Mixing
Finding an interpolating second order polynomial going through a given point
Below is a problem that I made up and my attempt to solve it. I am fairly sure that my answer is wrong. Please note that when I write $D_A$ I mean the first partial derivative of $D$ with respect to $A$. I have a similar meaning for $D_B$. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
Problem:
Given the points $(0,1), (1,3), (2,5), (3,7), (4,4)$ find a second order
interpolating polynomial, of the form $f(x) = Ax^2 + Bx + C$, such that the point $(4,4)$ is on the curve and the following is minimized:
$$ (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2 $$
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
begin{align*}
D_A &= 2(A+B+C-3) + 8( 4A + 2B + C - 5) + 18( 9A + 3B + C - 7) \
D_A &= (2 +32+182)A + (2+16+54)B + (2+8+18)C - 6 - 40 - 126 \
D_A &= 216A + 72B + 28C - 172 \
D_A &= 216A + 72B + 28(4-16A-4B) - 172 \
D_A &= -232A - 40B + 112 - 172 \
D_A &= -120A - 12B - 60 \
D_B &= 3(A+B+C-3) + 4(4A+2B+C-5) + 6(9A+3B+C-7) \
D_B &= (3+16+54)A + (3+8+18)B + (3+4+6C) - 9 - 20 - 54 \
D_B &= 73A + 29B + 13C - 83 \
D_B &= 73A + 29B + 13(4 - 16A - 4B) - 83 \
D_B &= 73A + 29B + 52 - 208A - 52B - 83 \
D_B &= -135A - 23B - 31 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
-120A - 12B - 60 &= 0 \
-135A - 23B - 31 &= 0 \
60A + 6B + 30 &= 0 \
30A + 3B + 15 &= 0 \
3B &= -30A - 15 \
B &= -10A - 5 \
135A + 23B + 31 &= 0 \
135A + 23( -10A - 5) + 31 &= 0 \
135A - 230A - 115 + 31 &= 0 \
-95A - 84 &= 0 \
A &= -frac{84}{95} \
end{align*}
Now we solve for $B$ and then $C$.
begin{align*}
B &= -10left( -frac{84}{95} right) - 5 = 10left( frac{84}{95} right) - 5\
B &= frac{840}{95} - 5 \
B &= frac{ 365 } {95} \
B &= frac{73}{9} \
C &= 4 - 16left( -frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + 16left( frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + frac{1344}{95} - frac{292}{9} = frac{ 4(95)(9) + 1344(9) - 292(95)}{9(95)} \
C &= frac{ 15516 - 27740}{9(95)} \
C &= frac{ -12224}{855} \
end{align*}
Hence our interpolating polynomial is:
$$ f(x) = -left( frac{84}{95} right)x^2 + left( frac{73}{9} right) - frac{ 12224}{855} $$
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Based upon the comments of Siong Thye Goh I have updated my post. However, my
answer is still wrong. I am hoping that somebody can tell me where I went wrong
and/ or correct it.
Thanks,
Bob
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
Now we need to find $D_A$ and $D_B$. Observe that $C_A = -16$ and $C_B = -4$.
begin{align*}
D_A &= 2C_A(C-1) + 2( 1+ C_A)(A + B + C- 3) + 2(4 + C_A)( 4A + 2B + C - 5) \
&+ 2(9 + C_A)( 9A + 3B + C - 7) \
D_A &= 2(-16)(C-1) + 2( 1 - 16)(A + B + C- 3) + 2(4 - 16)( 4A + 2B + C - 5) \
&+ 2(9 - 16)( 9A + 3B + C - 7) \
D_A &= -32(C-1) + 2( -15 )(A + B + C- 3) + 2( -12 )( 4A + 2B + C - 5) \
&+ 2(-7)( 9A + 3B + C - 7) \
D_A &= -32(C-1) - 30(A + B + C- 3) - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -32C + 32 - 30A - 30B -30C + 90 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B -62C + 122 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B - 62C + 122 - 96A - 48B - 24C + 120 - 126A - 42B - 14C + 98 \
D_A &= -252A - (30 + 48 + 42)B -(62 + 24 + 14)C + 340 \
D_A &= -252A - 120B - 100C + 340 \
D_A &= -252A - 120B - 100(4 - 16A - 4B) + 340 \
D_A &= -252A - 120B - 400 + 1600A +400B + 340 \
D_A &= 1358A + 280B - 60 \
end{align*}
begin{align*}
D_B &= 2C_B (C - 1) + 3(1 +C_B)(A + B + C- 3) + 2(2 + C_B)(4A + 2B + C - 5) \
&+ 2(3+C_B)( 9A + 3B + C - 7) \
D_B &= -8 (C - 1) + 3(-3)(A + B + C- 3) + 2(-2)(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9(A + B + C- 3) - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9A - 9B - 9C + 27 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17C + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17(4 - 16A - 4B) + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 68 - 16(17)A + 68B + 35 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) - 2( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 16A - 8B - 4C + 20 -18A - 6B - 2C + 14 \
D_B &= -43A + (-9 + 68 -8 - 6)B - 6C - 33 + 20 + 14 \
D_B &= -43A + 45B - 6C + 1 \
D_B &= -43A + 45B - 6( 4 - 16A - 4B ) + 1 \
D_B &= -43A + 45B - 24 + 96A + 24B + 1 \
D_B &= 53A + 69B - 23 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
1358A + 280B - 60 &= 0 \
53A + 69B - 23 &= 0 \
679A + 140B - 30 &= 0 \
140B &= 30 - 679A \
B &= frac{30}{140} - frac{679A}{140} \
B &= frac{3}{14} - frac{679A}{140} \
53A + 69left( frac{3}{14} - frac{679A}{140} right) - 23 &= 0 \
53A + frac{69(3)}{14} - frac{35987A}{140} - 23 &= 0 \
53A - frac{35987A}{140} - frac{115}{14} &= 0 \
-frac{28567A}{140} &= frac{115}{14} \
-28567A &= 1150 \
A &= -frac{1150}{28567} \
B &= frac{3}{14} - frac{-679(1150)}{140(28567)} \
B &= frac{3}{14} + frac{679(1150)}{140(28567)} \
end{align*}
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Hence, I am confident that my solution is wrong.
statistics multivariable-calculus
asked Jan 1 at 1:57
BobBob
902514
add a comment |
Below is a problem that I made up and my attempt to solve it. I am fairly sure that my answer is wrong. Please note that when I write $D_A$ I mean the first partial derivative of $D$ with respect to $A$. I have a similar meaning for $D_B$. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
Problem:
Given the points $(0,1), (1,3), (2,5), (3,7), (4,4)$ find a second order
interpolating polynomial, of the form $f(x) = Ax^2 + Bx + C$, such that the point $(4,4)$ is on the curve and the following is minimized:
$$ (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2 $$
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
begin{align*}
D_A &= 2(A+B+C-3) + 8( 4A + 2B + C - 5) + 18( 9A + 3B + C - 7) \
D_A &= (2 +32+182)A + (2+16+54)B + (2+8+18)C - 6 - 40 - 126 \
D_A &= 216A + 72B + 28C - 172 \
D_A &= 216A + 72B + 28(4-16A-4B) - 172 \
D_A &= -232A - 40B + 112 - 172 \
D_A &= -120A - 12B - 60 \
D_B &= 3(A+B+C-3) + 4(4A+2B+C-5) + 6(9A+3B+C-7) \
D_B &= (3+16+54)A + (3+8+18)B + (3+4+6C) - 9 - 20 - 54 \
D_B &= 73A + 29B + 13C - 83 \
D_B &= 73A + 29B + 13(4 - 16A - 4B) - 83 \
D_B &= 73A + 29B + 52 - 208A - 52B - 83 \
D_B &= -135A - 23B - 31 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
-120A - 12B - 60 &= 0 \
-135A - 23B - 31 &= 0 \
60A + 6B + 30 &= 0 \
30A + 3B + 15 &= 0 \
3B &= -30A - 15 \
B &= -10A - 5 \
135A + 23B + 31 &= 0 \
135A + 23( -10A - 5) + 31 &= 0 \
135A - 230A - 115 + 31 &= 0 \
-95A - 84 &= 0 \
A &= -frac{84}{95} \
end{align*}
Now we solve for $B$ and then $C$.
begin{align*}
B &= -10left( -frac{84}{95} right) - 5 = 10left( frac{84}{95} right) - 5\
B &= frac{840}{95} - 5 \
B &= frac{ 365 } {95} \
B &= frac{73}{9} \
C &= 4 - 16left( -frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + 16left( frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + frac{1344}{95} - frac{292}{9} = frac{ 4(95)(9) + 1344(9) - 292(95)}{9(95)} \
C &= frac{ 15516 - 27740}{9(95)} \
C &= frac{ -12224}{855} \
end{align*}
Hence our interpolating polynomial is:
$$ f(x) = -left( frac{84}{95} right)x^2 + left( frac{73}{9} right) - frac{ 12224}{855} $$
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Based upon the comments of Siong Thye Goh I have updated my post. However, my
answer is still wrong. I am hoping that somebody can tell me where I went wrong
and/ or correct it.
Thanks,
Bob
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
Now we need to find $D_A$ and $D_B$. Observe that $C_A = -16$ and $C_B = -4$.
begin{align*}
D_A &= 2C_A(C-1) + 2( 1+ C_A)(A + B + C- 3) + 2(4 + C_A)( 4A + 2B + C - 5) \
&+ 2(9 + C_A)( 9A + 3B + C - 7) \
D_A &= 2(-16)(C-1) + 2( 1 - 16)(A + B + C- 3) + 2(4 - 16)( 4A + 2B + C - 5) \
&+ 2(9 - 16)( 9A + 3B + C - 7) \
D_A &= -32(C-1) + 2( -15 )(A + B + C- 3) + 2( -12 )( 4A + 2B + C - 5) \
&+ 2(-7)( 9A + 3B + C - 7) \
D_A &= -32(C-1) - 30(A + B + C- 3) - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -32C + 32 - 30A - 30B -30C + 90 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B -62C + 122 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B - 62C + 122 - 96A - 48B - 24C + 120 - 126A - 42B - 14C + 98 \
D_A &= -252A - (30 + 48 + 42)B -(62 + 24 + 14)C + 340 \
D_A &= -252A - 120B - 100C + 340 \
D_A &= -252A - 120B - 100(4 - 16A - 4B) + 340 \
D_A &= -252A - 120B - 400 + 1600A +400B + 340 \
D_A &= 1358A + 280B - 60 \
end{align*}
begin{align*}
D_B &= 2C_B (C - 1) + 3(1 +C_B)(A + B + C- 3) + 2(2 + C_B)(4A + 2B + C - 5) \
&+ 2(3+C_B)( 9A + 3B + C - 7) \
D_B &= -8 (C - 1) + 3(-3)(A + B + C- 3) + 2(-2)(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9(A + B + C- 3) - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9A - 9B - 9C + 27 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17C + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17(4 - 16A - 4B) + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 68 - 16(17)A + 68B + 35 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) - 2( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 16A - 8B - 4C + 20 -18A - 6B - 2C + 14 \
D_B &= -43A + (-9 + 68 -8 - 6)B - 6C - 33 + 20 + 14 \
D_B &= -43A + 45B - 6C + 1 \
D_B &= -43A + 45B - 6( 4 - 16A - 4B ) + 1 \
D_B &= -43A + 45B - 24 + 96A + 24B + 1 \
D_B &= 53A + 69B - 23 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
1358A + 280B - 60 &= 0 \
53A + 69B - 23 &= 0 \
679A + 140B - 30 &= 0 \
140B &= 30 - 679A \
B &= frac{30}{140} - frac{679A}{140} \
B &= frac{3}{14} - frac{679A}{140} \
53A + 69left( frac{3}{14} - frac{679A}{140} right) - 23 &= 0 \
53A + frac{69(3)}{14} - frac{35987A}{140} - 23 &= 0 \
53A - frac{35987A}{140} - frac{115}{14} &= 0 \
-frac{28567A}{140} &= frac{115}{14} \
-28567A &= 1150 \
A &= -frac{1150}{28567} \
B &= frac{3}{14} - frac{-679(1150)}{140(28567)} \
B &= frac{3}{14} + frac{679(1150)}{140(28567)} \
end{align*}
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Hence, I am confident that my solution is wrong.
statistics multivariable-calculus
asked Jan 1 at 1:57
BobBob
902514
add a comment |
Below is a problem that I made up and my attempt to solve it. I am fairly sure that my answer is wrong. Please note that when I write $D_A$ I mean the first partial derivative of $D$ with respect to $A$. I have a similar meaning for $D_B$. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
Problem:
Given the points $(0,1), (1,3), (2,5), (3,7), (4,4)$ find a second order
interpolating polynomial, of the form $f(x) = Ax^2 + Bx + C$, such that the point $(4,4)$ is on the curve and the following is minimized:
$$ (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2 $$
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
begin{align*}
D_A &= 2(A+B+C-3) + 8( 4A + 2B + C - 5) + 18( 9A + 3B + C - 7) \
D_A &= (2 +32+182)A + (2+16+54)B + (2+8+18)C - 6 - 40 - 126 \
D_A &= 216A + 72B + 28C - 172 \
D_A &= 216A + 72B + 28(4-16A-4B) - 172 \
D_A &= -232A - 40B + 112 - 172 \
D_A &= -120A - 12B - 60 \
D_B &= 3(A+B+C-3) + 4(4A+2B+C-5) + 6(9A+3B+C-7) \
D_B &= (3+16+54)A + (3+8+18)B + (3+4+6C) - 9 - 20 - 54 \
D_B &= 73A + 29B + 13C - 83 \
D_B &= 73A + 29B + 13(4 - 16A - 4B) - 83 \
D_B &= 73A + 29B + 52 - 208A - 52B - 83 \
D_B &= -135A - 23B - 31 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
-120A - 12B - 60 &= 0 \
-135A - 23B - 31 &= 0 \
60A + 6B + 30 &= 0 \
30A + 3B + 15 &= 0 \
3B &= -30A - 15 \
B &= -10A - 5 \
135A + 23B + 31 &= 0 \
135A + 23( -10A - 5) + 31 &= 0 \
135A - 230A - 115 + 31 &= 0 \
-95A - 84 &= 0 \
A &= -frac{84}{95} \
end{align*}
Now we solve for $B$ and then $C$.
begin{align*}
B &= -10left( -frac{84}{95} right) - 5 = 10left( frac{84}{95} right) - 5\
B &= frac{840}{95} - 5 \
B &= frac{ 365 } {95} \
B &= frac{73}{9} \
C &= 4 - 16left( -frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + 16left( frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + frac{1344}{95} - frac{292}{9} = frac{ 4(95)(9) + 1344(9) - 292(95)}{9(95)} \
C &= frac{ 15516 - 27740}{9(95)} \
C &= frac{ -12224}{855} \
end{align*}
Hence our interpolating polynomial is:
$$ f(x) = -left( frac{84}{95} right)x^2 + left( frac{73}{9} right) - frac{ 12224}{855} $$
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Based upon the comments of Siong Thye Goh I have updated my post. However, my
answer is still wrong. I am hoping that somebody can tell me where I went wrong
and/ or correct it.
Thanks,
Bob
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
Now we need to find $D_A$ and $D_B$. Observe that $C_A = -16$ and $C_B = -4$.
begin{align*}
D_A &= 2C_A(C-1) + 2( 1+ C_A)(A + B + C- 3) + 2(4 + C_A)( 4A + 2B + C - 5) \
&+ 2(9 + C_A)( 9A + 3B + C - 7) \
D_A &= 2(-16)(C-1) + 2( 1 - 16)(A + B + C- 3) + 2(4 - 16)( 4A + 2B + C - 5) \
&+ 2(9 - 16)( 9A + 3B + C - 7) \
D_A &= -32(C-1) + 2( -15 )(A + B + C- 3) + 2( -12 )( 4A + 2B + C - 5) \
&+ 2(-7)( 9A + 3B + C - 7) \
D_A &= -32(C-1) - 30(A + B + C- 3) - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -32C + 32 - 30A - 30B -30C + 90 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B -62C + 122 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B - 62C + 122 - 96A - 48B - 24C + 120 - 126A - 42B - 14C + 98 \
D_A &= -252A - (30 + 48 + 42)B -(62 + 24 + 14)C + 340 \
D_A &= -252A - 120B - 100C + 340 \
D_A &= -252A - 120B - 100(4 - 16A - 4B) + 340 \
D_A &= -252A - 120B - 400 + 1600A +400B + 340 \
D_A &= 1358A + 280B - 60 \
end{align*}
begin{align*}
D_B &= 2C_B (C - 1) + 3(1 +C_B)(A + B + C- 3) + 2(2 + C_B)(4A + 2B + C - 5) \
&+ 2(3+C_B)( 9A + 3B + C - 7) \
D_B &= -8 (C - 1) + 3(-3)(A + B + C- 3) + 2(-2)(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9(A + B + C- 3) - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9A - 9B - 9C + 27 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17C + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17(4 - 16A - 4B) + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 68 - 16(17)A + 68B + 35 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) - 2( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 16A - 8B - 4C + 20 -18A - 6B - 2C + 14 \
D_B &= -43A + (-9 + 68 -8 - 6)B - 6C - 33 + 20 + 14 \
D_B &= -43A + 45B - 6C + 1 \
D_B &= -43A + 45B - 6( 4 - 16A - 4B ) + 1 \
D_B &= -43A + 45B - 24 + 96A + 24B + 1 \
D_B &= 53A + 69B - 23 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
1358A + 280B - 60 &= 0 \
53A + 69B - 23 &= 0 \
679A + 140B - 30 &= 0 \
140B &= 30 - 679A \
B &= frac{30}{140} - frac{679A}{140} \
B &= frac{3}{14} - frac{679A}{140} \
53A + 69left( frac{3}{14} - frac{679A}{140} right) - 23 &= 0 \
53A + frac{69(3)}{14} - frac{35987A}{140} - 23 &= 0 \
53A - frac{35987A}{140} - frac{115}{14} &= 0 \
-frac{28567A}{140} &= frac{115}{14} \
-28567A &= 1150 \
A &= -frac{1150}{28567} \
B &= frac{3}{14} - frac{-679(1150)}{140(28567)} \
B &= frac{3}{14} + frac{679(1150)}{140(28567)} \
end{align*}
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Hence, I am confident that my solution is wrong.
statistics multivariable-calculus
asked Jan 1 at 1:57
BobBob
902514
Below is a problem that I made up and my attempt to solve it. I am fairly sure that my answer is wrong. Please note that when I write $D_A$ I mean the first partial derivative of $D$ with respect to $A$. I have a similar meaning for $D_B$. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob
Problem:
Given the points $(0,1), (1,3), (2,5), (3,7), (4,4)$ find a second order
interpolating polynomial, of the form $f(x) = Ax^2 + Bx + C$, such that the point $(4,4)$ is on the curve and the following is minimized:
$$ (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2 $$
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
begin{align*}
D_A &= 2(A+B+C-3) + 8( 4A + 2B + C - 5) + 18( 9A + 3B + C - 7) \
D_A &= (2 +32+182)A + (2+16+54)B + (2+8+18)C - 6 - 40 - 126 \
D_A &= 216A + 72B + 28C - 172 \
D_A &= 216A + 72B + 28(4-16A-4B) - 172 \
D_A &= -232A - 40B + 112 - 172 \
D_A &= -120A - 12B - 60 \
D_B &= 3(A+B+C-3) + 4(4A+2B+C-5) + 6(9A+3B+C-7) \
D_B &= (3+16+54)A + (3+8+18)B + (3+4+6C) - 9 - 20 - 54 \
D_B &= 73A + 29B + 13C - 83 \
D_B &= 73A + 29B + 13(4 - 16A - 4B) - 83 \
D_B &= 73A + 29B + 52 - 208A - 52B - 83 \
D_B &= -135A - 23B - 31 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
-120A - 12B - 60 &= 0 \
-135A - 23B - 31 &= 0 \
60A + 6B + 30 &= 0 \
30A + 3B + 15 &= 0 \
3B &= -30A - 15 \
B &= -10A - 5 \
135A + 23B + 31 &= 0 \
135A + 23( -10A - 5) + 31 &= 0 \
135A - 230A - 115 + 31 &= 0 \
-95A - 84 &= 0 \
A &= -frac{84}{95} \
end{align*}
Now we solve for $B$ and then $C$.
begin{align*}
B &= -10left( -frac{84}{95} right) - 5 = 10left( frac{84}{95} right) - 5\
B &= frac{840}{95} - 5 \
B &= frac{ 365 } {95} \
B &= frac{73}{9} \
C &= 4 - 16left( -frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + 16left( frac{84}{95} right) - 4 left(frac{73}{9} right) \
C &= 4 + frac{1344}{95} - frac{292}{9} = frac{ 4(95)(9) + 1344(9) - 292(95)}{9(95)} \
C &= frac{ 15516 - 27740}{9(95)} \
C &= frac{ -12224}{855} \
end{align*}
Hence our interpolating polynomial is:
$$ f(x) = -left( frac{84}{95} right)x^2 + left( frac{73}{9} right) - frac{ 12224}{855} $$
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Based upon the comments of Siong Thye Goh I have updated my post. However, my
answer is still wrong. I am hoping that somebody can tell me where I went wrong
and/ or correct it.
Thanks,
Bob
Answer:
Since the point $(4,4)$ must be on the curve, we have:
begin{align*}
f(4) &= 16A + 4B + C = 4 \
C &= 4 - 16A - 4B \
end{align*}
Let $D = (f(0) - 1)^2 + (f(1) -3)^2 + (f(2) - 5)^2 +( f(3) - 7)^2$. We need to find values for $A$ and
$B$ such that $D$ is minimized.
begin{align*}
(f(0) - 1)^2 &= ( C - 1)^2 \
(f(1) -3)^2 &= (A + B + C - 3)^2 = (A + B - 2)^2 \
(f(2) - 5)^2 &= ( 4A + 2B + C - 5)^2 \
(f(3) - 7)^2 &= (9A + 3B + C - 7)^2 \
D &= (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2 \
end{align*}
Now we need to find $D_A$ and $D_B$. Observe that $C_A = -16$ and $C_B = -4$.
begin{align*}
D_A &= 2C_A(C-1) + 2( 1+ C_A)(A + B + C- 3) + 2(4 + C_A)( 4A + 2B + C - 5) \
&+ 2(9 + C_A)( 9A + 3B + C - 7) \
D_A &= 2(-16)(C-1) + 2( 1 - 16)(A + B + C- 3) + 2(4 - 16)( 4A + 2B + C - 5) \
&+ 2(9 - 16)( 9A + 3B + C - 7) \
D_A &= -32(C-1) + 2( -15 )(A + B + C- 3) + 2( -12 )( 4A + 2B + C - 5) \
&+ 2(-7)( 9A + 3B + C - 7) \
D_A &= -32(C-1) - 30(A + B + C- 3) - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -32C + 32 - 30A - 30B -30C + 90 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B -62C + 122 - 24( 4A + 2B + C - 5) - 14( 9A + 3B + C - 7) \
D_A &= -30A - 30B - 62C + 122 - 96A - 48B - 24C + 120 - 126A - 42B - 14C + 98 \
D_A &= -252A - (30 + 48 + 42)B -(62 + 24 + 14)C + 340 \
D_A &= -252A - 120B - 100C + 340 \
D_A &= -252A - 120B - 100(4 - 16A - 4B) + 340 \
D_A &= -252A - 120B - 400 + 1600A +400B + 340 \
D_A &= 1358A + 280B - 60 \
end{align*}
begin{align*}
D_B &= 2C_B (C - 1) + 3(1 +C_B)(A + B + C- 3) + 2(2 + C_B)(4A + 2B + C - 5) \
&+ 2(3+C_B)( 9A + 3B + C - 7) \
D_B &= -8 (C - 1) + 3(-3)(A + B + C- 3) + 2(-2)(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9(A + B + C- 3) - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -8C + 8 - 9A - 9B - 9C + 27 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17C + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 17(4 - 16A - 4B) + 35 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 68 - 16(17)A + 68B + 35 - 4(4A + 2B + C - 5) \
&+ 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) + 2(3 - 4)( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 4(4A + 2B + C - 5) - 2( 9A + 3B + C - 7) \
D_B &= -9A - 9B - 272A + 68B - 33 - 16A - 8B - 4C + 20 -18A - 6B - 2C + 14 \
D_B &= -43A + (-9 + 68 -8 - 6)B - 6C - 33 + 20 + 14 \
D_B &= -43A + 45B - 6C + 1 \
D_B &= -43A + 45B - 6( 4 - 16A - 4B ) + 1 \
D_B &= -43A + 45B - 24 + 96A + 24B + 1 \
D_B &= 53A + 69B - 23 \
end{align*}
Now we need to solve th following system of two equations:
begin{align*}
1358A + 280B - 60 &= 0 \
53A + 69B - 23 &= 0 \
679A + 140B - 30 &= 0 \
140B &= 30 - 679A \
B &= frac{30}{140} - frac{679A}{140} \
B &= frac{3}{14} - frac{679A}{140} \
53A + 69left( frac{3}{14} - frac{679A}{140} right) - 23 &= 0 \
53A + frac{69(3)}{14} - frac{35987A}{140} - 23 &= 0 \
53A - frac{35987A}{140} - frac{115}{14} &= 0 \
-frac{28567A}{140} &= frac{115}{14} \
-28567A &= 1150 \
A &= -frac{1150}{28567} \
B &= frac{3}{14} - frac{-679(1150)}{140(28567)} \
B &= frac{3}{14} + frac{679(1150)}{140(28567)} \
end{align*}
However, the program R gets:
$$ f(x) = -0.9355 x^2 + 4.8323 x - 0.3613 $$
Hence, I am confident that my solution is wrong.
statistics multivariable-calculus
statistics multivariable-calculus
asked Jan 1 at 1:57
BobBob
902514
asked Jan 1 at 1:57
BobBob
902514
edited Jan 2 at 0:45
Bob
asked Jan 1 at 1:57
BobBob
902514
asked Jan 1 at 1:57
BobBob
902514
asked Jan 1 at 1:57
BobBob
902514
902514
add a comment |
add a comment |
1 Answer
1
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oldest
votes
You are treating $C$ as a constant but it is a function of $A$ and $B$.
We have $C = 4 - 16A - 4B$. That is $C_A=-16, C_B=-4$.
We have
$$D = (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2$$
Hence,
begin{align}D_A &=2C_A(C-1) +2(1+C_A)(A+B+C-3) + 2(4+C_A)( 4A + 2B + C - 5) + 2(9+C_A)( 9A + 3B + C - 7) \
end{align}
Similarly for $D_B$.
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
add a comment |
Your Answer
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1 Answer
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1 Answer
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You are treating $C$ as a constant but it is a function of $A$ and $B$.
We have $C = 4 - 16A - 4B$. That is $C_A=-16, C_B=-4$.
We have
$$D = (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2$$
Hence,
begin{align}D_A &=2C_A(C-1) +2(1+C_A)(A+B+C-3) + 2(4+C_A)( 4A + 2B + C - 5) + 2(9+C_A)( 9A + 3B + C - 7) \
end{align}
Similarly for $D_B$.
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
add a comment |
You are treating $C$ as a constant but it is a function of $A$ and $B$.
We have $C = 4 - 16A - 4B$. That is $C_A=-16, C_B=-4$.
We have
$$D = (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2$$
Hence,
begin{align}D_A &=2C_A(C-1) +2(1+C_A)(A+B+C-3) + 2(4+C_A)( 4A + 2B + C - 5) + 2(9+C_A)( 9A + 3B + C - 7) \
end{align}
Similarly for $D_B$.
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
add a comment |
You are treating $C$ as a constant but it is a function of $A$ and $B$.
We have $C = 4 - 16A - 4B$. That is $C_A=-16, C_B=-4$.
We have
$$D = (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2$$
Hence,
begin{align}D_A &=2C_A(C-1) +2(1+C_A)(A+B+C-3) + 2(4+C_A)( 4A + 2B + C - 5) + 2(9+C_A)( 9A + 3B + C - 7) \
end{align}
Similarly for $D_B$.
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
You are treating $C$ as a constant but it is a function of $A$ and $B$.
We have $C = 4 - 16A - 4B$. That is $C_A=-16, C_B=-4$.
We have
$$D = (C-1)^2 + ( A+B+C-3)^2 + (4A+2B+C-5)^2 + (9A+3B+C -7)^2$$
Hence,
begin{align}D_A &=2C_A(C-1) +2(1+C_A)(A+B+C-3) + 2(4+C_A)( 4A + 2B + C - 5) + 2(9+C_A)( 9A + 3B + C - 7) \
end{align}
Similarly for $D_B$.
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 1 at 3:47
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
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