Mathematics Dictionary
Dr. K. G. Shih
Partial Fractions
Subjects
Symbol Defintion
AL 10 00 |
- Outlines
AL 10 01 |
- Introduction
AL 10 02 |
- Rules
AL 10 03 |
- Find partial fraction of 1/(x^2-6*x+8)
AL 10 04 |
- Find partial fraction of 1/(1-x^3)
AL 10 05 |
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AL 10 06 |
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AL 10 07 |
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AL 10 08 |
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AL 10 09 |
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AL 10 10 |
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Answers
AL 10 01. Introduction
Why do we need partial fractions ?
One reason is to find integral.
Eample : Find
∫
dx/(x^2-6*x+8)
Sine x^2 - 6*x + 8 = (x-2)*(x-4)
Hence 1/(x^2 - 6*x + 8) = A/(x-2) + B/(x-4)
Then we can find the integral
∫
dx/(x^2-6*x+8) =
∫
(A*dx)/(x-2) +
∫
(A*dx)/(x-4)
Go to Begin
AL 10 02. Rules
The numerator in partial fraction is less one power than denominator
Example 1 : (A*x + B)/(a*x^2 + b*x + c)
Example 2 : (A*x^2 + B*x + C)/(a*x^3 + b*x^2 + c*x + d)
Go to Begin
AL 10 03. Example : Find partial fraction of 1/(x^2 - 6*x + 8)
Since x^2 - 6*x + 8 = (x-2)*(x-4)
Hence 1/(x^2 - 6*x + 8) = A/(x-2) + B/(x-4)
Both sides times (x^2 - 6*x + 8)
1 = A*(x-4) + B*(x-2)
1 = (A+B)*x - (4*A+2*B)
Hence A+B = 0 ............. (1)
4*A + 2*B =-1 ............. (2)
Solve (1) and (2) we have A = -1/2 and B = 1/2
Hence 1/(x^2 - 6*x + 8) = -1/(2*(x-2)) + 1/(2*(x-4))
Go to Begin
AL 10 04. Find partial fraction of 1/(1-x^3)
Since 1 - x^3 = (1-x)*(1+x+x^2)
Hence 1/(1-x^3) = A/(1-x) + (B*x+C)/(1+x+x^2)
Both sides times (1-x^3)
1 = A*(1+x+x^2) + (B*x+C)*(1-x)
1 = (A+C) + (A+B-C)*x + (A-B)*x^2
Hence A+C = 1 ....................... (1)
A+B-C = 0 ........................... (2)
A-B = 0 ............................. (3)
(2)+(3) we have 2*A-C = 0 ........... (4)
(1)+(4) we have 3*A = 1 and A = 1/3
Hence B = A = 1/3
Hence C = 1-A = 2/3
Hence 1/(1-x^3) = 1/(3*(1-x)) + (1*x+2)/(1+x+x^2)
Go to Begin
AL 10 05. Answer
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AL 10 06. Answer
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AL 10 07. Answer
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AL 10 08. Answer
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AL 10 09. Answer
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AL 10 10. Answer
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AL 10 00. Outlines
Rules
1. The denominator must have factors
2. For factor with power 2 : (A*x + B)/(a*x^2 + b*x + c)
3. For factor with power 3 : (A*x^2 + B*x + C)/(a*x^3 + b*x^2 + c*x + d)
Major application
Find intergal
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