What are some quickly convergent, easily calculated approximations for common functions for when you've forgotten a calculator to a test?

Question

I think it's nice not to rely too much on a calculator, whether it's forgotten or forbidden. Approximations can be useful on exams when you want a good guess at the answer to see if it's somewhat correct. Evaluating decimals are often better than $\dfrac{e^2}{\pi}-\sqrt3$.

For instance $e^2$ can be approximated as $\left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4} \approx 6$, but $e^2 = 7.238905...$ so it's not a very good approximation. So what are some nice convergent approximations for commonly used functions? I'm thinking mostly of functions like $\ln$, $\exp$, $\sin$, $\sin^{-1}/\arcsin$.

Answer 1

I don't agree that "Evaluating decimals are always better...", but

$$ x_{n+1} = \frac{1}{2}\left(x_n+\frac{x}{x_n}\right) $$ converges to $\sqrt{x}$ quite quickly.

Answer 2

I agree with @Hebrik that "decimals are better" is dubious. But:

$\sin(x) \approx x$ for small $x$; for not-small $x$, the half-angle formula (together with rules like $\sin(x) = \sin (\pi - x)$ get you to small $x$ fairly quickly.

And $\ln(1+x) \approx x$ for small $x$, and you can use $\ln(x/e) = \ln(x) - 1$ to rapidly reduce the size of $x$ until it's small. And if dividing by $e$ seems like a pain, work with $\log_2$, and convert to $\ln$ at the end via $$ \log_2(x) = \frac{\ln(x)}{\ln (2)} \approx \frac{10}{7} \ln(x) $$

Answer 3

$x_{n+1}=x_n+2 x_n (A-x_n^2)/(3 x_n^2+A)$ for $x_1>0<A,$ converges to $\sqrt A$ faster than $y_{n+1}=y_n+(A-y_n^2)/2 y_n.$

$\pi$ is close to $355/113=3+16/113=3+1/(7+1/16)=3.14159292.... $

$e^2$ is about $2.72^2=2.7^2+2(2.7)(0.02)=0.2^2=7.29+0.108+.0004....$. Use basic algebra to break a product into parts, using the Distributive law.

For $\log x,$ if have a good approximation for $\log y$ with $y/x$ close to $1$, then $\log x=\log y+\log (1+(x-y)/y)=\log y +(z-z^2/2+z^3/3-...)$ where $z=(x-y)/y.$

A quick method of approximate long division: Example:$ X=1/2.7376:$ $$X=0.3/0.82128.$$ Add $6\times 0.1$ to LHS and $6\times 2.7376$ to RHS ,giving $$X=0.36/0.985536.$$ Add $ 5\times 0.01$ to LHS and $5\times 0.026378$ to RHS giving $$X=0.366/0.999194.$$ The next step will show that $X$ is quite close to $0.3663$. The idea is to add (or subtract)a multiple of some LHS already obtained, to the LHS,and add (or subtract) the corresponding multiple to the RHS, getting the RHS closer to $1$.

An analogous method works for some approximate multiplication. e.g $$e^2=(2.71828...)^2=(0.90609...)(8.15484)=(1.00677 [approx])(7.34645...)$$ (For the next step multiply $7.34645$ by $(1-1/150)$ to get a good approximation to $e^2$).

Look at The Trachtenburg (Trachtenberg?) Speed System of Basic Mathematics. There was a Q similar to yours and several people recommended this book, and the OP was very impressed with it.I have it myself.