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Find x(2), x(3) from X(z)=(2z²+3z+12)/(z-1)⁴

Short answer

Expand X(z) in powers of z^{-1}: X(z)=(2z^{-2}+3z^{-3}+12z^{-4})/(1-z^{-1})^4. The coefficients give x(2)=2 and x(3)=11.

If Z[x(k)] = (2z² + 3z + 12)/(z - 1)⁴, find the value of x(2) and x(3).

Idea. By definition X(z) = x(0) + x(1)z⁻¹ + x(2)z⁻² + x(3)z⁻³ + … So x(k) is the coefficient of z⁻ᵏ in the expansion of X(z) in powers of 1/z (long division / binomial series).

Step 1: write in powers of w = z⁻¹.

X(z) = (2z² + 3z + 12) / (z − 1)⁴
     = (2z² + 3z + 12) / z⁴  ·  1/(1 − w)⁴
     = (2w² + 3w³ + 12w⁴) · 1/(1 − w)⁴

Step 2: binomial series.

1/(1 − w)⁴ = 1 + 4w + 10w² + 20w³ + 35w⁴ + …

Step 3: multiply.

coefficient of w⁰ : 0                        → x(0) = 0
coefficient of w¹ : 0                        → x(1) = 0
coefficient of w² : 2                        → x(2) = 2
coefficient of w³ : 2·4 + 3  = 11            → x(3) = 11
coefficient of w⁴ : 2·10 + 3·4 + 12 = 44     → x(4) = 44

Answer: x(2) = 2 and x(3) = 11.

✓ Answer checked by Claude Opus 5.5 (high effort)

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