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Find the indicated Maclaurin or Taylor series for the given function about the indicated point, and find the radius of convergence for the series.

sinx,x0=0

Short Answer

Expert verified

The Maclurin series for the function is f(x)=∑k=0∞(-1)k(2k+1)!x2k+1

Step by step solution

01

Given information

The function isf(x)=sinx

02

 Find the general of the Maclurin series of the function 

The Maclurin series at x0=0 for any function f with a derivative of order n is given by

f(x)=f(0)+f'(0)x+f''(0)2!x2+f'''(0)3!x3+f''''(0)4!x4+…

The function's general Maclurin series is

f(x)=∑n=0∞fn(0)n!xn
03

 Make a table of the Maclurin series for the function f(x)=sinx

n
fn(x)
fn(0)
fn(0)n!
0
sinx
0
0
1
cosx
1
1
2
-sinx
0
0
3
-cosx
-1
-13!
...
...
...
...
2k
(-1)ksinx
0
0
2k+1
(-1)kcosx
(-1)k
(-1)k1(2k+1)!
04

 Find the Maclurin series for the function f(x)=sinx 

The Maclurin series for the function f(x)=sinxis:

0+1·x+02!x2+(-1)3!x3+0x4+15!x5+06!x6+…

Or, we can write as:

f(x)=∑k=0∞(-1)k(2k+1)!x2k+1

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