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Partial Fractions: Where am I going wrong?

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(edited 4 years ago)
Original post by Apologetic Cube

Spoiler


Small mistake at the end, I think you flipped A, B around with C, D.

So you have 3x+4x2(x2+1)≡Ax+Bx2+Cx+Dx2+1\displaystyle \frac{3x + 4}{x^2 (x^2 + 1)} \equiv \frac{Ax + B}{x^2} + \frac{Cx + D}{x^2 + 1}

and A=3, B=4, C=−3, D=−4A = 3, \ B=4, \ C = -3, \ D = -4.

So:

3x+4x2(x2+1)≡3x+4x2−3x+4x2+1\displaystyle \frac{3x + 4}{x^2 (x^2 + 1)} \equiv \frac{3x + 4}{x^2} - \frac{3x + 4}{x^2 + 1} which you can simplify to get the textbook answer.
(edited 10 years ago)
Reply 2
Avatar for TenOfThem
TenOfThem
18
Original post by Apologetic Cube
Express (3x+4)/((x^2(x^2+1))
Just to clarify, that's 3x+4 divided by x squared multiplied by x squared plus one.

Acc​ording to the textbook, the answer is:

(3/x) + (4/x^2) - ((3x+4)/(x^2+1))

I get: (3x)/(x^2 + 1) + 4/(x^2 + 1) - 3/x - 4/(x^2)

Here's my working:

(3x + 4)/(x^2(x^2 + 1)) = (Ax + B)/(x^2) + (Cx + D)/(x^2 + 1)

(Ax + B)(x^2 + 1) + (Cx + D)(x^2) = 3x + 4

When x = 0:

B = 4

When x = i:

(Ci + D)(-1) = 3i + 4

C = -3, D = -4

When x = 1:

2(A + 4) - 7 = 7 => A = 3



Up to here is fine



So the partial fraction expansion is:

(3x + 4)/(x^2 + 1) + (-3x - 4)/(x^2)

= (3x)/(x^2 + 1) + 4/(x^2 + 1) - 3/x - 4/(x^2)


Then you simply mixed up where your A, B, C, and D were

Your working out gives the correct answer
Reply 3
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TenOfThem
18
Original post by Apologetic Cube



Can I ask, what exactly is meant when x = i? What does i represent? Could you possibly write out that step but without i, so I can see how C and D were found? This was copied from a website so it's not necessarily comparative to the curriculum I'm following.


i represents the imaginary value that is −1\sqrt{-1}
Original post by Apologetic Cube



Can I ask, what exactly is meant when x = i? What does i represent? Could you possibly write out that step but without i, so I can see how C and D were found? This was copied from a website so it's not necessarily comparative to the curriculum I'm following.

I take it you've not done complex numbers yet.

Basically i=−1  ⟹  i2=−1i = \sqrt{-1} \implies i^2 = -1.

So you have: 3x+4≡(Ax+B)(x2+1)+(Cx+D)x23x + 4 \equiv (Ax +B)(x^2 + 1) + (Cx + D)x^2

x=i:3i+4=(Ai+B)(i2+1)⏟=0+(Ci+D)i2⏟=−13i+4=−Ci−D\displaystyle x = i: 3i + 4 = (Ai + B)\underbrace{(i^2 + 1)}_{=0} + (Ci + D)\underbrace{i^2}_{=-1} \\ 3i + 4 = -Ci - D.

Now, you when you have an equation with complex numbers, you can equate real and imaginary parts. So if:

x+iy=u+iv  ⟹  x=u, y=vx + iy = u + iv \implies x = u, \ y = v.

And you can apply this to the above.
Original post by Apologetic Cube


Complex numbers are part of the AH course but we haven't covered them yet! So what value would I use instead of i, if I were to try and find C and D without the use of complex numbers?

Just plug in some simple values like x=1, −1x=1, \ -1, etc and you'll get simultaneous equations which you can solve.
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by Apologetic Cube


Complex numbers are part of the AH course but we haven't covered them yet! So what value would I use instead of i, if I were to try and find C and D without the use of complex numbers?



Use any other 2 values and solve the simultaneous equations for C and D

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