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What can I use to solve this?

  1. Jan 10, 2005 #1
    I've run into a calculation hurdle that I cannot cross in a problem I'm trying to solve.

    May I know what values of z satisfy the following equation:

    [tex]z^4 + 2z^3 - 5z + 1 = 0[/tex]

    I've never encountered this in mathematics class before.
     
  2. jcsd
  3. Jan 10, 2005 #2

    dextercioby

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    Would u please post the problem??Maybe you wouldn't get a quartic after all...

    Daniel.
     
  4. Jan 10, 2005 #3
    Neither do I, I can solve those that miss ax^3, never before for ax^2
     
  5. Jan 10, 2005 #4
    A ladder 3m long rests against a wall with one end a short distance from the base of the wall. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m wide. What is the maximum distance up the wall that the ladder can reach?
     
  6. Jan 10, 2005 #5

    dextercioby

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    The maximum height on the wall is attained when the ladder touches the storage box.
    The equations i have found read
    [tex] (a+1)^{2}+(b+1)^{2} =9 [/tex](1)
    [tex] ab=1 [/tex] (2)

    Then u find the eq.for "a"
    [tex] a^{2}+\frac{1}{a^{2}}+2(a+\frac{1}{a})=7 [/tex] (3)
    ,which can be put in the form
    [tex] u^{2}+2u-9 =0 [/tex] (3)

    ,where
    [tex] u=:a+\frac{1}{a} [/tex] (4)

    Solve for "u" and then for "a".Then add 1 to the "a" and find the final result.

    Daniel.
     
  7. Jan 10, 2005 #6
    Yes, that's how I solved it. I never thought of substituting [tex]a+\frac{1}{a}[/tex] for u. Very smart. Thanks a lot.
     
  8. Jan 10, 2005 #7

    Hurkyl

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    Use the symmetry of the problem . With dexter's notation, notice that whenever s is a solution, so is 1/s. So, we can take his quartic:

    z^4 + 2z^3 - 7z^2 + 2z + 1 = 0

    And write down its four factors:

    (z - s) (z - 1/s) (z - t) (z - 1/t) = 0

    If we group them into their symmetric pairs and multiply:

    (z^2 + az + 1) (z^2 + bz + 1)

    So one quadratic corresponds to the two solutions s and 1/s and the other to t and 1/t. In particular, notice, now, that our unknowns only have two possible values, instead of four, which means we should be able to determine them from a quadratic equation:

    Since the polynomials should be equal, we should expand and equate them:

    z^4 + (a+b)z^3 + (2+ab)z^2 + (a+b)z + 1
    =
    z^4 + 2z^3 - 7z^2 + 2z + 1 = 0

    Equating coefficients gives us a system of equations to solve.

    Once we have a and b, we can then solve for s and t and get the four possible solutions. (Two should be complex, I imagine)


    (Incidentally, I don't see how you got your equation, Recon)
     
  9. Jan 10, 2005 #8

    dextercioby

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    I multiple checked it.Can u prove your statement???I frankly doubt it... :wink:


    Daniel.
     
  10. Jan 10, 2005 #9

    Hurkyl

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    Another (similar) approach:

    We can arrange the factors into these pairs:

    (z - s)(z - t) and (z - 1/s)(z - 1/t)

    So each quadratic contains exactly one solution from each of the symmetric solutions (instead of both).

    Again, we break the quartic into quadratic factors:

    (z^2 + az + b) (z^2 + cz + d)

    But, in this case, we know that we can convert from one to the other by applying the symmetry. In particular,

    if (z^2 + az + b) = 0 then (1/z)^2 + c(1/z) + d = 0
    Rewriting the second gives:

    z^2 + (c/d)z + (1/d) = 0
    which must be the same polynomial as
    z^2 + az + b = 0

    Since they have the same solutions.

    So we can equate coefficients here, then multiply them to equate coefficients to the quartic polynomial. Again, we can solve for our unknowns, and then we know how to solve quadratics to get the answers to our problem.




    Lesson learned: use symmetry to break the problem into parts. We've seen two different ways to do it:

    (1) Each part corresponds to one of the different "orbits" under the symmetry. (e.g. {s, 1/s})

    (2) Each part consists of exactly one element from each "orbit". (e.g. {s, t})


    Then, we can solve the problem in two steps:

    (i) Determine the two subproblems.
    (ii) For each subproblem, get the corresponding answers.
     
  11. Jan 10, 2005 #10

    Hurkyl

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    Dexter gave a solution using a third approach. What precisely did he do, and why did it work? Is it really a different approach?


    The variable he defined, u, has the property that it is invariant under the symmetry. Notice that if you replace z with 1/z, u remains unchanged.

    Because of this invariance under symmetry, there are fewer solutions for u than there are for z, making them easier to find.


    Actually, in some sense, both of my approaches make use of invariants under symmetry. For instance, (z - s)(z - 1/s) is invariant under swapping s and 1/s. My second approach found two quadratics that are each invariant under swapping s and t.



    P.S. for a polynomial which is the same as its reverse, making dexter's solution is the standard way of getting a lower degree polynomial.


    (The reverse of a polynomial is the one with the coefficients in the opposite order. Equivalently, the multiplicative inverse of the roots of the polynomial are precisely the roots of the reverse. To make sense, of course, this assumes that 0 is not a root)
     
    Last edited: Jan 10, 2005
  12. Jan 10, 2005 #11

    Hurkyl

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    And just for fun, here's a trig approach:

    Let theta be the angle the ladder makes with the base. A is the distance from the box to the base, B is the distance from the box tot he top of the ladder.

    So, the length of the ladder is 1/sin theta + 1/cos theta = 3.

    With trig identities, and a change of variable, I can get a quadratic in sin phi.
     
  13. Jan 10, 2005 #12
    I was a bit sleepy at that time Dex. I did checked it just now. You are right. Sorry Dex!!

    ________________________________________________________________________

    "The beautiful mind goes faster than the hand"
     
  14. Jan 10, 2005 #13
    I got it the same way as dextercioby. I just made a mistake multiplying the right side of the equation with z^3 instead of z^2. I multiplied everything on the left side with z^2.

    EDIT: This is the equation I'm talking about.

    [tex]z^2 + 2z + \frac{2}{z} + \frac{1}{z^2} = 7[/tex]
     
    Last edited: Jan 10, 2005
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