You have made many, many measurements of a variable x whose errors are randomly

Question

You have made many, many measurements of a variable x whose errors are randomly distributed with a standard deviation sigma. The population mean (average) is mu. The Central Limit Theorem of calculus tells us that as the number of measurements N gets larger and larger the average of the N measurements of x with approach a Gaussian function y = 1/ sigma squareroot 2 pi e^-(x - mu)^2/2 sigma n^2 a) Find what the inflection points of the Gaussian are (give their formula) showing all of your work. b) Find the full width at half maximum (FWHM) - give the formula for the width of the Gaussian at the points where y = plusminus 0.5y_max.

Explanation / Answer

A random variable that is normally distributed with mean and standard deviation of has a probability density function of

f( x ) =1/ ( (2 ) )exp[-(x - )2/(22)].

Here we use the notation exp[y] = ey, where e is the mathematical constant approximated by 2.71828.

The first derivative of this probability density function is found by knowing the derivative for ex and applying the chain rule.

f’ (x ) = -(x - )/ (3 (2 ) )exp[-(x -) 2/(22)] = -(x - ) f( x )/2.

We now calculate the second derivative of this probability density function. We use the product rule to see that:

f’’( x ) = - f( x )/2 - (x - ) f’( x )/2

Simplifying this expression we have

f’’( x ) = - f( x )/2 + (x - )2 f( x )/(4)

Now set this expression equal to zero and solve for x. Since f( x ) is a nonzero function we may divide both sides of the equation by this function.

0 = - 1/2 + (x - )2 /4

To eliminate the fractions we may multiply both sides by 4

0 = - 2 + (x - )2

We are now nearly at our goal. To solve for x we see that

2 = (x - )2

By taking a square root of both sides (and remembering to take both the positive and negative values of the root

± = x -

From this it is easy to see that the inflection points occur where x = ± . In other words the inflection points are located one standard deviation above the mean and one standard deviation below the mean.

b)

y_max occut at mean

y = 0.5 y_max

solving we get

0.5 =exp[-(x -) 2/(22)]

solving we x = - *sqrt(log (4))

other point is x = + *sqrt(log (4))

width = 2* sqrt(2) **sqrt(log (2))

=2* *sqrt(log (4))

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