i have done partial fractions, but we are getting into series, and I need to use partial fractions to split it up, and just wondered if I did this one right, as its been a little while since ive done them. Thanks!
i have done partial fractions, but we are getting into series, and I need to use partial fractions to split it up, and just wondered if I did this one right, as its been a little while since ive done them. Thanks!
that is (technically) true. in fact, that is how we started:
$\displaystyle \frac {2n + 1}{n^2(n + 1)^2} = \frac {An + B}{n^2} + \frac {C(n + 1) + D}{(n + 1)^2} = \frac An + \frac B{n^2} + \frac C{n + 1} + \frac D{(n + 1)^2}$
this is just the rule, here's the trick. look at the degree of the "inside" function. that is, the function that is raised to the power, that is the function your numerator must be one degree less than. consider the illustration below
$\displaystyle \frac {f(x)}{[g(x)]^n} = \frac {p_1(x)}{g(x)} + \frac {p_2(x)}{[g(x)]^2} + \cdots + \frac {p_n(x)}{[g(x)]^n}$
in the above, $\displaystyle f(x)$ is of lower degree than $\displaystyle g(x)$, and each $\displaystyle p_n(x)$ is a general polynomial function that is ONE degree less than $\displaystyle g(x)$.
so, for example, if it were $\displaystyle \frac {2n + 1}{n^2(n^2 + 1)^2}$, you would write
$\displaystyle \frac {2n + 1}{n^2(n^2 + 1)^2} = \frac An + \frac B{n^2} + \frac {Cn + D}{n^2 + 1} + \frac {En + F}{(n^2 + 1)^2}$
in the last two fractions, the "inside" function was a quadratic, so the numerators had to be linear.
if it were $\displaystyle \frac {2n + 1}{n^2(n^3 + 1)^2}$, then you would write
$\displaystyle \frac {2n + 1}{n^2(n^3 + 1)^2} = \frac An + \frac B{n^2} + \frac {Cn^2 + Dn + E}{n^3 + 1} + \frac {Fn^2 + Gn + H}{(n^3 + 1)^2}$
in the last two fractions, the "inside" function was a cubic, so the numerators had to be quadratic
got it?