Question

If f(x) = e^ax then show that f(0), Δf(0), Δ²f(0) are in G.P

Answer 1

Given f(x) = e^ax

f(0) = e° = 1  .......(1)

Δf(x) = `"e"^("a"(x + "h")` – eax

= `"e"^("a"x+"ah")` – eax

= e^ax. e^ah – e^ax

= e^ax (e^ah – 1)

Δf(0) = e° (e^ah – 1)

= (e^ah – 1)   ........(2)

Δ²f(x)= Δ[Δf(x)]

= `Δ["e"^("a"(x + "h")) –  "e"^("a"x)]`

`["e"^("a"(x + "h" + "h")) – "e"^("a"(x + "h"))] – ["e"^("a"(x + "h")) –  "e"^("a"x)]`

= `"e"^("a"(x + 2"h")) –  "e"^("a"(x + "h")) –  "e"^("a"(x + "h")) +  "e"^("a"x)`

Δ²f(0) = Δ[Δf(x)]

= e^a(2h) – e^a(h) – e^a(h) + e⁰

= e^2ah – e^ah – e^ah + 1

= (e^ah)² – 2e^ah + 1

= [e^ah – 1]²   ..........(3)

From (1), (2) and (3)

[t₂]² = [Δf(0)]² = (e^ah – 1)²

t₁ × t₃ = f(0) × Δ²f(0)

= (1)(e^ah – 1)² = (e^ah – 1)²

⇒ [Δf(0)]² = f(0) × Δ²f(0)

∴ f(0), Δf(0), Δ²f(0) an is G.P.