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Using interpolants to solve a polynomial.

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riddle
#1
Sep14-10, 02:37 PM
P: 38
1. The problem statement, all variables and given/known data

Show that a root of the equation x3 - 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 = (x1 - 2) [tex]\div[/tex] (3-x1)
to end up with x1 = 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 = (3-x1) [tex]\div[/tex] (x1 - 2)

I'm utterly confused. Please help.
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LCKurtz
#2
Sep14-10, 05:18 PM
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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 x1 looks OK, and it is between 2 and 3. Did you check whether you got lucky and x1 is your root? If it isn't, then how do you know whether the root is in [2, x1] or in [x1, 3]? You have to know that in order to know which interval to do the next step with.
riddle
#3
Sep14-10, 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.

LCKurtz
#4
Sep14-10, 09:05 PM
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Using interpolants to solve a polynomial.

Quote Quote by riddle View Post
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.
Your first approximation for x1 of 35/16=2.1875 is correct. The value of the function at that point is negative so you want to use x = 2.1875 and x = 3 for your calculation of x2. I get 2.250619230 for it.
riddle
#5
Sep15-10, 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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