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Real solutions

  1. Dec 27, 2014 #1
    1. The problem statement, all variables and given/known data
    To find number of real solutions of:
    ##\frac{1}{x-1}## ##+\frac{1}{x-2}## + ##\frac{1}{x-3}## + ##\frac{1}{x-4}## =2


    2. Relevant equations
    It will form a 4th degree polynomial equation.

    3. The attempt at a solution
    The real solutions could be 0 or 2 or 4 as complex solutions always exist in pairs.
    It is a 4th degree polynomial equation , the real roots could be 4 but how?
     
  2. jcsd
  3. Dec 27, 2014 #2

    Stephen Tashi

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    The given equation is not a 4th degree polynomial equation. When you do multiplications on it to form a 4th degree polynomial equation, you may create an equation that has a different solution set than the original equation. For example, the original equation does not have solutions x = 1, x = 2, x =3 or x = 4 since such values require division by zero. (For example, for [itex] \frac {1}{x-2} [/itex] to have a defined value, we cannot have [itex] x = 2 [/itex].

    Are you studying functions that have asymptotes? Perhaps you are expected to sketch a graph of the function [itex] f(x) = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} - 2 [/itex]. The question asked for the number of roots ( the number of places the graph of the function crosses the x-axis) and not the particular values of the roots.
     
  4. Dec 27, 2014 #3
    Sketching graph of that function seems difficult. I am not studying functions that have asymptotes. I only require number of real roots. Calculus can be used here I think but not able to apply. I know that the number of real roots maybe 0 or 2 or 4.
     
  5. Dec 27, 2014 #4

    BvU

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    If you find the combined sketch difficult, start with the first term. Then the second, then estimate what 1st + 2nd looks like. You will see the pattern quickly -- and have overcome your aversion to things that seem difficult as well :)
     
  6. Dec 27, 2014 #5
    I was trying and found out that whenever value of x is negative y is always negative. So, there is no crossing of x axis when x -ve, so no real root till that point. Now the problem arises that around x=5 y value is coming +ve. So there is a crossing but there may be more. Can anyone help in that?
     
  7. Dec 27, 2014 #6

    Stephen Tashi

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    As I said before, you don't know that.

    Let h be a small positive number, say h = 1/100,000. What is the approxomate value of the function at x = 1 + h ?

    At that value, all terms except [itex] \frac{1}{x-1} [/itex] are of moderate size. The term [itex]\frac{1}{x-1} [/itex] has value 100,000. Now consider the value of the function at x = 2 - h.
     
  8. Dec 27, 2014 #7
    At x = 2 - h , the term [itex] \frac{1}{x-2} [/itex] would be -100,000. It means the graph crossed x axis one time there.
    Yeah , it seems I got it a bit.
    Then, at x = 2 + h the value again becomes 100,000 second crossing
    Then at 3-h , -100,000 third crossing
    At 3+h fourth crossing,
    4-h fifth crossing
    4+h sixth crossing
    Then at value between 5 and 6 seventh crossing and after that no crossing.
    But from this the answer is coming 7.
    I know the answer is 4 from calculators.
    So, what is my mistake?
     
  9. Dec 27, 2014 #8

    Stephen Tashi

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    If you know the function was postive for some x < 2 then you could say there has been a crossing. But was f(x) ever positive for x < 2 ?

    You have the basic idea. But to determine a crossing you must consider what happens between two places where f(x) "blows up", not just "at" one place.

    Considering x = 1 + h and x = 2-h is what tells you that the graph crosses the x-axis between x = 1 and x = 2.
     
  10. Dec 27, 2014 #9

    Ray Vickson

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    I think you are mis-understanding what is meant by "crossing". When the graph of y = f(x) jumps from y = -∞ to y = +∞ as x passes from left to right through x = 1 (or x = 2, 3, or 4) we do not regard that as a "crossing", because there is no value of x in the interval (1-h,1+h) that gives f(x) = 0 exactly. Of course, if you draw the graph of y = f (x) and put in a vertical line at x = 1 (or x = 2, etc.) that vertical line does pass through the x-axis; but the line is not actually part of the graph.

    Anyway, x-axis crossings are not the issue in this problem: y = 2 crossings are what matter.
     
  11. Dec 27, 2014 #10

    LCKurtz

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    @Ray Vickson

    Hey Ray, have you ever seen this with Maple? I tried to plot the given function and the line y=2 and got this output:

    10 + BB BB BB B B
    + BB BB BB B B
    8 + BB BB BB B B
    + BB BB B B B B
    6 + BB BB B B B B
    + BB B B B B B BB
    4 + BB B B B BB B BB
    + B B B BB B B B BB
    AAAAAAAAA2A*AAAAAAAAA*A*AAAAAAA*AA*AAAAAA*AA**AAAAA*AAAAAA******AAAAAAAAA
    + B BB B BB B BB B BBBBBBBBB
    -+-+-+-+--+-+-+-+-+-+-*-+**-+-+-*-+-**+-+-*-+-+**-+-*-+-+-+-+-+-+-+--+-+-+-
    + B BB B B B BB B
    -1BBBBBBBB + 1 BB 2 BB 3 BB 4 5 6
    -2B*BBB B B B B B B B
    + BB B B B B B B B
    -4 + BB B B B B B BB
    + B B B B B B BB
    -6 + BB B B B BB BB
    + B B B B BB BB
    -8 + B B BB BB BB
    + B B BB BB BB
    -10 + B B BB BB BB


    It's obviously some sort of ASCII output in a very rough shape of the graph. I'm using Maple 13.02 but I don't know how it got in this mode or how to get it out of it.
     
    Last edited: Dec 27, 2014
  12. Dec 27, 2014 #11
    Sorry Ray as I am not able to get it. Can you please elaborate.
    It is seeming a hard puzzle to me what you said Stephen. Isn't it true that for x= 1 + h that is x < 2 f(x) is positive?
     
  13. Dec 27, 2014 #12

    Stephen Tashi

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    I didn't mean to imply that f(x) was never positive for x < 2. I meant that you must check that x was positive for some x < 2 before you conclude that there was a place where x = 0 between that place and x = 2 - h.

    An example of what Ray is saying is that x = 2 is not a place where the graph crosses the x-axis, even though f(2-h) is negative and f(2+h) is positive. The value at x = 2 is a "singularity". It is a place where the graph does not exist.
     
  14. Dec 27, 2014 #13
    So, how now to go for it?
     
  15. Dec 27, 2014 #14
  16. Dec 27, 2014 #15

    Stephen Tashi

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    For example, think about what happens between x = 1+h and x = 2 -h instead of thinking of something between x = 2-h and x = 2 + h.

    After that think about what happens "at the ends". What happens between when x is very small and x = 1-h? What happens between x = 4+h and when x becomes very large.
     
  17. Dec 27, 2014 #16
    Please can you refer my above post?
    For my better understanding with diagram.
     
  18. Dec 27, 2014 #17

    Ray Vickson

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    No, I have never seen it, and when I give the following instructions (in Maple 11 on an HP Probook) everything works just fine---it even puts in the vertical line:
    f:=1/(x-1):
    pflot(f,x=-1..3,y=-10..10);

    To plot it without the vertical line I just said
    plot(f,x=-1..3, y=-10..10,discont=true);

    Similarly,
    g = add(1/(x-i),i=1..4): plot(g, x=-1..6, y=-10..10);
    produces a graph with 4 vertical lines at x = 1,2,3,4.
     
  19. Dec 27, 2014 #18

    Stephen Tashi

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    The plot in your link isn't an informative plot. It doesn't clearly what happens between x = 0 and x = 5 and that interval is the important part of the graph. (I don't use Wolfram, so I can't advise you how to get a clearer plot. Can you plot only the interval between x = 0 and x = 5? Perhaps it will show up better that way.)
     
  20. Dec 27, 2014 #19

    Ray Vickson

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    If you want to understand what is happening, forget about trying to use Wolfram Alpha, and just make simple sketches done manually. (Yes, I am serious!) For ##f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)##, you can understand what happens near ##x = 1## just by looking at the simple function ##1/(x-1)##. Slightly to the left of ##x = 1## this function is large negative and slightly to the right it is large positive. Basically, it jumps from ##y = -\infty## to ##y = +\infty## as ##x## passes through 1, from left to right. The other terms ##1/(x-2) + 1/(x-3) + 1/(x-4)## will just change things a bit near ##x = 1## but will not alter in any way the jump behavior at ##x = 1##. (You should thing about why that is true!)

    Similarly, you can understand what happens near ##x = 2, 3, 4## just by looking at the simpler functions ##1/(x-2), \; 1/(x-3), \; 1/(x-4)##.

    So, just to the right of ##x = 1## the function is large positive, and just to the left of ##x = 2## it is large negative, which means that between ##x=1## and ##x=2## there is a root of your equation ##f(x) = 2##. Furthermore, you should think about why there is just one, single root in the interval ##(1,2)##.

    Keep going like that for other ##x##-intervals.
     
  21. Dec 27, 2014 #20

    SammyS

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    Try this in Wolfram Alpha.

    http://m.wolframalpha.com/input/?i=...1/(x-3)+++1/(x-4)-2,+from+x=0+to+x=7&x=4&y=10

    That takes x from 0 to 7.

    If you simply plot ##\
    f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)\
    ## without subtracting 2, you may get some idea regarding the methods Steven and Ray are suggesting.
     
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