I request you to kindly help me to understand the point of conflict that I struck at while deriving an alternative proof of Sin x expansion (apart from Taylor series approach). I have mentioned the steps of proof and also the point of paradox below. I would like to apologize in advance, in case you find difficulty in understanding the proof and would be happy to clarify any confusion or ambiguity in the steps of proof.
Please find the derivation of alternative proof of $\sin x = \sum^{\infty}_{k=0} {(-1)^k}\frac{x^{2k+1}}{(2k+1)!}$ expansion, below.
Please refer to this diagram link for understanding Section 1
Section 1
First I will prove that $\sin x = \lim _{dx\to 0}\sum^{\infty}_{n=0} dx*\cos ndx $
Consider sector of circle “A6 O A0”
angle “A6 O A0” = x radian.
Arc “A6 A0” = $(1 Unit)*(x Radian) = x Unit$ (Arc length = radius * angle in radian)
Now let us divide $\angle$ 'A6 O A0' into N(= 6 in diagram) equal parts, with each part = dx. Diving the angle will also divide arc "A6 A0" in N equal parts
Consider length “A6 B6”
$\sin x$ = “A6 B6” = “A6 C6” + “A5 C5” + “A4 C4” + “A3 C3” + “A2 C2” + “A1 C1” -- equation 1
We know that
“A6 A5” = “A5 A4” = “A4 A3” = “A3 A2” = “A2 A1” = “A1 A0” = dx unit -- equation 2
Let us consider example of right angle Triangle “O A4 B4”
->Angle “B4 O A4” = 4dx radian
->Then we can say that angle “C4 A4 A3” = 4dx radian as well (because line “O A4” perpendicular to “A4 A3”)
->Hence length “A4 C4” = “A4 A3”*$\cos 4dx = dx*\cos 4dx$ (“A4 A3” = dx unit, from equation 2)
Base on above example, we can say the following
“A6 C6” = $dx*\cos 6dx$, “A5 C5” = $dx*\cos 5dx$,
“A4 C4” = $dx*\cos 4dx$, “A3 C3” = $dx*\cos 3dx$,
“A2 C2” = $dx*\cos 2dx$, “A1 C1” = $dx*\cos 1dx$ -- equation 3
Hence,
$sin x = dx*\cos 6dx + dx*\cos 5dx + dx*\cos 4dx + dx*\cos 3dx + dx*\cos 2dx + dx*\cos 1dx$, From equation 1 and 2 and as per diagram where number of division, N = 6 )
Since in diagram, I have divided angle x into N=6 parts to make it easy to understand.
But actually, angle “A6 O A0” = x radian should be divided into N equal parts such that following conditions should hold
$\lim _{N\to \infty }$
$\lim _{dx\to 0 }$
x radian $= N * dx$
Based on this
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left(dx*\cos 1dx + dx*\cos 2dx + dx*\cos 3dx + dx*\cos 4dx + dx*\cos 5dx + dx*\cos 6dx + dx*\cos 7dx + \cdots + dx*\cos Ndx\right)$
Or
$\sin x = \lim _{N\to \infty }\lim _{dx\to 0}\sum^{N}_{n=1} dx*\cos ndx $
Hence Section 1 Proved
Section 2
In this section, I will further breakdown, transform and restructure equation $\sin x = \lim _{N\to \infty }\lim _{dx\to 0}\sum^{N}_{n=1} dx*\cos ndx $, in an attempt to achieve the desired result of Taylor's series expansion of Sin x
We know that, (Refer Link for more details on this formula)
$\cos n\theta = \sum_{r=0}^{2r \leqslant n}(-1)^r\binom{n}{2r}\left(\cos\theta\right)^{n-2r}{\left(\sin\theta\right)^{2r}}$
if $\theta = \lim _{dx\to 0} dx$, then above formula will become,
$\cos ndx = \sum_{r=0}^{2r \leqslant n}(-1)^r\binom{n}{2r}dx^{2r}$ -- equation 4,
because, $\lim _{dx\to 0} \cos dx = 1$ and $\lim _{dx\to 0} \frac{\sin dx}{dx} = 1$
In Section 1, we have already proved that
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left(dx*\cos 1dx + dx*\cos 2dx + dx*\cos 3dx + dx*\cos 4dx + dx*\cos 5dx + dx*\cos 6dx + dx*\cos 7dx + \cdots + dx*\cos Ndx\right)$
Using equation 4, we will get
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left(\sum_{r=0}^{2r \leqslant 1}(-1)^r\binom{1}{2r}dx^{2r} + \sum_{r=0}^{2r \leqslant 2}(-1)^r\binom{2}{2r}dx^{2r} + \sum_{r=0}^{2r \leqslant 3}(-1)^r\binom{3}{2r}dx^{2r} + \sum_{r=0}^{2r \leqslant 4}(-1)^r\binom{4}{2r}dx^{2r} + \sum_{r=0}^{2r \leqslant 5}(-1)^r\binom{5}{2r}dx^{2r} + \cdots + \sum_{r=0}^{2r \leqslant N}(-1)^r\binom{N}{2r}dx^{2r}\right)*dx$
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left({\Bigl [\binom{1}{0}dx^0\Bigr]}+{\Bigl[\binom{2}{0}dx^0-\binom{2}{2}dx^2\Bigr]}+{\Bigl[\binom{3}{0}dx^0-\binom{3}{2}dx^2\Bigr]}+{\Bigl[\binom{4}{0}dx^0-\binom{4}{2}dx^2+\binom{4}{4}dx^4\Bigr]}+{\Bigl[\binom{5}{0}dx^0-\binom{5}{2}dx^2+\binom{5}{4}dx^4\Bigr]}+\cdots+{\Bigl[\binom{m}{0}dx^0-\binom{m}{2}dx^2+\binom{m}{4}dx^4+\cdots\Bigr]} \right)*dx$
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left(-{\binom{0}{0}dx^0}+{\sum_{n=0}^{N}(-1)^{\frac{0}{2}}\binom{n}{0}dx^0}+{\sum_{n=2}^{N}(-1)^{\frac{2}{2}}\binom{n}{2}dx^2}+{\sum_{n=4}^{N}(-1)^{\frac{4}{2}}\binom{n}{4}dx^4}+{\sum_{n=6}^{N}(-1)^{\frac{6}{2}}\binom{n}{6}dx^6}+\cdots+{\sum_{n=(m-1)}^{N}(-1)^{\frac{(m-1)}{2}}\binom{n}{m-1}{dx^\left(m-1\right)}}+\cdots+{\sum_{n=(N-2)}^{N}(-1)^{\frac{(N-2)}{2}}\binom{n}{N-2}{dx^\left(N-2\right)}}+{\sum_{n=N}^{N}(-1)^{\frac{N}{2}}\binom{n}{N}{dx^N}}\right)*dx$
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left(-{\binom{0}{0}dx^1}+{\sum_{n=0}^{N}(-1)^{\frac{0}{2}}\binom{n}{0}dx^1}+{\sum_{n=2}^{N}(-1)^{\frac{2}{2}}\binom{n}{2}dx^3}+{\sum_{n=4}^{N}(-1)^{\frac{4}{2}}\binom{n}{4}dx^5}+{\sum_{n=6}^{N}(-1)^{\frac{6}{2}}\binom{n}{6}dx^7}+\cdots+{\sum_{n=(m-1)}^{N}(-1)^{\frac{(m-1)}{2}}\binom{n}{m-1}{dx^m}}+\cdots+{\sum_{n=(N-2)}^{N}(-1)^{\frac{(N-2)}{2}}\binom{n}{N-2}{dx^\left(N-1\right)}}+{\sum_{n=N}^{N}(-1)^{\frac{N}{2}}\binom{n}{N}{dx^\left(N+1\right)}}\right)$
In above equation we can ignore 1st term $-{\binom{0}{0}dx^1}$ because $\lim _{dx\to 0 }$
$\sin x$ = $\lim _{N\to \infty }\lim _{dx\to 0}\left({\sum_{n=0}^{N}(-1)^{\frac{0}{2}}\binom{n}{0}dx^1}+{\sum_{n=2}^{N}(-1)^{\frac{2}{2}}\binom{n}{2}dx^3}+{\sum_{n=4}^{N}(-1)^{\frac{4}{2}}\binom{n}{4}dx^5}+{\sum_{n=6}^{N}(-1)^{\frac{6}{2}}\binom{n}{6}dx^7}+\cdots+{\sum_{n=(m-1)}^{N}(-1)^{\frac{(m-1)}{2}}\binom{n}{m-1}{dx^m}}+\cdots+{\sum_{n=(N-2)}^{N}(-1)^{\frac{(N-2)}{2}}\binom{n}{N-2}{dx^\left(N-1\right)}}+{\sum_{n=N}^{N}(-1)^{\frac{N}{2}}\binom{n}{N}{dx^\left(N+1\right)}}\right)$
we know that x $= N * dx$, then after re-writing above equation, we will get
$\sin x$ = $\lim _{N\to \infty }\left((-1)^0\frac{\sum_{n=0}^{N}\binom{n}{0}}{N^1}x^1+(-1)^1\frac{\sum_{n=2}^{N}\binom{n}{2}}{N^3}x^3+(-1)^2\frac{\sum_{n=4}^{N}\binom{n}{4}}{N^5}x^5+(-1)^3\frac{\sum_{n=6}^{N}\binom{n}{6}}{N^7}x^7+\cdots+(-1)^k\frac{\sum_{n=2k}^{N}\binom{n}{2k}}{N^\left(2k+1\right)}{x^\left(2k+1\right)}+\cdots+(-1)^{\frac{(N-2)}{2}}\frac{\sum_{n=(N-2)}^{N}\binom{n}{N-2}}{N^\left(N-1\right)}{x^\left(N-1\right)}+(-1)^{\frac{N}{2}}\frac{\sum_{n=N}^{N}\binom{n}{N}}{N^\left(N+1\right)}{x^\left(N+1\right)}\right)$
Let us consider the kth term in above sequence
kth term $=(-1)^k\frac{\sum_{n=2k}^{N}\binom{n}{2k}}{N^\left(2k+1\right)}{x^\left(2k+1\right)}$
Kth term can be re-written as
kth term $=(-1)^k\frac{1}{N^\left(2k+1\right)}\Biggl(\sum_{n=0}^{(N-2k)}\frac{(2k+n)!}{n!}\Biggr)\frac{x^\left(2k+1\right)}{2k!}$
$\sin x$ = $\lim _{N\to \infty }\left( (-1)^0\frac{1}{N^1}\Biggl(\sum_{n=0}^{N}\frac{n!}{n!}\Biggr)\frac{x^1}{0!}+ (-1)^1\frac{1}{N^3}\Biggl(\sum_{n=0}^{(N-2)}\frac{(2+n)!}{n!}\Biggr)\frac{x^3}{2!}+ (-1)^2\frac{1}{N^5}\Biggl(\sum_{n=0}^{(N-4)}\frac{(4+n)!}{n!}\Biggr)\frac{x^5}{4!}+ (-1)^3\frac{1}{N^7}\Biggl(\sum_{n=0}^{(N-6)}\frac{(6+n)!}{n!}\Biggr)\frac{x^7}{6!}+ \cdots+ (-1)^k\frac{1}{N^\left(2k+1\right)}\Biggl(\sum_{n=0}^{(N-2k)}\frac{(2k+n)!}{n!}\Biggr)\frac{x^\left(2k+1\right)}{2k!}+ \cdots+ (-1)^{\frac{(N-2)}{2}}\frac{1}{N^\left(N-1\right)}\Biggl(\sum_{n=0}^{2}\frac{(N-2+n)!}{n!}\Biggr)\frac{x^\left(N-1\right)}{(N-2)!}+ (-1)^{\frac{N}{2}}\frac{1}{N^\left(N+1\right)}\Biggl(\sum_{n=0}^{0}\frac{(N+n)!}{n!}\Biggr)\frac{x^\left(N+1\right)}{N!}\right)$
Section 3 - Paradox
Let $P(k) = \lim_{N\to \infty} \Biggl [\frac{1}{N^\left(2k+1\right)}\Biggl(\sum_{n=0}^{(N-2k)}\frac{(2k+n)!}{n!}\Biggr)\Biggr]$
Then
$\sin x$ = $
(-1)^0P(0)\frac{x^1}{0!}+
(-1)^1P(1)\frac{x^3}{2!}+
(-1)^2P(2)\frac{x^5}{4!}+
(-1)^3P(3)\frac{x^7}{6!}+
\cdots+
(-1)^kP(k)\frac{x^\left(2k+1\right)}{2k!}+
\cdots\cdots\infty$ -- Equation 5
$\sin x = \sum^{\infty}_{k=0} {(-1)^k}P(k)\frac{x^{2k+1}}{2k!}$
As per Taylor's Series expansion
$\sin x$ = $
(-1)^0\frac{x^1}{1!}+
(-1)^1\frac{x^3}{3!}+
(-1)^2\frac{x^5}{5!}+
(-1)^3\frac{x^7}{7!}+
\cdots+
(-1)^k\frac{x^\left(2k+1\right)}{2k+1!}+
\cdots\cdots\infty$
$\sin x = \sum^{\infty}_{k=0} {(-1)^k}\frac{x^{2k+1}}{(2k+1)!}$
Paradox :- I need to evaluate expression P(k) for all k, $0 \leqslant k \leqslant \infty$ Ideally, P(k) should be equal to $ \frac{1}{2k+1} $ to get the result of this alternative proof to be equal to actual Taylor's Series expansion of Sin x. But when I attempt to evaluate P(k) (I may be wrong in evaluating it), I am getting P(k) equal to 1 irrespective of value of k. (I can provide steps on how I am attempting to evaluate P(k), if needed)
(Kindly Note :- I used below formula to evaluate P(k). I found it in Wikipedia for gamma function, in definition section. Click here to open Wikipedia link)
$\lim _{n\to \infty }{\frac {n!\;(n+1)^{m}}{(n+m)!}}=1$
Kindly provide your suggestions !