Functional Equations: Substitution · Little Blue Book, Visualized

What this move does

In a functional equation $x$ is “arbitrary”, so you can replace it by any expression (inside the domain) and the equation still holds. The substitution method uses this to create new equations:

If the equation has the form $f (φ (x)) = g (x)$ : let $t = φ (x)$ , solve for $x$ , and you get $f (t)$ directly.
If both $f (x)$ and $f (φ (x))$ appear, and iterating $φ$ a few times brings you back to $x$ (for example $\frac{1}{x}$ and $- x$ come back after two steps, $\frac{1}{1 - x}$ after three): replace $x$ by $φ (x)$ again and again until $f (x)$ appears once more. Now there are as many equations as unknowns, so eliminate.
Substitute the $f$ you found back into the original equation to check it.

This uses the iteration from the previous chapter: the period of $φ$ decides how many substitutions you need.

Predict first

$f (x) + 2 f (\frac{1}{x}) = 3 x$ ( $x \neq = 0$ ). What is $f (2)$ ?

Choose an answer first; then you can see the result.

Work through one: a substitution of period 3

Example: f(x) + f(1/(1−x)) = x (x ≠ 0, 1)

Step 1 / 5

One equation, two unknowns
Equation ① contains $f (x)$ and $f (\frac{1}{1 - x})$ . With only one equation, you cannot solve it.

$f (x) + f (\frac{1}{1 - x}) = x$
Substitute once
Once more: back home
Eliminate
Substitute back

The simplest case: one substitution

When the equation contains only $f (some expression)$ , one substitution is enough: let $t = 2 x + 1$ ; then $x = \frac{t - 1}{2}$ , and substitute that into the right side.

Press “Put

t =

the inside”. Check:

(2 x + 1)^{2} - (2 x + 1) = 4 x^{2} + 2 x

Another example, with a symmetry: $2 f (x) - f (- x) = x^{2} + 3 x$ . Two substitutions of $- x$ bring you back, and elimination gives $f (x) = x^{2} + x$ .

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: repeated substitution always closes up the system

$φ (x) = \frac{1}{x}$ : one substitution brings you back, and the two equations are exactly enough.

Practice

Exercise 1Basic

$f (x + 1) = x^{2} + 3 x$ . Find $f (x)$ .

Exercise 2Basic

$2 f (x) - f (- x) = 3 x + 1$ . Find $f (x)$ .

Exercise 3Advanced

$f (x) - 2 f (\frac{1}{x}) = x$ ( $x \neq = 0$ ). Find $f (3)$ .

Exercise 4Challenge

$f (x) + f (\frac{1}{1 - x}) = 2 x$ ( $x \neq = 0, 1$ ). Find $f (3)$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Forgetting the range of the new variable

f (x^{2}) = x + 1

for every real

x

. Let

t = x^{2}

, so

x = t

and

f (t) = t + 1

What went wrong? Think first, then open

Diagnosis $x$ can be positive or negative, but $t = x^{2}$ can only be non-negative, and both $x = \pm t$ must satisfy the equation. $x = 2$ gives $f (4) = 3$ , while $x = - 2$ gives $f (4) = - 1$ : a contradiction.

This equation has no solution: no such function exists. When you substitute, state the range of the new variable clearly, and check with every value of

x

that can occur.

“The substitution must be reversible, and the range must be stated”

Exhibit

Stopping once you have a formula, without substituting back

Elimination in the system gives a formula for

f (x)

, so that is the function we want.

What went wrong? Think first, then open

Diagnosis Elimination only proves: “if $f$ satisfies the original equation, then $f$ can only be this formula.” That is a necessary condition. Sometimes the formula is undefined at some points, or does not satisfy the original equation at all.

Substitute the

f

you found back into the original equation (the “Substituted back” meter on the workbench does exactly that), and state the domain, for example

x \neq = 0, 1

“Elimination gives a candidate; substituting back makes it an answer”