Page 96 - 71 the abc of dft_opt

192 APPENDIX A. MATH BACKGROUND
A.2 Properties of the δ-function

We review briefly the useful properties of δ-function. The δ-function is an infinitely narrow,
infinitely tall peak at x = 0, chosen so that its area is exactly 1. Thus we can write several
Appendix A limiting formulas:
δ(x) = lim (Θ(x + a/2) − Θ(x − a/2)) /a,
a→0
1
Math background = lim √ exp(−x /γ )
2
2
γ→0 γ π
1 γ
= lim (A.5)
2
γ→0 π x + γ 2
A.1 Lagrange multipliers
where Θ(x) is the step function (= 0 for x < 0, 1 for x > 0), and the δ-function looks like
(with a finite width) what’s shown in Fig X.
To illustrate how Lagrange multipliers work, we do a simple example on a function of two
variables, instead of functionals. The usefulness of the δ function is that, for any reasonably smooth function:
1
Suppose we wish to maximize the function f(x, y) = x + y subject to the constraint dx f(x)δ(x) = f(0) (A.6)
2
2
x + y = 1. The inelegant approach is to simply solve the constraint equation for y as a
√ assuming the integration interval includes the origin. This can be used to extract the value
2
function of x, finding y = 1 − x , and substitute into f, yielding
of the function at any point x 0 :
√ 1
˜
f(x) = x + 1 − x 2 (A.1) dx f(x)δ(x − x 0 ) = f(x 0 ) (A.7)
Then we find the extrema of f by differentiation: again assuming the integration interval includes the point x 0 .
Another handy formula is the integral over the δ-function:
df ˜ x 1
= 1 − √ (A.2) x # #
dx 1 − x 2 dx δ(x ) = Θ(x) (A.8)
−∞
√ √
which we set to zero, finding x = 1/ 2. Then y = 1/ 2 also, yielding the maximum which is just the step function, i.e., you get zero unless the interval includes the δ-function,
√
f = 2. when you get 1. Another way to see this is that the first formula of Eq. (A.5) implies that
But a much more elegant way is to introduce a Lagrange multiplier. Write the constraint the δ function is the derivative of the step function.
2
2
as g(x, y) = 0, where g = x + y − 1. Then we define The Fourier transform of the δ-function is
1 ∞ dp
δ(x) = exp(ipx) (A.9)
h = f − µ g (A.3) −∞ 2π
i.e., it’s just a constant.
where µ is a constant, independent of x and y. We then minimize h freely, with no restriction
between x and y. Write Finally, one can show:
- δ(x − x i )
δ(f(x)) = (A.10)
∂h ∂h i,f(x i )=0 |f (x i )|
#
= 1 − 2µ x = 1 − 2µ y (A.4)
∂x ∂y where f (x) = df/dx.
#
and set both to zero, yielding (x, y) = (1, 1)/(2µ). Then we enforce the condition g = 0, Lastly, with some care, 1
√ δ (x)f(x) = −f (0) (A.11)
#
#
yielding µ = ±1/ 2, and choose the positive sign for a maximum. This method avoids
differentiating square roots, etc. This is especially important if we cannot solve the constraint assuming the interval includes 0.
equation analytically. In fact, its not strictly a function at all, but properly called a distribution.
191