- #1
PFuser1232
- 479
- 20
Suppose we have a rational function ##P## defined by:
$$P(x) = \frac{f(x)}{(x-a)(x-b)}$$
This is defined for all ##x##, except ##x = a## and ##x = b##.
To decompose this function into partial fractions we do the following:
$$\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Multiplying both sides by ##(x-a)(x-b)##:
$$f(x) = A(x-b) + B(x-a)$$
Which again holds whenever ##x## does not equal ##a## or ##b##.
To find ##A## or ##B##, we set ##x## equal to ##a## or ##b##, which is confusing. Didn't we state earlier that the rational function, and all steps that took us from the first equation to the last equation, are only valid when ##x## does not equal ##a## or ##b##?
$$P(x) = \frac{f(x)}{(x-a)(x-b)}$$
This is defined for all ##x##, except ##x = a## and ##x = b##.
To decompose this function into partial fractions we do the following:
$$\frac{f(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Multiplying both sides by ##(x-a)(x-b)##:
$$f(x) = A(x-b) + B(x-a)$$
Which again holds whenever ##x## does not equal ##a## or ##b##.
To find ##A## or ##B##, we set ##x## equal to ##a## or ##b##, which is confusing. Didn't we state earlier that the rational function, and all steps that took us from the first equation to the last equation, are only valid when ##x## does not equal ##a## or ##b##?