Found 3 solutions by solver91311, greenestamps, ikleyn:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Six points uniquely determine a 5th-degree polynomial function. (In general, n points uniquely determine an (n-1)th degree polynomial function).
The general form of a 5th-degree polynomial function is:
\ =\ ax^5\ +\ bx^4\ +\ cx^3\ +\ dx^2\ +\ ex\ +\ f)
So if \ =\ 0)
when 
, then
And if \ =\ 3)
when 
, then
Similarly:
Use Cramer's Rule to solve the 6X6 system. Easiest is to put your matrices into Excel and use the MDETERM function to calculate the required determinants.
For example, your coefficient determinant and Da determinant will be:
| 1 1 1 1 1 1|
| 32 16 8 4 2 1|
D = | 243 81 27 9 3 1|
| 1024 256 64 16 4 1|
| 3125 625 125 25 5 1|
| 7776 1296 216 36 6 1|
| 0 1 1 1 1 1|
| 3 16 8 4 2 1|
Da = | 16 81 27 9 3 1|
| 45 256 64 16 4 1|
| 96 625 125 25 5 1|
| 175 1296 216 36 6 1|
and so on...
You can also find an online matrix calculator that will do Gauss-Jordan reduction for you. The only alternative to getting the exact answer by solving the 6X6 linear system is to use a "curve of best fit" calculator that will handle a 5th-degree polynomial model. These generally result in inexact but very precise coefficients.
John
My calculator said it, I believe it, that settles it
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!
Use the method of finite differences to find the required degree of the polynomial. 6 points define a unique polynomial of degree 5; but it is possible a polynomial of lower degree will define the given points.
0 3 16 45 96 175
3 13 29 51 79
10 16 22 28
6 6 6
The third row of differences is constant; that means the points can be defined by a polynomial of degree 3:
t(n) = an^3+bn^2+cn+d
Substitute n = 1, 2, 3, and 4 to get four equations in the coefficients a, b, c, and d and solve the system. Note it will always be easy to see what the next step should be in solving the system.
a + b + c + d = 0
8a + 4b + 2c + d = 3
27a + 9b + 3c + d = 16
64a + 16b + 4c + d = 45
Comparing successive pairs of equations...:
7a + 3b + c = 3
19a + 5b + c = 13
37a + 7b + c = 29
Again comparing successive pairs of equations...:
12a + 2b = 10
18a + 2b = 16
And comparing those two equations...:
6a = 6
a = 1
Then back substitute the known values to find the others:
12(1)+2b = 10
2b = -2
b = -1
7(1)+3(-1)+c = 3
c = -1
1+(-1)+(-1)+d = 0
d = 1
We have a=1, b=-1, c=-1 and d=1; the polynomial of degree 3 that defines the given numbers is
t(n) = n^3-n^2-n+1
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.
I became interesting what will happen if I solve the system of equations derived by @solver91311
. =
Will I get the answer obtained by @greenestamps ?
So, I used an online solver http://matrix.reshish.com/gauss-jordanElimination.php , which employs the Gauss-Jordan elimination.
I filled the extended matrix and clicked the "Solve" button.
Below is what I got:
A = 0
B = 0
C = 1
D = -1
E = -1
F = 1
- exactly and precisely what @greenestamps got.
So, both methods produce the same solution.
If to think one more minute, then it becomes clear that it can not be in other way,
because the matrix of the system is NON-SINGULAR (is NOT DEGENERATED) and allows ONLY ONE SOLUTION.
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