The problem involves finding a polynomial function of degree n-1 that equals 1 at a specific point xi and equals 0 at other distinct points xj, where j is not equal to i. The context is rooted in polynomial functions and their properties, particularly focusing on interpolation and the behavior of polynomials at given points.
The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided insights into constructing polynomials for specific cases, while others seek clarification on the generalization to arbitrary n. There is a recognition of the need to manipulate polynomial forms to achieve the desired properties.
Participants express confusion regarding the definitions and implications of polynomial degrees, particularly in relation to the problem's requirements. There is a focus on understanding how to derive a polynomial of degree n-1 from a higher degree polynomial and the conditions under which certain polynomials are zero at specified points.
slider142 said:It gives you the hint that if you want a polynomial that is 0 at A, B, and C, the polynomial (x - A)(x - B)(x - C) does nicely. It wants you to find a polynomial that is 0 at a bunch of points (xj's) except for the particular point xi, where you have to come up with something that will make it 1 there.
emyt said:thanks for the reply, what does the polynomial of n-1 degree imply?
thanks
slider142 said:It doesn't imply anything really. It is giving you a limit of n-1 binomial factors, or less if you increase the degree of one or more factors.
emyt said:this is one of the main things I didn't understand about the question. so would a polynomial of the nth degree be the number of binomial terms present?
I have an answer in mind but I don't know how to get a polynomial of n-1th degree
thanks a lot for the responses
slider142 said:When you multiply out the expression, what is the monomial with the highest power? That power is the degree of the polynomial.
Obviously, if you simply use the nth degree polynomial (x - x1)...(x - xn), it will be 0 for each xj including the distinct xi. So what we want to start with is (x - x1)...(x - xn) where x - xi is not included. Can we do something to this (n-1)th degree polynomial to get 1 when x = xi?
emyt said:ohh, I see.. taking out a factor makes it to the n-1th degree, makes sense
(x-x0) / (x1-x0) would be 1 when x = x1 I and 0 when x = x0, i just didn't know how that would be a polynomial of the n-1th degree
slider142 said:This is an excellent solution for the case n=2, where you provide an (n-1)=1st degree polynomial that is 1 for x = x1 and 0 when x = x0. Can you provide one for n=3, and then generalize to all n?
emyt said:hang on, I'm still a bit confused.. how is that a case of n=2?
sorry.. I'm a bit slow
thanks
slider142 said:Replace n in your original question with 2:
"If x1, x2 are distinct numbers, find a polynomial function fi of degree 2-1=1 which is 1 at xi and 0 at xj for j =/ i (not equal)."
emyt said:okay, I see that (x-xj)/(xi-xj) is just a degree 1 polynomial. how come we consider the polynomial (x-xj)...(xjn)? where does it say that we should consider that polynomial so (x-xj)/(xi-xj) would be degree n-1?
slider142 said:I'm not sure what you're saying. What is xjn? The polynomial you name is of degree n - 1 = 2 - 1 = 1, so it fits the solution to the problem if they only wanted to see the case n = 2.
emyt said:okay, so the polynomial (x - x1)... (x-xn) will be zero when x = xj. But how come this is our nth degree polynomial? Where does it tell me to use (x - x1)... (x-xn) as the nth degree polynomial so that (x - xj) / (xi- xj) is n-1th degree?