Maclaurin, Taylor & Binomial Series Expansion Assignment Solution

This document presents a detailed solution to an assignment focusing on Maclaurin, Taylor, and Binomial series expansions. It begins by deriving the Maclaurin series for cosh(x) and then applies Taylor’s Inequality to estimate the error bound when approximating cosh(x) with a 4th order Maclaurin series. The solution also assesses the series' convergence using Taylor's Inequality. Further, the assignment delves into the sinc (sine cardinal) function, utilizing Maclaurin series to validate its special definition at x=0. The Maclaurin series for sin(t^2) is derived and used to approximate the Fresnel-sine integral. Finally, the solution covers the binomial series expansion for √1−x^2, applying it to evaluate the elliptic integral E(k). This comprehensive solution provides step-by-step calculations and explanations for each problem.

Order Id 814870
1. .
i) MacLaurin series for cosh(x)
cosh ( x )= ex+ e−x
2
MacLaurin series of cosh ( x ) = ex+ e−x
2 is defined as:
¿ e0+ e−0
2 +
d
dx ( ex +e−x
2 ) ( 0 )
1 ! x+
d2
d x2 ( e x+ e−x
2 ) ( 0 )
2 ! x2+
d3
d x3 ( ex+e−x
2 ) ( 0 )
2! x3 +…
e0+ e−0
2 =1
d
dx ( ex +e− x
2 ) ( 0 ) =0 d2
d x2 ( ex+ e−x
2 ) ( 0 )=1 d3
d x3 ( ex+ e−x
2 ) ( 0 )=0 d4
d x4 ( ex +e− x
2 ) ( 0 )=1
d5
d x5 ( ex+ e−x
2 ) ( 0 )=0 d6
d x6 ( ex+ e−x
2 ) ( 0 ) =1 d7
d x7 ( ex+ e−x
2 ) ( 0 )=0 d8
d x8 ( ex+ e−x
2 ) ( 0 ) =1
¿ 1+ 0
1 ! x+ 1
2 ! x2 + 0
3 ! x3+ 1
4 ! x4 + 0
5 ! x5 + 1
6 ! x6 + 0
7 ! x7 + 1
8 ! x8 +…¿ 1+ 1
2! x2+ 1
4 ! x4 + 1
6 ! x6 + 1
8 ! x8 +…
Thus the series representation is;
∑
n=0
∞ x2n
( 2 n ) !
ii) Taylor’s inequality says
|R4 ( x )|=|f ( x ) −T 4 ( x )|≤ M
( 4 +1 ) ! x4 +1T 4 ( x )=1+ 1
2 ! x2 + 1
4 ! x4
f 5 ( x )= d5
d x5 ( ex−e−x
2 )= 1
2 ( ex−e−x ) f 5 ( 1 ) =M = e1−e−1
2
The Taylor’s inequality becomes:

|R4 ( x )|≤ e1−e−1
2×5 ! |x5
| |R4 ( x )|≤ e1−e−1
2×5 ! |15
|= e1 −e−1
240 ≈ 0.009793343
So the approximation is accurate to within e1 −e−1
240 ≈ 0.009793343
iii) From part (ii), the convergence of cosh(x) can be determined by ratio test, thus:
lim
n ⟶ ∞ ( an+1
an ) = lim
n ⟶ ∞
( x2 (n+1)
( 2(n+1) ) !
x2 n
( 2 n ) ! )=0
Therefore the series converges for all x
2. sinc (sine cardinal) function
sinc ( x ) = { 1 for x=0
sin ( x )
x Otherwise
MacLaurin series of cosh ( x ) = ex+ e−x
2 is defined as:
¿ sinc(x)+
d
dx ( sinc ( x) ) ( 0 )
1 ! x+
d2
d x2 ( sinc(x ) ) ( 0 )
2 ! x2+
d3
d x3 ( sinc( x ) ) ( 0 )
2 ! x3+ …
sin ( x )
x = { 1 x =0
sin ( x )
x x ≠ 0
d
dx ( sinc( x) ) =
{ 0 x=0
xcos ( x )−sin ( x )
x2 x ≠ 0
d2
d x2 ( sinc( x ) )=
{ −1
3 x=0
( 2−x2 ) sin ( x )−2 x cos (x )
x3 x ≠ 0
Using the first and second order:
sinc ( x ) =1− x2
3! + x4
5 ! −… sinc ( x ) =∑
n=0
∞
( −1 ) n x2 n
( 2 n+1 ) !
Evaluating the limit of sinc(x) as x approaches 0

lim
n ⟶ 0
sinc ( x ) = lim
n ⟶ 0 ( sin ( x )
x )= lim
n ⟶ 0 (1− x2
3! + x4
5 ! … )=1
3. MacLaurin series for sin(s^2)
¿ sin ( t2 ) +
d
dx ( sin ( t2 ) ) ( 0 )
1 ! x+
d2
d x2 ( sin ( t2 ) ) ( 0 )
2 ! x2+
d3
d x3 ( sin ( t2 ) ) ( 0 )
2 ! x3+ …
sin ( 0 ) =0
d
dx (sin ( t2 ) ) ( 0 ) =0 d2
d x2 ( sin ( t2 ) ) ( 0 ) =2 d3
d x3 ( sin ( t2 ) ) ( 0 )=0 d4
d x4 (sin ( t2 ) ) ( 0 )=0 d5
d x5 ( sin ( t2 ) ) ( 0 )=0
d6
d x6 ( sin ( t2 ) ) ( 0 )=−120 d7
d x7 ( sin ( t2 ) ) ( 0 )=0 d8
d x8 ( sin ( t2 ) ) ( 0 )=0 d9
d x9 ( sin ( t2 ) ) ( 0 ) =0
d10
d x10 (sin ( t2 ) ) ( 0 )=30240
¿ 1+ 0
1 ! x+ 2
2 ! x2 + 0
3 ! x3+ 0
4 ! x4 + 0
5 ! x5 +−120
6 ! x6+ 0
7 ! x7 + 0
8 ! x8 + +0
9! x9 + 30240
10 ! x10 …
¿ t 2− 1
6 t6 + 1
120 t10 …The Fresnel-sine integral:
∫
0
x
sin ( t2 ) dt=∫
0
x
( t2− 1
6 t6 + 1
120 t10 … … ) ¿ x3
3 − x7
42 + x11
1320 −…
4. Binomial series for 1
√1−x2
1
√1−x2 = ( 1−x2 )
−1
2
1
√1−x2 =1+ −1
2 (−x2 ) +
−1
2 (−3
2 ) (−x2 )2
2 ! +
−1
2 (−3
2 )(−5
2 ) (−x2 )3
3 ! +… ¿ 1+ x2
2 + 3 x4
8 +5 x6
16 + …
Now to evaluate E (k) let x2=k2 sin2 ( t ) ⟹ x=ksin(t)E ( k ) =∫
0
π
2
1
√1−k2 sin2 ( t ) dt

In equation 1+ x2
2 + 3 x4
8 +5 x6
16 + …replace x with ksin ( t )
E ( k ) =∫
0
π
2
( 1+ ( ksin ( t ) )
2
2 + 3 ( ksin ( t ) ) 4
8 + 5 ( ksin ( t ) )
6
16 +… ) dt¿ ¿ ¿¿ π
2 + π k 2
8 + 9 π k 4
128 + 25 πk6
1024 +…
¿ π
2 ∑
n=0
∞ ( 2 n−1 ) 2 k2 n
23n