Decomposing Polynomials into Partial Fractions: A Detailed Guide

Decomposing a polynomial like (frac{x^9 1}{x - 1^{10}}) into partial fractions can be a complex process. In this guide, we will walk through the steps to break down this polynomial into a sum of simpler fractions, using algebraic techniques and a systematic approach.

Introduction to Partial Fractions

Partial fractions are a way of decomposing a rational function into a sum of simpler rational functions. This technique is particularly useful in calculus, algebra, and engineering for simplifying integrals, solving differential equations, and analyzing systems.

The Problem at Hand

We are given the polynomial (frac{x^9 1}{x - 1^{10}}). Our goal is to decompose this into partial fractions of the form:

[frac{x^9 1}{x - 1^{10}} frac{A}{x - 1} frac{B}{(x - 1)^2} frac{C}{(x - 1)^3} frac{D}{(x - 1)^4} frac{E}{(x - 1)^5} frac{F}{(x - 1)^6} frac{G}{(x - 1)^7} frac{H}{(x - 1)^8} frac{I}{(x - 1)^9} frac{J}{(x - 1)^{10}}]

Substitution and Simplification

First, we let (x - 1 t). Then, the expression becomes:

[frac{x^9 1}{(x - 1)^{10}} frac{t^9 1}{t^{10}}]

Next, we rewrite the fraction as:

[frac{t^9 1}{t^{10}} frac{1}{t} frac{1}{t^2} frac{1}{t^3} frac{1}{t^4} frac{1}{t^5} frac{1}{t^6} frac{1}{t^7} frac{1}{t^8} frac{1}{t^9} frac{2}{t^{10}}]

Substituting back (t x - 1), we get:

[frac{x^9 1}{(x - 1)^{10}} frac{1}{x - 1} frac{9}{(x - 1)^2} frac{36}{(x - 1)^3} frac{84}{(x - 1)^4} frac{126}{(x - 1)^5} frac{126}{(x - 1)^6} frac{84}{(x - 1)^7} frac{36}{(x - 1)^8} frac{9}{(x - 1)^9} frac{2}{(x - 1)^{10}}]

Detailed Coefficient Calculation

The coefficients (A, B, C, D, E, F, G, H, I, J) are calculated by equating the expanded form of the right-hand side to the left-hand side and comparing coefficients of (x^9, x^8, x^7, ldots, x^0).

Step-by-Step Calculation

1. **For (x^9):** [1 A] Thus, (A 1).

2. **For (x^8):** [0 -9A - B] Since (A 1), we get (B 9).

3. **For (x^7):** [0 36A - 8B - C] Substituting (A 1) and (B 9), we get (C 36).

4. **For (x^6):** [0 -84A 28B - 7C - D] Substituting (A 1), (B 9), and (C 36), we get (D 84).

5. **For (x^5):** [0 126A - 56B 21C - 6D E] Substituting (A 1), (B 9), (C 36), and (D 84), we get (E 126).

6. **For (x^4):** [0 -126A 70B - 35C 15D - 5E F] Substituting (A 1), (B 9), (C 36), (D 84), and (E 126), we get (F 126).

7. **For (x^3):** [0 84A - 56B 35C - 20D 10E - 4F G] Substituting (A 1), (B 9), (C 36), (D 84), (E 126), and (F 126), we get (G 84).

8. **For (x^2):** [0 -36A 28B - 21C 15D - 10E 6F - 3G H] Substituting (A 1), (B 9), (C 36), (D 84), (E 126), (F 126), and (G 84), we get (H 36).

9. **For (x):** [0 9A - 8B 7C - 6D 5E - 4F 3G - 2H I] Substituting (A 1), (B 9), (C 36), (D 84), (E 126), (F 126), (G 84), and (H 36), we get (I 9).

10. **For (x^0):** [1 2J] Thus, (J 1).

Final Answer

The polynomial (frac{x^9 1}{x - 1^{10}}) decomposes into:

[frac{x^9 1}{x - 1^{10}} frac{1}{x - 1} frac{9}{(x - 1)^2} frac{36}{(x - 1)^3} frac{84}{(x - 1)^4} frac{126}{(x - 1)^5} frac{126}{(x - 1)^6} frac{84}{(x - 1)^7} frac{36}{(x - 1)^8} frac{9}{(x - 1)^9} frac{2}{(x - 1)^{10}}]

Conclusion

Decomposing polynomials into partial fractions is a critical mathematical skill. By following the steps outlined here, you can solve similar problems with confidence. Whether you are working on advanced calculus or engineering problems, understanding partial fractions will prove invaluable.