly distributed about a particular number.
What is that number? Compare with the expression for ##x_n##.

There is really no details on the problem as it was not really homework.But I assumed n=1 to N and each were equally likely.

I do not understand your approach.

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I do not understand your approach.
You have asked if what you have done is correct. This suggests to me that you don't know how to check or that you do not understand the subject.

I am suggesting you sketch a graph of how x and x^2 varies with n and use the symmetry of that graph, along with your understanding of what "expectation value" means, to check to see if your answers are plausible. If the symmetry is easy (it is) then it is easy to tell if you are right or not.

i.e. you have got E[x]=0 - so sketch x vs n and see if x=0 is a plausible average value.

But if you don't understand "expectation value" then that won't make much sense to you: this is part of what I'm trying to find out.

i.e. There is a complication in the formula if N is a small number.

say if ##N=2##
then ##x_1=2\cos(\pi)=0, x_2=2\cos(2\pi)=2\implies E[x]=(x_1+x_2)/2=(0+2)/2=1## ...

It can also make a difference where you start your sum from.
If you start from ##n=1##, then the expectation works out as above, but if you start from ##n=0##:

##E[x]=(x_0+x_1+x_2)/3 = (2+0+2)/3=4/3##

... notice: in both cases: ##(E[x])^2\neq 0##

So: is N a small number or a large number? (see the definition of the expectation value)
Details matter - what is the original problem statement?

1 person
Thanks I understand it much better now.

## 1. What does it mean to prove E[x]=0 for x=cos(2n*pi/n)?

Proving E[x]=0 for x=cos(2n*pi/n) means to show that the expected value of the random variable x, defined as the cosine of 2n*pi/n, is equal to zero. In other words, it is a mathematical way of demonstrating that the average value of x is zero.

## 2. Why is it important to prove E[x]=0 for x=cos(2n*pi/n)?

Proving E[x]=0 for x=cos(2n*pi/n) is important because it helps us understand the behavior of the random variable x and its likelihood of taking certain values. It also allows us to make predictions and draw conclusions about the data set that x is derived from.

## 3. What is the significance of using cos(2n*pi/n) as the random variable in this proof?

Cos(2n*pi/n) is a commonly used function in probability and statistics because it produces a periodic curve that can represent a variety of real-world phenomena. It is also a simple and convenient function to work with mathematically, making it a popular choice for proving statistical concepts.

## 4. How can we prove that E[x]=0 for x=cos(2n*pi/n)?

There are several methods for proving E[x]=0 for x=cos(2n*pi/n), depending on the specific context and assumptions. One approach could be using the definition of expected value, which involves summing the products of the possible values of x and their corresponding probabilities. Another approach could be using properties of integrals and trigonometric identities to simplify the calculation.

## 5. Are there any real-world applications of proving E[x]=0 for x=cos(2n*pi/n)?

Yes, there are many real-world applications of proving E[x]=0 for x=cos(2n*pi/n). For example, in signal processing, the expected value of a periodic signal can be used to determine its average frequency. In economics, expected value calculations are used to analyze risk and make investment decisions. In physics, expected values are used to calculate the average energy of a quantum system.

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