Taylor Series beyond 2-3 terms?

Question

im looking to understand the tangent taylor series, but im struggling to understand how to use long division to divide one series (sine) into the other (cosine). I also can't find examples of the Tangent series much beyond X^5 (wikipedia and youtube videos both stop at the second or third term), which is not enough for me to see any pattern. (x^3/3 + 2x^5/15 tells me nothing).

Wiki says Bernouli Numbers which i plan on studying next, but seriously, i could really use an example of tangent series out to 5-6 just to get a ballpark of what's going on before i start plug and pray. If someone can explain why the long division of the series spits out x^3/3 instead of x^3/3x^2, that would help too,

because I took x^3/6 divided by x^2/2 and got 2x^3/6x^2, following the logic that 4/2 divided by 3/5 = 2/0.6 or 20/6. So I multiplied my top and bottom terms for the numerator, and my two middle terms for the denominator (4x5)/(2x3) = correct.

But when i do that with terms in the taylor series I'm doing something wrong. does that first x from sine divided by that first 1 from cosine have anything to do with it?

Completely lost.

Answer 1

$$\tan(x) = x+{\frac{1}{3}}{x}^{3}+{\frac{2}{15}}{x}^{5}+{\frac{17}{315}}{x}^{7}+ {\frac{62}{2835}}{x}^{9}+{\frac{1382}{155925}}{x}^{11}+{\frac{21844}{ 6081075}}{x}^{13}+\ldots$$

EDIT: Long division:

$$ \matrix{& & x &+ \frac{x^3}{3} &+ \frac{2 x^5}{15} &+ \frac{17 x^7}{315}&+ \ldots\cr& &---&---&---&---&--- \cr 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \ldots & | & x &- \frac{x^3}{6} &+ \frac{x^5}{120} &- \frac{x^7}{5040} &+ \ldots\cr & & x &- \frac{x^3}{2} &+ \frac{x^5}{24} &- \frac{x^7}{720} &+ \ldots\cr & & ---&---&---&---&---\cr & & &\frac{x^3}{3} &- \frac{x^5}{30} &+ \frac{x^7}{840} &+ \ldots\cr & & & \frac{x^3}{3} & - \frac{x^5}{6} & + \frac{x^7}{72} &+\ldots\cr & & & --- & --- & --- & ---\cr & & & & \frac{2 x^5}{15} & - \frac{4 x^7}{315} & +\ldots\cr & & & & \frac{2 x^5}{15} & - \frac{2 x^7}{30} & +\ldots\cr & & & & --- & --- & ---\cr & & & & & \frac{17 x^7}{315} & + \ldots }$$

Answer 2

Write $\frac{\sin x}{x}=\frac{\tan x}{x}\cos x$ as a power series in $x^2$, with $\frac{\tan x}{x}=t_0+t_1 x^2+t_2 x^4+\cdots$. Equating coefficients of powers of $x^2$ one by one gives $1=t_0,\,-\frac{1}{6}=-\frac{t_0}{2}+t_1,\,\frac{1}{120}=\frac{t_0}{24}-\frac{t_1}{2}+t_2$ etc. Write down as many of those as you like. Thus $t_0=1,\,t_1=\frac{1}{3},\,t_2=\frac{2}{15}$ etc.

Answer 3

You might find it conceptually easier to set up the identity of power series and compare the first few coefficients, and solve. This is algebraically equivalent to long division, though the order of some of the arithmetic operations is somewhat rearranged.

Write the desired Taylor series at $x = 0$ as $$\tan x \sim a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots .$$ Since $\tan$ is odd, all of the coefficients of the even terms vanish, i.e., $0 = a_0 = a_2 = a_4 = \cdots$. (This insight isn't necessary---we'd recover this fact soon anyway---but it does make the next computation easier.)

Replacing the functions in $$\cos x \tan x = \sin x$$ with their Taylor series gives $$\left(1 - \frac{1}{2!} x^2 + \frac{1}{4!} x^4 - \cdots\right)(a_1 x + a_3 x^3 + a_5 x^5 \cdots) = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \cdots .$$

Now, comparing the coefficients of the terms $x, x^3, x^5, \ldots$, on both sides respectively gives $$\begin{align*} a_1 &= 1 \\ a_3 - \tfrac{1}{2} a_1 &= -\tfrac{1}{6} \\ a_5 - \tfrac{1}{2} a_3 + \tfrac{1}{24} a_1 &= \tfrac{1}{120} \\ & \,\,\vdots \end{align*}$$ and successively solving and substituting gives $$\tan x \sim x + \tfrac{1}{3} x^3 + \tfrac{2}{15} x^5 + \cdots .$$ Of course, it's straightforward (if eventually tedious) to compute as many terms as you want this way.

An efficient proof of the formula you mentioned involving the Bernoulli numbers for the general coefficient is given in this answer.

Answer 4

My impression is that it's kind of backwards, in a numerical sense, to think about the coefficients of the $\tan$ series in terms of the Bernoulli numbers because it's simple and numerically stable to calculate the $\tan$ coefficients directly and in fact provides a reasonable method for computing the Bernoulli numbers given the formula in @RobJohn's post. Since $y(x)=\tan x$ is an odd function of $x$ analytic at $x=0$, $$y=\sum_{n=0}^{\infty}a_nx^{2n+1}$$ Then $y^{\prime}=\sec^2x=\tan^2x+1=y^2+1$ so $$\sum_{n=0}^{\infty}(2n+1)a_nx^{2n}=1+\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ia_jx^{2i+2j+2}=1+\sum_{n=1}^{\infty}\left(\sum_{i=0}^{n-1}a_ia_{n-i-1}\right)x^{2n}$$ The constant term reads $$a_0=1$$ The terms in $x^{4n}$ are $$a_{2n}=\frac1{4n+1}\sum_{i=0}^{2n-1}a_ia_{2n-i-1}=\frac2{4n+1}\sum_{i=0}^{n-1}a_ia_{2n-i-1}$$ While the terms in $x^{4n+2}$ are $$a_{2n+1}=\frac1{4n+3}\sum_{i=0}^{2n}a_ia_{2n-i}=\frac1{4n+3}\left(a_n^2+2\sum_{i=0}^{n-1}a_ia_{2n-i}\right)$$ The numerical stability arises because all terms in the formulas for $a_n$ have the same sign.

Answer 5

An alternative and straightforward method is: $$\begin{align}y&=\tan x \ (=0)\\ y'&=\frac{1}{\cos^2 x}=1+\tan^2x=1+y^2 \ (=1)\\ y''&=2yy'=2y(1+y^2)=2y+2y^3 \ (=0) \\ y'''&=2y'+6y^2y'=2+8y^2+6y^4 \ (=2)\\ y^{(4)}&=16yy'+24y^3y'=16y+40y^3+24y^5 \ (=0)\\ y^{(5)}&=16+120y^2y'+120y^4y'=16+136y^2+240y^4+120y^6 \ (=16)\\ y^{(6)}&=272yy'+960y^3y'+720y^5y'=272y+1232y^3+1680y^5+720y^7 \ (=0)\\ y^{(7)}&=272y'+3696y^2y'+8400y^4y'+5040y^6y'=272+3968y^2+\cdots \ (=272)\end{align}$$ Hence: $$\begin{align}\tan x&=0+\frac{1}{1!}x+\frac{0}{2!}x^2+\frac{2}{3!}x^3+\frac{0}{4!}x^4+\frac{16}{5!}x^5+\frac{0}{6!}x^6+\frac{272}{7!}x^7+\cdots\\ &=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\cdots\end{align}$$ Note: You can continue as far as you want, though the computation gets tedious. WA shows the expansion to many more terms (press on "More terms" button).