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Homework Help: Spivak's Calculus polynomial question

  1. Jul 24, 2008 #1
    1. The problem statement, all variables and given/known data

    (This problem is from the Spivak 2nd Ed. I had to translate it from spanish since my book is in spanish)

    If [tex]x_1, \ldots, x_n[/tex] are different numbers, find a polynomial function [tex]f_i[/tex] of [tex]n-1[/tex] degree that takes value 1 on [tex]x_i[/tex] and 0 in [tex]x_j[/tex] for [tex]j \neq i[/tex]. Indication: the product of every [tex](x-x_j)[/tex] for [tex]j \neq i[/tex] is 0 if [tex]j \neq i[/tex].

    \prod_{j=1}^{n} (x-x_{j})

    2. Relevant equations

    [tex]\prod_{j=1}^{n} (x-x_{j})[/tex]

    3. The attempt at a solution

    So far... Well so basically I stated all the known and unknown but I can't seem to get
    past that. So here's what I have...

    There is a set of [tex]x_1, \ldots, x_n[/tex]
    [tex]f_{i}[/tex] is of [tex]n-1[/tex] degree.

    There's a function such

    [tex]f_{i}(x) = a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \ldots + a_{1}x + a_{0}[/tex]

    There's a pair [tex](x_{i}, f_{i}(x_{i})[/tex] such that

    [tex]f_{i}(x_{i}) = a_{n-1}x_{i}^{n-1} + a_{n-2}x_{i}^{n-2} + \ldots + a_{1}x_{i} + a_{0} = 1[/tex]

    And there's also a pair [tex](x_{j}, f_{i}(x_{j}))[/tex] such that

    [tex]f_{i}(x_{j}) = a_{n-1}x_{j}^{n-1} + a_{n-2}x_{j}^{n-2} + \ldots + a_{1}x_{j} +a_{0} = 0[/tex]

    But I can't seem to connect the indication with the whole problem... any help? Oh, and I posted it in calculus but I am not quiet sure if this belongs in precalculus forum instead. I am sorry if this doesn't belong here.

    Thanks for any advice in advance.
  2. jcsd
  3. Jul 24, 2008 #2
    Try [tex]\prod_{k \neq i} (x-x_k)[/tex], a polynomial of degree n - 1, and divide this expression by an appropriate constant (try a product of factors). For more information, look up: Lagrange interpolation polynomials.
  4. Dec 19, 2009 #3
    First of all, I know that this thread is very old, but since I am working on this exact problem I assume it is better not to create a new thread. (+ it shows that I did a search :) )

    Here's my attempt:

    [tex] f_i(x) = \prod^n_{\frac{j=1}{j\neq i}} \frac{x-x_j}{x_i-x_j} [/tex]

    The next part of this problem is as follows.
    Find a polynomial function [tex]f[/tex] of degree [tex]n-1[/tex] such that [tex]f(x_i)=a_i[/tex], where [tex]a_1,...a_n[/tex] are given numbers. (You should use the function [tex]f_1[/tex] from the first part. The formula you will obtain is called the "Lagrange interpolation formula".

    [tex] f_i(x) = a_i\prod^n_{\frac{j=1}{j\neq i}} \frac{x-x_j}{x_i-x_j} [/tex]
    since the function from the previous problem was [tex]1[/tex] at [tex]x_i[/tex]

    I would appreciate it if someone could take a look at this and tell me if it is correct.
  5. Dec 19, 2009 #4
    Close. It should be:

    [tex]\Sigma[/tex] aifi(x)

    where f(x) is your function above

    you need a function such that fi(xi) = ai where xi = x1,x2,....,xn and ai = a1,a2,....,an.

    The LATEX subscripting doesn't seem to work :S
    Last edited: Dec 19, 2009
  6. Dec 20, 2009 #5
    Ok, so my new function is,

    [tex] f(x) = \sum^n_{i=1}a_if_i(x) [/tex]

    if I put [tex] x_i [/tex] into this new function, I would get:

    [tex] f(x_i) = a_1+a_2+...+a_n [/tex]

    The problem asks for a function where [tex] f(x_1)=a_i [/tex]. Does this imply a sum over the [tex]a_i[/tex]'s?
    By the way, I am in no way saying your answer is wrong. It fits nicely with my google search on lagrange interpolating polynomial :)

  7. Dec 20, 2009 #6


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    Science Advisor

    No. AT xi, fi will be 1, every other fn will be 0. [itex]f(x_i)= a_1(0)+ a_2(0)+ \cdot\cdot\cdot+ a_i (1)+ \cdot\cdot\cdot+ a_n(0)= a_i[/itex].

  8. Dec 20, 2009 #7
    This is why your function is wrong:

    You need to find a single function that gives you certain numbers whenever you input certain numbers.

    So at f(x1) you should get a1, at f(x2), you should get a2.. at f(xn) you should get an.

    With your function, aifi(x) where fi(x) is the quotient function shown earlier - you cannot get anything other than 0 and whatever ai was supposed to be, unless you keep changing your function.
    Last edited: Dec 20, 2009
  9. Dec 20, 2009 #8
    Ah, now I get it :)
    Thank you both very much!
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