Question

I don't have a specific problem. Can someone explain partial decomposition fractions please?

Answer 1

Mathematical: if you have a rational function with multiple "zeros" (more than 2) in the denominator, then the partial fraction decomposition allows you to express the same fraction as sums of rational functions each with lesser number of zeros in the denominator

Answer 2

reply man

Answer 3

replay peter

Answer 4

http://tutorial.math.lamar.edu/Classes/Alg/PartialFractions.aspx

Answer 5

It's a decomposition which may help you later in integration.

Answer 6

Or other places. Laplacoian, Z-transform, etc...
By this decompoition, the analysis you'd do will be on functions of a smaller degree.

Answer 7

its a way to undo fractions that have been added together

Answer 8

in many cases, its easier to work on the parts of a rational expression, as opposed to the "simplified" version

Answer 9

Nonono, I know what they are, I need you to explain how to do them.

Answer 10

factor the denominator so that you can break the rational expression apart into seperate pieces.

then define each new numerator according to the degree of the denominators, and start zeroing out each part and equate it to the results of the original numerator

Answer 11

the way we combine fractions tends to produce a product for the denominator and a summation for the numerator
\[\frac ab +\frac cd=\frac{ad+bc}{bd}\]
by factoring the denominator we reveal the denominators that we "combined" to begin with

Answer 12

the rest of it is coming up with a system of equation that systematically determines the numerators that we would have began with

Answer 13

\[\frac{2x+3}{(x-4)(x^2+3)}=\frac{A}{x-4}+\frac{Bx+C}{x^2+3}\]

try to redo the numerators

\[\frac{2x+3}{(x-4)(x^2+3)}=\frac{A(x^2+3)}{(x-4)(x^2+3)}+\frac{(Bx+C)(x-4)}{(x-4)(x^2+3)}\]

this leaves us with comparing numberators

\[2x+3=A(x^2+3)+(Bx+C)(x-4)\]

Answer 14

we know when x=4, we are left with\[2(4)+3=A(4^2+3)\]and can solve for A

Answer 15

unless we go complex values, we would have to try a different setup for Bx+C, but we already know A and we also know that for any x values, the sides have to equate to be equal

Answer 16

\[2x+3=\frac{11}{19}(x^2+3)+(Bx+C)(x-4)\]
\[2x+3=\frac{11}{19}x^2+\frac{33}{19}+Bx^2-(4B-C)x-4C\]
\[0x^2+2x+3=(B+\frac{11}{19})x^2-(4B-C)x-(4C+\frac{33}{19})\]
equte coefficients

Answer 17

\[B+\frac{11}{19}=0\\C-4B=2\\-4C-\frac{33}{19}=3\]

Answer 18

with any luck, i didnt mismath anything :)