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 cos x
f(x)=cosx
Using x=0, the given equation function becomes
- f(0)=cos(0)=1
Now taking the derivatives of the given function and using x=0, we have
- f′(x)=–sinx,f′(0)=–sin(0)=0
- f”(x)=–cosx,f”(0)=–cos(0)=–1
- f”′(x)=sinx,f”′(0)=sin(0)=0
- f(iv)(x)=cosx,f(iv)(0)=cos(0)=1
- f(v)(x)=–sinx,f(v)(0)=–sin(0)=0
- f(vi)(x)=–cosx,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 (cos x) = \(\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\)
Answer
Maclaurian series of cos x
f (cos x) = \(\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\ldots\)
