Assignment of Mathematics Solved PDF

Running Head: MATH
MATH
Name
Institute of Institute of Affiliation
Date

MATH 2
Question 1
General formula
xn+1=xn − xn
1
7
1
7 xn
−6
7
= xn−7 xn
= - 6xn
=4.33333 ̄3
x2= 2.94214333
x3= 2.07695292
x4= 1.61645303
x5= 1.46034889
x6= 1.44247296
x7= 1.4422496
x8= 1.44224957
Question 2
S ( t )= M
1+r ( t−t0 )2
a. The rate ....
( 1+ r ( t−t0 ) 2
) S ( t )=M

MATH 3
S ( t )+ S ( t ) r ( t−t0 )2= M
r = M −s ( t )
S ( t ) ( t−t0 )2
b. What time is spread highest,
For spread to be maximum,
r ( t−t0 )2 → 0
1.5( t-1)2 −→ 0
Opening up the bracket
t2−2 t+1=0
t= 1
c. Point of inflection
T(t-t0) -to(t-t0)
t2−2 tt 0−t o2
S ( t )= M
1+r ( t2−2 tt 0−t o2 )
S' ( t ) =−2 Mr(t−t 0)
(r ( t−t 0 ) 2 +1)
S' ' ( t )= 8 m r2 ( t −t 0 )2
(r ( t−t0 )2 +1 )3 − 2 mr
( r ( t−t0 )2 +1 )2 (where m, r and t0 are constants)
= 8∗2500∗1.52
( t−t0 )2
(1.5 ( t−1 )2 +1 )3 − 2∗2500∗1.5
(1.5∗( t−1 )2 +1 )2

MATH 4
Solving the above as follows
0 = 4∗1.5 ( t−1 ) 2
1.5 ( t−1 ) 2+1 −1
= 6 t2−12t +6
1.5t2−3 t+ 1.5+1 −1
=6 t2−12 t+6−1.5 t2 +3 t−2.5
= 4.5t2−9 t+ 3.5
So t= 1.471. t= 0.285
Since at t= 1.471. t= 0.285, S' ' ( t )=0, they are point of inflection
Question 3
Interpolating the above points using the LaGrange to obtain the equations
G ( t ) =40. x−0
40−30 . x−10
40−135 . x−15
40−200 . x −20
40−180 . x−25
40−70 . x−30
40−150 +
135. x−0
135−30 . x−5
135−40 . x−15
135−200 . x −20
135−180 . x−25
135−70 . x−30
135−150 +
200. x−0
200−30 . x −5
200−40 . x−10
200−135 . x−20
135−180 . x −25
135−70 . x−30
135−150
180. x
180−30 . x−5
180−40 . x−10
180−135 . x−15
180−200 . x−25
180−70 . x−30
180−150
70. x
70−30 . x−5
70−40 . x−10
70−135 . x−15
70−200 . x−20
70−180 . x−30
70−150
150. x
150−30 . ( x−5 )
150−40 . x−10
150−135 . x −15
150−200 . x−20
150−180 . x−25
150−70