Depending on what kind of curve you want fitted will determine the line of best fit. I found a pretty good polynomial model for your data. It is:
This has been bugging me for quite some time now, I have many uses for this, but that's off-topic.
How do I find out the equation of a curve using the givin points?
I have the points for the curve.
(1, 25)
(2, 40)
(3, 60)
(4, 85)
(5, 115)
(6, 150)
(7, 190)
(8, 235)
(9, 285)
(10, 340)
I should also mention that I just got into geometry (I'm on my freshman high school year) and my math is a little rough over the break. This is actually something I've been trying to figure out for me, not for the school or any homework. Well, technically homework because of work at home, but not for the school.
I should also mention that I've tried the methods at which I found while searching on how.
My answer was way off. Let's just leave it at that.
Thank you,
-Tyler W.
Fair question, what kind of curve are you looking to fit? My example is polynomial of order 2. You could do this for a high order or even fit a stright line or exponential model.
For a polynomial model you propose a solution of type
Then you can pick three of the points in your series and substitute them into this model to find a,b & c.
for example you could pick
(1, 25), (5, 115) and (10, 340)
giving
and
then solve this system.
Using different points will produce different models.
I've returned after eating out tonight and trying out everything at which you said. The part I'm stuck at right now is the solving the system of polynomial equations.
I attempted it and tried a method to find 3 variables from 3 equations to see if that's how you solve a system of polynomial equations.
I might of done the math wrong, I tried my equation from the system that you showed me and it was horribly off.
If it isn't too much trouble, may you please explain to me the method for solving a system of polynomial equations?
Here's what I gotten
-Tyler W.
EDIT: I think I may have found the issue here. I did the square root of 11.25 to simplify to .
When I try it out for the first number it's fine. I will report back when I've tested it.
EDIT2: Nope, resulted in failure after 1.
pickslides is a very good method.
The methods I use:
Plot the data and determine the best linear fit
(least squares)
Then a log plot of the data.
Then a log.log plot.
(& several other types of plot)
With 10 data points you can get a perfect fit with a polynomial of degree 9.
You can use the method of differences to determine the approximate degree of polynomial you require.
(1, 25)
(2, 40)
(3, 60)
(4, 85)
(5, 115)
(6, 150)
(7, 190)
(8, 235)
(9, 285)
(10, 340)
first difference
25,40: 15
40,60: 20
60,85: 25
...
285,340:55
2nd difference
15-20: 5
20-25: 5
25-30: 5
3rd difference
5-5:0
etc.
So, by the looks of the method of differences, I require a polynomial of degree 0? Like I said, just starting geometry, and I really loathe polynomials. looks like finding the equation of a curve is my arch-enemy.
So what pickslides was saying, I just keep trying different points until I get a good equation that works out?
-Tyler W.
NO.by the looks of the method of differences, I require a polynomial of degree 0?
You have first differences & 2nd differences & all 3rd differences are zero, so you will need an equation NO higher than 2nd degree.
ONLY if you want to.So what pickslides was saying, I just keep trying different points until I get a good equation that works out?
Try this
from the y values:
difference from one Y value to the next Y value
first differences
y1 - y0 = d10 = ? (it is not given, so we don't know)
y2 - y1 = d21 = 40-25 = 15
y3 - y2 = d32 = 60-40 = 20
y4 - y3 = d43 = 85-60 = 25
y5 - y4 = d54 = 115-85 = 30
y6 - y5 = d65 = 150-115 = 35
You only need to do this until you see what's occurring.
2nd DIFFERENCES (difference between 1st diffs)
d32-d21 = dif3221 = 20-15 = 5
d43-d32 = dif4332 = 25-20 = 5
d54-d43 = dif5443 = 30-25 = 5
d65-d54 = dif6554 = 35-30 = 5
3rd Differences will all be zero
That means you will have an equation of degree 2
THIS FORM:
y = ax^2 + bx + c
from above
the difference between y1 & y0
will be the difference between
y2 and y1 (which is 15) MINUS (since we're going backwards) the 2nd difference (which is 5)
which makes
y1 - y0 = 10
y1 = 25
y1 - 10 = y0 = 15
We know that y0 = 15; when x=0 , y=15; We have the constant!
y = ax^2 + bx + 15
--
solely for ease of compuation
reduce all of the y values by 15
v0 = 0
v1 = 10
v2 = 25
v3 = 45
v4 = 70
etc
work out some values for the x's
at x=1: v1 =
at x=2: v2 =
we have two equations with 2 unknowns
10 = a + b ;thus 10-a = b
25 = 4a + 2b
substituting
25 = 4a + 2(10-a)
25 = 4a + 20 - 2a
25 = 2a + 20
5 = 2a
a =2.5
since 10 = a + b
10 = 2.5 + b
7.5 = b
We have all the coefficients
and can write the equation:
which is exactly the equation that pickslides supplied in his first post.
The procedure he gave in his other post is an excellent method of deriving the polynomial.
.