Results 1 to 4 of 4

Thread: Polynomial

  1. #1
    Junior Member Hardwork's Avatar
    Joined
    Sep 2010
    Posts
    36

    Polynomial

    Question: If $\displaystyle x_{1}, ... , x_{n}$ are distinct numbers, find a polynomial $\displaystyle f_{i}$ of degree $\displaystyle n-1$ which is $\displaystyle 1$ at $\displaystyle x_{i}$ and 0 at $\displaystyle x_{j}$ for $\displaystyle j \ne i$. Hint: the product of all $\displaystyle (x-x_{i})$ for $\displaystyle j \ne i$ is zero at $\displaystyle x_{j}$ if $\displaystyle j \ne i$. .

    Confusion: The part of the question that I don't understand is the 'which is $\displaystyle 1$ at $\displaystyle x_{i}$ and 0 at $\displaystyle x_{j}$ for $\displaystyle j \ne i$.' What are $\displaystyle x_{i}$ and $\displaystyle x_{j}$? Does it mean that for any two numbers we choose from the sequence $\displaystyle x_{1}, ..., x_{n}$, we get $\displaystyle f_{i}$ as 0, for one, and 1, for the other? And how would you go about finding this polynomial?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    The polynomial is...

    $\displaystyle \displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})} $

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Hardwork's Avatar
    Joined
    Sep 2010
    Posts
    36
    Quote Originally Posted by chisigma View Post
    The polynomial is...

    $\displaystyle \displaystyle P(x)= \frac{\prod_{i=1, i\ne j}^{n} (x-x_{i})} {\prod_{i=1, i\ne j}^{n} (x_{j}-x_{i})} $

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Can you do a bit of explanation, please, if you don't mind? For example, it's not apparent to me that it's of degree $\displaystyle n-1$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    Quote Originally Posted by Hardwork View Post
    The part of the question that I don't understand is the 'which is $\displaystyle 1$ at $\displaystyle x_{i}$ and 0 at $\displaystyle x_{j}$ for $\displaystyle j \ne i$.' What are $\displaystyle x_{i}$ and $\displaystyle x_{j}$?
    The problem asks to find, for every $\displaystyle i=1,\dots,n$, a polynomial $\displaystyle f_i$ with the given property, which is a total of $\displaystyle n$ polynomials.

    For example, it's not apparent to me that it's of degree $\displaystyle n - 1$.
    In the denominator, you have a product of numbers. In the numerator, you have a product of $\displaystyle n-1$ expressions of the form $\displaystyle x-c$ for various constants $\displaystyle c$.

    This is similar to the interpolation polynomial in the Lagrange form (the link uses notation $\displaystyle \ell_j$ instead of $\displaystyle f_j$).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Oct 23rd 2011, 06:36 AM
  2. Replies: 1
    Last Post: Feb 24th 2011, 06:46 PM
  3. Replies: 1
    Last Post: Dec 15th 2009, 07:26 AM
  4. [SOLVED] dividing polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 3rd 2008, 02:00 PM
  5. dividing a polynomial by a polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Aug 2nd 2005, 12:26 AM

Search Tags


/mathhelpforum @mathhelpforum