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In Exercises 49–56 find the Taylor series for the specified function and the given value of x0.

56.sin3x,Ï€6

Short Answer

Expert verified

The Taylor series of the functionsin3xatx=π6ispn(x)=∑k=0∞(-1)k+132k(2k)!(x-π6)2k

Step by step solution

01

Step 1. Given information.

We have given that f(x)=sin3xandx0=Ï€6.

02

Step 2. Table of the Taylor series

Any function fwith a derivative of order n, the Taylor series at x=Ï€6is given by,

Pn(x)=f(Ï€6)+f'(Ï€6)(x-Ï€6)+f''(Ï€6)2!(x-Ï€6)2+f'''(Ï€6)3!(x-Ï€6)3+....

So, let us find the derivatives of the given function and construct a table of the Taylor series for the function f(x)=sin3xat π6.

03

Step 3. Required Taylor series 

Therefore, the Taylor series of the function f(x)=sin3xat x=Ï€6.

Pn(x)=1+0·(x-π6)+-322!(x-π6)2+03!(x-π6)3+...+(-1)k32k(2k)!(x-π6)2k+0(2k+1)!(x-π6)2k+1+...

otherwise, it can be written as

Pn(x)=∑k=0∞(-1)k+132k(2k)!(x-π6)2k

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