# How to find the expectation value of cos x

• DEEPTHIgv
In summary, the probability distribution function is not the same as the wavefunction in QM. This does not read like a quantum theory question to me, just straight probability. But the given f(x) is not a valid pdf; its integral over the full range is not 1. Seems like it should be 4e-4x. Anyway, to answer your question, all you care about at infinity is that the trig functions are bounded. Look at the other factor.
DEEPTHIgv

## Homework Statement

If x is a continuous variable which is uniformly distributed over the real line from x=0 to x -> infinity according to the distribution f (x) =exp(-4x) then the expectation value of cos 4x is?

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Request

## Homework Equations

the expectation value of any function g is, <g>=∫Ψ*gΨdx

## The Attempt at a Solution

Here,
Ψ=f(x)
since f is independent of time
∴Ψ*=f(x)
now, for <cos 4x>=∫exp(-4x)cos(4x)exp(-4x)dx
let <cos4x>=I=∫cos(4x)exp(-8x)dx
rest of the solution if in the picture attached
my main doubt is how to put the limit infinity in sin and cos

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The limits ##\lim_{x\rightarrow \infty}\cos{4x}##, ##\lim_{x\rightarrow \infty}\sin{4x}## do not exist, however here you have the limits

##\lim_{x\rightarrow \infty}e^{-8x}\cos{4x}## , ##\lim_{x\rightarrow \infty}e^{-8x}\sin{4x}## which are zero, because ##lim_{x\rightarrow \infty}e^{-8x}=0## and the functions ##\sin{4x}, \cos{4x}## are bounded. (It is a well known lemma that the limit of the product of a bounded function with another function that has limit zero, is also zero).

DEEPTHIgv
DEEPTHIgv said:

## Homework Statement

If x is a continuous variable which is uniformly distributed over the real line from x=0 to x -> infinity according to the distribution f (x) =exp(-4x) then the expectation value of cos 4x is?

Follow· 01
Request

## Homework Equations

the expectation value of any function g is, <g>=∫Ψ*gΨdx

## The Attempt at a Solution

Here,
Ψ=f(x)
since f is independent of time
∴Ψ*=f(x)
now, for <cos 4x>=∫exp(-4x)cos(4x)exp(-4x)dx
let <cos4x>=I=∫cos(4x)exp(-8x)dx
rest of the solution if in the picture attached
my main doubt is how to put the limit infinity in sin and cos
The probability distribution function is not the same as the wavefunction in QM.

This does not read like a quantum theory question to me, just straight probability. But the given f(x) is not a valid pdf; its integral over the full range is not 1. Seems like it should be 4e-4x.
Anyway, to answer your question, all you care about at infinity is that the trig functions are bounded. Look at the other factor.

Beaten by Δ2 and ehild because I spent time figuring out the answer, and I do not believe it is 1/2. Also, there is still the problem that the given pdf is not valid.

Edit: the reason I got a different answer is that I used cos(x), as stated in the title, instead of cos(4x), as stated in post #1.

Last edited:
DEEPTHIgv and Delta2
That was my concern also, but my subconscious thought was that he meant wave function, though the problem statement clearly states probability density function.

DEEPTHIgv
Delta2 said:
the problem statement clearly states probability density function.
Except, as I posted, it fails the totality of 1 criterion.

haruspex said:
Except, as I posted, it fails the totality of 1 criterion.
Yes it should have been with a factor 4 in front (as you already said), probably a typo by the OP.

DEEPTHIgv said:
solution if in the picture attached
You may find it a bit simpler to spot that a likely integral of ##e^{-\lambda x}\cos(\mu x)## is the form ##e^{-\lambda x}(A\cos(\mu x)+B\sin(\mu x))## then differentiate that to find A and B.

ehild said:
The probability distribution function is not the same as the wavefunction in QM.

Could you please explain or suggest a source from which I can understand the difference.

haruspex said:
This does not read like a quantum theory question to me, just straight probability. But the given f(x) is not a valid pdf; its integral over the full range is not 1. Seems like it should be 4e-4x.
Anyway, to answer your question, all you care about at infinity is that the trig functions are bounded. Look at the other factor.

Beaten by Δ2 and ehild because I spent time figuring out the answer, and I do not believe it is 1/2. Also, there is still the problem that the given pdf is not valid.

It is possible.
The above question is from a previous year entrance test.

Delta2 said:
That was my concern also, but my subconscious thought was that he meant wave function, though the problem statement clearly states probability density function.

Isn't the expectation value supposed to be ∫f* cos4x f dx

Delta2 said:
Yes it should have been with a factor 4 in front (as you already said), probably a typo by the OP.
I have posted the question just as it was given.

DEEPTHIgv said:
Isn't the expectation value supposed to be ∫f* cos4x f dx
That is correct if ##f## is given as a wave function. However the problem states that ##f## is a probability density function hence the expectation value of the variable ##g(x)## is simply ##\int g(x)f(x)dx##.

To unify the two cases, when ##f## is the wave function then the probability density function is ##|f|^2=f^{*}f##

Last edited:
DEEPTHIgv
Delta2 said:
That was my concern also, but my subconscious thought was that he meant wave function, though the problem statement clearly states probability density function.

Delta2

## 1. How do I define the expectation value of cos x?

The expectation value of cos x is defined as the average value of cos x over a large number of measurements or trials. It is calculated by taking the integral of cos x multiplied by the probability density function and integrating over the entire range of possible values.

## 2. What is the significance of finding the expectation value of cos x?

The expectation value of cos x provides a measure of the central tendency of a set of data or a probability distribution. It is useful in analyzing and predicting outcomes in various scientific and mathematical fields, such as quantum mechanics, statistics, and signal processing.

## 3. What are the steps to finding the expectation value of cos x?

The steps to finding the expectation value of cos x are as follows:

• 1. Determine the probability distribution function (PDF) for the variable x.
• 2. Multiply the PDF by cos x to get the product of the two functions.
• 3. Integrate the product over the entire range of possible values for x.
• 4. The result of the integration is the expectation value of cos x.

## 4. Can the expectation value of cos x be negative?

Yes, the expectation value of cos x can be negative. This indicates that the values of cos x in the data set or probability distribution are more likely to be negative than positive. However, it is also possible for the expectation value to be zero or positive.

## 5. How is the expectation value of cos x related to the uncertainty principle?

The expectation value of cos x is related to the uncertainty principle in that it provides information about the spread or uncertainty in the values of cos x. A smaller expectation value indicates a more tightly clustered set of values, while a larger expectation value suggests a wider spread of values. This is in line with the uncertainty principle, which states that the more precisely the position of a particle is known, the less precisely its momentum can be known and vice versa.

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