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

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 +...`

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

For...

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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 +...`

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

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

`f(x) =sinh(x)`

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

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

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

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

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

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

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

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

`f^2(0) = sinh(0)=0`

`f^3(0) = cosh(0)=1`

`f^4(0) = sinh(0)=0`

`f^5(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`

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

`= 1/(1!)x+1/(3!)x^3+1/(5!)x^5+...`

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

The Maclaurin series is `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` for the function `f(x)=sinh(x)` .

To determine the interval of convergence for the Maclaurin series: `sum_(n=0)^oo x^(2n+1)/((2n+1)!)` , 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+1)/((2n+1)!)` , we have:

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

Then,

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

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

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

`=x^((2n+1)+2)/((2n+3)!)`

`=(x^(2n+1)*x^2)/((2n+3)(2n+2)((2n+1)!))`

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+1)*x^2)/((2n+3)(2n+2)((2n+1)!))*((2n+1)!)/x^(2n+1)|`

Cancel out common factors: `x^(2n+1) and ((2n+1)!)` .

`lim_(n-gtoo)|x^2/((2n+3)(2n+2))|`

Evaluate the limit.

`lim_(n-gtoo)|x^2/((2n+3)(2n+2))| = |x^2|lim_(n-gtoo)|1/((2n+3)(2n+2))|`

`=|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+1)/((2n+1)!)` is absolutely converges for all `x` .

Interval of convergence: `-ooltxltoo` .

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