How do you model a quartic function?

I

itank2828

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Joined
Oct 6, 2012
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ok, so I have these four co-ordinates:
1. (0,197)
2. (5, 204)
3. (8, 218)
4. (11, 225)
5. (12, 236)

Which are on a graph, and I want to find a model function to them which is a quartic function of the form: y = ax4 + bx3 + cx2 + dx + e

I put the co-ordinates into the general formula to get the following equations:
∴ e=197

204= 625a+125b+25c +5d+197

218= 4096a+512b+64c +8d+197

225= 14641a+1331b+121c +11d+197

231= 20736a+1728b+144c +12d+197

Now I need to know how the **** I can find a,b,c and d!!
Please help me - I need to complete this maths task very soon!!
 
itank2828 said:
ok, so I have these four co-ordinates:
1. (0,197)
2. (5, 204)
3. (8, 218)
4. (11, 225)
5. (12, 236)

Which are on a graph, and I want to find a model function to them which is a quartic function of the form: y = ax4 + bx3 + cx2 + dx + e

I put the co-ordinates into the general formula to get the following equations:
∴ e=197

204= 625a+125b+25c +5d+197

218= 4096a+512b+64c +8d+197

225= 14641a+1331b+121c +11d+197

231= 20736a+1728b+144c +12d+197

Now I need to know how the **** I can find a,b,c and d!!
Please help me - I need to complete this maths task very soon!!
Click to expand...

There are several ways to solve system of linear equations. Namely,

Gauss elimination

Matrix inversion

etc.

Those can be done by hand or by computer (using software like Excel).

What method/s have you been taught?
 
Help me!

Subhotosh Khan said:
There are several ways to solve system of linear equations. Namely,

Gauss elimination

Matrix inversion

etc.

Those can be done by hand or by computer (using software like Excel).

What method/s have you been taught?
Click to expand...

None of them, but by myself I tried to research using the Reduced Row Echelon Form. I will give you an attachment of what I came up with, but i know it was wrong as it didn't fit in the graph. Btw I'm not allowed to use technology to find the answer.

Please, I have a deadline very shortly on this I really to do it quickly!!
I can't post attachment here for some reason do you have an email or something I can send it to? I Need Urgent help man please help me!!
 
itank2828 said:
None of them, but by myself I tried to research using the Reduced Row Echelon Form. I will give you an attachment of what I came up with, but i know it was wrong as it didn't fit in the graph. Btw I'm not allowed to use technology to find the answer.

Please, I have a deadline very shortly on this I really to do it quickly!!
I can't post attachment here for some reason do you have an email or something I can send it to? I Need Urgent help man please help me!!
Click to expand...
Use the method of substitution. BE VERY CAREFUL with the arithmetic.

You use one equation to solve for one unknown in terms of the others. Example

\(\displaystyle 204 = 625a + 125b + 25c + 5d + 197 \implies 5d = 7 - 625a - 125b - 25c \implies d = \dfrac{7}{5} - 125a - 25b - 5c.\)

Now you substitute that value of d into the remaining equations.

Example

\(\displaystyle 218 = 4096a + 512b + 64c + 8d + 197 \implies 21 = 4096a + 512b + 64c + 8\left(\dfrac{7}{5} - 125a - 25b - 5c\right) \implies\)

\(\displaystyle 21= 4096a - 1000a + 512b - 200b + 64c - 40c + \dfrac{56}{5} \implies \dfrac{105 - 56}{5} = \dfrac{49}{5} = 3096a + 312b + 24c \implies\)

\(\displaystyle 49 = 15480a + 1560b + 120c.\)

Notice that d has been eliminated from this equation by the substitution.

Now solve for c in the equation above.

You now have two equations left with a, b, c, and d.

Eliminate c and d from one of them by substitution. Now use that equation to solve for b.

Now eliminate b, c, d from the last equation.

Now you are down to one equation in one unknown, which you solve by normal means.

This is a ton of arithmetic so make sure you check that the answers that you get work in the original equations.
 
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JeffM said:
Use the method of substitution. BE VERY CAREFUL with the arithmetic.

You use one equation to solve for one unknown in terms of the others. Example

\(\displaystyle 204 = 625a + 125b + 25c + 5d + 197 \implies 5d = 7 - 625a - 125b - 25c \implies d = \dfrac{7}{5} - 125a - 25b - 5c.\)

Now you substitute that value of d into the remaining equations.

Example

\(\displaystyle 218 = 4096a + 512b + 64c + 8d + 197 \implies 21 = 4096a + 512b + 64c + 8\left(\dfrac{7}{5} - 125a - 25b - 5c\right) \implies\)

\(\displaystyle 21= 4096a - 1000a + 512b - 200b + 64c - 40c + \dfrac{56}{5} \implies \dfrac{105 - 56}{5} = \dfrac{49}{5} = 3096a + 312b + 24c \implies\)

\(\displaystyle 49 = 15480a + 1560b + 120c.\)

Notice that d has been eliminated from this equation by the substitution.

Now solve for c in the equation above.

You now have two equations left with a, b, c, and d.

Eliminate c and d from one of them by substitution. Now use that equation to solve for b.

Now eliminate b, c, d from the last equation.

Now you are down to one equation in one unknown, which you solve by normal means.

This is a ton of arithmetic so make sure you check that the answers that you get work in the original equations.
Click to expand...


Hey thanks a lot man!! I decided to go for a quadratic, but really appreciate all the help!!
 
itank2828 said:
Hey thanks a lot man!! I decided to go for a quadratic, but really appreciate all the help!!
Click to expand...
Why did you decide "to go for a quadratic?

A quadratic is of the form \(\displaystyle y= ax^2+bx+ c\) so is determined by 3 equations, equivalent to requiring that its graph pass through three points. That means that a quadratic will NOT, in general, pass through five given points. A quartic, of the form \(\displaystyle y= ax^4+ bx^3+ cx^2+ dx+ e\) has five coefficients so can be determined by the five points you are given.
 
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HallsofIvy said:
Why did you decide "to go for a quadratic?

A quadratic is of the form \(\displaystyle y= ax^2+bx+ c\) so is determined by 3 equations, equivalent to requiring that its graph pass through three points. That means that a quadratic will NOT, in general, pass through five given points. A quartic, of the form \(\displaystyle y= ax^4+ bx^3+ cx^2+ dx+ e\) has five coefficients so can be determined by the five points you are given.
Click to expand...

Well, my deadline is tommorow so I have no time to complete the quartic. I already spent more than 4 hours trying to find the quartic (before I got the answers) and it has gone to waste, but at least I can finish my task now :/
 
Well, if you only concerned with giving an answer quickly, and not about whether that answer is right, why not just give something like "y= 204"?
 
HallsofIvy said:
Well, if you only concerned with giving an answer quickly, and not about whether that answer is right, why not just give something like "y= 204"?
Click to expand...
It's not like that HallsofIvy... I'm trying to be as accurate as possible with the time I have. Besides, the graph I'm trying to modell a function for has 12 points, not 5. I just took a sample, so even a quartic won't fit in well
 
lookagain said:
You have five sets of co-ordinates.
Click to expand...

No, I gave you 5 sets of co-ordinates to work with. I have 12 sets of co-ordinates.

I have 12 co-ordinates, with 2/3 empty.

0
1
2
3
4
5
6
7
8
9
10
11
12
197
203

198
204
212
216
218
224
223
225
236
 
Halls

As usual, we have very little context for the problem. If it is to find an equation that generates 12 specific points, a polynomial of degree 11 is one type of function that will do so, and there is not necessarily any quartic that will do so. If, however, the problem is to find a model that gives a good approximation, a linear model or a quadratic model may be quite robust. I would certainly not propose initially a polynomial of degree 11, which would guarantee a perfect fit, because there would be no degrees of freedom.

In an applied problem, I would start by looking at the scatter diagram of all twelve data points and trying to perceive patterns therein, then think about what, if anything, can be conjectured about the nature of the generating process, and finally test a number of potentially relevant functions for goodness of fit. If I saw the possibility of two or three local extrema, I'd be excited about trying a quartic. If I saw only one, I'd be inclined to start with a quadratic.

Without understanding more about the actual problem posed, I do not think we can be sure that a quartic is the best solution. Given the five points provided, I have an intuition that the coeffecient of the quartic term would not be statistically different from zero: y is not increasing rapidly.
 
If you plot the points, you'll see that the first two points are out-liers.

With the rest of the points a best fit quadratic would be

y = -0.5357x2 + 11.869x + 158.93 .......................................R2 = 0.9872

Best fit quartic is:

y = 0.0076x4 - 0.2172x3 + 1.697x2 + 2.149x + 174.03.............R2 = 0.9875

No significant improvement in quality of fit.
 
itank2828 said:
No, I gave you 5 sets of co-ordinates to work with. I have 12 sets of co-ordinates.

I have 12 co-ordinates, with 2/3 empty.
Click to expand...

What "2/3 empty?" You gave (presented) us with five sets of co-ordinates.
You state that you have 12 sets of co-ordinates.

12 (total that you have) - 5 (given to us to work with) = 7 left


\(\displaystyle \dfrac{7}{12} \ \ne \ \dfrac{2}{3}\)


Am I misunderstanding you?



Edit: But with positions 0 to 12, you have 13 potential sets of co-ordinates.

Also, if "2/3" is supposed to represent the values for box 2 and box 3 to be empty, then state that.
And then don't use the fraction, 2/3.
 
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