# 2013_Question2

(i)

We have:

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

so:

kf(1 - t) - k^2f(t) = k(1 - t)

Adding this to the identity in the question:

(1 - k^2)f(t) = k - (k - 1)t

Hence:

\displaystyle f(t) = \frac {k + (1 - k)t} {1 - k^2}
(iia)

Suppose there existed such a function f.

We have:

f(t) - f(1 - t) = t

and:

f(1 - t) - f(t) = 1 - t

0 = 1

Clearly nonsense so there exists no such f.

(b)

Since we have:

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

and:

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

We must have:

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

for all t.

We have:

(2(1 - t) - 1)^3 = (2 - 2t - 1)^3 = (1 - 2t)^3 = -(2t - 1)^3

So we’re reassured g satisfies the condition in (b)

Since the LHS is a cubic, it make sense to start by trying f cubic, ie. trying f of the form f(t) = At^3 + Bt^2 + Ct

We then have:

\begin{align*}f(t) - f(1 - t) &= f(t) + f(t - 1) \\ &= At^3 + Bt^2 + Ct + A(t - 1)^3 + B(t - 1)^2 + C(t - 1) \\ &= At^3 + Bt^2 + Ct + A(t^3 - 3t^2 + 3t - 1) + B(t^2 - 2t + 1) + Ct - C \\ & = 2A t^3 + (2B - 3A)t^2 + (2C + 3A - 2B)t + (B - C - A)\end{align*}

and:

(2t - 1)^3 = 8t^3 - 12t^2 + 6t - 1

Comparing coefficients we have a solution of A = 4, B = 0, C = -3. So a solution is f(t) = 4t^3 - 3t.

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