Always the same problem comes on me, I will be mad

[opeth_35]

Please help for this probability function !

 

[I like Serena]

You seem to have left out a number of steps.
Let's start with the original statement of the problem.

Could it be something like:

$$A = \int_0^{\infty} e^{-2ax^2} dx$$

Calculate A?

 

[opeth_35]

yes, as you said, I have tried to solve this problem and I wrote in the paper that I've found value... I've checked again and again,I found the same value like in the paper.. ??:(

 

[I like Serena]

yes, as you said, I have tried to solve this problem and I wrote in the paper that I've found value... I've checked again and again,I found the same value like in the paper.. ??:(

Well, I'll give you a quick answer, and a slow question.

In your notes you write: $$1 = A^2 \frac 1 {4a} \int_0^{\pi/2}d\theta$$

This is wrong. It should be: $$A^2 = \frac 1 {4a} \int_0^{\pi/2}d\theta$$

Applying this should give you the same answer as your teacher.

The slow question is: what happened in the first step?
You move here from a single integral over x to a double integral over r and theta.
Do you know what the line of reasoning is here?
Note that this is the most important step in the reasoning.

 

[opeth_35]

actually,I d be appreciate that, If you explain you asked question. What did you want to ask me about the line of reasoning?

 

[I like Serena]

actually,I d be appreciate that, If you explain you asked question. What did you want to ask me about the line of reasoning?

I'll give you the step.
It is:

$$A^2 = \int_0^{\infty} e^{-2ax^2} dx \cdot \int_0^{\infty} e^{-2ax^2} dx <br /> = \int_0^{\infty} e^{-2ax^2} dx \cdot \int_0^{\infty} e^{-2ay^2} dy<br /> = \int_0^{\infty} \int_0^{\infty} e^{-2ax^2} e^{-2ay^2} dxdy<br /> = \int_0^{\infty} \int_0^{\infty} e^{-2a(x^2+y^2)} dxdy$$

This is a surface integral in cartesian coordinates over the first quadrant.
Now we shift to polar coordinates over the first quadrant, which means that r goes from 0 to infinity and theta goes from 0 to pi/2.
Also where in cartesian coordinates we have dxdy as an infinitesimal surface element, in polar coordinates we have r.dtheta.dr.

In other words:

$$A^2 = \int_0^{\infty} \int_0^{\infty} e^{-2a(x^2+y^2)} dxdy<br /> = \int_0^{\frac {\pi} 2} \int_0^{\infty} e^{-2a(r^2)} r dr d\theta$$

Does this make sense to you?

 

[opeth_35]

thank you for your reasonable explanation about that, Let me I want to ask you other question about this kind of probability functions.. You were talking about quadrant and you said: First quadrant and for this quadrant while the boundries go from 0 to +infinite. we need to use pi/2 while we compute for theta funtion. If we are using for the boundries go from -infinite to +infinite, we should use the boundries goes from 0 to 2pi which this is fourth quadrant I think. I know that. But,
So, Is there any function which we will use second and third quadrant about in this kind of questions. I am wondering.. like the boundries of pi and 3*pi/2..
I hope, I explained what I want to ask,,:)

 

[I like Serena]

thank you for your reasonable explanation about that, Let me I want to ask you other question about this kind of probability functions.. You were talking about quadrant and you said: First quadrant and for this quadrant while the boundries go from 0 to +infinite. we need to use pi/2 while we compute for theta funtion. If we are using for the boundries go from -infinite to +infinite, we should use the boundries goes from 0 to 2pi which this is fourth quadrant I think. I know that. But,
So, Is there any function which we will use second and third quadrant about in this kind of questions. I am wondering.. like the boundries of pi and 3*pi/2..
I hope, I explained what I want to ask,,:)

It's possible of course, but it seems unlikely to me.
Note that it's just a "trick" to be able to calculate the original integral from 0 to infinity.
Furthermore, in probability theory we're usually interested in the range from -infinity to +infinity.