Functions that turn adding into multiplying
Tonight's idea
Here is a model. Something grows steadily: savings, or a colony of bacteria. Let \(G(t)\) be the factor it is multiplied by over \(t\) years. Growing for \(s\) years and then for \(t\) more is the same as growing for \(s + t\) years, so
\(G(s + t) = G(s)\,G(t)\): adding the times multiplies the factors.
Nobody gave you a formula for \(G\). You were given a rule it obeys. That is a functional equation: an equation whose unknown is a whole function, not a number. Your job: squeeze out everything the rule forces, and notice what it does not force.
The same rule runs things bigger than numbers. Turning the plane by angle \(s\) and then by angle \(t\) is one turn by \(s + t\), so the rotation matrices obey \(R(s + t) = R(s)\,R(t)\). A family of actions, one for each length of time, where doing \(s\) and then \(t\) equals doing \(s + t\), is called a one-parameter semigroup. It is how mathematicians describe a system that evolves steadily in time.
Watch one: \(f(x + y) = f(x)\,f(y)\) and \(f(1) = 2\)
Suppose \(f\) is defined for every real number, and the rule holds for all real \(x\) and \(y\). You may choose any \(x\) and \(y\) you like.
Whole numbers. Take \(x = y = 1\): \(f(2) = f(1)\,f(1) = 4\). Take \(x = 2\), \(y = 1\): \(f(3) = f(2)\,f(1) = 8\).
Zero. Take \(x = 1\), \(y = 0\): \(f(1) = f(1)\,f(0)\), that is \(2 = 2 f(0)\), so \(f(0) = 1\). Dividing by \(f(1)\) was allowed only because \(f(1) = 2\) is not zero. The function that is 0 everywhere also obeys the rule, and for it \(f(0) = 0\).
Negatives. Take \(x = 1\), \(y = -1\): \(f(0) = f(1)\,f(-1)\), that is \(1 = 2 f(-1)\), so \(f(-1) = \tfrac12\).
So far \(f\) agrees with \(2^x\) at \(x = -1, 0, 1, 2, 3\), and you never used a formula.
Your turn
- Keep the \(f\) from the example. (a) Use \(f(x)\,f(-x) = f(0)\) to explain why \(f(x)\) is never 0. (b) Write \(x = \tfrac{x}{2} + \tfrac{x}{2}\) and explain why \(f(x)\) is never negative. (c) Now find \(f\!\left(\tfrac12\right)\), and say where you needed (a) and (b).
- A different rule: \(g(x + y) = g(x) + g(y)\) for all real \(x\) and \(y\), and \(g(1) = 3\). Find \(g(2)\), \(g(0)\), \(g(-1)\) and \(g\!\left(\tfrac13\right)\).
- A third rule: \(h(xy) = h(x) + h(y)\) for all positive \(x\) and \(y\). (a) Show that \(h(1) = 0\). (b) Show that \(h(8) = 3h(2)\). (c) If you are also told \(h(2) = 1\), find \(h(8)\), \(h\!\left(\tfrac12\right)\) and \(h(\sqrt2)\).
- A made-up savings account doubles your money every year and grows steadily in between. As a model, its growth factor obeys \(G(s + t) = G(s)\,G(t)\) with \(G(1) = 2\). (a) A friend says half a year must multiply your money by 1.5. Use the rule to show that is wrong. (b) Find the true half-year factor, and what \(\$1000\) becomes after six months, to the nearest cent.
Stretch: every fraction
Keep the \(f\) from the example. Let \(n\) be a positive whole number and \(m\) any whole number. Show that \(f\!\left(\tfrac1n\right) = 2^{1/n}\), the positive \(n\)th root of 2, and then that \(f\!\left(\tfrac mn\right) = 2^{m/n}\). Does your argument tell you \(f(\sqrt2)\)?
Answers
- (a) \(f(x)\,f(-x) = f(x + (-x)) = f(0) = 1\). If \(f(x)\) were 0 the left side would be 0, not 1. (b) \(f(x) = f\!\left(\tfrac x2\right) f\!\left(\tfrac x2\right) = f\!\left(\tfrac x2\right)^2\), and the square of a real number is never negative. With (a), \(f(x) \gt 0\) for every \(x\). (c) \(f\!\left(\tfrac12\right)^2 = f(1) = 2\), so \(f\!\left(\tfrac12\right)\) is \(\sqrt2\) or \(-\sqrt2\). Because \(f\) is positive, \(f\!\left(\tfrac12\right) = \sqrt2 \approx 1.414\).
- \(g(2) = g(1) + g(1) = 6\). From \(g(1) = g(1) + g(0)\), \(g(0) = 0\). From \(g(0) = g(1) + g(-1)\), \(g(-1) = -3\). Adding three copies of \(\tfrac13\) gives \(g(1) = 3\,g\!\left(\tfrac13\right)\), so \(g\!\left(\tfrac13\right) = 1\). All four agree with \(g(x) = 3x\). This rule turns adding into adding.
- (a) Take \(x = y = 1\): \(h(1) = h(1) + h(1)\), so \(h(1) = 0\). (b) \(h(4) = h(2 \cdot 2) = 2h(2)\), then \(h(8) = h(4 \cdot 2) = h(4) + h(2) = 3h(2)\). (c) \(h(8) = 3\). Since \(h(2) + h\!\left(\tfrac12\right) = h(1) = 0\), \(h\!\left(\tfrac12\right) = -1\). Since \(h(\sqrt2) + h(\sqrt2) = h(2) = 1\), \(h(\sqrt2) = \tfrac12\). This rule turns multiplying into adding. It is the rule of logarithms, and these values match \(\log_2\).
- (a) Two half-years make one year, so \(G\!\left(\tfrac12\right)^2 = G(1) = 2\). But \(1.5^2 = 2.25\), which is more than doubling. (b) A growth factor is positive, so \(G\!\left(\tfrac12\right) = \sqrt2 \approx 1.4142\), and \(\$1000\) becomes about \(\$1414.21\). The growth is about 41.4% in each half-year, not 50%.
Stretch. Adding \(n\) copies of \(\tfrac1n\) and using the rule \(n - 1\) times gives \(f\!\left(\tfrac1n\right)^n = f(1) = 2\). By exercise 1, \(f\!\left(\tfrac1n\right)\) is positive, so it is the positive \(n\)th root, \(2^{1/n}\). For \(m \gt 0\), adding \(m\) copies of \(\tfrac1n\) gives \(f\!\left(\tfrac mn\right) = \left(2^{1/n}\right)^m = 2^{m/n}\). For \(m = 0\), \(f(0) = 1 = 2^0\). For \(m \lt 0\), \(f\!\left(\tfrac mn\right) f\!\left(-\tfrac mn\right) = 1\), so \(f\!\left(\tfrac mn\right) = 1 / 2^{-m/n} = 2^{m/n}\). It does not tell you \(f(\sqrt2)\): these steps only ever reach fractions, and \(\sqrt2\) is not one. If you also know that \(f\) is continuous (its graph has no jumps), then \(f(x) = 2^x\) for every real \(x\). Without some extra condition like that, it is known that wild solutions exist: they equal \(2^x\) at every fraction but not at every real number.