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

**Vendetta** · Sep 29th 2013, 10:38 PM

roots of an equation ... any help would be appreciated thanks in advance

**Prove It** · Sep 29th 2013, 11:01 PM

What have you tried so far?

**zzephod** · Sep 30th 2013, 05:32 AM

Originally Posted by Prove It

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)

.

**Vendetta** · Sep 30th 2013, 05:59 AM

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??

**zzephod** · Sep 30th 2013, 06:21 AM

Originally Posted by Vendetta

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 $x=-1+a \varepsilon +O(\varepsilon^2)$ ...

**HallsofIvy** · Sep 30th 2013, 06:28 AM

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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