Mth 1002 Powers Series of Functions from Known Function

CALCULUS 2 NAME: ________________________________

LAB 31 11.10 TAYLOR & MACLAURIN SERIES Lab Time: ____________ Date: __________

Basic Definitions:

a
Using the known Geometric Series Formula:  ar
n0
n
 a  ar  ar 2  ar 3  ... 
1 r
which holds when r  1,

new series formulas can be created by a suitable replacement of terms.


a
a  ar  ar  ar  ...  ar and making the replacements a 1
2 3 n
1. Using the formula
1 r n0


and r  x , create a new series formula for 1 and give its interval of converegence.
1 x


1
1  1x  1x 2  1x 3  ...  1x n
1 x n 0



1
1  x  x  x  ...  x
2 3 n

1 x n 0


r 1  x 1   1  x 1
Interval :   1,1


1
2. Replace the x-term in your formula from the previous problem with   x to create a formula for .
1 x


1 1
 1   x    x  2   x 3  ...    x  n
1  x 1   x  n 0

1  x  x 2  x 3  ...    1 x n
n

n0



3. Integrate your formula from the previous problem with term-by-term to create a formula for ln  x  1 .
(Don’t forget to find the constant “C” by plugging in x 0 )


ln x  1 x 11 dx  1  x  x2  x3  ... dx C  x   1
2 x 2  13 x3  1
4 x 4  ...


ln  x  1 C  x  1
2
x 2  13 x3  1
4
x 4  ...
x  0  ln 0  1 C  0  0  13 0 
3 2
1
2
1
4 04  ...
ln 1 C  0  0  0  0  ...
0 C
 ln x 1  0  x  1
2
2 3
x  13 x  1
4
4
x  ...

x4  ...    1n
n1
So : ln  x  1  x  1
2
2
x  x  1
3
3 1
4 xn
n 1



1
4. Replace the x-term in the formula 1  x  x 2  x3  ...  x n
1 x n 0

1
with   x2  to create a formula for .
1 x 2