Taylor Series Problems

Here are four summations you need.  You are going to evaluate them for the x-values shown below and up to n=30.

Starting Equations

Taylor1(x) = 
Taylor2(x) = 
Taylor3(x) = 
Taylor4(x) = 

Expected Output / Test Cases

Taylor1(double x)
x = 1      2.7182818284590455
x = 2      7.389056098930649
x = 3      20.08553692318766
x = 4      54.598150033144265
x = 5      148.41315910257657
x = 6      403.4287934927351
x = 7      1096.6331584284578
x = 8      2980.957987041728
x = 9      8103.083927575384
x = 10    22026.46579480671

Taylor2(double x)
x = 0           1.0
x = PI/4      0.7071067811865475
x = PI/2      4.2539467343847745E-17
x = 3PI/4    -0.7071067811865474
x = PI          -1.0000000000000002
x = 5PI/4     -0.707106781186548
x = 3PI/2     1.3033660283077339E-15
x = 7PI/4     0.7071067811865466
x = 2PI        0.9999999999999902

Taylor3(double x)
x = 0         0.0
x = PI/4     0.7071067811865475
x = PI/2     1.0000000000000002
x = 3PI/4   0.7071067811865477
x = PI         .3280566969799443E-16
x = 5PI/4    -0.7071067811865474
x = 3PI/2    -1.0000000000000002
x = 7PI/4    -0.7071067811865495
x = 2PI        3.3009043599749695E-15

Taylor4(double x)
x = -(1/2)      0.6666666666666667
x = -(1/3)      0.7500000000000002
x = -(1/4)      0.8
x = -(1/5)      0.8333333333333334
x = 0             1.0
x = (1/5)       1.25
x = (1/4)       1.3333333333333333
x = (1/3)       1.5
x = (1/2)        2.0