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Homework Help: 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

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    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

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

    fresh_42

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    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

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    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

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    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

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    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

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    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.
     
  22. Sep 30, 2016 #21
    I think modern wolfram alpha shows the input in variable forms like fraction line form of the division as a picture
     
  23. Sep 30, 2016 #22

    did you get the same result as post 14 from me
     
  24. Sep 30, 2016 #23

    fresh_42

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    Yes.

    This way one can see, whether the input was correct or not. However, one shouldn't get used to it in the sense that "everybody knows what is meant", because it is a total unnecessary source of potential mistakes.

    We often read things like "e^t^2/2" where it is not clear without additional knowledge that ##e^{\frac{1}{2}t^2}## is meant and not ##\frac{1}{2}e^{t^2}## or ##e^t \, ##.
     
  25. Oct 1, 2016 #24

    SammyS

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    What I get was s3 - s2 - s +1
     
  26. Oct 1, 2016 #25
    ¨now we have competition between the answers it seems:biggrin:

    looks like that comes out at the same factorization so it looks like what I got,
     
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