Give the Maclaurin series for sin x.

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plumingagcqr · Accepted Answer

Step 1: Maclaurin series explanationA 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&#039; general equation isf(x)=f(x0)+f&#039;(x0)(x-x0)+f”(x0)2!(x-x0)2+f”&#039;(x0)3!(x-x0)3+…..Step 2: Maclaurin series for sin x.Given, fx=sinxUsing x=0, the given equation function becomesf(0)=sin(0)=0Now taking the derivatives of the given function and using x=0, we have  1.      f    ′    (  0  )  =  cos  ⁡  (  0  )  =  1    2.  f  &quot;  (  0  )  =  −  sin  ⁡  (  0  )  =  −  0  =  0    3.      f    ‴    (  0  )  =  sin  ⁡  (  0  )  =  0    4.      f    ⁗    (  0  )  =  sin  ⁡  (  0  )  =  0Thus, we get the series as,f(x)=f(0)+xf&#039;(0)+x22!f&#039;&#039;(0)+x33!f&#039;&#039;&#039;(0)+x44!f&#039;&#039;&#039;&#039;(0)+….Putting the values in the preceding series yieldssinx=1-x22!+x44!-x66!+…Therefore, the Maclaurin series for sin x is 1-x22!+x44!-x66!+….

Give the Maclaurin series for sin x.

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