# Two Mathematical Analysis questions

*label*Mathematics

*timer*Asked: Feb 3rd, 2019

*account_balance_wallet*$10

### Question Description

Hello

Here are only two simple questions in analysis marked in red in the pictures attached below,

Please answer the questions showing all the solution details and steps so i can understand every single step,

I prefer handwriting solutions,

Thanks in advance,

### Unformatted Attachment Preview

## Tutor Answer

The solutions are ready. I explained every step but feel free to ask for more explanations at any point.I believe your teacher knows your handwriting so you cannot use mine directly. Also I believe equation editor writes symbols better than me:)docx and pdf files are identical

oh forgot the attachments

12. The equation of the least squares line is π¦ = ππ₯ + π.

(a) The point of this line which is directly above or below a point (π₯π , π¦π ) has the same xcoordinate, π₯π . Therefore its y-coordinate must satisfy the equation of the line, i.e. its ycoordinate is πππ + π.

(b) the squared distance between the points (π₯π , π¦π ) and (π₯π , ππ₯π + π) is

2

π

(π₯π β π₯π )2 + (π¦π β (ππ₯π + π)) = (ππ β (πππ + π)) .

π

(c) The sum of these squares is obviously π(π, π) = βππ=π(ππ β (πππ + π)) .

Its partial derivatives are:

π

ππ(π, π)

π

2

=β

[(π¦π β (ππ₯π + π)) ] =

ππ

π=1 ππ

π

π

π

= β 2(π¦π β (ππ₯π + π)) [(π¦π β (ππ₯π + π))] = β 2(π¦π β (ππ₯π + π)) β (β1) =

ππ

π=1

π=1

π

(ππ β (πππ + π)),

= βπ β

π=π

π

ππ(π, π)

π

2

=β

[(π¦π β (ππ₯π + π)) ] =

ππ

π=1 ππ

π

π

π

= β 2(π¦π β (ππ₯π + π))

[(π¦π β (ππ₯π + π))] = β 2(π¦π β (ππ₯π + π)) β (βπ₯π ) =

ππ

π=1

π=1

π

= βπ β

(ππ β (π...

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