FWHM for a quadratic

This post contains a derives a result I needed recently. The derivation is simple but a little tedious, so I wanted to save it in case I need it again.

Full width half maximum

A common way to measure the width of a function peak in a function f(x) is to find the place x₀ where f takes on its maximum, and find two points, x_-1 to the left and x₁ to the right, where f drops to half its peak value, i.e.

f₁ = f(x₀) / 2.

The width of the peak is then defined to be the distance between these two points:

FWHM = x₁ - x_-1

where FWHM stands for full width half maximum." I've mentioned FWHM a few times before, such as here.

It's also useful sometimes to find the full width at k times the maximum for values of k other than 1/2 and so we'll solve the more general problem.

Quadratic case

Now suppose f is a quadratic function

f(x) = ax^2 + bx + c.

where a is not zero. We want to find the FWHM of f, and more generally find the distance between two points where f takes on values k times its maximum (or minimum).

Taking the derivative of f shows that the vertex of the parabola occurs when

2ax + b = 0

and so

x₀ = -b/(2a).

and

f(x₀) = c - b^2/(4a).

Now we have to find two solutions to

f(x) = k f(x₀)

which means

ax^2 + bx + c - k(c - b^2/(4a)) = 0.

This is a quadratic equation with constant term

c = c - k(c - b^2/(4a))

and so from the quadratic formula, the difference between the two roots is

( b^2 - 4ac ) / |a| = ( b^2 - 4a(1-k)c - kb^2 ) / |a|

When k = 1/2, this reduces to

FWHM = (b^2/2 - 2ac) / |a|.

Examples

Let's try this on a couple examples to see if this checks out.

Maximum example

Let

f(x) = 20 - (x - 2)^2 = -x^2 + 4x +16

Clearly the maximum is 20 and occurs at x = 2. The quadratic formula shows the two places where f takes half its maximum value are

x = 2 10

and so the FWHM equals 210.

If we use the formula for FWHM above we get

( 4^2/(2) + 32) = 40 = 210.

Minimum example

Let's do another example, this time looking for where a convex parabola takes on twice its minimum value. So here we set k = 2 and so the expression

( b^2 - 4ac ) / |a| = ( b^2 - 4a(1-k)c - kb^2 ) / |a|

above reduces to

(4ac - b^2) / |a|.

Let

f(x) = 3x^2 + 2x + 1

Then the minimum of f occurs at -1/3 and the minimum equals 2/3. We want to find where f equals 4/3. The quadratic formula shows this occurs at

(-1 2)/3

and so the distance between these two points is 22 / 3.

If we plug a = 3, b = 2, and c = 1 into

(4ac - b^2) / |a|

we get the same result

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