does anybody know where this "1" comes from?

Printable View

- Jan 5th 2013, 02:03 PMcxz7410123matching the terms in partial fractions
does anybody know where this "1" comes from?

- Jan 5th 2013, 02:20 PMskeeterRe: matching the terms in partial fractions
expanding the RHS yields the quadratic terms $\displaystyle 2Au^2 + Cu^2 = (2A+C)u^2$

note the coefficient of $\displaystyle u^2$ on the LHS is 1 , hence by matching coefficients $\displaystyle 2A+C = 1$ - Jan 5th 2013, 02:20 PMDevenoRe: matching the terms in partial fractions
u

^{2}= (1)(u^{2}), that is, in the polynomial u^{2}, the coefficient of u^{2}is 1 (this is usually suppressed when we write polynomials). - Jan 5th 2013, 02:56 PMcxz7410123Re: matching the terms in partial fractions
Thank you very much.

Your answer rises me an other problem, why in the second scan there is this "0" and not 3?.

Attachment 26483 - Jan 5th 2013, 03:13 PMskeeterRe: matching the terms in partial fractions