Partial Fraction Paradox

[James10492]

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.


[math]\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}[/math]
[math]u^2 = A(u-2) + B(u+2)[/math]
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

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

so,

[math]\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)} [/math]

but now,


[math]\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}[/math]
????

 

[topsquark]

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:
[imath]\dfrac{u^2}{u^2 - 4} = 1 + \dfrac{4}{u^2 - 4}[/imath].

Now do a partial fraction decomposition on [imath]\dfrac{4}{u^2 - 4}[/imath].

-Dan

 

[James10492]

Ah thank you that was what I was not taking into account. Mind expanded! Thank you again sir.

 

[Dr.Peterson]

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.


[math]\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}[/math]
[math]u^2 = A(u-2) + B(u+2)[/math]
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

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

so,

[math]\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)} [/math]

but now,


[math]\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}[/math]
????

Observe that [imath]A = -1, B = 1[/imath] does not make [imath]u^2 = A(u-2) + B(u+2)[/imath] an identity. On substitution, the equation becomes [imath]u^2 = -1(u-2) + 1(u+2) = 4[/imath]. This happens because you didn't equate coefficients of [imath]u^2[/imath], 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 [imath]Au+B[/imath] and [imath]Cu+D[/imath]. And then you would have too few equations rather than too many.

 

[BigBeachBanana]

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.


[math]\frac{u^2}{u^2 - 4} = \frac{u^2}{(u+2)(u-2)} = \frac{A}{(u+2)} + \frac{B}{(u-2)}[/math]
[math]u^2 = A(u-2) + B(u+2)[/math]
let u = 2,

then 4 = 4 B, so B = 1

let u = -2,

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

so,

[math]\frac{A}{(u+2)} + \frac{B}{(u-2)} =\frac{1}{(u -2)} - \frac{1}{(u+2)} [/math]

but now,


[math]\frac{1}{u-2} - \frac{1}{u+2} \neq \frac{u^2}{u^2-4}[/math]
????

On a side note, there's a trick to deal with such fractions. Time saver.
[math]\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}[/math]