Deriving power series for $\sin x$ without using Taylor's Theorem or $\exp z$
Starting with defining $(\cos t, \sin t)$ from the unit circle, is it possible to derive the power series for $\sin(t)$:
$$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$$
Note: I will be answering my question, I hope this doesn't offend anyone. If is any issue with the proof, I am grateful for any improvement.
$\endgroup$6 Answers
$\begingroup$Another method is using the inequality
$$\sin(x)\leq x$$
Integrate this between $0$ and $t$
$$-\cos(t)+\cos(0)\leq t^2/2$$ $$1-t^2/2\leq \cos(t)$$
Integrate this between $0$ and $x$:
$$x-x^3/3!\leq \sin(x)$$
Repeat this and you will get a lower and upper bound, which are just partial sums of the Taylor series. Note that you directly get the taylor series for $\cos(x)$.
$\endgroup$ 3 $\begingroup$I thought that you might want to derive the series without calculus. From angle addition formulas we have $$\sin(n-1)x=\sin nx\cos x-\cos nx\sin x$$ $$\sin(n+1)x=\sin nx\cos x+\cos nx\sin x$$ Adding, we get $$\sin(n+1)x+\sin(n-1)x=2\sin nx\cos x$$ And the key identity $$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$ So we can show that $$\sin2x=2\sin x\cos x-0=2\sin x\cos x$$ $$\sin3x=2(2\sin x\cos x)\cos x-\sin x=4\sin x(1-\sin^2x)-\sin x=3\sin x-4\sin^3x$$ This implies $$\sin x=3\sin\frac x3-4\sin^3\frac x3\tag{1}$$ And for a for a first-quadrant angle $x\le\frac{\pi}2$, $$\sin x\ge\sin\frac x3(4-4\sin^2\frac{\pi}6)=2\sin\frac x3\tag{2}$$ At this point we will try to find the series for $\sin^{-1}x=x+bx^3+cx^5+\dots$. We know the leading term is $x$ and since $x=3\left(\frac x3\right)$ we start from eq $(1)$ with $$\sin x=3\left(\sin\frac x3-\frac43\sin^3\frac x3\right)$$ So far so good. Now for the next term: $$\begin{align}\sin x+b\sin^3x&=3\sin\frac x3-4\sin^3\frac x3+b\left(3\sin\frac x3-4\sin^3\frac x3\right)^3\\ &=3\sin\frac x3+(27b-4)\sin^3\frac x3-108b\sin^5\frac x3+144b\sin^7\frac x3-64b\sin^9\frac x3\\ &=3\left(\sin\frac x3+b\sin^3\frac x3\right)+O\left(\sin^5\frac x3\right)\end{align}$$ Comparing coefficients of $\sin^3\frac x3$ on the last two lines we get $27b-4=3b$ or $b=\frac16$. Then $$\sin x+\frac16\sin^3x=3\sin\frac x3+\frac12\sin^3\frac x3-18\sin^5\frac x3+24\sin^7\frac x3-\frac{32}3\sin^9\frac x3$$ Let's get one more term: $$\begin{align}\sin x+\frac16\sin^3x+c\sin^5x&=3\sin\frac x3+\frac12\sin^3\frac x3-18\sin^5\frac x3+24\sin^7\frac x3\\ &-\frac{32}3\sin^9\frac x3+c\left(3\sin\frac x3-4\sin^3\frac x3\right)^3\\ &=3\sin\frac x3+\frac12\sin^3\frac x3+\left(243c-18\right)\sin^5\frac x3+\left(24-1620c\right)\sin^7\frac x3\\ &+\left(4320c-\frac{32}3\right)\sin^9\frac x3-5760c\sin^{11}\frac x3+3840c\sin^{13}\frac x3-1024c\sin^{15}\frac x3\\ &=3\left(\sin\frac x3+\frac16\sin^3\frac x3+c\sin^5\sin^5\frac x3\right)+O\left(\sin^7\frac x3\right)\end{align}$$ Again comparing coefficients of the $\sin^5\frac x3$ terms in the last two lines we find that $243c-18=3c$, so $c=\frac3{40}$ . Then $$\begin{align}\sin x+\frac16\sin^3x+\frac3{40}\sin^5x&=3\left(\sin\frac x3+\frac16\sin^3\frac x3+\frac3{40}\sin^5\frac x3\right)+\frac{\sin^7\frac x3}{30}\left(-149-184\cos^2\frac x3\right.\\ &\left.-864\cos^4\frac x3+576\cos^6\frac x3-2304\cos^8\frac x3\right)\end{align}$$ As can be seen, given enough patience and accurate algebra we could derive as many terms as we wished of the series for $\sin^{-1}(x)$ in this fashion. Applying inequality $(2)$ $n$ times, we find that $$\sin x\ge2^n\sin\frac x{3^n}$$ For first quadrant angles $x$, also $$0<149+184\cos^2\frac x3+864\cos^4\frac x3-576\cos^6\frac x3+2304\cos^8\frac x3<3501$$ so for such angles, $$\begin{align}-\frac{389}{10}\frac{3^n}{2^{7n}}\sin^7x&\le3^{n-1}\left(\sin\frac x{3^{n-1}}+\frac16\sin^3\frac x{3^{n-1}}+\frac3{40}\sin^5\frac x{3^{n-1}}-\frac x{3^{n-1}}\right)\\ &-3^n\left(\sin\frac x{3^n}+\frac16\sin^3\frac x{3^n}+\frac3{40}\sin^5\frac x{3^n}-\frac x{3^n}\right)\le0\tag{3}\end{align}$$ Since $$\frac{3^n}{2^{7n}}=\frac{\frac{3^n}{2^{7n}}}{1-\frac3{2^7}}-\frac{\frac{3^{n+1}}{2^{7(n+1)}}}{1-\frac3{2^7}}$$ We can sum inequality $(3)$ from $n=1$ to $n=N$ to get $$\begin{align}-\frac{1167}{1250}\sin^7x&\le-\frac{24896}{625}\cdot\frac3{128}\sin^7x+\frac{24896}{625}\cdot\frac{3^{N+1}}{2^{7N+7}}\sin^7x\\ &\le\sin x+\frac16\sin^3x+\frac3{40}\sin^5x-x\\ &-3^N\left(\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}+\frac3{40}\sin^5\frac x{3^N}-\frac x{3^N}\right)\le0\end{align}$$ Since the $N$-dependent part is $$f(x,N)=x\cdot\frac{\sin\frac x{3^N}+\frac16\sin^3\frac x{3^N}+\frac3{40}\sin^5\frac x{3^N}-\frac x{3^N}}{\frac x{3^N}}$$ And $$\lim_{N\rightarrow\infty}f(x,N)=0$$ For any first quadrant angle $x$, it follows that for any first quadrant angle $x$, $$-\frac{1167}{1250}\sin^7x\le\sin x+\frac16\sin^3x+\frac3{40}\sin^5x-x\le0$$ The knowledge we obtained from that limit above we could have gotten more fundamentally without actually taking a limit, so we have the first $3$ terms of the series for $\sin^{-1}(x)$ from algebra.
Now we want to work backwards to find the series for $\sin x$. $$\begin{align}\sin x&=x+px^3+qx^5+O(x^7)\\ &=\sin^{-1}(\sin x)+p\left(\sin^{-1}(\sin x)\right)^3+q\left(\sin^{-1}(\sin x)\right)^5+O\left(\left(\sin^{-1}(\sin x)\right)^7\right)\\ &=\sin x+\left(p+\frac16\right)\sin^3x+\left(q+\frac p2+\frac3{40}\right)\sin^5x+O(\sin^7x)\end{align}$$ Equating coefficients of like powers of $\sin x$, we find that $$p=-\frac16$$ $$q=\frac p2-\frac3{40}=\frac1{120}$$ Hopefully you can see how to work out the series for $\sin x$ with just algebra and trigonometry. How many terms would you have to derive like this before it would be easier to learn enough calculus to write down the Taylor series directly?
$\endgroup$ 2 $\begingroup$I think this works:
Starting with defining $(cos(t), sin(t))$ from the unit circle, it can be shown by elementary means that $sin(t)$ satisfies $f''=-f$. Also by observation $f(0)=0$.
Consider a power series $f(x) = a_0 + a_1 x + a_2 x^2 + ... $ with the same two constraints, we have $a_0=0$ and, by differentiating twice and comparing coefficients, $(k+2)(k+1)a_{k+2} = -a_k$.
It can be immediately seen that for even $k$, $a_k=0$.
Also that:
$a_3 = -\frac{1}{(3)(2)}a_1$ = -$\frac{a_1}{3!}$
$a_5 = -\frac{1}{(5)(4)}a_3 = +\frac{a_1}{5!}$
... etc.
So $f(x) = a_1 (x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...)$ which (by the Ratio Test) converges.
$a_1$ must have value 1 if $f$ is to behave similarly to $sin$ in the neighbourhood of 0 (as $sin(x) \to x$ as $x \to 0$).
So we have $sin(x)$ and $f(x)$ both satisfying $f''=-f$ and having value 0 at $x=0$.
EDIT:
However, as pointed out in the comments, there might be more than one function satisfying this property.
If the function can be written as a power series, we have found it! It must be $f$ with $a_1=1$.
But maybe there is some weird function that can't be written as a power series... so this proof is not quite complete. Can anyone put the last nail in?
$\endgroup$ 5 $\begingroup$One approach is to use the differential equation $$y'' + y = 0$$ satisfied by $y = \sin x$. The differential equation has unique solution $$y = y(0)\cos x + y'(0)\sin x$$ Now consider the power series $$f(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$$ then $f(x)$ is defined for all $x$ (because the series is convergent for all $x$) and by differentiating the series twice we can see that $f''(x) + f(x) = 0$. Hence $f(x) = f(0)\cos x + f'(0)\sin x$ and noting that $f(0) = 0, f'(0) = 1$ we get $f(x) = \sin x$.
The above proof uses the derivatives of $\sin x, \cos x$ and differentiation of power series.
$\endgroup$ $\begingroup$By definition of the unit circle and $\cos(t),\sin(t)$, we have the following lemma :
If $g : \mathbb{R} \to \mathbb{R}^2$, $g(t) = (x(t),y(t))$ is a smooth parametrized curve in $\mathbb{R}^2$ such that $\ \|g(t)\|= 1$ for every $t$, then $g(t)$ moves on the unit circle.
Moreover, if $\|g'(t)\|=1$ for every $t$ then $g(t) = (\cos(t+a),\pm\sin(t+a))$ for some $a$.
Let $$x(t) = \sum_{k=0}^\infty (-1)^k \frac{t^{2k}}{(2k)!}, \qquad\qquad y(t) = \sum_{k=0}^\infty (-1)^k \frac{t^{2k+1}}{(2k+1)!}$$both series converge and define smooth functions of $t \in \mathbb{R}$.
Consider the function $g : \mathbb{R} \to \mathbb{R}^2$ $$g(t) = (x(t),y(t))$$
By differentiating the power series term by term, we get $x'(t) = -y(t)$, $y'(t) = x(t)$, so that $$\frac{d}{dt}\|g(t)\|^2 = \frac{d}{dt}[x(t)^2+y(t)^2] = 2 x'(t) x(t) + 2 y'(t)y(t) = -2 y(t) x(t)+2 x(t) y(t) = 0$$
thus $$\|g(t)\|^2 = \|g(0)\|^2 = 1$$
and the same for $\|g'(t) \|^2 = \|g(t)\|^2$.
Hence the lemma applies and $$g(t) = (\cos(t+a),\pm\sin(t+a))$$
obviously $a=0$ since $g(0) = (1,0)$, and $g(t) = (\cos(t),\sin(t))$ since $g'(0) = (0,1)$.
$\endgroup$ $\begingroup$We want to start with a statement in the form: For any $n\in\mathbb{Z}^+$ there is some $c_n\in\mathbb{R}^+$ such that for any $x\in\left(-\frac{\pi}4,\frac{\pi}4\right)$ $$\left|\sin x-\sum_{k=1}^nb_kx^{2k-1}\right|\le c_n|x|^{2n+1}$$ And $$\left|\cos x-\sum_{k=0}^{n-1}a_kx^{2k}\right|\le c_n|x|^{2n}$$ And $$a_k=\frac{(-1)^k}{(2k)!}\text{, }b_k=\frac{(-1)^{k-1}}{(2k-1)!}$$ We already established the result for $\sin x$ for $n\le3$ in our other answer and the result for $\cos x$ for $n=1$ follows because $$1-\cos x=2\sin^2\frac x2\le\frac12x^2$$ If true for $n$, then $$\begin{align}\cos x&=1-2\sin^2\frac x2=1-2\left(\sum_{k=1}^n\frac{(-1)^{k-1}}{(2k-1)!}\frac{x^{2k-1}}{2^{2k-1}}+f_n(x)x^{2k+1}\right)^2\\ &=\sum_{k=0}^{n-1}\frac{(-1)^k}{(2k)!}x^{2k}-\frac2{(2n)!}\sum_{k=1}^n\frac{(-1)^{k-1}}{2^{2k-1}(2k-1)!}\frac{(2n)!(-1)^{n-k+1-1}}{2^{2n-2k+2-1}(2n-2k+2-1)!}x^{2n}+F_n(x)x^{2n+2}\end{align}$$ We got the first sum from the induction hypothesis, the second sum is the collection of coefficients of $x^{2n}$ and the last term reflects the fact that all the other terms have at least $x^{2n+2}$ in them. We can spot the middle sum as a difference of binomial series: $$a_n=-\frac2{(2n)!}\sum_{k=1}^n\frac{(-1)^{k-1}}{2^{2k-1}(2k-1)!}\frac{(2n)!(-1)^{n-k+1-1}}{2^{2n-2k+2-1}(2n-2k+2-1)!}x^{2n}$$ $$=\frac{(-1)^n}{(2n)!}\left[\left(\frac12+\frac12\right)^{2n}-\left(\frac12-\frac12\right)^{2n}\right]=\frac{(-1)^n}{(2n)!}$$ So that gives us $a_n$ and if we worked out $\max\left|F_n(x)x^{2n+2}\right|$ we would have one bound for $c_{n+1}$. Then $$\begin{align}\sin x&=2\sin\frac x2\cos\frac x2\\ &=2\left(\sum_{k=1}^n\frac{(-1)^{k-1}}{(2k-1)!}\frac{x^{2k-1}}{2^{2k-1}}+b_{n+1}\frac{x^{2n+1}}{2^{2n+1}}+f_{n+1}(x)x^{2n+3}\right)\left(\sum_{k=0}^n\frac{(-1)^k}{(2k)!}\frac{x^{2k}}{2^{2k}}+F_n(x)x^{2n+2}\right)\\ &=\sum_{k=1}^{n+1}\frac{(-1)^{k-1}}{(2k-1)!}x^{2k-1}+G_n(x)x^{2n+3}\\ &+\left[\frac{b_{n+1}}{2^{2n}}+\frac2{(2n+1)!}\sum_{k=1}^n\frac{(-1)^{k-1}}{2^{2k-1}(2k-1)!}\frac{(2n+1)!(-1)^{n-k+1}}{2^{2n-2k+2}(2n-2k+2)!}\right]x^{2n+1}\end{align}$$ Since the stuff in the square brackets must be $b_{n+1}$ and the sum is again almost a difference of binomial series, $$b_{n+1}=\frac{b_{n+1}}{2^{2n}}+\frac{(-1)^n}{(2n+1)!}\left[\left(\frac12+\frac12\right)^{2n+1}-\left(\frac12-\frac12\right)^{2n+1}-\frac1{2^{2n}}\right]=\frac{(-1)^n}{(2n+1)!}$$ Ooh, I was hoping for more mathematical rigor than that, but I didn't properly foresee the situation in that last step. I hope it is evident, however, that the series for $\sin x$ and $\cos x$ follow from formal multiplication of series as in the two steps shown above so someone working before Newton-Leibniz calculus like Madhava could have derived the series algebraically.
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