Request for (quadratic?) equation?

Question

I'm trying to produce an equation that will provide the results, below. The first column is "x". The second column, "y", is the result I'm looking to produce with the equation.

So, for example, when x = 23, the equation should produce an answer of 92. When x = 12, the equation should produce a result of 840. When x = 4, the equation should produce 1080... and so on. Here is the full list of values:

01104
11104
21100
31092
41080
51064
61044
71020
8992
9960
10924
11884
12840
13792
14740
15684
16624
17560
18492
19420
20344
21264
22180
2392
240

..can anyone help me with this one?

Thanks a whole whole whole bunch!

Jamie

Okay, let me try the two columns again, for some reason the first column (starting at 0 and going up to 24 in increments of "1") didn't appear:

01104
11104
21100
31092
41080
51064
61044
71020
8992
9960
10924
11884
12840
13792
14740
15684
16624
17560
18492
19420
20344
21264
22180
2392
240

0 1104
1 1104
2 1100
3 1092
4 1080
5 1064
6 1044
7 1020
8 992
9 960
10 924
11 884
12 840
13 792
14 740
15 684
16 624
17 560
18 492
19 420
20 344
21 264
22 180
23 92
24 0

Answer 1

Checking by method of differences

. x ............. f(x) ...... Δ1 ...... Δ2
. 0 .......... 1104
. 1 .......... 1104 ........ 0
. 2 .......... 1100 ........ 4 ....... -4
. 3 .......... 1092 ........ 8 ....... -4
. 4 .......... 1080 ...... 12 ....... -4
. 5 .......... 1064 ...... 16 ....... -4
. 6 .......... 1044 ...... 20 ....... -4
. 7 .......... 1020 ...... 24 ....... -4
. 8 ............ 992 ...... 28 ....... -4
. 9 ............ 960 ...... 32 ....... -4
10 ............ 924 ...... 36 ....... -4
11 ............ 884 ...... 40 ....... -4
12 ............ 840 ...... 44 ....... -4
13 ............ 792 ...... 48 ....... -4
14 ............ 740 ...... 52 ....... -4
15 ............ 684 ...... 56 ....... -4
16 ............ 624 ...... 60 ....... -4
17 ............ 560 ...... 64 ....... -4
18 ............ 492 ...... 68 ....... -4
19 ............ 420 ...... 72 ....... -4
20 ............ 344 ...... 76 ....... -4
21 ............ 264 ...... 80 ....... -4
22 ............ 180 ...... 84 ....... -4
23 .............. 92 ...... 88 ....... -4
24 ................ 0 ...... 92 ....... -4

So the relation is quadratic (as Δ2 is constant which imples the 2nd derivative is constant)

So the relation is of the form

f(x) = ax^2 + bx + c

Now f(0) = c = 1104
f(1) = a + b + c = 1104
So a + b + 1104 = 1104
ie a + b = 0

f(2) = 4a + 2b + c = 1100
4a + 2b + 1104 = 1100
So 4a + 2b = -4

ie 2a + 2a + 2b = 2a + 2(a + b) = -4
So 2a + 0 = -4
ie a = -2 and b = 2

Thus f(x) = -2x^2 + 2x + 1104
= -2(x^2 - x - 552)
= -2(x - 24)(x + 23)

Test f(10) = -2*(10 - 24)(10 + 23)
= -2 * (-14) * (33)
= 28 * 33
= 924 YESSSSSSSSSSS!!

Answer 2

Are you treating experimental data? If so, then there is a special branch of maths for it. I don’t know how this method is mentioned in English, but in my tongue something like regression analysis or method of minimal squares. The matter is:
If you get a set of experimental data Y for a set of values of factor X, then you may correlate them like: S=sum[for j=0 until n] of (y(xj)-yj)^2, S being minimum for a set of parameters B for expected function y(x). In most cases you may assume that
y(x)=bn*x^n+ … +b2*x^2+b1*x1+b0 – that is polynomial of nth power. For your data I’d suggest n>3. Your task is to find this set of optimal coefficients B.
How it must be done? Find derivatives dS/d(bj) and assume each being = to Zero, thus you’ll receive a system of n+1 linear equations with n+1 unknown bj, hence you’ll have your particular approximation y(x). I’ve given you just some important hints and tips. Now you better read a handbook on this stuff to know not less important details!

Answer 3

I only see one column. These problems are solved by a method called "linear regression". Your equation is not linear, but you can linearize it by making the substitution z = √y. Then you take the square root of each of the y values and get a linear function for that. You can search for formulas for linear regression or find more info here http://en.wikipedia.org/wiki/Linear_regression

Also, Excel has several linear regression functions.

You can get more information on curve fitting here http://en.wikipedia.org/wiki/Curve_fitting

Answer 4

-2x^2+2x+1104