I understand that Taylor series expansion for $\sin(x)$ is derived as follow: $$ \sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-... $$ Now, what exactly is the first, second, and third term? Is the first term just $\sin(x) = x$? Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
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$\begingroup$In the Taylor expansion at $0$ of the function $\sin(x)$, the even powers of $x$, i.e. the "missing" terms, are zero because $\sin(x)$ is an odd function: $$ \begin{align}\sin(x)&=\sum_{k=0}^{\infty}\frac{D^k(\sin(x))_{x=0}}{k!}\cdot x^k \\&=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!}\cdot x^2+\frac{-\cos(0)}{3!}\cdot x^3+ \frac{\sin(0)}{4!}\cdot x^4+\frac{\cos(0)}{5!}\cdot x^5+o(x^5) \\&=\underbrace{0+x+0\cdot x^2-\frac{x^3}{3!}+0\cdot x^4+\frac{x^5}{5!}}_{\text{polynomial expansion up to the $5$-th term }}+o(x^5). \end{align}$$
$\endgroup$ 6 $\begingroup$If you know Euler's formula $e^{ix} = \cos(x)+i\sin(x)$, the power series $e^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$, and don't mind using complex expressions to get real results, then $$e^{ix}-e^{-ix} =(\cos(x)+i\sin(x))-(\cos(-x)+i\sin(-x)) =2i\sin(x) $$ so $$\sin(x) =-i(e^{ix}-e^{-ix})/2 ,$$ and $$e^{ix}-e^{-ix} =\sum_{n=0}^{\infty} \dfrac{(ix)^n-(-ix)^n}{n!} =\sum_{n=0}^{\infty} ((i^n)-(-i)^n)\dfrac{x^n}{n!} $$ and, with period 4, $$(i^n)-(-i)^n =1-1, i-(-i), -1-(-1), -i-(i) =0, 2i, 0, -2i $$ so $-i/2$ times this is $0, 1, 0, -1$.
The power series is therefore $$x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{x^7}{5040}... =\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} .$$
$\endgroup$ 2 $\begingroup$Terms are the members of a summation (whatever the formula) and they are enumerated in the order of their appearance (left to right, starting from first)
$x$ is the first term,
$-\dfrac{x^3}{3!}$ is the second term,
$\dfrac{x^5}{5!}$ is the third term.
When speaking of a polynomial, a term is said of the $n^{th}$ degree depending on the power of the variable. So here there are no terms of even degree and
$x$ is the first degree term,
$-\dfrac{x^3}{3!}$ is the third degree term,
$\dfrac{x^5}{5!}$ is the fifth degree term.
Now a Taylor expansion is written up to a remainder term, with as many terms as you like. The word order is used and equals the highest degree. So you can say
$\sin(x)=x+r_1(x)$ is the first order expansion,
$\sin(x)=x-\dfrac{x^3}{3!}+r_3(x)$ is the third order expansion,
$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+r_5(x)$ is the fifth order expansion.
Similarly, for the cosine you would have
First term $1$, second term $-\dfrac{x^2}2$, third term $\dfrac{x^4}{4!}$, and zero$^{th}$ order development/expansion $1$, second order $1-\dfrac{x^2}2$, fourth order $1-\dfrac{x^2}2+\dfrac{x^4}{4!}$.
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