IX.B. MacLaurin's Polynomials and Taylor's Theory.

Exercises IX.B: Solutions for odd problems 1-7.

In exercises 1-10 , find Pn(x,f(x)) for the specified function and n.

1. f(x) = sin(x) ; n = 7

 k f(k)(x) f(k)(0) 0 sin(x) 0 1 cos(x) 1 2 -sin(x) 0 3 -cos(x) -1 4 sin(x) 0 5 cos(x) 1 6 -sin(x) 0 7 -cos(x) -1
So P7(x,sin(x)) =  x - x3/6  +  x 5/120 - x 7/5040.

3. f(x) = sin(2x) ; n = 7

 k f(k)(x) f(k)(0) 0 sin(2x) 0 1 2cos(2x) 2 2 -4sin(2x) 0 3 -8cos(2x) -8 4 16sin(2x) 0 5 32cos(2x) 32 6 -64sin(2x) 0 7 -128cos(2x) -128
So P7(x,sin(2x)) =  2x - 8x3/6  +  32x 5/120 -128 x 7/5040
= 2x - 4x3/3  +  4x 5/15 - 8x 7/315.

5. f(x) = e 2x ; n = 6 and n = 7

 k f(k)(x) f(k)(0) 0 exp(2x) 1 1 2exp(2x) 2 2 4exp(2x) 4 3 8exp(2x) 8 4 16exp(2x) 16 5 32exp(2x) 32 6 64exp(2x) 64 7 128exp(2x) -128
So P6(x,exp(2x)) = 1 + 2x + 4x2/2  + 8x3/6 + 16x4/24 + 32x 5/120 + 64x 6/720
and
P7(x,exp(2x)) = 1 + 2x + 4x2/2  + 8x3/6 + 16x4/24 + 32x 5/120 + 64x 6/720 +128 x 7/5040

7. f(x) = ln(1+x) ; n = 4

 k f(k)(x) f(k)(0) 0 ln(1+x) 0 1 1/(1 + x) 1 2 -1/(1+x)2 -1 3 2/(1+x)3 2 4 -6/(1+x)4 -6
So P4(x,ln(1+x)) =  x - x2/2  + x3/3  - x4/4