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In Exercises 49–56 find the Taylor series for the specified function and the given value of x0. Note: These are the same functions and values as in Exercises 41–48.

51.sinx,Ï€

Short Answer

Expert verified

The Taylor series for the function f(x)=sinx at x=π isPn(x)=∑k=0∞(−1)k+1(2k+1)!(x−π)2k+1

Step by step solution

01

Step 1. Given data 

We have the functionf(x)=sinx

02

Step 2. Table of the taylor series 

Any function fwith a derivative of order n, the taylor series at x=Ï€is given by,

Pn(x)=f(Ï€)+f′(Ï€)(x−π)+f′â¶Ä²(Ï€)2!(x−π)2+f′â¶Ä²â€²(Ï€)3!(x−π)3+f′â¶Ä²â€²â¶Ä²(Ï€)4!(x−π)4+…

we can write the general of the Taylor series of the function fis,

Pn(x)=∑k=0∞fk(x0)k!(x−x0)n

So, let us first construct the table of the Taylor series for the function f(x)=sinxat x=Ï€

nfn(x)


0sinx
00
1cosx
-1-1
2-sinx
00
3-cosx
113!
.
.
.
.
.
.
.
.
.
.
.
.
2k(−1)ksinx
00
2k+1(−1)kcosx
(−1)k+1
(−1)k+11(2k+1)!
.
.
.
.
.
.
.
.
.
.
.
.
03

Step 3. Taylor series for the  f(x)=sinx

The Taylor series for the function f(x)=sinxat x=Ï€is

Pn(x)=0+−1⋅(x−π)+02!(x−π)2+13!(x−π)3+⋯+0(2k)!(x−π)2k+(−1)k+1(2k+1)!(x−π)2k+1+⋯

Or we can write this asPn(x)=∑k=0∞(−1)k+1(2k+1)!(x−π)2k+1

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