   Chapter 11, Problem 45RE

Chapter
Section
Textbook Problem

# Find the Taylor series of f(x) = sin x at a = π/6.

To determine

To find: The Taylor series for f(x)=sinx centered at π6 .

Explanation

Result used:

If f has a power series expansion at a , f(x)=n=0f(n)(a)n!(xa)n , f(x)=f(a)+f(a)1!(xa)+f(a)2!(xa)2+f(a)3!(xa)3+

Calculation:

Consider the function f(x)=sinx centered at a=π6 .

Obtain the first four nonzero terms of the series as follows,

The function f(x)=sinx at a=π6 is computed as follows,

f(π6)=sin(π6)=12

That is, f(π6)=12 .

The first derivative of f(x) at a=π6 is computed as follows,

f(x)=ddx(sinx)=cosx

f(x)=cosx (1)

Substitute at π6 .for x,

f(π6)=cos(π6)=32

That is, f(π6)=32 .

The second derivative of f(x) at a=π6 is computed as follows,

f(2)(x)=d2dx2(f(x))=ddx(f(x))=ddx(cosx)    (by equation(1))

f(2)(x)=sinx (2)

Substitute at π6 .for x,

f(2)(π6)=sin(π6)

That is, f(2)(π6)=12

The third derivative of f(x) at a=π6 is computed as follows,

f(3)(x)=d3dx3(f(x))=ddx(f(2)(x))=ddx(sinx)   (by equation(2))

f(3)(x)=cosx (3)

Substitute at π6 .for x,

f(3)(π6)=cos(π6)=32

That is, f(3)(π6)=32

The fourth derivative of f(x) at a=π6 is computed as follows,

f(4)(x)=d4dx4(f(x))=ddx(f(3)(x))=ddx(cosx)   (by equation(3))=sinx

Substitute at π6 .for x,

f(4)(π6)=sin(π6)=12

That is, f(4)(π6)=12 .

Tabulate the computed values as follows,

 n fn(x) fn(π6) 0 sinx 12 1 cosx 32 2 −sinx −12 3 −cosx −32 4 sinx 12

By the Result stated above, the power series expansion at a=π6 is computed as follows,

f(x)=n=0f(n)(π6)n!(xπ6)n

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