The Minimization of the Quadratic Mean of an Integral Dose

Abstract

In this work, we define the optimal dose as a combination of the projections on orthogonal axes of the absorbed dose and an integer multiple of the integral dose. Here, we show that such optimal dose minimizes the mean square of the total absorbed dose subject to certain conditions of integration. We prove that there is a unique minimizer.

Appendices

Appendix A: Proof of Lemma 3.1

Integrate the function d with respect to x ₁,…,x _i−1, x _1+i ,…,x _n .

by Equality (1).

Appendix B: Proof of Lemma 3.2

We expand the function d ².

Integrate with respect to x ₁,…,x _n .

Appendix C: Proof of Theorem 3.1

By Equality (2) and Lemma 3.1, by integrating the difference f−d with respect to all x’s but x _i we get zero.

$$ \int_0\sp1\cdots\int_0 \sp1(f-d) (x_1,\ldots ,x_n)\,dx_1\cdots dx_{i-1}\,d x_{1+i}\cdots d x_n=0. $$

(3)

Multiply by p _i , and add up with respect to the index i.

$$\sum_{i=1}\sp n p_i(x_i)\int _0\sp1\cdots\int_0\sp1(f-d) (x_1, \ldots , x_n)\,d x_1\cdots d x_{i-1}\,d x_{1+i}\cdots d x_n=0. $$

Integrate with respect to x _i .

(4)

Integrate with respect to x _i equality (3), and multiply by a constant.

Subtract this equality from Equality (4).

Multiply by 2. Suppose that the function f ² is integrable, and add the integral of the sum of the squares of the function d and the difference (f−d).

Therefore,

$$\int_0\sp1\cdots\int_0 \sp1d(x_1, \ldots , x_n)\sp2\,d x_1\cdots d x_n\le\int_0\sp1\cdots\int_0 \sp1f(x_1, \ldots , x_n)\sp2\,d x_1\cdots d x_n. $$

Equality occurs only if the function f is equal to the function d.