MHB Least squares method : approximation of a cubic polynomial

[mathmari]

Hey!

I want to determine an approximation of a cubic polynomial that has at the points $$x_0=-2, \ x_1=-1, \ x_2=0 , \ x_3=3, \ x_4=3.5$$ the values $$y_0=-33, \ y_1=-20, \ y_2=-20.1, \ y_3=-4.3 , \ y_4=32.5$$ using the least squares method.

So we are looking for a cubic polynomial $p(x)$ such that $$\sum_{i=0}^4\left (p(x_i)-y_i\right )^2$$ is minimal, right? (Wondering) Let $p(x)=a_3x^3+a_2x^2+a_1x+a_0$. Then we get the following sum:
\begin{align*}\sum_{i=0}^4\left (p(x_i)-y_i\right )^2=&\left (-8a_3+4a_2-2a_1+a_0+33\right )^2+\left (-a_3+a_2-a_1+a_0+20\right )^2+\left (a_0+20.1\right )^2 \\ & +\left (27a_3+9a_2+3a_1+a_0+4.3\right )^2 +\left (42.875a_3+12.25a_2+3.5a_1+a_0-32.5\right )^2\end{align*}

Now we calculate the partial derivatives in respect to each $a_i$ and then we set these equal to $0$. In that we get a system of equations.

Let this sum be $S$.

We get the following partial derivatives:

\begin{align*}\frac{\partial{S}}{\partial{a_0}}=&2\left (-8a_3+4a_2-2a_1+a_0+33\right )+2\left (-a_3+a_2-a_1+a_0+20\right )\\ & +2\left (a_0+20.1\right ) +2\left (27a_3+9a_2+3a_1+a_0+4.3\right ) +2\left (42.875a_3+12.25a_2+3.5a_1+a_0-32.5\right )\\ = &89.8 + 10 a_0 + 7 a_1+ 52.5 a_2 + 121.75 a_3\end{align*}

\begin{align*}\frac{\partial{S}}{\partial{a_1}}=&2\left (-8a_3+4a_2-2a_1+a_0+33\right )\cdot 2+2\left (-a_3+a_2-a_1+a_0+20\right )\cdot (-1) \\ & +2\left (27a_3+9a_2+3a_1+a_0+4.3\right )\cdot 3 +2\left (42.875a_3+12.25a_2+3.5a_1+a_0-32.5\right )\cdot 3.5 \\ = & -109.7 + 15 a_0 + 36.5 a_1 + 153.75 a_2 + 432.125 a_3\end{align*}

\begin{align*}\frac{\partial{S}}{\partial{a_2}}=&2\left (-8a_3+4a_2-2a_1+a_0+33\right )\cdot 4+2\left (-a_3+a_2-a_1+a_0+20\right ) \\ & +2\left (27a_3+9a_2+3a_1+a_0+4.3\right )\cdot 9 +2\left (42.875a_3+12.25a_2+3.5a_1+a_0-32.5\right )\cdot 12.25\\ = & -414.85 + 52.5 a_0 + 121.75 a_1 + 496.125 a_2 + 1470.44 a_3 \end{align*}

\begin{align*}\frac{\partial{S}}{\partial{a_3}}=&2\left (-8a_3+4a_2-2a_1+a_0+33\right )\cdot (-8)+2\left (-a_3+a_2-a_1+a_0+20\right )\cdot (-1) \\ & +2\left (27a_3+9a_2+3a_1+a_0+4.3\right )\cdot 27 +2\left (42.875a_3+12.25a_2+3.5a_1+a_0-32.5\right )\cdot 42.875\\ = & -3122.68 + 121.75 a_0 + 496.125 a_1 + 1470.44 a_2 + 5264.53 a_3\end{align*}

So, we get the system:
\begin{align*}& \ \ \ \ 89.8 + 10 a_0 + 7 a_1+ 52.5 a_2 + 121.75 a_3=0 \\ &-109.7 + 15 a_0 + 36.5 a_1 + 153.75 a_2 + 432.125 a_3=0 \\ &-414.85 + 52.5 a_0 + 121.75 a_1 + 496.125 a_2 + 1470.44 a_3=0 \\ &-3122.68 + 121.75 a_0 + 496.125 a_1 + 1470.44 a_2 + 5264.53 a_3=0\end{align*}

Then we get the solution: \begin{equation*}a_0\approx -21.6611, \ a_1\approx -7.08875, \ a_2\approx -2.0.6057, \ a_3\approx 2.33768\end{equation*}

The cubic polynomial that we are looking for is \begin{equation*}p(x)=2.33768x^3-2.0.6057x^2-7.08875x-21.6611 \end{equation*}
Doing the above in an other way I got an other result, but I don't know why and which is the correct one. (Wondering)

Using the least square method and the normal equation we get that the solution that we are looking for is the vector $\vec{a}$ such that $A \vec{a}=b \Rightarrow A^TA\vec{a}=A^Tb$.

So we get the following:
\begin{equation*}
\begin{bmatrix}
5 & 3.5 & 26.25 & 60.875 \\
3.5 & 26.25 & 60.875 & 248.0625 \\
26.25 & 60.875 & 248.0625 & 735.21875 \\
60.875 & 248.0625 & 735.21875 & 2632.265625
\end{bmatrix}
\begin{bmatrix}
a_0 \\
a_1 \\
a_2 \\
a_3 \\
\end{bmatrix}=
\begin{bmatrix}
-44.9 \\
186 \\
207.425 \\
1561.3375 \\
\end{bmatrix}
\end{equation*}

Solving this one we get the polynomial $$p(x)=-23.0087-12.2424x-2.7837x^2+3.0565x^3$$ This is not the same polynomial as the one that we get at the first method. And for the second polynomial the condition $p(0)=-20.1$ is not true. (Wondering)

 

[I like Serena]

Hey mathmari!

I checked the solution $a=(A^T A)^{-1} A^T y$, and found the same solution that you did.
If I graph it, I get:
View attachment 8788
https://octave-online.net/

Spoiler

xx=[-2 -1 0 3 3.5]';
yy=[-33 -20 -20.1 -4.3 32.5]';
A=[ones(5,1) xx xx.^2 xx.^3];
ATA = A'*A
ATy = A'*yy
a = ATA\ATy
plot(xx,yy)
hold on
plot(xx, A*a)

That looks pretty close doesn't it, even though $p(0)$ is slightly off? (Wondering)

And yes, your solution with the derivatives set to zero should have the same result. (Worried)EDIT: Now I've also plotted it with the $a$ you found with the derivatives:
View attachment 8789

Spoiler

a_diff = [-21.6611, -7.08875, -2.06057, 2.33768]'
plot(xx, A*a_diff)

Hmm... that's not the same is it?
The first one does look better though.
Perhaps there is a mistake? (Wondering)EDIT 2:
If I calculate the residu $\|Aa-y\|$ in both cases, I get:

>> norm(A*a-yy)
ans =  5.2292
>> norm(A*a_diff-yy)
ans =  7.5340

In other words, the solution with the derivatives did not find the lowest possible residu.
It confirms that there must be a calculation mistake. (Worried)

 

[mathmari]

I checked my calculations again and I saw that I had a wrong sign. Now I get \begin{equation*}p(x)=3.05677x^3-2.78419x^2-12.2439x-23.0084 \end{equation*} which is approximately the same as the second one. Do we have have to get the exactly same polynomial or is it correct that we get approximately the same one? But what about the value of the polynomial at $x=0$ ? Shouldn't it be as the given value? (Wondering)

 

[I like Serena]

I checked my calculations again and I saw that I had a wrong sign. Now I get \begin{equation*}p(x)=3.05677x^3-2.78419x^2-12.2439x-23.0084 \end{equation*} which is approximately the same as the second one. Do we have have to get the exactly same polynomial or is it correct that we get approximately the same one? But what about the value of the polynomial at $x=0$ ? Shouldn't it be as the given value?

Good!
The first 4 significant digits of each coefficient are the same.
I think it means that intermediate results of the numbers have been rounded to perhaps 5 or 6 significant digits.
The results should be exactly the same, but of cause they are sensitive to rounding.
Anyway, they are close enough that I believe that the calculations are correct now. (Thinking)And no, the value of polynomial is not expected to be the same as the given value at $x=0$.
It is an approximation after all.
We can only fit a cubic polynomial perfectly through 4 points.
Since 5 points were given, it is normal that every point is slightly off from the given value.
The squared sum of the differences is merely as low as possible. (Nerd)