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

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marizi eNotes educator| Certified Educator

Maclaurin series is a special case of Taylor series that is centered at c=0. The expansion of the function about 0 follows the formula:

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

 or

`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +(f^5(0))/(5!)x^5+...`

To determine the Maclaurin series for the given function `f(x)=cosh(x)` , we may apply the formula for Maclaurin series.

To list `f^n(x),`  we may follow the derivative formula for hyperbolic trigonometric functions:  `d/(dx) cosh(x) = sinh(x)` and 

`d/(dx) sinh(x) = cosh(x). `

`f(x) =cosh(x) `

`f'(x) = d/(dx) cosh(x)= sinh(x) `

`f^2(x) = d/(dx) sinh(x)= cosh(x)`

`f^3(x) = d/(dx) cosh(x)= sinh(x)`

`f^4(x) = d/(dx) sinh(x)=cosh(x) `

`f^5(x) = d/(dx) cosh(x)=sinh(x)`

`f^6(x) = d/(dx) sinh(x)=cosh(x)`

Note: When n= even then `f^n(x)=cosh(x)`.

When n= odd then `f^n(x)=sinh(x)`.

Plug-in `x=0` on each `f^n(x)` , we get:

`f'(0) =cosh(0)=1`

`f'(0) =sinh(0)=0`

`f^2(0) =cosh(0)=1`

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

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

 `f^5(0) =sinh(0)=0`

 `f^6(0) =cosh(0)=1`

Plug-in the values on the formula for Maclaurin series, we get:

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

`= 1+0/(1!)x+1/(2!)x^2+0/(3!)x^3+1/(4!)x^4 +0/(5!)x^5+1/(6!)x^6+ ...`

`=1+0+1/(2!)x^2+0+1/(4!)x^4+0+1/(6!)x^6+ ...`

`=1+/(2!)x^2+1/(4!)x^4+1/(6!)x^6+ ...`

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

The Maclaurin series is `sum_(n=0)^oo x^(2n)/((2n)!)` for the function `f(x)=cosh(x)` .

To determine the interval of convergence for the Maclaurin series: `sum_(n=0)^oo x^(2n)/((2n)!)` , we may apply Ratio Test.  

In Ratio test, we determine the limit as: `lim_(n-gtoo)|a_(n+1)/a_n| = L` .

The series converges absolutely when it satisfies `Llt1` .

In the Maclaurin series: `sum_(n=0)^oo x^(2n)/((2n)!)` , we have:

`a_n=x^(2n)/((2n)!)`

Then,

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

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

           `=x^(2n+2)/((2n+2)!)`

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

Applying the Ratio test, we set-up the limit as:

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

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

Cancel out common factors: `x^(2n)` and `(2n)!` .

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

Evaluate the limit.

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

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

                                `= |x^2|*0`

                                ` =0`

The `L=0` satisfies `Llt1` for all` x` . Thus, the Maclaurin series: `sum_(n=0)^oo x^(2n)/((2n)!)` is absolutely converges for all `x` .

Interval of convergence: -`ooltxltoo` .