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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.

53.lnx,3

Short Answer

Expert verified

The Taylor series for the function f(x)=lnxat x=3isPn(x)=ln3+13(x−3)+∑k=2∞(−1)k+11⋅2⋅3⋯(k−1)3kk!(x−3)k

Step by step solution

01

Step 1. Given data

We have the functionf(x)=lnx

02

Step 2. Table of the taylor series 

Any function fwith a derivative of order n, the taylor series at x=3is given by,

Pn(x)=f(3)+f′(3)(x−3)+f′â¶Ä²(3)2!(x−3)2+f′â¶Ä²â€²(3)3!(x−3)3+f′â¶Ä²â€²(3)4!(x−3)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)=lnxat x=3.

nfn(x)
fn(3)
fn(3)n!
0lnx
ln3
ln3
11x
13
13
2-1x2
-132
-1322!
31â‹…2x3
1â‹…233
1â‹…2333!
4−1⋅2⋅3x4
−1⋅2⋅334
−1⋅2⋅3344!
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k(−1)k+11⋅2⋅3⋯(k−1)xk
(−1)k+11⋅2⋅3⋯(2k−3)3k
(−1)k+11⋅2⋅3⋯(k−1)3kk!
03

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

The Taylor series for the function f(x)=lnxat x=3isPn(x)=ln3+13⋅(x−3)+−1322!(x−3)2+1⋅2333!(x−3)3+−1⋅2⋅3344!(x−3)4+…+(−1)k+11⋅2⋅3⋯(k−1)3kk!(x−3)k

Or we can write this asPn(x)=ln3+13(x−3)+∑k=2∞(−1)k+11⋅2⋅3⋯(k−1)3kk!(x−3)k

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