Mathematics Assignment: Sequence, Series, Complex Numbers Solutions

This document presents a comprehensive solution to a mathematics homework assignment. It covers a range of topics including sequences and series, where problems involving geometric progressions, recurrence systems, and determining the nth term are addressed. The solution also delves into complex numbers, tackling problems related to finding roots, solving quadratic equations, and applying De Moivre's theorem. Furthermore, the assignment explores Taylor series, including finding the series for given functions and determining the intervals of validity. The solutions are detailed, providing step-by-step explanations and calculations to facilitate understanding of the concepts and problem-solving techniques. Graphical representations are included where appropriate to aid visualization of the functions and their approximations.

Ans.1. (a) Here the first term is a= 2.6 and Common Ratio (r) =?
We have, ar= 4.68, ar^2 =8.424, ar^3= 15.1632
r=ar/a= 4.68/2.6= 1.8
r=ar^2/ar=8.424/4.68=1.8
r=ar^3/ar^2=15.1632/8.424=1.8
Therefore the First term is 2.6 and Common Ratio= 1.8
The recurrence System is ( a∗r(n−1) )=¿2.6*(1.8)^(n-1) ; n=1,…..
(b) The closed form for this sequence is 2.6*(1.8)^(n-1) ; n=1,…..
(c) Tenth Term= a*r^(10-1) = 2.6*(1.8)^(9)=515.734
(d)
Let the number be ‘n’.
a*r^(n-1) ≤ 39000
or, 2.6*(1.8)^(n-1)≤39000
or, (1.8)^(n-1)≤15000
or, (n-1)ln(1.8)≤ln(15000)
or, n-1≤16.35
or, n≤17.35
Since ‘n’ should be a natural number, so it should be n=17
Ans.2. The sequence converges to -26
Reason:
lim
n → ∞
{52∗( 0.91 )n−26 }
¿ lim
n → ∞
{52∗( 0.91 )n }−26
=0-26 =-26
Ans.3. It is an Arithmetic Progression
Here 1st Term a= 6 (From Feb 2016)
And Common Difference = 4
Number of terms = 14
Therefore, the required number of people=
12+ ∑
n =1
14
{6+ ( n−1 )∗4 }

= 12 + ∑
n =1
14
{ 6 } + ∑
n =1
14
{ ( n−1 )∗4 }
= 12+6*14 + ∑
n =1
14
{ n∗4 } - ∑
n =1
14
{ 4 }
=12 +14*6+4*14*15/2-14*4
=460
Ans.4. ∑
n=11
38
{ n3 −3∗n2 }
= ∑
n =1
38
{n3−3∗n2 } - ∑
n =1
11
{n3−3∗n2 }
= ∑
n =1
38
{n3 } - ∑
n =1
38
{3∗n2 } - ∑
n =1
11
{n3 } + ∑
n =1
11
{3∗n2 }
=(38*39/2)^2 -3*38*39*77/6 – (11*12/2)^2 + 3*11*12*23/6
=549081-57057- 4356+ 1518
=489186
Ans.5. We have the 8th term as
14C8 *((3*a^2)^8)*(9*a^3)^(-6)
Therefore the required coefficient is
14C8*3^(-4)
Ans.8.(a) We have f(x)=( 1
4 + x)^(-1/4)
Therefore differentiating both sides w.r.t. x we get,
f (x′ )=(-1/4)( 1
4 + x)^(-5/4)
or, g(x)=(-4)f′(x)= ( 1
4 + x)^(-5/4)
Now, we are having
f(x)=√2 -√2x+ 5 √ 2
2 x^2 - 15 √ 2
2 x^3 +….
Differentiating, both sides w.r.t. x and multiplying by -4 we get,
g(x)=(-4)f′(x)= 4√2+20√2x -+90√2 x^2+…. Which is the required Taylor Series for g(x)= ( 1
4 + x)^(-5/4).

(b)
We have f(x)=( 1
4 + x)^(-1/4)
Therefore integrating both sides w.r.t. x we get,
F(x)=(4/3)( 1
4 + x)^(3/4)
or, h(x)=F(x)=(4/3) ( 1
4 + x)^(3/4)
Now, we are having
f(x)=√2 -√2x+ 5 √ 2
2 x^2 - 15 √ 2
2 x^3 +….
Integrating, both sides w.r.t. x and multiplying by ¾ we get,
h(x)=F(x)= ¾ √2x – 3/8 √2 x^2 + 5/3 √2 x^3 – 15/8 √2 x^4 , which is the required Taylor Series for
h(x)= ( 1
4 + x)^(3/4).
Ans.9.
The required equation is:
z^2 – (3/4 +2i +3/4 -2i)z + (3/4 + 2i)(3/4-2i)=0
or, z^2 –(6/4)z + (73/16)=0
Ans.10.
z^4 = -2√3-2i
= -2(√3 + i)
=-4( √ 3
2 + i
2)
=(2i)^2 (cos(π/6)+sin(π/6))
=((2i)^2)*(e^(iπ/6))
Or, z^2 = ±(2*i)*(e^(iπ/12))
Or, z= ±(√(2*i))*(e^(iπ/24))

Ans.6. (a)
(b)
(d) q(x) is better approximation than p(x) as q(x) gives finer approximations than p(x).
Ans. No. 6(c) is done at the end
Ans.7. (a)
(i)
f(0)= 1 f ′(0)=0
f′′(0)=- 5 *log(e) f′′′(0)=0
f4(0)= ((5* log (e))^2)/2 f5(0)=0
f6(0)= - ((5* log (e))^3)/6 f7(0)=0
f8(0) = ((5* log (e))^4)/24

Therefore, the required Taylor Series is:
T(x)= 1 - 5 *log(e) * x^2 + ((5* log (e))^2)/2 * x^4 - ((5* log (e))^3)/6 * x^6 + ((5* log (e))^4)/24 *
x^8+ ……
(ii) The interval of validity for the series is
|-5*x^2|≤1
or, |x^2|≤1/5
or, - 1
√ 5 ≤ x ≤ 1
√ 5
(b) (i) 5
2 x +3 = 5
2 x +5−2 = 5
2(x−1)+5 = 1
2(x−1)/5+1 =
1
1+ 2 ( x−1 )
5
(ii) f(1)= 1
f′(1)= -(2/5)
f′′(1)= (4/25)
f′′′(1)= -(8/125)
Therefore the required Taylor Polynomial is
T(x)= 1 -(2/5) * (x-1) + (4/25) * (x-1)^2 - (8/125) * (x-1)^3 + ……
(iii)
The interval of validity is:
| 2(x−1)
5 |≤1
or, |x-1|≤ 5
2
or, -−5
2 ≤x-1≤ 5
2
or, -−3
2 ≤x≤ 7
2
Ans.11.
L.H.S.=

(sinΘ)^4= ( 1
2∗i )^4 (e^(i*Θ)- e^(-i*Θ) )^4
= 1
16 ( (e^(2*i* Θ))^2+ (e^(-2*i* Θ))^2 + 2^2 – 2* (e^(2*i* Θ))*2 – 2* (e^(-2*i* Θ)) +6)
= 1
16( (e^(4*i* Θ))+ (e^(-4*i* Θ) +6 – 4 * (e^(2*i* Θ)) – 4* (e^(-2*i* Θ)))
= 1
16∗{ei∗4∗θ +e−i∗4∗θ−4∗ei∗2∗θ−4∗e−i∗2∗θ +6 }
R.H.S.= 1
8 (cos(4*Θ)- 4*cos(2*Θ)+ 3)
= 1
8 {1
2 ( ei∗4∗Θ +e−i∗4∗Θ )− 4
2 ( ei∗2∗Θ+ e−i∗2∗Θ ) +3 }
=
1
8 ∗1
2 {ei∗4∗θ +e−i∗4∗θ−4∗ei∗2∗θ−4∗e−i∗2∗θ +6 }
= 1
16∗{ei∗4∗θ +e−i∗4∗θ−4∗ei∗2∗θ−4∗e−i∗2∗θ +6 }
L.H.S.= R.H.S.
Ans.6. (c)
Graph of f:

Graph of p:

Graph of q: