`f(x)=cosx ` Prove that the Maclaurin series for the function converges to the function for all x

Maclaurin series is a special case of Taylor series which is centered at `c=0` . We follow the formula:

`f(x)=sum_(n=0)^oo (f^n(0))/(n!)x^n`

or

`f(x) = f(0)+ f'(0)x +(f^2(0))/(2!)x^2 +(f^3(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +...`

To list the `f^n(x)` , we may apply derivative formula for trigonometric functions: 

`d/(dx) sin(x) = cos(x)` and `d/(dx)cos(x) = -sin(x).`

`f(x)=cos(x)`

`f'(x)=d/(dx)cos(x) = -sin(x)`

`f^2(x)=d/(dx) -sin(x) = -cos(x)`

`f^3(x)=d/(dx) -cos(x)= - (-sin(x))= sin(x)`

`f^4(x)=d/(dx)d/(dx) sin(x) = cos(x)`

Plug-in ` x=0` , we get:

`f(0)=cos(0) =1`

...

Maclaurin series is a special case of Taylor series which is centered at `c=0` . We follow the formula:

`f(x)=sum_(n=0)^oo (f^n(0))/(n!)x^n`

or

`f(x) = f(0)+ f'(0)x +(f^2(0))/(2!)x^2 +(f^3(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +...`

To list the `f^n(x)` , we may apply derivative formula for trigonometric functions: 

`d/(dx) sin(x) = cos(x)` and `d/(dx)cos(x) = -sin(x).`

`f(x)=cos(x)`

`f'(x)=d/(dx)cos(x) = -sin(x)`

`f^2(x)=d/(dx) -sin(x) = -cos(x)`

`f^3(x)=d/(dx) -cos(x)= - (-sin(x))= sin(x)`

`f^4(x)=d/(dx)d/(dx) sin(x) = cos(x)`

Plug-in ` x=0` , we get:

`f(0)=cos(0) =1`

`f'(0) = -sin(0)=0`

`f^2(0) = -cos(0)=-1`

`f^3(0)=sin(0)=0`

`f^4(0)= cos(0) =1`

Note: `cos(0)= 1` and `sin(0)=0` .

Plug-in the` f^n(0)` values on the formula for Maclaurin series, we get:

`cos(x) =sum_(n=0)^oo (f^n(0))/(n!)x^n`

              `=1 +0*x+(-1)/(2!)x^2+(0)/(3!)x^3+(1)/(4!)x^4+...`

               `=1 +0-1/2x^2+0/6x^3 +1/24x^4+...`

              `=1 +0-1/2x^2+0 +1/24x^4+...`

               `=1 -1/2x^2 +1/24x^4+...`

                `= sum_(n=0)^oo ((-1)^n x^(2n))/((2n)!)`

To determine the interval of convergence, we apply Ratio test.

In ratio test, we determine a limit as `lim_(n-gtoo)| a_(n+1)/a_n| =L` where `a_n!=0` for all `ngt=N` .

The series `sum a_n` is a convergent series when `L lt1` .

From the Maclaurin series of cos(x) as `sum_(n=0)^oo ((-1)^n x^(2n))/((2n)!)` , we have:

`a_n= ((-1)^n x^(2n))/((2n)!)` then `1/a_n=((2n)!) /((-1)^n x^(2n))`

Then, `a_(n+1) =(-1)^(n+1) x^(2(n+1))/((2(n+1))!)`

                      `=(-1)^(n+1) x^(2n+2)/((2n+2)!)`

                      ` =(-1)^n*(-1)^1 (x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)`

                      ` = ((-1)^n*(-1)x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)`

We set up the limit `lim_(n-gtoo)| a_(n+1)/a_n|` as:

`lim_(n-gtoo) |a_(n+1)/a_n| =lim_(n-gtoo) |a_(n+1) * 1/a_n |`

` =lim_(n-gtoo) |((-1)^n*(-1)^1* x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)*((2n)!) /((-1)^n x^(2n))|`

Cancel out common factors: `(-1)^n, (2n)!, and x^(2n)` , the limit becomes;

`lim_(n-gtoo) |(-x^2)/((2n+2)(2n+1))|`

Evaluate the limit.

`lim_(n-gtoo) |- x^2/((2n+2)(2n+1))|=|-x^2/2| lim_(n-gtoo) 1/((2n+2)(2n+1))`

                                           `=|-x^2/2|*1/ oo `

                                           `=|-x^2/2|*0 `

                                            `=0`

The `L=0` satisfy the `Llt1` for every `x` .

Therefore, Maclaurin series of `cos(x)` as `sum_(n=0)^oo (-1)^n x^(2n)/((2n)!)` converges for all x.

Interval of convergence:` -ooltxltoo` .