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Dividing power series (2): More complicated examples

This thread will include more intricate long division problems. We start by deriving the power series for tan(x) and cot(x). It is easy to note that tan(x) and cot(x) can be obtained by using their very definition. tan(x)=sin(x)cos(x) and cot(x)=cos(x)sin(x). We first expand in series and then divide the two power series using the scheme of long division. This method, while simple, do not reveal the relationship between these two functions' coefficients and the Bernoulli numbers. We will illuminate this point in another thread where we seek to obtain the power series for the generating function of Bernoulli numbers. For now, we contend to just obtain the power series for tan(x) and cot(x).

Series representation of tan(x)

Thus tan(x)=x+x33+2x515+17x7315+62x92835+1382x11155925...

Series representation of cot(x)

The function is not defined at x=0, since sin(0)=0, division by zero is undefined. The function doesn't have a Maclaurin series centered at 0. Nevertheless, we can still expand it into what is called the Laurent series. The Laurent series is a power series that includes negative degree.

Thus cot(x)=1xx3x3452x5945x747252x993555...

Series representation of sec(x)

We use the identity sec(x)=1cos(x) to derive the Maclaurin series for sec(x).

Thus sec(x)=1+1x22!+5x424+61x6720+277x88064+50521x103628800...

Long division may give us the series, but it doesn't hint at the fact that sec(x) is one of the two generating functions of Euler numbers. In fact, since we reduce fractions to their simplest form, this obscure the general formula for the coefficients of sec(x). The series of sec(x) is as followed:

n=0(1)nE2nx2n(2n)!

The first few Euler numbers are:
E0=1, E2=1, E4=5, E6=61, E8=1385, E10=50521, E12=2702765...

Thus we can rewrite the series as followed:

sec(x)=1+1x22!+5x44!+61x66!+1385x88!+50521x1010!...

Series representation of tanh(x) and sech(x)

Using the same method above, we can derive the series for tanh(x) and sech(x). The difference is that instead of being a positive term series, they become alternating series. The series for tanh(x) is:

tanh(x)=xx33+2x51517x7315+62x928351382x11155925...

The series for sech is:

sech(x)=1x22!+5x42461x6720+277x8806450521x103628800...

n=1(1)nE2nx2n(2n)!

Series representation of xsin(x)

At x=0, the function has the form 00. Using L'Hopital rule to find the limit limx0xsin(x), we have limx01cos(x), which is 1. So the function is defined when x=0. By long division, we have:

Thus xsin(x)=1+x23!+7x4360+31x615120+127x8604800+73x103421440...

Series representation of 1sin(x)=csc(x)

It is now easy to obtain the series representation for csc(x), we only have to divide it by x, the series is:

Thus csc(x)=1x+x3!+7x3360+31x515120+127x7604800+73x93421440...

Again, the function is not defined at x=0, so there is no Maclaurin series for csc(x). What we obtain is the Laurent series.

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