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Homework Help: Cardano formula

  1. Dec 19, 2013 #1

    bobie

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    1. The problem statement, all variables and given/known data
    Can we apply the formula to the equation

    x^3 -px +q = 0 ?

    If not what is the best way to find x ?

    Thanks for your help
     
  2. jcsd
  3. Dec 19, 2013 #2

    SteamKing

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    I don't see why not.
     
  4. Dec 19, 2013 #3

    bobie

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    I've been trying for days to solve this to no avail, can you show me where I go wrong, please?

    .9³ - 400*.9 = 359.271 (359../2 = 179.6355)

    ³√ (√(179..²+ 400³/27) - 179..) = 11.1075117

    ³√ (-√(179..²+ 400³/27) - 179..) = -12.0388862 + 11.1075117 = .896

    I have tried to change plus to minus around to no avail
    Thanks, Steam king
     
  5. Dec 19, 2013 #4

    SteamKing

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    You mistake me for a clairvoyant. Please post the original equation you are trying to solve. Your previous post makes no sense to me.
     
  6. Dec 19, 2013 #5

    bobie

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    x³ -400 x + 359.271 = 0
    x= 0.9
    applying cardano (q/2 = 179.6355)
    ± √(q/2² + p³/27 ) -q/2
    I get x= 0.896....
     
    Last edited: Dec 19, 2013
  7. Dec 19, 2013 #6

    NascentOxygen

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    It looks like one q/2 needs a sign change, the 400 should be -400, and the resulting square root of a negative leads to complex numbers. (You'll then take into account that each complex number has 3 cube roots.)

    I expect that eventually you'll have a handful of candidates, one of which will be the 0.9 you are looking for. :smile:

    Good luck!
     
    Last edited: Dec 19, 2013
  8. Dec 19, 2013 #7

    HallsofIvy

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    If you mean that Cardano's formula says that the root is "± √(q/2² + p³/27 ) -q/2", you are wrong. That isn't Cardano's formula! You have p and q switched and have forgotten a cube root.

    If a and b are any two real numbers then
    [itex](a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3[/itex] and
    [itex]-3ab(a+ b)= -3a^3b- 3ab^2[/itex]

    so [itex](a+ b)^3- 3ab(a+ b)= a^3+ b^3[/itex]

    If we let x= a+ b, m= 3ab, and [itex]n= a^3+ b^3[/itex] then [itex]x^3- mx= n[/itex].

    Now, suppose we know m and n. Can we find a and b and so x?

    Yes. From m= 3ab, b= m/3a so [itex]a^3+ b^3= a^3+ m^3/3^3a^3= n[/itex]. Multiply through by [itex]a^3[/itex] to get
    [itex](a^3)^2- (m/3)^3= n(a^3)[itex] or [itex](a^3)^2- n(a^3)- (m/3)^3= 0[/itex], a quadratic equation in [itex]a^3[/itex].

    The quadratic formula gives
    [tex]a^3= \frac{n\pm\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]

    Since [itex]a^3+ b^3= n[/itex], [itex]b^3= n- a^3[/itex] so
    [tex]b^3= \frac{n\mp\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]

    Take the cube roots to find a and b, then x= a+ b.
     
  9. Dec 19, 2013 #8

    bobie

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    that is what I did in post 3:

    ³√ (-√(179.271²+ 400³/27) - 179.271) = -12.00294772 +
    ³√ (√(179.271²+ 400³/27) - 179.271) = +11.10838241 = 0.89..
     
    Last edited: Dec 19, 2013
  10. Dec 19, 2013 #9

    bobie

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    I have tried all sign changes including
    -x^3 + 400 x - 359.271 =0
    but no 0.9 as a result
    There are all real roots, wolfram gives no imaginary root
    http://www.wolframalpha.com/input/?i=x^3-400x+359.271=0
     
    Last edited: Dec 19, 2013
  11. Dec 19, 2013 #10

    NascentOxygen

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  12. Dec 19, 2013 #11

    NascentOxygen

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    And in post #6 I pointed out your mistake, that 400 should be -400.

    Wolframalpha is not telling you the whole story. You need to break the evaluation of t1/3 + u1/3 into 2 parts, carefully writing down the 3 roots for t1/3 and then for u1/3. Then perform the final addition/s manually. Out of the 6 answers, you will see one is 0.9
     
    Last edited: Dec 19, 2013
  13. Dec 19, 2013 #12

    Ray Vickson

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    Note that ##(0.9)^3 - 400 \, (0.9) = -539.271## (just by striaght calculation), so x = 0.9 is a solution of the equation ##x^3 - 400 x + 539.271 = 0.##
     
  14. Dec 19, 2013 #13
    I'm noticing that in all of bobie's posts, he is working with the number 359.271... That might be where his error is.

    Edit: Nevermind, the 5 and 3 were mixed up in your post. Sorry.
     
  15. Dec 19, 2013 #14

    Ray Vickson

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    Yes, you are right: it should have been 359.271.
     
  16. Dec 20, 2013 #15

    bobie

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    I tried all sorts of equation with -p
    x³ - 33x - 18 =0
    18/2 = 9, 33/3 = 11
    the solutions are:
    http://m.wolframalpha.com/input/?i=solve+z^3+-+33z+-18+=0&x=8&y=7
    [tex]x = 6 ; -5.4495 ; 0.55 05[/tex]
    [tex]a^3 = \frac{18}{2}\pm\sqrt{\left(\frac{18}{2}\right)^2+ \left(\frac{33}{3}\right)^3} → 9 + \sqrt{81+1331} = 46.5766 [/tex][tex]a = \sqrt[3]{46.5766} = 3.597956362 [/tex]
    [tex]b^3 = 9 - \sqrt{81+ 1331} = - 28.5766[/tex][tex]b= \sqrt[3]{-28.5766} = -3.05729111[/tex][tex]x = a + b = 3.5979 -3.05729 = 0. 54 06[/tex]
    changing 9 to -9 gives the same result with inverted sign - 0.54 06
    where do I go wrong, please?
    or the formula cannot be applied when p < 0 ?
     
    Last edited: Dec 20, 2013
  17. Dec 20, 2013 #16

    ehild

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    It is easier to apply some iterative method.

    Try xk+1=(xk3+359.271)/400

    ehild
     
  18. Dec 20, 2013 #17

    bobie

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    probably to get a negative p we should start from
    [itex](a- b)^3= a^3- 3a^2b- 3ab^2+ b^3[/itex]
    and two such numbers do not exist
     
  19. Dec 20, 2013 #18

    NascentOxygen

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    You are making exactly the same mistake as I pointed out yesterday. :cry: The Cardano formula should contain (-33/3)^3 but for some reason you keep messing up the sign of the term to be cubed.

    Maybe you have written down the formula incorrectly? A readable treatment can be found here: http://www.math.cornell.edu/~dwh/courses/M122-S00/supplements/cardano.html

    I can confirm that when you apply the formula correctly, the Cardano method gives the root you are hoping for, viz., 6
    Here's a good start: http://m.wolframalpha.com/input/?i=solve+a%3D-33%3B+b%3D-18%3B+t%3D-b%2F2+%2B+sqrt%28%28b%2F2%29%5E2+%2B+%28a%2F3%29%5E3%29%3B+w%3Dt%5E1%2F3&x=0&y=0
     
    Last edited: Dec 20, 2013
  20. Dec 21, 2013 #19

    bobie

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    I found b, is that correct?
    [tex]a^3 = \frac{18}{2}\pm\sqrt{\left(\frac{18}{2}\right)^2- \left(\frac{33}{3}\right)^3} → 9 + \sqrt{81-1331} = 46.5766 [/tex][tex]a = 3+\sqrt{2} i [/tex]
    [tex]b^3 = 9 - \sqrt{81- 1331} = 3- \sqrt{2} i [/tex]
    I am not familiar with complex numbers but I suppose that the opposite √2 cancel out
    [tex] a + b = 3+3 + (+ √2i+ - √2i) =0 = 6 [/tex]
    Is that correct?
    but I found b through wolfram, how can I find it by myself? is it too difficult?

    Thank you all for your kind help
     
  21. Dec 21, 2013 #20

    SteamKing

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    The roots of polynomials with integer coefficients can be complex. In such cases where there are complex roots, the roots will occur in pairs and be conjugates of one another, i.e. a + bi and a - bi, where a, b are real numbers.
     
  22. Dec 22, 2013 #21

    NascentOxygen

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    Looks better. (I can't see how "= 46.5766" comes into the picture, though.)

    If you aren't up with complex numbers, then you'll have to be like those in Cardano's day and restrict your use of the method to those cases where the value under the square root evaluates as non-negative.

    Wolframalpha makes maths more interesting, it gives you an enticing glimpse of what lies over the next crest. :smile: I don't think you could determine the cube root of a complex number easily manually.
    Enjoy your study! http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  23. Dec 22, 2013 #22

    bobie

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    Thanks for your kind help,
    I have noticed that it is not difficult, the square root of a negative √-n is just the root of the positive
    plus i, if √n =a, [itex]\sqrt[3]{n}[/itex]= b , then √-n = ai, and [itex]\sqrt[3]{-n}[/itex] just
    b/2 [itex]\pm[/itex] b/2 * i√3.
    Isn't there a hard-and-fast rule to get the cube root of m + √-n?
     
    Last edited by a moderator: May 6, 2017
  24. Dec 22, 2013 #23

    bobie

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    Thanks for the advice.
    Could you show me how to do that on a pocket calculator or tell me how many operations it requires? it must be surely slower, but how much slower
     
  25. Dec 23, 2013 #24

    ehild

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    Start with an easy trial xo value, say, xo=0
    Type in the trial value, take the cube.
    Add 359.271 (and store 359.271) . Divide the result by 400.
    Getting the result (x1) take the cube, add the retrieved value (359.271) divide the result by 400.... and so on.

    xo=0

    x1=359.271/4=0.898. Substituting back:
    x2=0.89999
    substituting back again
    x3=0.8999999, x4=0.9. It took a few seconds with my stone-age pocket calculator.

    If you start with xo=1
    x1=0.9006775
    x2=0.9000041
    x3=0.9

    If you have one root, divide the original equation by (x-0.9). That results a quadratic equation.

    ehild
     
    Last edited: Dec 23, 2013
  26. Dec 23, 2013 #25

    bobie

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    Thanks, ehild, that was vere quick!
    Merry Christmas to everybody!
    :smile:http://imageshack.us/scaled/landing/109/holly1756.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
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