Functional Equations

Solve functional equations by clever substitution and by recognising the three standard families: additive (kx), logarithmic (k log x) and exponential (aˣ), plus the swap-and-solve technique.

Builds on: Part 18 · Periodic Functions.

Video coming soon

Families

kx, k log x, aˣ

f(x + y) = f(x) + f(y) gives kx; f(xy) = f(x) + f(y) gives k log x; f(x + y) = f(x)f(y) gives aˣ.

Three family curves. — 3x, log₂x and 2ˣ.

Table of families. — Substitute 0, −x, 1, 1/x.

First facts

What substitution reveals

Additive: f(0) = 0 and f is odd. Logarithmic: f(1) = 0 and f(1/x) = −f(x). Exponential: f(0) = 1, f(−x) = 1/f(x), f > 0.

Swap and solve

f(x) + 2f(1/x) = 3x

Replace x by 1/x to get a second equation; solving the pair gives f(x) = 2/x − x.

Swap-and-solve steps. — Two equations, two unknowns.

Worked examples

From the video

1. f(x + y) = f(x) + f(y), f(1) = 3. f(5)?

15.

2. f(xy) = f(x) + f(y), f(2) = 1. f(8)?

3. f(x + y) = f(x)f(y), f(1) = 2. f(5)?

32.

4. f(x) + 2f(1/x) = 3x. Find f.

f(x) = 2/x − x.

JEE-style question

Your turn

f(x + y) = f(x) f(y) for all real x, y, and f(1) = 3. Then f(1) + f(2) + … + f(n) equals:

(a)3(3ⁿ − 1)/2

(b)3ⁿ − 1

(c)3n(n + 1)/2

(d)(3ⁿ − 1)/2

Show the answer and the traps

f(x) = 3ˣ, so the sum is 3 + 9 + … + 3ⁿ = 3(3ⁿ − 1)/2, a geometric series. Option (a).

Products of values signal exponentials; sums of values signal kx.

Watch out

Common mistakes

Confusing the additive and exponential familiesSums of values: kx; products of values: aˣ.

Forgetting the zero function in f(x + y) = f(x)f(y)f(0) = 0 gives f ≡ 0; otherwise f(0) = 1.

Substituting without a planStart with 0, −x, 1 and 1/x.

Practice

Try these

1. f(x + y) = f(x) + f(y), f(2) = 10. Find f(7).

35.

2. f(xy) = f(x) + f(y), f(3) = 2. Find f(81) and f(1/9).

8 and −4.

3. 2f(x) + f(1 − x) = x². Find f(x).

f(x) = (x² + 2x − 1)/3.