Series representation of tan(x)
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)=1x−x3−x345−2x5945−x74725−2x993555...
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)=x−x33+2x515−17x7315+62x92835−1382x11155925...
The series for sech is:
sech(x)=1−x22!+5x424−61x6720+277x88064−50521x103628800...
∞∑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 limx→0xsin(x), we have limx→01cos(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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