How do you find the equation of a Cubic function given 5 points? (no zeros)?

[srizen]

Homework Statement

what the questions asks is that i need to find the equation of a polynomial with these given points:
1,1
2,-3
3,5
4,37
5,105i know that one way to solve is by creating 5 equations then solve for ax^3+bx^2+cx+d using the elimination/substitution method.
however is there another, much easier way of doing this question?

Homework Equations

ax^3+bx^2+cx+d

The Attempt at a Solution

1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

fixed, yes, my mistake

 

[eumyang]

1= a+b+c+d
-3=8a+4b+3c+d
5= 64a+16b+5c+d
37=64a+16b+5c+d
108=125a+25b+5c+d

Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

Also, you don't need the last equation, because there are 4 unknowns.

As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).

 

[srizen]

Three of the equations are wrong. They should be
1= a+b+c+d
-3=8a+4b+2c+d
5= 27a+9b+3c+d
37=64a+16b+4c+d
108=125a+25b+5c+d

fixed, i typed the equations too fast

 

[SammyS]

...

As for other methods, there's the finite difference method, but I don't think it will help for this particular problem (because you already told us that this is a cubic).

True, but I got the coefficients fairly quickly using a difference method.

Actually, after playing around a bit with this, I got the result with two different difference methods.

 

[srizen]

True, but I got the coefficients fairly quickly using a difference method.

Actually, after playing around a bit with this, I got the result with two different difference methods.

what exactly is the difference method?
i ask because I've tried this question, and i kept getting it wrong.

 

[eumyang]

what exactly is the difference method?
i ask because I've tried this question, and i kept getting it wrong.

A couple of results from searching via Google:
http://www.jimloy.com/algebra/finite.htm
http://www.math-mate.com/chapter37.shtml

 

[SammyS]

what exactly is the difference method?
i ask because I've tried this question, and i kept getting it wrong.

Make a table of differences

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline <br /> \quad i \quad &amp; \quad x_i \quad &amp; \quad f(x_i) \quad &amp;\quad (\Delta^1)_i \quad &amp;\quad (\Delta^2)_i \quad &amp; \quad (\Delta^3)_i \quad &amp; \quad (\Delta^4)_i \quad \\ <br /> \hline <br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 1 &amp; 1 &amp; 1 &amp; -4 &amp; 12 &amp; 12 &amp; 0 \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 2 &amp; 2 &amp; -3 &amp; 8 &amp; 24 &amp; &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 3 &amp; 3 &amp; 5 &amp; &amp; &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 4 &amp; 4 &amp; 37 &amp; &amp; -- &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 5 &amp;5 &amp; 105 &amp; -- &amp; -- &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> \hline \end{array}

Where: (\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,

(\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i

etc.

See if you can fill in the rest.

If f(x) is truly a cubic function then the Δ³ column will all be the same.

Fill out a similar Table for g(x) = x³ . The Δ³ column will all be 6's.

What do you suppose that means about the x³ coefficient of f(x) ?

 

[srizen]

Make a table of differences

\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline <br /> \quad i \quad &amp; \quad x_i \quad &amp; \quad f(x_i) \quad &amp;\quad (\Delta^1)_i \quad &amp;\quad (\Delta^2)_i \quad &amp; \quad (\Delta^3)_i \quad &amp; \quad (\Delta^4)_i \quad \\ <br /> \hline <br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 1 &amp; 1 &amp; 1 &amp; -4 &amp; 12 &amp; 12 &amp; 0 \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 2 &amp; 2 &amp; -3 &amp; 8 &amp; 24 &amp; &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 3 &amp; 3 &amp; 5 &amp; &amp; &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 4 &amp; 4 &amp; 37 &amp; &amp; -- &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> 5 &amp;5 &amp; 105 &amp; -- &amp; -- &amp; -- &amp; -- \\<br /> &amp; &amp; &amp; &amp; &amp; &amp; \\<br /> \hline \end{array}

Where: (\Delta^1)_i=f(x_{i+1})-f(x_{i})\,,

(\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i

etc.

See if you can fill in the rest.

If f(x) is truly a cubic function then the Δ³ column will all be the same.

Fill out a similar Table for g(x) = x³ . The Δ³ column will all be 6's.

What do you suppose that means about the x³ coefficient of f(x) ?

OMG! i love you forever, i had no idea this method existed, i already solved through almost an hour of writing matrices, with this i solved it in 3 minutes. thank you!