399: For 'Independent Variable'-Value Pairs Data, Choosing Origin-Passing Approximating Line with Least Value Difference Squares Sum Equals Projecting Values Vector to Independent Variables Vector Line

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A description/proof of that for 'independent variable'-value pairs data, choosing origin-passing approximating line with least value difference squares sum equals projecting values vector to independent variables vector line

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of projection of vector to line.

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Target Context

• The reader will have a description and a proof of the proposition that for any 'independent variable'-value pairs data, choosing the origin-passing approximating line with the least value difference squares sum equals projecting the values vector to the independent variables vector line in the ${\mathbb{R}}^n$ space.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any 'independent variable'-value pairs data, $(x_1,y_1),(x_2,y_2),...,(x_n,y_n)$, choosing the origin-passing approximating line, $y=sx$, such that the value difference squares sum, $S:=\sum _i(y_i-s{x}_i{)}^2$, is the least equals projecting the values vector, $(y_1,y_2,...,y_n)$, to the independent variables vector line, $t(x_1,x_2,...,x_n)$ in the ${\mathbb{R}}^n$ space, what which means is that $s$ equals the length of the projection.

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2: Proof

Let us determine $s$. Let us take $\frac{dS}{ds}=0$. $\frac{dS}{ds}=\sum _i(-2(y_i-s{x}_i)x_i)=0$. $\sum _i(y_i{x}_i)=s\sum _i{x}_i^2$. $s=(\sum _i(y_i{x}_i))(\sum _i{x}_i^2{)}^{-1}$.

Let us take the projection. The inner product of $(x_1,x_2,...,x_n)$ and $(y_1,y_2,...,y_n)$ is $\sum _i(y_i{x}_i)=(\sum _i{x}_i^2{)}^{2^{-1}}(\sum _i{y}_i^2{)}^{2^{-1}}cos\theta$ where $\theta$ is the angle between the vectors, and the length of the projection is $l=(\sum _i{y}_i^2{)}^{2^{-1}}cos\theta =(\sum _i{y}_i^2{)}^{2^{-1}}\sum _i(y_i{x}_i)((\sum _i{x}_i^2{)}^{2^{-1}}{)}^{-1}((\sum _i{y}_i^2{)}^{2^{-1}}{)}^{-1}=\sum _i(y_i{x}_i)((\sum _i{x}_i^2{)}^{2^{-1}}{)}^{-1}$.

So, $l=s$, which is what this proposition means.

The projection is $(\sum _i(y_i{x}_i))((\sum _i{x}_i^2{)}^{2^{-1}}{)}^{-1}((\sum _i{x}_i^2{)}^{2^{-1}}{)}^{-1}(x_1,x_2,...,x_n)=(\sum _i(y_i{x}_i))((\sum _i{x}_i^2{)}^{2^{-1}}{)}^{-2}(x_1,x_2,...,x_n)=(\sum _i(y_i{x}_i))(\sum _i{x}_i^2{)}^{-1}(x_1,x_2,...,x_n)$.

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References

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