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Fraction calculation, can't find mistake

  1. Sep 30, 2016 #1
    1. The problem statement, all variables and given/known data


    I did some homework and wolfram alpha afterwards gave different result than what I have. Apparently I must have done some error, but I couldn't find error after I doublechecked about two times. So I decided to see if you guys can find an error. And what is the correct answer also. I was pulling my hair out after doing this problem for some time :nb)


    $$ \frac{\frac{1}{s}-s}{\frac{1}{s}-\frac{1}{s-1}} ~~~~~~ $$

    2. Relevant equations


    3. The attempt at a solution

    here we expanded
    ##(\frac{1}{s}-\frac{s^2}{s})/(\frac{s-1}{s(s-1)}-\frac{s}{s(s-1)})##


    here I tried to get the plusses and minuses right, but I'm not sure I succeeded
    ##[\frac{1}{s}+\frac{-s^2}{s}] / [\frac{s-1}{s(s-1)}+\frac{-s}{s(s-1)}]##

    ##(\frac{1-s^2}{s}) / [\frac{(s-1)-s}{s(s-1)}]##

    ##[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ]##


    ##\frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}##


    ## \frac{(s-1)^2(s+1)(s)}{-s}##

    ##(-1)(s-1)^2(s+1)##
     
  2. jcsd
  3. Sep 30, 2016 #2

    fresh_42

    Staff: Mentor

    What is ##(1-s^2)## factored?
     
  4. Sep 30, 2016 #3

    SammyS

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    Looks OK.

    What's the problem you're having?
     
  5. Sep 30, 2016 #4
    basically ##(1-s^2) = (1^2-s^2)##

    so that is (1+s) (1-s)

    = 1-s +s -s^2
    = 1- s^2
     
  6. Sep 30, 2016 #5

    fresh_42

    Staff: Mentor

    Yes.
     
  7. Sep 30, 2016 #6
    can you bold print my mistake in quote for example
     
  8. Sep 30, 2016 #7

    fresh_42

    Staff: Mentor

    You wrote
    ##[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ] = \frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}##
    But ## (s-1)(s+1) \neq (1-s)(1+s)##
     
  9. Sep 30, 2016 #8
    I think I got it right this time.

    (s-1)(s+1)

    = s^2 +s -s -1

    ##=~~ (+s^2~~-1)##

    the correct part said that the factors were as follows

    ## (1-s) (1+s) = 1 +s -s -s^2##
    ##= -s^2 +1##

    the latter case is the "counter number" to the other number (multiplied by minus one)
     
  10. Sep 30, 2016 #9

    SammyS

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    Oh!

    not quite OK !

    sign mistake
     
  11. Sep 30, 2016 #10

    fresh_42

    Staff: Mentor

    The other way around. Until ##1-s^2## everything was fine, and by the way, afterwards, too. You simply changed the sign by substituting ##1-s^2 = (s-1)(s+1)## which is wrong by ##-1##. You should have used ##1-s^2=(1-s)(1+s)## or ##1-s^2=(-1)(s-1)(s+1)##
     
  12. Sep 30, 2016 #11
    looks like the tricky math prob got under your skin at first also !! xD
     
  13. Sep 30, 2016 #12

    fresh_42

    Staff: Mentor

    By the way. Me, too. I only found out as I calculated it on my own. At first sight I assumed a false input in Wolfram.
     
    Last edited: Sep 30, 2016
  14. Sep 30, 2016 #13
    ok i will try to post the new answer in full, but it will be tedious because of Latex code
     
  15. Sep 30, 2016 #14
    ##\frac{1/s-s}{1/s-[1/(s-1)]}##


    ##\frac{(1-s^2)/s}{(-1)/[s(s-1)]}##


    ##\frac{(1-s)(1+s)}{s}*\frac{s(s-1)}{-1}##


    ##\frac{(1-s)(1+s)(s)(s-1)}{(-1)(s)}##


    ##\frac{(-1)(s-1)(s)(s-1)(s+1)}{-1s}##

    ##\frac{(s-1)^2(s)(s+1)}{s}##

    ##(s-1)^2(s+1)##
     
  16. Sep 30, 2016 #15

    SammyS

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    What does Wolfram give?

    Did I somehow miss that in one of these posts?
     
  17. Sep 30, 2016 #16

    I think I did a wrong factorization for the term ##(1^2 -s^2)##


    evidently wolfram alpha did not become fooled by that.

    this skewed my result from early on. I tried to use the memory formula for the ##(a^2-b^2)## for finding factors
     
  18. Sep 30, 2016 #17

    fresh_42

    Staff: Mentor

    No. The OP hasn't said and I haven't done. Far too many parenthesis to type in.
     
  19. Sep 30, 2016 #18
    code for wolfram alpha


    && inside these &&

    && (1/s-s)/(1/s-1/(s-1)) &&
     
  20. Sep 30, 2016 #19
    looks like I did the problem correctly as post # 14 will show the same alternative form result as wolfram alpha.

    ##
    (1-s)(1-s^2)
    ##

    or

    ##
    (s-1)^2(s+1)
    ##
     
  21. Sep 30, 2016 #20

    fresh_42

    Staff: Mentor

    I wouldn't have relied on this ambiguous inline coding. I prefer to use parenthesis to avoid false interpretations.
    Concerning the sign error all of us blundered into: it happens. Nothing to worry about.
     
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