Home » Blog » Find a formula for the area of an ordinate set under a similarity transform

Find a formula for the area of an ordinate set under a similarity transform

From Example 4 of Section 2.3 of Apostol, we know that given a nonnegative, integrable function f on an interval [a,b], and the area of the ordinate set of f is S, then if we define a function g(x) = k f(x/k), we have,

    \[ a(kS) = \int_{kb}^{kb} g(x) \, dx = k \int_{ka}^{kb} f(x/k) \, dx = k^2 \int_a^b f(x) \, dx. \]

Generalize this formula and prove the result.

To generalize this, we proceed as follows.

Let f be a nonnegative integrable function on [a,b], and let S be the ordinate set of f. If we apply a transformation under which we multiply the x-coordinate of each point (x,y) on the graph of f by a constant k > 0 and each y-coordinate by a constant j> 0, then we obtain a new function g were a point (x,y) is on g if and only if \left( \frac{x}{k}, \frac{y}{j} \right) is on f. Then,

    \[ \frac{y}{j} = f \left( \frac{x}{k} \right) \quad \implies \quad y = j \cdot f \left( \frac{x}{k} \right) \quad \implies \quad g(x) = j \cdot f \left( \frac{x}{k} \right). \]

Letting jkS denote the ordinate set of g:

    \begin{align*}    a(S) = \int_a^b f(x) \,dx \quad \implies \quad a(jkS) &= \int_{ka}^{kb} g(x) \, dx \\   &= j \cdot \int_{ka}^{kb} f \left( \frac{x}{k} \right) \, dx \\   &= jk \cdot \int_a^b f (x) \, dx \\   &= jk \cdot a(S). \end{align*}


  1. Anonymous says:

    I think it would be better to prove that the ordinate set of the new graph is measurable first. To do this, the solution (the part from a(S) to jk . a(S)) only needed to be written in a reverse order

    • Richard Sullivan says:

      No, [ka, kb] is correct. The horizontal scale factor is k. The interval [ka, kb] is the horizontal dimension (along the x-axis) of the ordinate set of g, after the ordinate set of f is scaled horizontally by k and vertically by j.

      • Daniel says:

        Richard, why do you assume that “the horizontal scale factor is k”? The problem statement says nothing about scaling [a, b] by the same factor in both ends, and you could perfectly scale the left-end by ka and the right end by jb, yielding the interval [ka, jb].

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):