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Math Help - I need to submit this in 30/9/2013 before 12am

  1. #1
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    I need to submit this in 30/9/2013 before 12am

    roots of an equation ... any help would be appreciated thanks in advance
    Attached Thumbnails Attached Thumbnails I need to submit this in 30/9/2013 before 12am-advanced-algebra.png  
    Last edited by Vendetta; September 29th 2013 at 10:41 PM.
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    Re: I need to submit this in 30/9/2013 before 12am

    What have you tried so far?
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    Re: I need to submit this in 30/9/2013 before 12am

    Quote Originally Posted by Prove It View Post
    What have you tried so far?
    Posting his entire assignment here 5.5hrs before the deadline by the look of it (and not giving a time zone for the deadline! looks like a 11:00 UT deadline to me)

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    Re: I need to submit this in 30/9/2013 before 12am

    i know how to solve the question if he didn't write that ~=0 the root would be -0.322 i took 2 points -0.5 and zero and i did the bisection method one is positive and one is negative so c=( a+b)/2 and i keep iterating ... but what does he mean ~= 0 is that suppose to mean the equation is now x^3+1??
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    Re: I need to submit this in 30/9/2013 before 12am

    Quote Originally Posted by Vendetta View Post
    i know how to solve the question if he didn't write that ~=0 the root would be -0.322 i took 2 points -0.5 and zero and i did the bisection method one is positive and one is negative so c=( a+b)/2 and i keep iterating ... but what does he mean ~= 0 is that suppose to mean the equation is now x^3+1??
    You have double posted this question, in the other thread I point out that when \varepsilon=0 the real root is -1 so when \varepsilon \ne 0 you expand the root as a power series x=-1+a \varepsilon +O(\varepsilon^2) ...
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    Re: I need to submit this in 30/9/2013 before 12am

    If you are given a problem like this, surely you have learned some approximation methods. This looks like a "perturbation method" would work. Let x= x_0+ \epsilon x_1+ \epsilon^2x_2+ \cdot\cdot\cdot. Then x^3= x_0^3+\epsilon(3x_0^2x_1)+ \epsilon^2(3x_0x_1^2+ 3x_0^2x_2)+ \epsilon^3(x_1^3+ 2x_0^2x_3+ 6x_0x_1x_2) and 3\epsilon x=\epsilon x_0+ \epsilon^2x_1+ \epsilon^3x_2 to the third power in \epsilon. The "zeroth" power of \epsilon in the equation would be x_0^3+ 1= 0 which has x_0= -1 as it only real root. The "first" power of [tex]\epsilon[/b] would be 3x_0^2x_1+ x_0= 0 or, since x_1= -1, 3x_1- 1= 0 so that x_1= 1/3. Do the same to find x_2 and x_3.

    Have you not seen anything like that? Are you sure you have been attending the right class? You seem to be saying that you have a whole series of problems on topics you have never seen before!
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