In exercises 1-10 , find P_n(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