Question

real analysis (1 sec typing up prob.)

Answer 1

For f, a real valued function on the real line, define the function Δf by

Δf(x) = f(x + 1) - f(x).

For n ≥ 2, define \[\Delta ^{n}f\]
recursively by

\[\Delta ^{n} = \Delta(\Delta ^{n-1}f)\]

Prove that Δ^n f = 0 if and only if f has the form

\[f(x) = a _{0}(x) + a _{1}(x)x + . . . + a _{n-1}(x)x ^{n-1}\]

where a0, a1, . . . a_n-1 are periodic functions of period 1

Answer 2

an example i worked out , let \[f(x) = \sin(2 \pi x) + \sin(2 \pi x) * x\]

working out Δ^2 gives:

(((sin(2*pi*(x+2)) + sin(2*pi*(x+2)) * (x+2)) - (sin(2*pi*(x+1)) + sin(2*pi*(x+1)) * (x+1))) - ((sin(2*pi*(x+1)) + sin(2*pi*(x+1)) * (x+1)) - (sin(2*pi*(x)) + sin(2*pi*(x)) * (x))))

which is equal to 0

Δ will always output 0 for periodic functions of 1
its the recursive Δ function giving me problem proving. hm.