- #1
ko_kidd
- 21
- 0
[tex]\frac{7}{3s^{2}(3s+1)}[/tex]
Can this be decomposed, and how?
Can this be decomposed, and how?
ko_kidd said:I have one more problem.
Would this:[tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]
simplify to something like
[tex]\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}[/tex] = [tex]\frac{87}{(x)(x^{2}+13x+38)}[/tex]
ko_kidd said:[tex]\frac{7}{3s^{2}(3s+1)}[/tex]
Can this be decomposed, and how?
symbolipoint said:I'd say, yes; it can be decomposed; without my first relearning the method and trying to decompose to partial fractions. Your denominators might be [tex] \[
3s^2
\]
[/tex] and [tex] \[
3s + 1
\]
[/tex]
HallsofIvy said:With that "s2", you are going to need both 1/s and 1/s2.
[tex]\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}[/tex]
Multiplying through by the common denominator, [itex]7= As(3s+1)+ B(3s+1)+ Cs^2[/itex]. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.
Partial fraction decomposition is a method used to break down a rational function into simpler fractions. This is done by finding the individual fractions that, when added together, are equal to the original function.
Partial fraction decomposition is used when dealing with rational functions, particularly when integrating them. It allows for easier integration by breaking down the function into simpler fractions.
The first step in partial fraction decomposition is to factor the denominator of the rational function. Then, using the factors, we can write the rational function as a sum of simpler fractions with unknown coefficients. These coefficients can be solved for using algebraic manipulation and solving a system of equations.
The purpose of partial fraction decomposition is to simplify a rational function into simpler fractions. This allows for easier integration, as well as a better understanding of the overall function.
Yes, there are limitations to partial fraction decomposition. It can only be used on rational functions with distinct linear factors in the denominator. If the function has repeated or irreducible quadratic factors, additional techniques may be needed to fully decompose the function.