Partial Fraction Paradox

James10492

Junior Member
Joined
May 17, 2020
Messages
50
Hi all, this is driving me nuts, I am probably doing something dumb but would appreciate it if someone could point out what. Thanks in advance.


u2u24=u2(u+2)(u2)=A(u+2)+B(u2)\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}
u2=A(u2)+B(u+2)u^2 = A(u-2) + B(u+2)
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

then 4 = -4 A, so A = -1

so,

A(u+2)+B(u2)=1(u2)1(u+2)\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)}

but now,


1u21u+2u2u24\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}
????
 
To use partial fractions you need a fraction that has a lower degree in the numerator than in the denominator. So you first need to do a division:
u2u24=1+4u24\dfrac{u^2}{u^2 - 4} = 1 + \dfrac{4}{u^2 - 4}.

Now do a partial fraction decomposition on 4u24\dfrac{4}{u^2 - 4}.

-Dan
 
Ah thank you that was what I was not taking into account. Mind expanded! Thank you again sir.
 
Hi all, this is driving me nuts, I am probably doing something dumb but would appreciate it if someone could point out what. Thanks in advance.


u2u24=u2(u+2)(u2)=A(u+2)+B(u2)\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}
u2=A(u2)+B(u+2)u^2 = A(u-2) + B(u+2)
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

then 4 = -4 A, so A = -1

so,

A(u+2)+B(u2)=1(u2)1(u+2)\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)}

but now,


1u21u+2u2u24\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}
????
Observe that A=1,B=1A = -1, B = 1 does not make u2=A(u2)+B(u+2)u^2 = A(u-2) + B(u+2) an identity. On substitution, the equation becomes u2=1(u2)+1(u+2)=4u^2 = -1(u-2) + 1(u+2) = 4. This happens because you didn't equate coefficients of u2u^2, which are respectively 1 and 0. The full system of equations is inconsistent.

If you were to do this without making a proper fraction, you would need to make the numerators Au+BAu+B and Cu+DCu+D. And then you would have too few equations rather than too many.
 
Hi all, this is driving me nuts, I am probably doing something dumb but would appreciate it if someone could point out what. Thanks in advance.


u2u24=u2(u+2)(u2)=A(u+2)+B(u2)\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}
u2=A(u2)+B(u+2)u^2 = A(u-2) + B(u+2)
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

then 4 = -4 A, so A = -1

so,

A(u+2)+B(u2)=1(u2)1(u+2)\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)}

but now,


1u21u+2u2u24\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}
????
On a side note, there's a trick to deal with such fractions. Time saver.
u2u24=u24+4adding 0u24=u24u24+4u24=1+4u24\frac{u^2}{u^2 - 4} =\frac{u^2\overbrace{-4+4}^{\text{adding 0}}}{u^2-4}=\frac{u^2-4}{u^2-4}+\frac{4}{u^2-4}=1+\frac{4}{u^2-4}
 
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