How to trace equation from values of f(x) using Interpolation formulae












1














I have the following data, representing the values of a function at given points:



$f(1)=0$



$f(2)=5$



$f(3)=12$



$f(4)=21$



$f(5)=32$



I want to know the exact quadratic equation which was used above to get values of $f(x)$. How can I get to know it?



I remember that, at school, we used Newton's interpolation and backward/forward interpolation formula. But all I see on the internet regarding the above methods is that we can find $f(x)$ for any $x$ given the above table, eg. we can find $f(4.5)$ by using above table. However I need to trace the original quadratic formula: is there any way to do so?
(The answer should be $f(x)=x^2+2x-3$.)






functions interpolation







share|cite|improve this question
























  • You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
    – lulu
    Dec 1 '18 at 13:37


















1














I have the following data, representing the values of a function at given points:



$f(1)=0$



$f(2)=5$



$f(3)=12$



$f(4)=21$



$f(5)=32$



I want to know the exact quadratic equation which was used above to get values of $f(x)$. How can I get to know it?



I remember that, at school, we used Newton's interpolation and backward/forward interpolation formula. But all I see on the internet regarding the above methods is that we can find $f(x)$ for any $x$ given the above table, eg. we can find $f(4.5)$ by using above table. However I need to trace the original quadratic formula: is there any way to do so?
(The answer should be $f(x)=x^2+2x-3$.)






functions interpolation







share|cite|improve this question
























  • You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
    – lulu
    Dec 1 '18 at 13:37
















1












1








1







I have the following data, representing the values of a function at given points:



$f(1)=0$



$f(2)=5$



$f(3)=12$



$f(4)=21$



$f(5)=32$



I want to know the exact quadratic equation which was used above to get values of $f(x)$. How can I get to know it?



I remember that, at school, we used Newton's interpolation and backward/forward interpolation formula. But all I see on the internet regarding the above methods is that we can find $f(x)$ for any $x$ given the above table, eg. we can find $f(4.5)$ by using above table. However I need to trace the original quadratic formula: is there any way to do so?
(The answer should be $f(x)=x^2+2x-3$.)






functions interpolation







share|cite|improve this question















I have the following data, representing the values of a function at given points:



$f(1)=0$



$f(2)=5$



$f(3)=12$



$f(4)=21$



$f(5)=32$



I want to know the exact quadratic equation which was used above to get values of $f(x)$. How can I get to know it?



I remember that, at school, we used Newton's interpolation and backward/forward interpolation formula. But all I see on the internet regarding the above methods is that we can find $f(x)$ for any $x$ given the above table, eg. we can find $f(4.5)$ by using above table. However I need to trace the original quadratic formula: is there any way to do so?
(The answer should be $f(x)=x^2+2x-3$.)






functions interpolation




functions interpolation






share|cite|improve this question















share|cite|improve this question













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edited Dec 1 '18 at 13:39









Daniele Tampieri

1,7261619




1,7261619










asked Dec 1 '18 at 13:31









help

84




84












  • You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
    – lulu
    Dec 1 '18 at 13:37




















  • You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
    – lulu
    Dec 1 '18 at 13:37


















You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
– lulu
Dec 1 '18 at 13:37






You can just solve for the coefficients; any three values gives you three linear equations in three unknowns. Or you can use Lagrange Interpolation.
– lulu
Dec 1 '18 at 13:37












2 Answers
2






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oldest

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1














We can use Lagrange Interpolation as mentioned in the comments by @lulu.



First, we find the Lagrange Polynomials for the five points:



$$L_0(x) = dfrac {(x - x_1) (x - x_2) (x-x _3) (x - x_4)} {(x_0 -
x_1 ) (x_0 - x_2)(x_0 - x_3) (x_ 0 -x_4)} \
L_1(x) = dfrac {(x - x_0) (x - x_2) (x-x _3) (x - x_4)} {(x_1 -x_0 ) (x_1 - x_2)(x_1 - x_3) (x_1 -x_4)} \
L_2(x) = dfrac {(x - x_0) (x - x_1) (x-x _3) (x - x_4)} {(x_2 -
x_0) (x_2 - x_1)(x_2 - x_3) (x_2 -x_4)}\
L_3(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_4)} {(x_3 -
x_0 ) (x_3 - x_1)(x_3 - x_2) (x_3 -x_4)} \
L_4(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_3)} {(x_4 -
x_0 ) (x_4- x_1)(x_4 - x_2) (x_4 -x_3)}$$



The final result is given by



$$f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) + f(x_3) L_3(x) + f(x_4) L_4(x)
= x^2+2 x-3$$



We can verify that $f(x)$ produces
$$f(1)=0, f(2)=5, f(3)=12, f(4)=21, f(5)=32$$






share|cite|improve this answer































    0














    For a potentially less computationally intense method (also included in lulu's comment), you can solve a system of three equations.



    You know that the general form of a quadratic equation is $$f(x)=ax^2+bx+c,$$ and since there are three unknowns ($a$, $b$, and $c$), you can plug in values from three points for $f(x)$ and $x$ and solve the resulting system of equations, drawing on the idea that any three points in the plane uniquely determine a parabola. For example, using the first three points you list, you would obtain the following system:



    $$begin{align}
    0&=a+b+c \
    5&=4a+2b+c \
    12&=9a+3b+c
    end{align}$$



    This does indeed give the desired result of $a=1$, $b=2$, $c=-3$.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      1














      We can use Lagrange Interpolation as mentioned in the comments by @lulu.



      First, we find the Lagrange Polynomials for the five points:



      $$L_0(x) = dfrac {(x - x_1) (x - x_2) (x-x _3) (x - x_4)} {(x_0 -
      x_1 ) (x_0 - x_2)(x_0 - x_3) (x_ 0 -x_4)} \
      L_1(x) = dfrac {(x - x_0) (x - x_2) (x-x _3) (x - x_4)} {(x_1 -x_0 ) (x_1 - x_2)(x_1 - x_3) (x_1 -x_4)} \
      L_2(x) = dfrac {(x - x_0) (x - x_1) (x-x _3) (x - x_4)} {(x_2 -
      x_0) (x_2 - x_1)(x_2 - x_3) (x_2 -x_4)}\
      L_3(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_4)} {(x_3 -
      x_0 ) (x_3 - x_1)(x_3 - x_2) (x_3 -x_4)} \
      L_4(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_3)} {(x_4 -
      x_0 ) (x_4- x_1)(x_4 - x_2) (x_4 -x_3)}$$



      The final result is given by



      $$f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) + f(x_3) L_3(x) + f(x_4) L_4(x)
      = x^2+2 x-3$$



      We can verify that $f(x)$ produces
      $$f(1)=0, f(2)=5, f(3)=12, f(4)=21, f(5)=32$$






      share|cite|improve this answer




























        1














        We can use Lagrange Interpolation as mentioned in the comments by @lulu.



        First, we find the Lagrange Polynomials for the five points:



        $$L_0(x) = dfrac {(x - x_1) (x - x_2) (x-x _3) (x - x_4)} {(x_0 -
        x_1 ) (x_0 - x_2)(x_0 - x_3) (x_ 0 -x_4)} \
        L_1(x) = dfrac {(x - x_0) (x - x_2) (x-x _3) (x - x_4)} {(x_1 -x_0 ) (x_1 - x_2)(x_1 - x_3) (x_1 -x_4)} \
        L_2(x) = dfrac {(x - x_0) (x - x_1) (x-x _3) (x - x_4)} {(x_2 -
        x_0) (x_2 - x_1)(x_2 - x_3) (x_2 -x_4)}\
        L_3(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_4)} {(x_3 -
        x_0 ) (x_3 - x_1)(x_3 - x_2) (x_3 -x_4)} \
        L_4(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_3)} {(x_4 -
        x_0 ) (x_4- x_1)(x_4 - x_2) (x_4 -x_3)}$$



        The final result is given by



        $$f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) + f(x_3) L_3(x) + f(x_4) L_4(x)
        = x^2+2 x-3$$



        We can verify that $f(x)$ produces
        $$f(1)=0, f(2)=5, f(3)=12, f(4)=21, f(5)=32$$






        share|cite|improve this answer


























          1












          1








          1






          We can use Lagrange Interpolation as mentioned in the comments by @lulu.



          First, we find the Lagrange Polynomials for the five points:



          $$L_0(x) = dfrac {(x - x_1) (x - x_2) (x-x _3) (x - x_4)} {(x_0 -
          x_1 ) (x_0 - x_2)(x_0 - x_3) (x_ 0 -x_4)} \
          L_1(x) = dfrac {(x - x_0) (x - x_2) (x-x _3) (x - x_4)} {(x_1 -x_0 ) (x_1 - x_2)(x_1 - x_3) (x_1 -x_4)} \
          L_2(x) = dfrac {(x - x_0) (x - x_1) (x-x _3) (x - x_4)} {(x_2 -
          x_0) (x_2 - x_1)(x_2 - x_3) (x_2 -x_4)}\
          L_3(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_4)} {(x_3 -
          x_0 ) (x_3 - x_1)(x_3 - x_2) (x_3 -x_4)} \
          L_4(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_3)} {(x_4 -
          x_0 ) (x_4- x_1)(x_4 - x_2) (x_4 -x_3)}$$



          The final result is given by



          $$f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) + f(x_3) L_3(x) + f(x_4) L_4(x)
          = x^2+2 x-3$$



          We can verify that $f(x)$ produces
          $$f(1)=0, f(2)=5, f(3)=12, f(4)=21, f(5)=32$$






          share|cite|improve this answer














          We can use Lagrange Interpolation as mentioned in the comments by @lulu.



          First, we find the Lagrange Polynomials for the five points:



          $$L_0(x) = dfrac {(x - x_1) (x - x_2) (x-x _3) (x - x_4)} {(x_0 -
          x_1 ) (x_0 - x_2)(x_0 - x_3) (x_ 0 -x_4)} \
          L_1(x) = dfrac {(x - x_0) (x - x_2) (x-x _3) (x - x_4)} {(x_1 -x_0 ) (x_1 - x_2)(x_1 - x_3) (x_1 -x_4)} \
          L_2(x) = dfrac {(x - x_0) (x - x_1) (x-x _3) (x - x_4)} {(x_2 -
          x_0) (x_2 - x_1)(x_2 - x_3) (x_2 -x_4)}\
          L_3(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_4)} {(x_3 -
          x_0 ) (x_3 - x_1)(x_3 - x_2) (x_3 -x_4)} \
          L_4(x) = dfrac {(x - x_0) (x - x_1) (x-x _2) (x - x_3)} {(x_4 -
          x_0 ) (x_4- x_1)(x_4 - x_2) (x_4 -x_3)}$$



          The final result is given by



          $$f(x) = f(x_0) L_0(x) + f(x_1) L_1(x) + f(x_2) L_2(x) + f(x_3) L_3(x) + f(x_4) L_4(x)
          = x^2+2 x-3$$



          We can verify that $f(x)$ produces
          $$f(1)=0, f(2)=5, f(3)=12, f(4)=21, f(5)=32$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 16:24

























          answered Dec 1 '18 at 14:11









          Moo

          5,53131020




          5,53131020























              0














              For a potentially less computationally intense method (also included in lulu's comment), you can solve a system of three equations.



              You know that the general form of a quadratic equation is $$f(x)=ax^2+bx+c,$$ and since there are three unknowns ($a$, $b$, and $c$), you can plug in values from three points for $f(x)$ and $x$ and solve the resulting system of equations, drawing on the idea that any three points in the plane uniquely determine a parabola. For example, using the first three points you list, you would obtain the following system:



              $$begin{align}
              0&=a+b+c \
              5&=4a+2b+c \
              12&=9a+3b+c
              end{align}$$



              This does indeed give the desired result of $a=1$, $b=2$, $c=-3$.






              share|cite|improve this answer


























                0














                For a potentially less computationally intense method (also included in lulu's comment), you can solve a system of three equations.



                You know that the general form of a quadratic equation is $$f(x)=ax^2+bx+c,$$ and since there are three unknowns ($a$, $b$, and $c$), you can plug in values from three points for $f(x)$ and $x$ and solve the resulting system of equations, drawing on the idea that any three points in the plane uniquely determine a parabola. For example, using the first three points you list, you would obtain the following system:



                $$begin{align}
                0&=a+b+c \
                5&=4a+2b+c \
                12&=9a+3b+c
                end{align}$$



                This does indeed give the desired result of $a=1$, $b=2$, $c=-3$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  For a potentially less computationally intense method (also included in lulu's comment), you can solve a system of three equations.



                  You know that the general form of a quadratic equation is $$f(x)=ax^2+bx+c,$$ and since there are three unknowns ($a$, $b$, and $c$), you can plug in values from three points for $f(x)$ and $x$ and solve the resulting system of equations, drawing on the idea that any three points in the plane uniquely determine a parabola. For example, using the first three points you list, you would obtain the following system:



                  $$begin{align}
                  0&=a+b+c \
                  5&=4a+2b+c \
                  12&=9a+3b+c
                  end{align}$$



                  This does indeed give the desired result of $a=1$, $b=2$, $c=-3$.






                  share|cite|improve this answer












                  For a potentially less computationally intense method (also included in lulu's comment), you can solve a system of three equations.



                  You know that the general form of a quadratic equation is $$f(x)=ax^2+bx+c,$$ and since there are three unknowns ($a$, $b$, and $c$), you can plug in values from three points for $f(x)$ and $x$ and solve the resulting system of equations, drawing on the idea that any three points in the plane uniquely determine a parabola. For example, using the first three points you list, you would obtain the following system:



                  $$begin{align}
                  0&=a+b+c \
                  5&=4a+2b+c \
                  12&=9a+3b+c
                  end{align}$$



                  This does indeed give the desired result of $a=1$, $b=2$, $c=-3$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 16:14









                  Robert Howard

                  1,9161822




                  1,9161822






























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