2
$\begingroup$

The double angle formulas (101) and (102), combined with the best linear approximations at $0$ (103) and (104) can be used to estimate $\sin$ or $\cos$ by expanding an expression $n$ times and shrinking the angle to $\frac{\theta}{2^n}$ and then applying the linear approximation.

$$ \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \tag{101} $$ $$ \cos(\theta) = \cos^2\left(\frac{\theta}{2}\right) - \sin^2\left(\frac{\theta}{2}\right) \tag{102} $$

$$ \sin(\theta) \approx \theta \tag{103} $$ $$ \cos(\theta) \approx 1 \tag{104} $$

It seems to converge slowly, maybe adding a bit of accuracy or less per iteration. How quickly does it converge to the true value of $\sin(\theta)$ ?

Here's a sample illustrating the relatively slow convergence.

$$ \begin{array}[cc] \;n & \sin_n(1) \\ 0 & 1.0 \\ 1 & 0.9375 \\ 2 & 0.89233017 \\ 3 & 0.86740446 \\ 4 & 0.8545371 \\ 5 & 0.84802556 \\ 6 & 0.84475327 \\ 7 & 0.8431133 \\ \end{array} $$

$\endgroup$
1
$\begingroup$

The error decays like a constant multiple of $2^n$, approximately halving at each step.

To calculate the speed, we first put this estimate into a more conventional form. Note that if we use the complex form $e^{i\theta}=\cos\theta+i\sin\theta$, this estimate becomes $e^{i\theta}\approx \left(1+\frac{i\theta}{2^n}\right)^{2^n}$, so $$\sin_n(1) = \operatorname{Im}\left(\left(1+\frac{i}{2^n}\right)^{2^n}\right)=\operatorname{Im}\left(\exp\left(2^n\ln\left(1+\frac{i}{2^n}\right)\right)\right)$$ Estimate that logarithm with the power series; $\ln\left(1+\frac{i}{2^n}\right)=\frac{i}{2^n}+\frac{1}{2\cdot 2^{2n}}+O\left(\frac1{2^{3n}}\right)$.
Multiply by $2^n$ and exponentiate for $(\cos 1+i\sin 1)\cdot \left(1+\frac1{2\cdot 2^n}+O(2^{-2n})\right)\cdot (1+O(2^{-2n}))$. We extract the imaginary part of that for $$\sin_n(1)=\sin 1+\frac1{2^n}\cdot\frac{\sin 1}{2}+O(2^{-2n})$$ There it is - the precise decay rate of the error, including the constant factor $\frac{\sin 1}{2}$.

That doesn't quite match your reported data, being off by a factor of $2$. Looking at that data, it looks like you've underreported the number of steps for each estimate by $1$. (This is now corrected in the question.) I get $$\sin 1=2\sin\frac12\cos \frac12\approx 2\cdot\frac12\cdot 1=1$$ for $n=1$ step and $$\sin 1= 2\sin\frac12\cos\frac12=4\sin\frac14\cos\frac14(\cos^2\frac14-\sin^2\frac14)\approx 4\cdot \frac14\cdot (1-\frac1{16})=0.9375$$ for $n=2$ steps.

$\endgroup$

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.