will also be true that f1(n) will contain a second series f2(n), similarly related to f1(n); and so on without end.
We have illustrated this truth. We now need to develope it for any and every series of f(n), however arbitrary. Consider, then, the values of f(n) as a part of the original number-series. These values of f(n) form an image or representative of the whole number-series in such wise that if r be a whole number appropriately chosen, some one value of f(n), say the value that corresponds to the number p in the original series, or, in other words, the pth value of f(n), is r. But since f(n) images the whole of the original number-series, it must contain, as a part of itself, a representation of its own self as it is in that number-series. In this representation, f1(n), there is again a first member, a second member, and so on.
Now we can indeed speak of the series f1(n) as “derived from” f(n) by a second and relatively new operation. But, as a fact, the very operation which defines the series f(n) already predetermines f1(n), and no really second, or new operation is needed. For if every whole number has its correspondent, or “image,” in f(n), then, for that very reason, every separate “image,” being, by hypothesis, a whole number, has again, in f(n), its own image; and this image again its own image, and so on without end. Merely to observe these images of images, already present in f(n), is to observe, in succession, the various members of the series f1(n). The law of the formation of f1(n) is already determined, then, when f(n) is written, no matter how arbitrary f(n) itself may be.
In particular, let p be any whole number, and suppose that, according to the original self-representation of the numbers, f(p) = r. Then r also will have its image in the series f(n). Let that image be called f(r). Then f(r) = f(f(p)), is at once defined as f1(p), that is, as that value which f1(n) takes when n = p, or as the image of the image of p. It is easy to see that f1(p) is the pth value, in serial order, of the series f1(n). At the same time, since f1(p) = f(r), and since f(r) occupies,
