Results 1 to 6 of 6

Thread: Polynomial Interpolation

  1. #1
    Junior Member
    Joined
    Sep 2017
    From
    Astral Plane
    Posts
    36

    Polynomial Interpolation

    (3) For each of the mysterious functions below, use the given properties to find a,b,c.

    (a) f(x) = ax^2+bx=c, where f(1)=3 f`(0) = -1 f`(1) = 2.

    Now, I think I'm comfortable doing this with 3 points of f(x), but I'm confused as to how to do this with 1 or more points of f`(x).
    Normally, I would create an augmented matrix and plug the points in, but again I don't know what to do with the f`(x) points.

    (b) g(x) = asin(x) + bcos(x) +c where, g(0) = 2 g(pi)=1 g`(0)=-2

    (c) h(x) a +b(e^x) + c(e^-x) where, h(1)=e^2+1 , h`(1) = e^2 - 1 integral of h(x)dx = e^2 from 0 to 1


    Don't even know how to start with b and c...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,245
    Thanks
    2665

    Re: Polynomial Interpolation

    a) $f(x) = a x^2 + b x + c$

    $f^\prime(x) = 2 a x + b$

    plug in the values given and we end up with 3 equations and since you mentioned matrix I will write them as

    $\begin{pmatrix}
    1 &1 &1 &| &3 \\
    0 &1 &0 &| &-1 \\
    2 &1 &0 &| &2
    \end{pmatrix}\begin{pmatrix}a \\ b \\ c \end{pmatrix} = \begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}$

    use your favorite form of matrix reduction to solve this for $\begin{pmatrix}a &b &c \end{pmatrix}$

    (b) and (c) use essentially the same method.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2017
    From
    Astral Plane
    Posts
    36

    Re: Polynomial Interpolation

    Would the general matrix look like this:

    x^2 x c | y1
    2x c 0 | y2
    2x c 0 | y3?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,966
    Thanks
    2921
    Awards
    1

    Re: Polynomial Interpolation

    Quote Originally Posted by MrJank View Post
    (3) For each of the mysterious functions below, use the given properties to find a,b,c.

    (a) f(x) = ax^2+bx=c, where f(1)=3 f`(0) = -1 f`(1) = 2.

    Now, I think I'm comfortable doing this with 3 points of f(x), but I'm confused as to how to do this with 1 or more points of f`(x).
    Normally, I would create an augmented matrix and plug the points in, but again I don't know what to do with the f`(x) points.
    (b) g(x) = asin(x) + bcos(x) +c where, g(0) = 2 g(pi)=1 g`(0)=-2
    (c) h(x) a +b(e^x) + c(e^-x) where, h(1)=e^2+1 , h`(1) = e^2 - 1 integral of h(x)dx = e^2 from 0 to 1
    I would find the inverse of the matrix.
    $\begin{pmatrix}
    1 &1 &1 \\
    0 &1 &0 \\
    2 &1 &0 \end{pmatrix}\begin{pmatrix}a \\ b \\ c \end{pmatrix} = \begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}$ See Here
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,926
    Thanks
    3098

    Re: Polynomial Interpolation

    Personally, I dislike using matrices for small sets of equations like this.

    (a) f(x) = ax^2+b+ c, where f(1)=3 f`(0) = -1 f`(1) = 2. (You have "(a) f(x) = ax^2+bx=c" but I presume that second "=" is a typo.)

    We are given that $\displaystyle f(x)= ax^2+ bx+ c$. So we know that $\displaystyle f'(x)= 2ax+ b$.

    We are told that f(1)= 3. Okay, $\displaystyle f(1)= a(1^2)+ b(1)+ c= a+ b+ c= 3$.
    We are told that f'(0)= -1. Okay, $\displaystyle f'(0)= 2a(0)+ b= b= -1$.
    We are told that f'(1)= 2. Okay, $\displaystyle f'(1)= 2a(1)+ b= 2a+ b= 2$
    So we have the three equation a+ b+ c= 3, b= -1, and 2a+ b= 2. Since b= -1, 2a+ b= 2a- 1= 2 so 2a= 3 and a= 3/2. Then a+ b+ c= 3/2- 1+ c= 1/2+ c= 2 so c= 2- 1/2= 3/2.
    $\displaystyle f(x)= (3/2)x^2- x+ 3/2$.


    (b) g(x) = asin(x) + bcos(x) +c where, g(0) = 2 g(pi)=1 g`(0)=-2
    Now g'(x)= a cos(x)- b sin(x)

    g(0)= a sin(0)+ b cos(0)+ c= b+ c= 2.
    g'(pi)= a cos(pi)- b sin(pi)= -a= 1.
    g'(0)= a cos(0)- b sin(0)= -a= -2.
    The last two equations say "a= -1" and "a= -2". Those are impossible. There is no function of that form satisfying these conditions.


    (c) h(x)= a +b(e^x) + c(e^-x) where, h(1)=e^2+1 , h`(1) = e^2 - 1 integral of h(x)dx = e^2 from 0 to 1.
    With h(x)= a+ be^x+ ce^(-x), h'(x)= be^x- ce^(-x) and the integral is
    $\displaystyle \int_0^1 a+ be^x+ c^{-x} dx= \left[ax+ be^x- ce^{-x}\right_0^1= (a+ be- \frac{c}{e})- (b- c)$.

    The condition that $\displaystyle h(1)= e^2+ 1$ gives the equation $\displaystyle a+ be+ \frac{c}{e}= e^2+ 1$.
    The condition that $\displaystyle h'(1)= e^2- 1$ gives the equation $\displaystyle be- \frac{c}{e}= e^2- 1$.
    The condition that $\displaystyle \int_0^1 h(x) dx= e^2$ gives the condition that $\displaystyle a+ be- \frac{c}{e}- b+ c= e^2$.
    Again, we have three equations to solve for a, b, and c.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    933
    Thanks
    212

    Re: Polynomial Interpolation

    Quote Originally Posted by MrJank View Post
    Would the general matrix look like this:

    x^2 x c | y1
    2x c 0 | y2
    2x c 0 | y3?
    You are trying to find a, b and c, so you need three equations involving a, b and/or c


    $\displaystyle f(x) = ax^2 + bx +c$
    So, $\displaystyle f'(x) = 2ax +b$


    Now $\displaystyle f(1)= a+b+c=3$

    and $\displaystyle f'(0)=2a*0+b=b=-1$

    and $\displaystyle f'(1)=2a*1+b=2a+b=2$


    So there are your 3 equations. Now you need to solve them simultaneously (using matrices if you want, or just algebraic elimination method)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. interpolation and polynomial approximation
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Nov 26th 2013, 08:45 AM
  2. lagrange interpolation polynomial
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jun 21st 2011, 10:46 PM
  3. Polynomial interpolation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Apr 10th 2010, 08:01 PM
  4. Polynomial Interpolation
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: Oct 21st 2009, 09:19 PM
  5. Polynomial Interpolation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Oct 9th 2009, 08:08 AM

/mathhelpforum @mathhelpforum