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

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Homework Help Overview

The discussion revolves around finding the equation of a cubic polynomial given five specific points. Participants explore various methods to derive the polynomial coefficients based on the provided coordinates.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up equations based on the polynomial form ax^3 + bx^2 + cx + d, with some questioning the necessity of all five equations given there are only four unknowns. Others mention alternative methods, such as finite differences, and share their experiences with these approaches.

Discussion Status

The conversation includes attempts to clarify the equations derived from the points and explores the finite difference method. Some participants express confusion about the method and seek further clarification, while others report success using it.

Contextual Notes

There is an indication that some participants have struggled with the problem, leading to a discussion about the validity of the methods and the assumptions made regarding the cubic nature of the function.

srizen
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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
 
Last edited:
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srizen said:
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).
 
Last edited:
eumyang said:
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
 
eumyang said:
...

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.
 
SammyS said:
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.
 
srizen said:
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

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

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

[itex](\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i[/itex]

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) ?
 
SammyS said:
Make a table of differences

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

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

[itex](\Delta^2)_i=(\Delta^1)_{i+1}-(\Delta^1)_i[/itex]

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!
 

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