[Calculus] Represent the function 7/1-3x as a power series

Question

Represent the function 7/1-3x as a power series f(x)= ∑∞_n=0 c_nxⁿ

Compute the first few coefficients of this power series:

C₀₌

C₁₌

C₂₌

C₃₌

C₄₌

Answer 1

7 / (1 - 3x) has the form a / (1 - r) which is the sum of the geometric series ∑_{(n=0 to ∞)} arⁿ

So, 7 / (1 - 3x) = 7 + 21x + 63x² + ...

The series converges only when -1 < 3x < 1 (that is, when -1/3 < x < 1/3)

Answer 2

The key here is to use the geometric power series formula: 1/(1-r) = ∑_n=0^∞rⁿ .

Here, we will use r = 3x, and so we get

7/(1-3x) = 7 * 1/(1-r)

= 7*∑_n=0^∞(3x)ⁿ

= 7 ∑_n=0^∞ 3ⁿ xⁿ

= ∑_n=0^∞ (7•3ⁿ) xⁿ

The coefficients of the power series are the numbers that are multiplied against xⁿ, that is, 7•3ⁿ. So, we get

n=0 ⇒ C₀ = 7•3⁰ = 7

n=1 ⇒ C₁ = 7•3¹ = 21

n=2 ⇒ C₂ = 7•3² = 7 • 9 = 63

n=3 ⇒ C₃ = 7•3³ = 7 • 27 = 189

n=4 ⇒ C₄ = 7•3⁴ = 7 • 81 = 567