Advanced Mathematics for Scientists: Coursework 1311 Assignment

This document provides a comprehensive solution to an advanced mathematics assignment, addressing several key concepts. The solution begins by determining critical points for a given function, classifying them as maxima, minima, or saddle points using partial derivatives. It then explores vector calculus, calculating the divergence of a vector field and identifying locations of convergence. The assignment further delves into the curl of a vector field, determining points where the curl is zero. Finally, the solution analyzes a periodic function, graphing it over a specified interval, deriving its Fourier series representation, and evaluating the function at a specific point. This assignment is a great resource for students studying advanced mathematics and calculus.

1)
Partial derivatives
f x=−2sin 2 x=0=¿ x= π
2 ,− π
2 , 0
f y=2 ( y−1 ) =0=¿ y =1
The critical points are
( π
2 , 1) , ( −π
2 , 1 ) , ( 0,1 )
f xx =−4 cos 2 x
f yy=2
f xy =0
At ( π
2 , 1),
f xx =4
AC−B2 >0
Similarly at (−π
2 , 1 ),
f xx =4
AC−B2 >0
Similarly at (0, 1)
f xx =−4
AC−B2 <0
( π
2 , 1 ) , ( −π
2 , 1 ) arethe local minimum points
( 0,1 ) is a saddle point
2)
∇ .⃗ f = ∂ f (x , y , z )
∂ x i. i+ ∂ f (x , y , z)
∂ y j . j+ ∂ f (x , y , z)
∂ z k . k
∇ .⃗ f =x2 +x+ ( 4 x +6 )=0=¿ x2 +5 x+6=0
( x +5 ) ( x+1 ) =0=¿ x=−5∨−1
3)

∇ ×⃗ f =( ∂
∂ x i+ ∂
∂ y j+ ∂
∂ z k )×(3 i+5 xz j+ y x2 k )
∇ ×⃗ f = ( x2−5 x ) i+ ( 0−2 xy ) j+ ( 5 z−0 ) k
∇ ×⃗ f = ( x2−5 x ) i−2 xy j+5 z k
( x2−x )=0=¿ x=0∨5
2 xy=¿ x∨ y=0
5 z=0=¿ z=0
The points where the curl are zero are
(0, y, 0) or (5, 0, 0) or (0, 0, 0)
4)
The x-axis represent the x values from −8 ≤ x ≤ 8 and the y axis represent the corresponding
f ( x )
f ( x ) = { 0−2< x< 0
2− x 0< x <2
Since it has a period of 4
f ( x ) = { 0−6< x ←4
2− x−4 < x←2
f ( x )= { 0−10<x ←8
2− x−8< x ←6
f ( x ) = { 0 2< x <4
2− x 4 < x <6
f ( x )= { 0 6< x <8
2− x 8< x <10
But since it is given that we can plot the graph between −8 ≤ x ≤ 8
So, the graph looks like,

a0= 1
4 ∫
0
2
( 2−x ) cos ( 0 πx
2 )dx= 1
4 ∫
0
2
( 2−x ) dx= 1
4 ( 4−2 )= 1
2
an= 1
2∫
0
2
( 2−x ) cos ( nπx
2 )dx= 1
2 ∫
0
2
2 cos ( nπx
2 )dx− 1
2 ∫
0
2
xcos ( nπx
2 )dx
¿ 0−1
2 ∫
0
2
xcos ( nπx
2 )dx=−1
2 ( 2
nπ )2
¿
bn=1
2∫
0
2
( 2−x ) sin ( nπx
2 )dx=1
2 ∫
0
2
2sin ( nπx
2 )dx −1
2 ∫
0
2
xsin ( nπx
2 )dx
¿ ¿
¿ ( 1− ( −1 ) n
( 2
nπ +1 ) )
f ( x ) 1
2 +∑
n=1
∞
[ −2
n2 π2 ¿ ( ( −1 ) n −1 ) cos nπx
2 +
( 1− ( −1 ) n
( 2
nπ + 1 ) ) sin nπx
2 ]¿
f ( 2 ) =1
2 +∑
n=1
∞
[ −2
n2 π 2 ¿ ( ( −1 ) n −1 ) cos 2 nπ
2 +
( 1− ( −1 ) n
( 2
nπ +1 ) ) sin 2 nπ
2 ]¿
f ( 2 ) =1
2 +∑
n=1
∞
[ −2
n2 π 2 ¿ ( ( −1 ) n −1 ) cos nπ ]¿

f ( 2 ) =1
2 +∑
n=1
∞
[ −2
n2 π 2 ¿ ( (−1 )n −1 ) (−1 )n ]¿