# [SOLVED]Method of least square: initial position & velocity

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

A point is moving linearly with constant velocity $v$ and the movement is $x=a+vt$.
The below information is given:

Find the initial position $a$ and the velocity using the method of least square.

Could you give me a hint how we use this method here? Couldn't we use the data of the matrix and get a $2\times 2$ linear system?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
A point is moving linearly with constant velocity $v$ and the movement is $x=a+vt$.
The below information is given:

Find the initial position $a$ and the velocity using the method of least square.

Could you give me a hint how we use this method here?
Hey mathmari !!

The method of least squares is that we minimize the sum of the squared deviations $\sum\limits_{i=1}^n (x_i - x(t_i))^2$, isn't it?

We should be able to find formulas for it somewhere, either in matrix form or in formula form.
Or we can derive those formulas ourselves.
Or we can use a tool like Excel to do it for us.

Couldn't we use the data of the matrix and get a $2\times 2$ linear system?
Which $2\times 2$ linear system do you mean?

#### mathmari

##### Well-known member
MHB Site Helper
The method of least squares is that we minimize the sum of the squared deviations $\sum\limits_{i=1}^n (x_i - x(t_i))^2$, isn't it?

We should be able to find formulas for it somewhere, either in matrix form or in formula form.
Or we can derive those formulas ourselves.
Or we can use a tool like Excel to do it for us.
Ah I think now I have an idea how we have to apply the method in this case.

From the matrix we have the exact values of $x_i$'s and from the formula of the movement we get an approximation $x(t_i)$, or not?

So we get \begin{align*}\sum\limits_{i=1}^5 (x_i - x(t_i))^2&=(x_1 - x(t_1))^2+(x_2 - x(t_2))^2+(x_3 - x(t_3))^2+(x_4 - x(t_4))^2+(x_5 - x(t_5))^2 \\ & =(-11 - [a-2v])^2+(-5 - [a-v])^2+(1 - a)^2+(5 - [a+v])^2+(10 - [a+2v])^2\end{align*}

#### HallsofIvy

##### Well-known member
MHB Math Helper
What do you mean by "the initial position"? I would think that would be x(0) but the table itself tells you that x(0)= 1. There is no calculation at all required for that!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ah I think now I have an idea how we have to apply the method in this case.

From the matrix we have the exact values of $x_i$'s and from the formula of the movement we get an approximation $x(t_i)$, or not?

So we get \begin{align*}\sum\limits_{i=1}^5 (x_i - x(t_i))^2&=(x_1 - x(t_1))^2+(x_2 - x(t_2))^2+(x_3 - x(t_3))^2+(x_4 - x(t_4))^2+(x_5 - x(t_5))^2 \\ & =(-11 - [a-2v])^2+(-5 - [a-v])^2+(1 - a)^2+(5 - [a+v])^2+(10 - [a+2v])^2\end{align*}
Yep.
We can minimize that with respect to $a$ and $v$ can't we?

What do you mean by "the initial position"? I would think that would be x(0) but the table itself tells you that x(0)= 1. There is no calculation at all required for that!
The value of $x(0)$ given in the table is a measurement.
The approximation of $a$ is the approximation of $x(0)$ based on a linear least square interpolation.
The question asks for $a$, which should be a more accurate version of the position at time $t=0$.

#### mathmari

##### Well-known member
MHB Site Helper
Yep.
We can minimize that with respect to $a$ and $v$ can't we?
For that we use the gradient, don't we?

\begin{align*}S(a,v)=&(-11 - [a-2v])^2+(-5 - [a-v])^2+(1 - a)^2+(5 - [a+v])^2+(10 - [a+2v])^2 \\ =&11^2 +22[a-2v]+ [a-2v]^2+5^2 +10[a-v]+ [a-v]^2+1 -2a+ a^2+5^2 -10[a+v]+ [a+v]^2\\ & +10^2 -20[a+2v]+ [a+2v]^2 \\ =&121 +22a-44v+ a^2-4av+4v^2+25 +10a-10v+ a^2-2av+v^2+1 -2a+ a^2\\ &+25 -10a-10v+ a^2+2av+v^2+100 -20a-40v+ a^2+4av+4v^2 \\ =& 5a^2+10v^2-104v+272 \end{align*}

The partial derivatives are:
\begin{align*} & \frac{\partial{S}}{\partial{a}}= 10a \\ & \frac{\partial{S}}{\partial{v}} = 20v-104 \end{align*}

We set these partial derivatives equal to zero and we get \begin{align*} & \frac{\partial{S}}{\partial{a}}= 0 \Rightarrow 10a=0 \Rightarrow a=0 \\ & \frac{\partial{S}}{\partial{v}} = 0 \Rightarrow 20v-104=0 \Rightarrow v=\frac{104}{20} \Rightarrow v=5.2 \end{align*}

Is everything correct?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We set these partial derivatives equal to zero and we get \begin{align*} & \frac{\partial{S}}{\partial{a}}= 0 \Rightarrow 10a=0 \Rightarrow a=0 \\ & \frac{\partial{S}}{\partial{v}} = 0 \Rightarrow 20v-104=0 \Rightarrow v=\frac{104}{20} \Rightarrow v=5.2 \end{align*}

Is everything correct?
The answer is correct so I think the intermediate steps are correct as well.

#### mathmari

##### Well-known member
MHB Site Helper
The answer is correct so I think the intermediate steps are correct as well.
How did you get the correct answer? Is there also an other method we can use? Or is there an online tool?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How did you get the correct answer? Is there also an other method we can use? Or is there an online tool?
I entered the data in Excel and used the [M]LINEST[/M] array function to get the result.

#### mathmari

##### Well-known member
MHB Site Helper
I entered the data in Excel and used the [M]LINEST[/M] array function to get the result.
Ahh! Interesting! Thank you!!