Review question

# When does $f$ satisfy the identity $f(t)-f(1-t)=g(t)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9798

## Solution

1. Let $k$ be a real number, $k \ne \pm 1$. The function $f(t)$ satisfies the identity $f(t) - kf(1-t)=t$ for all values of $t$. By replacing $t$ with $1-t$, determine $f(t)$.

We have that the function $f(t)$ satisfies the identity

$f(t) - kf(1-t)=t,$

and by replacing $t$ with $1-t$ we have

$f(1-t) -kf(t)=1-t.$

Rearranging the second identity and substituting into the first gives

$f(t) - k[kf(t) + 1-t]=t,$

and solving for $f(t)$ gives

$f(t)=\frac{k+(1-k)t}{1-k^2}.$

1. Consider the new identity $\begin{equation*} f(t) - f(1-t) = g(t). \label{eq:starrepeat}\tag{*} \end{equation*}$
1. Show that no function $f(t)$ satisfies $\eqref{eq:starrepeat}$ when $g(t)=t$.

Letting $g(t)=t$ gives

$f(t) - f(1-t) = t,$

and replacing $t$ with $1-t$ gives

$f(1-t)-f(t) = 1-t.$

Summing these two identities gives $0=1$, which is clearly a contradiction. So there is no function $f(t)$ that satisfies $\eqref{eq:starrepeat}$ when $g(t)=t$.

1. What condition must the function $g(t)$ satisfy for there to be a solution $f(t)$ to $\eqref{eq:starrepeat}$?

If there is a solution $f(t)$ to $\eqref{eq:starrepeat}$ then we have

$f(t) - f(1-t) = g(t)$

and

$f(1-t)-f(t)=g(1-t).$

Summing these identities gives

$g(t)+g(1-t)=0,$

so $g(t)$ must satisfy this condition for there to be a solution to $\eqref{eq:starrepeat}$.

1. Find a solution $f(t)$ to $\eqref{eq:starrepeat}$ when $g(t) = (2t-1)^3$.

### Approach 1

Letting $g(t)=(2t-1)^3$ means we require $f(t)-f(1-t)=8t^3-12t^2+6t -1.$

First let’s try $f_1(t)=t^3$. This gives

\begin{align*} f_1(t) - f_1(1-t) & = t^3 - (1-t)^3 \\ & = 2t^3 - 3t^2 + 3t-1.\\ \end{align*} Multiplying by $4$ gives $$$4[f_1(t) - f_1(1-t)] = 8t^3 -12t^2 +12t -4 \label{eq4f1}$$$

which is close to our required expression.

Now let $f_2(t)=t$, which gives $$$f_2(t) - f_2(1-t) = 2t-1. \label{eqf2}$$$ Subtracting $3$ times $\eqref{eqf2}$ from $\eqref{eq4f1}$ gives \begin{align*} 4[f_1(t) - f_1(1-t)] -3[f_2(t) - f_2(1-t)] &= 8t^3 -12t^2 +12t -4 -6t+3\\ &= 8t^3 - 12t^2 + 6t - 1 \end{align*}

as required. Hence, $f(t) = 4f_1(t)-3f_2(t) = 4t^3-3t$ is a solution.

This is just one solution; there are infinitely many solutions.

### Approach 2

Trying $f(t) = at^3+bt^2+ct+d$ seems like a sensible guess. Can we find constants $a,b,c$ and $d$ that will work?

We have $f(t)-f(1-t) = (at^3+bt^2+ct+d) -(a(1-t)^3+b(1-t)^2+c(1-t)+d)$ which on multiplying out gives

$f(t)-f(1-t)=2at^3 - 3at^2 + (3a + 2b + 2c)t - a - b - c.$

Now $(2t-1)^3 = 8t^3 - 12t^2 + 6t - 1$, so comparing coefficients, $a = 4, b=-1, c=-2, d=0$ is a solution that works.