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Question 94361: I need help with two different charts:
n 1 2 3 4 5
f(n) 0 4 20 54 112
n 1 2 3 4 5
f(n) 2 17 86 275 674
Need to find degree and equation for the polynomial
Don't really understand
Thank you
LLO
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! n 1 2 3 4 5
f(n) 0 4 20 54 112
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Its not linear because there is no constant difference in the f(n) line
Its not exponential because there is no constant ratio between consecutive terms.
Try a Quadratic Regression test and get: f(n) = 9n^2-26.6n+18.8 with R^2 = 99.9
which is very high.
I looked at Cubic and Quartic Regression test and got results that were worse.
------------------------
n 1 2 3 4 5
f(n) 2 17 86 275 674
Need to find degree and equation for the polynomial
-----------
Quadratic Regression gives this result:fn)=63.4n^2-220.37n+174.2
with R^2=0.992
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Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website! Soluton by Edwin McCravy:
`
Tutor Stanbon assumed you were taking statistics and were
looking for a regression equation. But you were not looking
for that. You want to find a polynomial f(n) that will have
those values exactly.
`
I need help with two different charts:
n 1 2 3 4 5
f(n) 0 4 20 54 112
To find the degree:
Make a difference table, by finding the
difference of each adjacent pair, writing
the differences until you find a line of
differences that are all the same.
n 1 2 3 4 5
f(n) 0 4 20 54 112
1 4 16 34 58
2 12 18 24
3 6 6
It took 3 lines of differences to find
a line on which all the numbers are the
same, so we know to assume a polynomial
of degree 3:
f(n) = An³ + Bn² + Cn + D
Since f(1) = 0,
f(1) = A(1)³ + B(1)² + C(1) + D = 0
f(1) = A + B + C + D = 0
So
Since f(2) = 4
f(2) = A(2)³ + B(2)² + C(2) + D = 4
f(2) = A(8) + B(4) + C(2) + D = 4
f(2) = 8A + 4B + 2C + D = 4
Since f(3) = 20
f(3) = A(3)³ + B(3)² + C(3) + D = 20
f(3) = A(27) + B(9) + C(3) + D = 2-
f(3) = 27A + 9B + 3C + D = 20
Since f(4) = 54
f(4) = A(4)³ + B(4)² + C(4) + D = 54
f(3) = A(64) + B(16) + C(4) + D = 54
f(3) = 64A + 16B + 4C + D = 54
So we have the system of equations:
A + B + C + D = 0
8A + 4B + 2C + D = 4
27A + 9B + 3C + D = 20
64A + 16B + 4C + D = 54
Do you know how to solve that system of
equations with matrices? If not post
again asking how. The solution is
A=1, B=0, C=-3, D=2
So
f(n) = An³ + Bn² + Cn + D so
f(n) = 1n³ + 0n² + (-3)n + 2
f(n) = n³ - 3n + 2
-------------------------------
n 1 2 3 4 5
f(n) 2 17 86 275 674
Need to find degree and equation for the polynomial
Make a difference table, by finding the
difference of each adjacent pair, writing
the differences until you find a line of
differences that are all the same.
n 1 2 3 4 5
f(n) 2 17 86 275 674
1 15 69 189 399
2 54 120 210
3 66 90
4 24
We had to make 4 lines this time because
we didn't come to a row that was all the
same, so we had to take it all the way to
4 lines, so the degree is 4.
f(n) = An4 + Bn³ + Cn² + Dn + E
Since f(1) = 2,
f(1) = A(1)4 + B(1)³ + C(1)² + D(1) + E = 2
f(1) = A + B + C + D + E = 2
Since f(2) = 17
f(2) = A(2)4 + B(2)³ + C(2)² + D(2) + E = 17
f(2) = A(16) + B(8) + C(4) + D(2) + E = 17
f(2) = 16A + 8B + 4C + 2D + E = 17
Since f(3) = 86
f(3) = A(3)4 + B(3)³ + C(3)² + D(3) + E = 86
f(3) = A(81) + B(27) + C(9) + 3D + E = 86
f(3) = 81A + 27B + 9C + 3D + E = 86
Since f(4) = 275
f(4) = A(4)4 + B(4)³ + C(4)² + D(4) + E = 275
f(4) = A(256) + B(64) + C(16) + 4D + E = 275
f(4) = 256A + 64B + 16C + 4D + E = 275
Since f(5) = 674
f(4) = A(5)4 + B(5)³ + C(5)² + D(5) + E = 674
f(4) = A(625) + B(125) + C(25) + 5D + E = 674
f(4) = 625A + 125B + 25C + 5D + E = 674
So we have the system of equations:
A + B + C + D + E = 2
16A + 8B + 4C + 2D + E = 17
81A + 27B + 9C + 3D + E = 86
256A + 64B + 16C + 4D + E = 275
625A + 125B + 25C + 5D + E = 674
Do you know how to solve that system of
equations with matrices? If not post
again asking how. The solution is
A=1, B=1, C=-4, D=5, E=-1
So
f(n) = An4 + Bn³ + Cn² + Dn + E
f(n) = 1n4 + 1n³ + (-4)n² + 5n - 1
f(n) = n4 + n³ - 4n² + 5n - 1
Edwin
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