Calculating FWHM and Standard Deviation Relationship for Gaussian Distribution

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In summary, the mathematical relationship between the FWHM and standard deviation for a Gaussian distribution is that the FWHM is equal to the square root of 2 times the standard deviation times the natural logarithm of 1/2. This can also be written as -2ln2 times the standard deviation. The "extra" factor of 2 in ln2 comes from the fact that there are two values of x for which N = A/2 and the difference between them gives the FWHM. Taking the natural logarithm of 1/2 is the same as taking the negative natural logarithm of 2.
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The question is: for a Gaussian distribution what is the mathematical relationship between the FWHM and the standard deviation.

The equations I'm using are:

[TEX]N(x) =\frac {\ a}{2}[/TEX]
[TEX]N(x) = Ae^- \frac {\ (x-x_2)^2}{2 sigma^2}[/tex]

I equated the equations and started to solve for x. I know you take the ln of both sides to get rid of e. But a few of my friends have 2ln2 in their answer...how do you get ln2? isn't it ln 1/2?
 
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The "extra" factor of 2 arises from the fact that there are two values of x for which N = A/2 and the difference between them gives you the FWHM.
 
  • #3
Sorry, I mean the 2 in ln2, not the factor 2

N(x) = A/2 ...the A cancels, giving you 1/2
then taking the ln of both sides would give you ln 1/2, right? not ln2

I'm getting [tex]sqrt(2*sigma*ln \frac {\1}{2}) = -(x-x_2)


Thanks.
 
  • #4
You do understand that [itex]\ln \frac {1}{2} = - \ln 2[/itex]?
 

1. What is the difference between ln and e?

ln and e are both mathematical constants, but they have different values. ln, also known as the natural logarithm, has a value of approximately 2.71828. e, also known as Euler's number, has a value of approximately 2.71828. ln is the inverse function of e, meaning that ln(e) = 1 and e^(ln(x)) = x.

2. How do you solve for x using ln and e?

To solve for x using ln and e, you will need to use logarithmic and exponential properties. For example, if you have an equation ln(x) = 3, you can rewrite it as e^(ln(x)) = e^3. Then, using the property e^(ln(x)) = x, you can solve for x by setting it equal to e^3. The same process applies for equations where x is in the exponent, such as e^(2x) = 10. You can rewrite it as ln(e^(2x)) = ln(10), and then use the property ln(e^(2x)) = 2x to solve for x.

3. Can you solve for x using ln and e in any type of equation?

No, ln and e can only be used to solve for x in exponential and logarithmic equations. These are equations where x is in the exponent or the argument of a logarithm, such as e^(2x) = 10 or ln(x) = 3. They cannot be used to solve for x in other types of equations, such as linear or quadratic equations.

4. Are there any restrictions when using ln and e to solve for x?

Yes, there are some restrictions when using ln and e to solve for x. For logarithmic equations, the argument of the logarithm must be positive. This means that if you have an equation ln(x) = 3, x must be a positive number. For exponential equations, the base (e) cannot be equal to 1, as any number raised to the power of 0 is equal to 1. Additionally, when using logarithmic properties, the argument of the logarithm cannot be 0 or negative.

5. Can you use ln and e to solve for multiple variables in an equation?

No, ln and e can only be used to solve for one variable at a time. If you have an equation with multiple variables, you will need to isolate the variable you want to solve for before using ln and e. However, if the equation only contains the variable you want to solve for and constants, then you can use ln and e to solve for that variable directly.

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