In Exercises 31–40 find the Maclaurin series for the specified function. Note: These are the same functions as in Exercises 21–30.

sin3x

Short Answer

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Ans: The Maclaurin series of the function f(x)=k=0(-1)k32k+1(2k+1)!x2k+1

Step by step solution

01

Step 1. Given information:

f(x)=sin3x

02

Step 2. Finding the general form  Maclaurin series:  

Since for any function fwith derivatives of all orders at the point x0=0, then the Maclurin series is

f(x)=f(0)+f'(0)x+f''(0)2!x2+f''(0)3!x3+f'''(0)4!x4+

Or, we can write the general form Maclurin series of the function f is

f(x)=n=0fn(0)n!xn
03

Step 3. Constructing the table of the Maclaurin series for the function :

So, let us first construct the table of the Maclaurin series for the functionf(x)=sin3x

nfn(x)fn(0)fn(0)n!0sin3x0013cos3x332-9sin3x003-27cos3x-27-273!·*··2k(-9)ksin3x0·2k+1(-1)k32k+1cos3x(-1)k32k+1(-1)k32k+1(2k+1)!········
04

Step 4. Finding the Maclaurin series :

Therefore, the Maclaurin series for the functionf(x)=sin3xis 0+3·x+02!x2+(-27)3!x3+04!x4+

Or, we can write as

f(x)=k=0(-1)k32k+1(2k+1)!x2k+1

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