-2
$\begingroup$

Question: Why does

$$ \sin\theta = \sum_{n=0}^{\infty} (-1)^n \frac{\theta^{2n+1}}{(2n+1)!} $$

converge to $\sin \theta$?

What I understand:

I understand that this formula comes from viewing $\sin \theta$ as a Maclaurin series.

I understand that this is the case because $\sin \theta$ can be viewed as a Taylor expansion:

$$ \sin x = \sum_{k=0}^{n} \frac{\sin^{(k)}(x)(x-0)^k}{k!} + \frac{\sin^{(n+1)}(c)(x-0)^n}{(n+1)!} $$

Moreover, I understand that since the derivative of $\sin \theta$ is at most $1$, that the error terms converge to $0$ (so that this series converges).

What I don't understand is why this series converges to the desired target, $\sin x$. Why is this the case?

NOTE: Other explanations on this site seem to only explain the steps I outlined above, but not why the Maclaurin series actually converges to $\sin \theta$ (unless I'm missing something).

$\endgroup$
  • $\begingroup$ Was perhaps the last sum meant to be $$\sum_{n=0}^k \frac{\sin^{(n)}(0)(x-0)^n}{n!}+\frac{\sin^{(k+1)}(c)(x-0)^{k+1}}{(k+1)!}\quad\quad ?$$ $\endgroup$ – user228113 Aug 24 '16 at 15:57
  • $\begingroup$ This answer of mine may help your understanding, as it illustrates the terms of the power series for sine and cosine (and secant and tangent) as lengths whose alternating sums "obviously" converge to the target values defined in the geometric sense. (To be clear: The convergence is visually "obvious"; that the lengths match the terms of the series is ... not. However, I link to a note that explains the latter.) $\endgroup$ – Blue Aug 24 '16 at 15:57
  • $\begingroup$ Reference the answers to this previous question: math.stackexchange.com/questions/185356/… $\endgroup$ – user363555 Aug 24 '16 at 16:00
  • 4
    $\begingroup$ What definition of the sine function are you applying here? $\endgroup$ – Mark Viola Aug 24 '16 at 16:37
  • 1
    $\begingroup$ In strongest possible agreement with @Dr.MV, I insist that your question can not even be answered until you have specified just what the sine function is for you. $\endgroup$ – Lubin Aug 24 '16 at 18:46
0
$\begingroup$

The thing that you do not tell us is: how do you define the function $\sin$? Most mathematicians define it precisely as the function $x \mapsto \sum \limits _{n=0} ^\infty (-1)^n \frac {x ^{2n+1}} {(2n+1)!}$, in which case your question has a trivial answer.

$\endgroup$
  • $\begingroup$ The question probably defines $\sin$ as the vertical displacement of a point on a unit circle after the segment from the center to the point has been rotationally translated $x$ radians. $\endgroup$ – GFauxPas Aug 24 '16 at 16:10
  • 3
    $\begingroup$ @Surb Surely, you must be joking. This is one of the more mature ways to define an analytic function. $\endgroup$ – Mark Viola Aug 24 '16 at 16:36
  • 2
    $\begingroup$ @Surb: In some cases, it makes sense. In particular, it makes a decent amount of sense to define $\sin z$ by series for a complex variable $z$. $\endgroup$ – Cameron Buie Aug 24 '16 at 16:37
  • $\begingroup$ @GFauxPas: In general, trigonometric definitions of the elementary functions are good during high-school. Once one enters the world of pure mathematics, other definitions are used (placing these functions in a different theoretical framework - not that the trigonometric one would be bad, but it is rather limited). $\endgroup$ – Alex M. Aug 24 '16 at 16:50
0
$\begingroup$

Part 1. For $n\geq 1$ we have $$ (\bullet ) \quad\sin y-\sin x=(n+1)!^{-1}(y-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(y-x)^j\sin^{(j)}x$$ where $c_n \in [x,y]\cup [y,x].$ (See footnote.) For $y=0$ this gives $$-\sin x=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(-x)^{j}\sin^{(j)}x$$ $$ \text {where }\quad c_n\in [0,x]\cup [x,0].$$ $$\text {Therefore }\quad\sin x=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}(-x)^{j-1}\sin^{(j)}x=$$ $$= R_n(x)+S_n(x)$$ $$ \text {where }\quad R_n(x)=(n+1)!^{-1}(-x)^{n+1}\sin^{(n+1)}c_n.$$ Since $R_n(x)\to 0$ as $n\to \infty$, we have $S_n(x)\to \sin x$ as $n\to \infty.$ That is, $$\sin x=\lim_{n\to \infty}S_n(x)=\sum_{j=1}^{\infty}j!^{-1}(-x)^{j-1}\sin^{(j)}x.$$

Part 2. When $n\geq 1,$ and $x=0$ in $(\bullet)$ we have $$\sin y=\sin y-\sin 0=(n+1)!^{-1}y^{n+1}\sin^{(n+1)}c_n+\sum_{j=1}^nj!^{-1}y^j\sin^{(j)}(0)=T_n(y)+U_n(y)$$ $$\text { where } \quad T_n(x)=(n+1)!^{-1}y^{n+1}\sin^{(n+1)}c_n.$$ Since $T_n(y)\to 0 $ as $n\to \infty,$ and since $\sin^{(j)}(0)=0$ when $j$ is even , while $\sin^{(j)}(0)=(-1)^{(j-1)/2}$ when $j$ is odd, we have $$\sin y=\lim_{j\to\infty}U_n(y)=\sum_{k=0}^{\infty}(2k+1)!^{-1}(-1)^k y^{2k+1}.$$

Footnote. $c_n\in [x,y]\cup [y,x]$ is just a convenient easy of saying that $c_n$ belongs to the closed interval from x to y.

Remarks. (1).It does not make sense to write $\sin x=R_n(x)+\sum_{n=1}^{\infty}B_n(x).$..... (2). There is an error in the summation for $\sin x$ in your Q.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.