I noticed that the following sine formula:
$$\sum_{r=0}^{a} \binom{a}{r} \sum_{m=0}^{r} \sum_{k=0}^{m} \binom{m}{k} \binom{r}{m} (-1)^{m-k} 2^{m-k} \sin(k) = \sin(a)$$
seem to hold for integer values of 'a' but only numerically up to 7.
8 and above, the formula only approximates to sin(a) and the accuracy becomes worse for larger values of 'a'
May there be a proper reason why? or might it just be a program error?
It is true, and is true for any function, not just $\sin$.
Here is my very messy derivation.
There is undoubtedly a simpler one, but this is mine.
A question about the bounds of accuracy for a sine addition formula
$\sum_{r=0}^{a} \binom{a}{r} \sum_{m=0}^{r} \sum_{k=0}^{m} \binom{m}{k} \binom{r}{m} (-1)^{m-k} 2^{m-k} \sin(k) = \sin(a) $
$\begin{array}\\ s_f(a) &=\sum_{r=0}^{a} \binom{a}{r} \sum_{m=0}^{r} \sum_{k=0}^{m} \binom{m}{k} \binom{r}{m} (-1)^{m-k} 2^{m-k} f(k)\\ &=\sum_{r=0}^{a} \binom{a}{r} \sum_{k=0}^{r}\sum_{m=k}^{r} \binom{m}{k} \binom{r}{m} (-1)^{m-k} 2^{m-k} f(k)\\ &=\sum_{k=0}^{a}\sum_{r=k}^{a} \binom{a}{r} \sum_{m=k}^{r} \binom{m}{k} \binom{r}{m} (-1)^{m-k} 2^{m-k} f(k)\\ &=\sum_{k=0}^{a}f(k)(-1)^{-k}2^{-k}\sum_{r=k}^{a} \binom{a}{r} \sum_{m=k}^{r} \binom{m}{k} \binom{r}{m} (-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}f(k)(-1)^{-k}2^{-k}\sum_{r=k}^{a} \binom{a}{r} \sum_{m=k}^{r} \dfrac{m!r!}{k!(m-k)!m!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}f(k)(-1)^{-k}2^{-k}\sum_{r=k}^{a} \binom{a}{r} \sum_{m=k}^{r} \dfrac{r!}{k!(m-k)!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}\dfrac{f(k)(-1)^{-k}2^{-k}}{k!}\sum_{r=k}^{a} \dfrac{a!}{r!(a-r)!}r! \sum_{m=k}^{r} \dfrac{1}{(m-k)!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}\dfrac{f(k)(-1)^{-k}2^{-k}}{k!}\sum_{r=k}^{a} \dfrac{a!}{(a-r)!} \sum_{m=k}^{r} \dfrac{1}{(m-k)!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}\dfrac{f(k)(-1)^{-k}2^{-k}}{k!}\sum_{r=k}^{a} \dfrac{a!}{(a-r)!(r-k)!} \sum_{m=k}^{r} \dfrac{(r-k)!}{(m-k)!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}\dfrac{f(k)(-1)^{-k}2^{-k}a!}{k!(a-k)!}\sum_{r=k}^{a} \dfrac{(a-k)!}{(a-r)!(r-k)!} \sum_{m=k}^{r} \dfrac{(r-k)!}{(m-k)!(r-m)!}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}f(k)(-1)^{-k}2^{-k}\binom{a}{k}\sum_{r=k}^{a} \binom{a-k}{r-k}\sum_{m=k}^{r} \binom{r-k}{r-m}(-1)^{m} 2^{m} \\ &=\sum_{k=0}^{a}f(k)(-1)^{-k}2^{-k}\binom{a}{k}\sum_{r=k}^{a} \binom{a-k}{r-k}\sum_{m=0}^{r-k} \binom{r-k}{r-(m+k)}(-1)^{m+k} 2^{m+k} \\ &=\sum_{k=0}^{a}f(k)\binom{a}{k}\sum_{r=k}^{a} \binom{a-k}{r-k}\sum_{m=0}^{r-k} \binom{r-k}{r-k-m}(-2)^{m}\\ &=\sum_{k=0}^{a}f(k)\binom{a}{k}\sum_{r=k}^{a} \binom{a-k}{r-k}(1-2)^{r-k}\\ &=\sum_{k=0}^{a}f(k)\binom{a}{k}\sum_{r=k}^{a} \binom{a-k}{r-k}(-1)^{r-k}\\ &=\sum_{k=0}^{a}f(k)\binom{a}{k}\sum_{r=0}^{a-k} \binom{a-k}{r}(-1)^{r}\\ &=\sum_{k=0}^{a}f(k)\binom{a}{k}(1-1)^{a-k}\\ &=f(a)\\ \end{array} $
since $(1-1)^{a-k}=0 $ if $a-k>0$ and $1$ if $a=k$.
