Question

How do you find the Maclaurin Series for `y= x / sin(x)`?

Answer 1

`f(x) = 1 + 1/6x^2+7/360x^4 + O(x^6)`

Explanation:

We have:

`f(x) = x/sinx`

First let us take a look at the graph and see how "well defined" the function is:

graph{x/sinx [-40, 40, -20, 20]}

The Maclaurin series can be expressed in the following way:

`f(x) = f(0) + (f'(0))/(1!) x + (f''(0))/(2!) x^2 + (f'''(0))/(3!) x^3 + (f^((4)))(0)/(4!) x^4 + ...`
`" "= sum_(n=0)^(∞) f^((n))(0)/(n!) x^n`

We also note from the graph that `f` is even, so we expect all odd powers of `x` in the series to vanish. So, Let us find the derivatives, and compute the values at `x=0`.

First Term

`f(x) = x/sinx`

We cannot evaluate `f(x)` when `x=0` as it is of an indeterminate form `0/0`, so must examine the limit of `f(x)` as `x rarr 0`, which we can evaluate using L'Hôpital's rule , we have:

`f(0) = lim_(x rarr 0) f(x)`
`" " = lim_(x rarr 0) x/sinx`
`" " = lim_(x rarr 0) 1/cosx`
`" " = 1`

Second Term:

`f(x) = x/sinx`

Using the quotient rule, we have:

`f'(x) = ( (sinx)(1) - (x)(cosx) ) / (sinx)^2`
`" " = ( sinx - xcosx ) / (sin^2x)`

Again, we cannot evaluate `f'(x)` at `x=0`, and again we can use L'Hôpital's rule, thus:

`f'(0) = lim_(x rarr 0) f'(x)`
`" " = lim_(x rarr 0) ( sinx - xcosx ) / (sin^2x)`
`" " = lim_(x rarr 0) ( cosx - (x)(-sinx) - (1)(cosx) ) / (2sinxcosx)`
`" " = lim_(x rarr 0) ( xsinx ) / (2sinxcosx)`
`" " = lim_(x rarr 0) ( x ) / (2cosx)`
`" " = 0`

Third Term:

`f'(x) = ( sinx - xcosx ) / (sin^2x)`

Using the quotient rule, and the product rule we have:

`f''(x) = ( (sin^2x)(xsinx) - (sinx-xcosx)(2sinxcosx) ) / (sin^2x)^2`
`" " = ( (xsin^2x - 2cosxsinx+2xcos^2x) ) / (sin^3x)`
`" " = ( (x - 2cosxsinx+xcos^2x) ) / (sin^3x)`

Again, we cannot evaluate `f''(x)` at `x=0`, and again we can use L'Hôpital's rule, thus:

`f''(0) = lim_(x rarr 0) f''(x)`
`" " = lim_(x rarr 0) ( (x - sin2x+xcos^2x) ) / (sin^3x)`

`" " = lim_(x rarr 0) ( (1 - 2cos2x+x(2cosx)(-sinx) + cos^2x) ) / ((3sin^2x)(cosx))`

`" " = lim_(x rarr 0) ( (1 - 2cos2x-xsin2x + cos^2x) ) / ((3sin^2x)(cosx))`

`" " = 2/3 lim_(x rarr 0) ( (1 - 2cos2x-xsin2x + cos^2x) ) / ((sinx)(sin2x))`

And a repeated application of L'Hôpital's rule gives:

`f''(0) = 2/3 lim_(x rarr 0) ( (4sin2x-x(2cos2x)-sin2x + 2cosx(-sinx)) ) / ((sinx)(2cos2x)+(cosx)(sin2x))`

`" " = 2/3 lim_(x rarr 0) ( (4sin2x-2xcos2x-sin2x - sin2x) ) / ((sinx)(2cos2x)+(cosx)(sin2x))`

`" " = 4/3 lim_(x rarr 0) ( sin2x-xcos2x ) / (2sinxcos2x+cosxsin2x)`

And a repeated application of L'Hôpital's rule gives:

`f''(0) = 4/3 lim_(x rarr 0) ( 2cos2x-x(-2sin2x)-cos2x ) / (2sinx(-2sinx)+(2cosx)(cosx)+cosx(2cos2x)+(-sinx)sin2x)`

`" " = 4/3 ( 2+0-1 ) / (2+2-0`
`" " = 1/3`

As you can see this is rapidly becoming tedious, although further terms can e calculated in a similar way.

Constructing the Maclaurin Series:

Using the above results we can construct the Maclaurin Series as follows:

`f(x) = 1 + (0)/(1!) x + (1/3)/(2!) x^2 + ...`
`" " = 1 + 1/6 x^2 + ...`

With the assistance of technology, further terms can be calculated readily, to get:

`f(x) = 1 + 1/6x^2+7/360x^4 + O(x^6)`

Answer 2

`x/sinx = 1 +x^2/6 + (7x^4)/360 +O(x^6)`

Explanation:

Start from the McLaurin series of `sinx`:

`sinx = sum_(n=0)^oo (-1)^nx^(2n+1)/((2n+1)!)`

Then:

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

Now note that if:

`f(x) = sum_(n=0)^oo a_nx^n`

and:

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

it follows that:

`(sum_(n=0)^oo a_nx^n)(sum_(n=0)^oo b_nx^n) = 1`

In our particular case `a_k = 0` and also `b_k=0` for `k` odd, because `sinx/x` and `x/sinx` are even functions:

`(a_0+a_2x^2+a_4x^4+...)(b_0+b_2x^2+b_4x^4+...) = 1`

so that expanding the product we have:

`a_0 b_0 = 1`

`a_0b_2+b_2a_0 = 0`

`a_0b_4 +a_2b_2+a_4b_0 = 0`

`...`

and solving for `b_k`:

`b_0 = 1/a_0`

`b_2 = -(b_0a_2)/a_0 = -a_2/a_0^2`

`b_4 = -(a_2b_2+a_4b_0)/a_0 = (a_2^2-a_0a_4)/a_0^3`

as:

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

`a_0 = 1 => b_0 =1`

`a_2 =-1/6 => b_2 = 1/6`

`a_4 = 1/120 => b_4 = 1/36-1/120 = (20-6)/720 = 7/360`

so:

`x/sinx = 1 +x^2/6 + (7x^4)/360 +O(x^6)`