Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fractional Reduction TROUBLE

  1. Nov 17, 2012 #1
    Hi, this seems like the right section to post, I have a few questions, three fractions that I know the answer for but I just cant figure out how they were arrived at (the middle step).
    The only way I can think of solving them is expanding it all out etc. which would result in a 100 term cubic polynomial, so I'm sure there is an easier way to get to the final fraction.
    If anyone can help me solve either of these three that would be great.
    Please see the pictures for the questions.

    P.S the 'j' is 'i', that is the square root of -1.


    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 17, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi toneboy1! :smile:

    for the first one, two hints:

    i] the denominator of the solution is a2 + b2, which factors as … ?

    ii] the first two denominators are almost identical, only some of the terms are multiplied by -1, so group together all the terms that are and all the terms that aren't :wink:
     
  4. Nov 17, 2012 #3
    Thanks for the reply, I appreciate it.

    Err...well I do remember a2+b2=(a+b)2-2ab
    and I can see the solution of the first is squared like that but I don't see how it was originally (a+b)2-2ab.

    Likewise I can see they're almost identical but I cant see what you mean by grouping the positive and negative ones....
     
  5. Nov 18, 2012 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi toneboy1! :smile:

    (just got up :zzz:)

    a2+b2 = a2 - (ib)2 = (a + ib)(a - ib) ? :wink:
     
  6. Nov 18, 2012 #5
    Well hopefully you've had your coffee now.

    Please, treat me like an idiot, I still don't see quite what your getting at.

    Thanks
     
  7. Nov 18, 2012 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    try factoring the denominator of the RHS (the a2 + b2 one) …

    then compare it with the LHS, and you may see what's going on :wink:
     
  8. Nov 18, 2012 #7
    Ok, applying '(a + ib)(a - ib)' to the RHS I got: (-w2+4jw-1-4w+j8)(-w2+4jw-1+4w-8j)...which doesn't mean a hell of a lot to me.

    But I only included the RHS answer to the question for varifications sake, it shouldn't be really necessary.
     
  9. Nov 18, 2012 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ok, for the LHS, write out separately the terms in the denominators which are the same for both denominators, and the terms which are minus each other …

    what do you get? :smile:
     
  10. Nov 19, 2012 #9
    You're the boss.
    Ok well than you'd get: 16j - 8jw - 8w on the denominator (?)

    Thanks
     
  11. Nov 19, 2012 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    uhh? :confused:

    if you expand each denonimator on the LHS, there are 9 terms which are the same apart possibly for a minus sign (5 are the same, 4 are minus)

    write out the ones that are the same, and then the ones that are minus :smile:
     
  12. Nov 19, 2012 #11
    what I previously posted was what I thought the result was, as the denominators I expanded to be:

    [-w2+4w+4jw-8j-1]-[-w2-4w+8j+4jw-1] = 16j - 8jw - 8w ...or was it -8w + 16j...

    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fractional Reduction TROUBLE
  1. Fractional Derivatives (Replies: 3)

  2. Integrating a fraction (Replies: 12)

  3. Partial fraction (Replies: 3)

Loading...