# Two Mathematical Analysis questions

Anonymous
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,

a80bf831_f32c_4438_9ba1_e0a99c475722.jpeg
ebde5d6d_b74a_4102_8d3a_12d3c6c67b7f.jpeg

### Unformatted Attachment Preview

Borys_S
School: University of Maryland

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¦ = ππ₯ + 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 πππ + π.
(b) the squared distance between the points (π₯π , π¦π ) and (π₯π , ππ₯π + π) is
2

π

(π₯π β π₯π )2 + (π¦π β (ππ₯π + π)) = (ππ β (πππ + π)) .
π

(c) The sum of these squares is obviously π(π, π) = βππ=π(ππ β (πππ + π)) .
Its partial derivatives are:
π
ππ(π, π)
π
2
=β
[(π¦π β (ππ₯π + π)) ] =
ππ
π=1 ππ
π
π
π
= β 2(π¦π β (ππ₯π + π)) [(π¦π β (ππ₯π + π))] = β 2(π¦π β (ππ₯π + π)) β (β1) =
ππ
π=1
π=1
π

(ππ β (πππ + π)),

= βπ β

π=π

π
ππ(π, π)
π
2
=β
[(π¦π β (ππ₯π + π)) ] =
ππ
π=1 ππ
π
π
π
= β 2(π¦π β (ππ₯π + π))
[(π¦π β (ππ₯π + π))] = β 2(π¦π β (ππ₯π + π)) β (βπ₯π ) =
ππ
π=1
π=1
π

= βπ β

(ππ β (π...

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Anonymous
awesome work thanks

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