Section 1.1

2a)

2b) 

2c)

2d)
 
 

4) Plot will be provided later.

6), , 

8), so limit = 1 as x->0, so set g(0)=1.

10) System:

,

so we get coefficients:

Section 1.2

2)

and to bound error we need find maximum of  on interval -, so find its derivative (the fourth derivative, set equal to zero, at it has a local max at x=0, , but comparing the endpoints, we see maximum is a ,  so 
 
 

4), b/c absolute value of cos(x) < 1. Using calculator, we need n>=4, or degree 2n-1 >=7.
 
 

6), and using the fact that on 0<x<1, e^x < 3 we get:

, and to make this less that 10^(-5) we need n>=9.

8) Algebra too complicated to type, basic cancellation is:

.
 
 

10) f(1)= 2, f'(1)=6, f''(1)=30, f'''(1)=120, f''''(1)=360, f'''''(1)=720, f''''''(1)=720. By Taylor's theorem , so .

12), by substitution.

, and following the work in the book (after equation 1.18). Using the MVT we get:

, this is our remainder term.
 
 

14)

Notice that in the log series, I have two different constants c in the remainder terms, but it turns out the are just opposites of each other.
 
 

16) The remainder term from question 12

, for 0<x<=1 we can bound it like: we know 0<c<1, so 0<c²<1 so 1<1+c²<2, so

, and if I want this less than 10^(-10), then I need 2n+3 > 10¹⁰. That is a HUGE number, not practical at all.