Give the Maclaurin series for sin x.

crubats4b5p

crubats4b5p

Answered question

2023-03-15

Give the Maclaurin series for sin x.

Answer & Explanation

plumingagcqr

plumingagcqr

Beginner2023-03-16Added 7 answers

Step 1: Maclaurin series explanation
A Maclaurin series is a function that has expansion series that gives the sum of derivatives of that function.
The Maclaurin series of a function f(x) up to order n may be found using series [f,x,0,n].
It is a special case of the Taylor series when x=0.
The Maclaurin series' general equation is
f(x)=f(x0)+f'(x0)(x-x0)+f(x0)2!(x-x0)2+f'(x0)3!(x-x0)3+..
Step 2: Maclaurin series for sin x.
Given, fx=sinx
Using x=0, the given equation function becomes
f(0)=sin(0)=0
Now taking the derivatives of the given function and using x=0, we have
1. f ( 0 ) = cos ( 0 ) = 1 2. f " ( 0 ) = sin ( 0 ) = 0 = 0 3. f ( 0 ) = sin ( 0 ) = 0 4. f ( 0 ) = sin ( 0 ) = 0
Thus, we get the series as,
f(x)=f(0)+xf'(0)+x22!f''(0)+x33!f'''(0)+x44!f''''(0)+.
Putting the values in the preceding series yields
sinx=1-x22!+x44!-x66!+
Therefore, the Maclaurin series for sin x is 1-x22!+x44!-x66!+.

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