Give the Maclaurian series for sin x

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.

General equation of Maclaurin series

\[\large f(x)=f(x_{0})+{f}'(x_{0})(x-x_{0})+\frac{{f}”(x_{0})}{2!}(x-x_{0})^{2}+\frac{{f}”'(x_{0})}{3!}(x-x_{0})^{3}+…..\]

The Maclaurin series formula is

\[\large f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(x_{0})}{n!}(x-x_{0})\]

Where, f(xo), f’(xo), f’‘(xo)……. are the successive differentials when xo = 0.

Maclaurian series of sin x

f(x)= sin x

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

  • f′(0)= cos(0)=1
  • f”(0)=–sin(0)=–0=0
  • f”′(0)=sin(0)=0
  • f(iv)(0)= sin(0)=o
  • f(v)(x)=–sinx,f(v)(0)=–sin(0)=0
  • f(v)(0)=cos(0)=1

f(x)=f(0)+xf′(0)+x2 / 2!f”(0)+x3 / 3!f”′(0)+x4 / 4!f(iv)(0)+⋯

Putting the values in the above series, we get

f (sin x) = \(\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\)

Answer

f(sin x)

$sin\;x$ $\sum_{k=0}^{\infty}(-1)^{2}=\frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\frac{x^{7}}{7!}+…..$

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