Operand n. Finding the equation is simply a matter of algebraic
elimination: from the two equations n" = n' + 1 and n' = n + 1,
eliminate n'. Substituting for n' in the first equation we get (with
brackets to show the derivation) n" = (n + 1) + 1, i.e. n" = n + 2.
This equation gives correctly the relation between operand (n)
and transform (n") when T 2 is applied, and in that way T 2 is speci-
Fied. For uniformity of notation the equation should now be re-writ-
ten as m' = m + 2. This is the transformation, in standard notation,
Whose single application (hence the single prime on m) causes the
Same change as the double application of T. (The change from n to
M is a mere change of name, made to avoid confusion.)
The rule is quite general. Thus, if the transformation is n' =
2n – 3, then a second application will give second transforms n"
that are related to the first by n" = 2n' – 3. Substitute for n', using
Brackets freely:
n" = 2(2n – 3) – 3
= 4n – 9.
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So the double application causes the same change as a single
application of the transformation m' = 4m – 9.
Higher powers. Higher powers are found simply by adding
symbols for higher transforms, n"', etc., and eliminating the sym-
Bols for the intermediate transforms. Thus, find the transforma-
tion caused by three applications of n' = 2n – 3. Set up the
Equations relating step to step:
n' = 2n – 3
n" = 2n' – 3
n"' = 2n" – 3
Take the last equation and substitute for n", getting
n"' = 2(2n' – 3) – 3
= 4n' – 9.
Now substitute for n':
n"' = 4(2n – 3) – 9
= 8n – 21.
So the triple application causes the same changes as would be
caused by a single application of m' = 8m – 21. If the original was
T, this is T3.
Ex. 1: Eliminate n' from n" = 3n' and n' = 3n. Form the transformation corre-
sponding to the result and verify that two applications of n' = 3n gives the
Same result.
Ex. 2: Eliminate a' from a" = a' + 8 and a' = a + 8.
Ex. 3: Eliminate a" and a' from a'" = 7a", a" = 7a', and a' = 7a.
Ex. 4: Eliminate k' from k" = – 3k' + 2, k' = – 3k + 2. Verify as in Ex.1.
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Ex. 5: Eliminate m' from m" = log m', m' = log m.
Ex. 6: Eliminate p' from p"=(p')2, p' =p2
Ex. 7: Find the transformations that are equivalent to double applications, on all
The positive numbers greater than 1, of:
(i) n' = 2n + 3;
(ii) n' = n2 + n;
(iii) n' = 1 + 2log n.
Ex. 8: Find the transformation that is equivalent to a triple application of
n' = – 3n – 1 to the positive and negative integers and zero. Verify as in
Ex. 1.
Ex. 9: Find the transformations equivalent to the second, third, and further
applications of the transformation n' = 1/(1 + n). (Note: the series discov-
Ered by Fibonacci in the 12th century, 1, 1, 2, 3, 5, 8, 13,... is extended by
taking as next term the sum of the previous two; thus, 3 + 5 = 8, 5 + 8 = 13,
8 + 13 = ......, etc.)
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A N I N T R O D UC T I O N T O C Y B E R NE T I C S
C H A NG E
Ex. 10: What is the result of applying the transformation n' = 1/n twice,
When the operands are all the positive rational numbers (i.e. all the
Fractions) ?
Ex. 11: Here is a geometrical transformation. Draw a straight line on paper and
Mark its ends A and B. This line, in its length and position, is the operand.
Obtain its transform, with ends A' and B', by the transformation-rule R: A' is
midway between A and B; B' is found by rotating the line A'B about A'
Through a right angle anticlockwise. Draw such a line, apply R repeatedly,
And satisfy yourself about how the system behaves.
*Ex. 12: (Continued). If familiar with analytical geometry, let A start at (0,0) and
B at (0,1), and find the limiting position. (Hint: Build up A’s final x-co-ordi-
Nate as a series, and sum; similarly for A’s y-co- ordinate.)
Transformation U, which will give a transform U(T(n)). Thus, if
They are
T: ↓
A b c d
B d a b
and U: ↓
A b c d
D c d b
Then T(b,) is d, and U(T(b)) is U(d), which is b. T and U applied in
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That order, thus define a new transformation, V, which is easily
Found to be
V: ↓
A b c d
C b d c
Notation. The notation that indicates the transform by the
addition of a prime (') is convenient if only one transformation is
Under consideration; but if several transformations might act on n,
the symbol n' does not show which one has acted. For this reason,
Another symbol is sometimes used: if n is the operand, and trans-
Formation T is applied, the transform is represented by T(n). The
Four pieces of type, two letters and two parentheses, represent one
Quantity, a fact that is apt to be confusing until one is used to it.
T(n), really n' in disguise, can be transformed again, and would be
Written T(T(n)) if the notation were consistent; actually the outer
brackets are usually eliminated and the T ’s combined, so that n"
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