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Math Help - hmmm... just wondering

  1. #1
    dan
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    hmmm... just wondering

    is it posible to solve a non-liner system...and how!!!
    for example
     x^2 +y^3 =10
     x^3 -y^4 =-6

    if any one has any thoughts on how you do this please let me know

    ~dan
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    Quote Originally Posted by dan
    is it posible to solve a non-liner system...and how!!!
    for example
     x^2 +y^3 =10
     x^3 -y^4 =-6

    if any one has any thoughts on how you do this please let me know

    ~dan
    Of course,
    x^3=y^4-6
    x=\sqrt[3]{y^4-6}
    Substitute,
    (y^4-6)^{2/3}+y^3=10
    Thus,
    (y^4-6)^{2/3}=(10+y^3)
    Cube both sides,
    (y^2-6)^2=(10+y^3)^3
    Expand and solve for the nasty nontic.

    Note, the solutions not necessarily can be expressed in algebraic terms.
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  3. #3
    dan
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    Quote Originally Posted by ThePerfectHacker

    (y^2-6)^2=(10+y^3)^3
    Expand and solve for the nasty nontic.

    i hate to ask...but how do you slolve for y in that equation.
    it seems i'v spent hours working on probelms of that type before....
    would i use the "binomal therem"?
    dan
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    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Of course,
    x^3=y^4-6
    x=\sqrt[3]{y^4-6}
    Substitute,
    (y^4-6)^{2/3}+y^3=10
    Thus,
    (y^4-6)^{2/3}=(10+y^3)
    Cube both sides,
    (y^2-6)^2=(10+y^3)^3
    Expand and solve for the nasty nontic.

    Note, the solutions not necessarily can be expressed in algebraic terms.
    using:

    x=\pm \sqrt{10-y^3}


    and substituting into

    <br />
x^3 -y^4 =-6<br />

    will give two octics - a much easier problem than a single nonic. ?

    RonL
    Last edited by CaptainBlack; September 5th 2006 at 01:35 PM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by dan
    i hate to ask...but how do you slolve for y in that equation.
    it seems i'v spent hours working on probelms of that type before....
    would i use the "binomal therem"?
    dan
    If you are lucky it will have a special form that will allow you to find
    all or some or the roots. Otherwise numerically.

    RonL
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    Grand Panjandrum
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    Quote Originally Posted by dan
    is it posible to solve a non-liner system...and how!!!
    for example
     x^2 +y^3 =10
     x^3 -y^4 =-6

    if any one has any thoughts on how you do this please let me know

    ~dan
    If you are happy with approximate solutions one could start garphically.

    These two equations represent curves in the x-y plane. Sketch them
    and your first approximation to the soution/s will be the point/s of
    intersection.

    Then they can be refined with an appropriate numerical procedure (say
    linearising about the approximate roots, or something like that).

    RonL
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  7. #7
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    Quote Originally Posted by dan
    i hate to ask...but how do you slolve for y in that equation.
    it seems i'v spent hours working on probelms of that type before....
    would i use the "binomal therem"?
    dan
    Not necessarily there is a general method.
    Yes, you do expand with binomial theorem then work with polynomial equation.
    (Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!")
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  8. #8
    dan
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    Quote Originally Posted by ThePerfectHacker
    (Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!")
    hahaha
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  9. #9
    dan
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    so if i had a polynomal. could i solve with newton??

    i dont so how you get a nontic though  (10 +y^3)^3 just because 3*3 is nine???

    some how i got from (y^2 -6)^2 =(10 +y^3)^3

    to  0= y^5 + 59y^4 +90y^3 + 12y^2 + 964

    is that right???
    dan
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  10. #10
    dan
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    Cool

    Quote Originally Posted by CaptainBlack
    using:

    x=\pm \sqrt{10-y^3}


    and substituting into

    <br />
x^3 -y^4 =-6<br />

    will give two octics - a much easier problem than an nonic.

    RonL

    mabye i'll try that
    Last edited by CaptainBlack; September 7th 2006 at 07:33 AM.
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  11. #11
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    Quote Originally Posted by dan

    some how i got from (y^2 -6)^2 =(10 +y^3)^3
    y^4-12y^2+36=1000+300y^3+30y^6+y^9
    Thus,
    y^9+30y^6-y^4+300y^3+12y^2+9964=0
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  12. #12
    dan
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    Quote Originally Posted by ThePerfectHacker
    y^4-12y^2+36=1000+300y^3+30y^6+y^9
    Thus,
    y^9+30y^6-y^4+300y^3+12y^2+9964=0
    yea...thanks i did something wrong...what else is new...
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  13. #13
    dan
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    so if
     f(y)=y^9 + 30y^6 - y^4 +300y^3 + 12y^2 +9964
    then
    f'(y)=y(9y^7 +180y^4 -4y^2 +900y +24)

    so could i use newton for here to approximate [tex] y [tex]
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  14. #14
    dan
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    by my calc  y \approx -3.13425651413923

    look right??
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  15. #15
    Grand Panjandrum
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    Quote Originally Posted by dan
    by my calc  y \approx -3.13425651413923

    look right??
    What do you get if you substitute this back into the original equations?
    Is there a x which with this as a value of y solves the equations?

    RonL

    (when I do this I get no consistent solution for x, but I could be doing the
    arithmetic wrongly )
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