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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.

role="math" localid="1649406173168" 49.cosx,Ï€2

Short Answer

Expert verified

The Taylor series for the functionf(x)=cosxatx=π2isPn(x)=∑k=0∞(−1)k+1(2k+1)!(x−π2)2k+1

Step by step solution

01

Step 1. Given data

We have the functionf(x)=cosx

02

Step 2. Table of the taylor series

Any function fwith a derivative of order n, the taylor series at x=Ï€2is given by,

Pn(x)=f(Ï€2)+f′(Ï€2)(x−π2)+f′â¶Ä²(Ï€2)2!(x−π2)2+f′â¶Ä²(Ï€2)3!(x−π2)3+f′â¶Ä²â€²â¶Ä²(Ï€2)4!(x−π2)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)=cosxat x=Ï€2

nfn(x)
fnπ2
fnπ2n!
0cosx
00
1-sinx
-1-1
2-cosx
00
3sinx
113!
.
.
.
.
.
.
.
.
.
.
.
.
2k(-1)kcosx
00
2k+1(-1)k+1sinx
(-1)k+1
(−1)k+11(2k+1)!
.
.
.
.
.
.
.
.
.
.
.
.
03

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

The Taylor series for the function f(x)=cosxat x=π2isPn(x)=0+−1⋅(x−π2)+02!(x−π2)2+13!(x−π2)3+04!(x−π2)4+⋯+0(2k)!(x−π2)2k+(−1)k+1(2k+1)!(x−π2)2k+1+⋯

Or we can write this asPn(x)=∑k=0∞(−1)k+1(2k+1)!(x−π2)2k+1

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