Using interpolants to solve a polynomial.by riddle Tags: interpolation, iteration, numerical analysis, polynomial 

#1
Sep1410, 02:37 PM

P: 38

1. The problem statement, all variables and given/known data
Show that a root of the equation x^{3}  3x  5 =0 lies in the interval [2,3], and then find the root using linear interpolation correct to one decimal place. 2. Relevant equations n/a 3. The attempt at a solution This is my first ever time using interpolants ( well at least in the sense of solving a polynomial. I think I've used them before... but instinctively.) This is what I did: 3 [tex]\div[/tex] 13 = (x_{1}  2) [tex]\div[/tex] (3x_{1}) to end up with x_{1} = 2.1875. I went on to get the third approximation only to find out that it was outside the range (less than 2). I checked what the solution said, and it said: 3 [tex]\div[/tex] 13 = (3x_{1}) [tex]\div[/tex] (x_{1}  2) I'm utterly confused. Please help. 



#2
Sep1410, 05:18 PM

HW Helper
Thanks
PF Gold
P: 7,175

The first thing your problem asked was to show that there was a root of your function between x = 2 and x = 3. Did you do that? And how did you do it?
Your x_{1} looks OK, and it is between 2 and 3. Did you check whether you got lucky and x_{1} is your root? If it isn't, then how do you know whether the root is in [2, x_{1}] or in [x_{1}, 3]? You have to know that in order to know which interval to do the next step with. 



#3
Sep1410, 06:03 PM

P: 38

Oh. I forgot to mention that I'd got the first part. Yes. The root is in between 2 and 3.
I just substituted "x" with 3 and 2 and saw that f(3) / f(2) = s. i.e., the sign changed, so there had to have had been a number which had f(x) = 0, i.e., the root of the equation. My first approximation is different from what the book says, the same with all my further approximations. I thought that it might have been a typo, but then I couldn't do the other questions either, and in the end I checked to see if the answers that the book gave worked, and they did. So theres got to be something wrong in what I'm doing. 



#4
Sep1410, 09:05 PM

HW Helper
Thanks
PF Gold
P: 7,175

Using interpolants to solve a polynomial. 



#5
Sep1510, 05:59 PM

P: 38

Ok. I'm getting the same thing.
I understand how this works now. But I'm still having trouble believing that the publishers messed up so bad. Can you just have a look at what they did. I'm really tired right now, and I wont be able to use the book for a week or two so I wont be able to look at it later. 


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