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

Printable View

- Sep 29th 2013, 10:38 PMVendettaI need to submit this in 30/9/2013 before 12am
roots of an equation ... any help would be appreciated thanks in advance :)

- Sep 29th 2013, 11:01 PMProve ItRe: I need to submit this in 30/9/2013 before 12am
What have you tried so far?

- Sep 30th 2013, 05:32 AMzzephodRe: I need to submit this in 30/9/2013 before 12am
- Sep 30th 2013, 05:59 AMVendettaRe: 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??

- Sep 30th 2013, 06:21 AMzzephodRe: I need to submit this in 30/9/2013 before 12am
You have double posted this question, in the other thread I point out that when $\displaystyle \varepsilon=0$ the real root is $\displaystyle -1$ so when $\displaystyle \varepsilon \ne 0$ you expand the root as a power series $\displaystyle x=-1+a \varepsilon +O(\varepsilon^2)$ ...

- Sep 30th 2013, 06:28 AMHallsofIvyRe: 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 $\displaystyle x= x_0+ \epsilon x_1+ \epsilon^2x_2+ \cdot\cdot\cdot$. Then $\displaystyle x^3= x_0^3+\epsilon(3x_0^2x_1)+ \epsilon^2(3x_0x_1^2+ 3x_0^2x_2)+ \epsilon^3(x_1^3+ 2x_0^2x_3+ 6x_0x_1x_2)$ and $\displaystyle 3\epsilon x=\epsilon x_0+ \epsilon^2x_1+ \epsilon^3x_2$ to the third power in $\displaystyle \epsilon$. The "zeroth" power of $\displaystyle \epsilon$ in the equation would be $\displaystyle x_0^3+ 1= 0$ which has $\displaystyle x_0= -1$ as it only real root. The "first" power of [tex]\epsilon[/b] would be $\displaystyle 3x_0^2x_1+ x_0= 0$ or, since $\displaystyle x_1= -1$, $\displaystyle 3x_1- 1= 0$ so that $\displaystyle x_1= 1/3$. Do the same to find $\displaystyle x_2$ and $\displaystyle x_3$.

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!