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 $\ln$, $\exp$, $\sin$, $\sin^{-1}/\arcsin$.

score 2 · Answer 1 · answered Dec 12 '14 at 14:14

2

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.

answered Dec 12 '14 at 14:14

Henrik supports the community

6,831

Changed to often, although I can't see why it would be better to use $\dfrac{e^2}{\pi}-\sqrt3$ in any context where you'd want to have an approximate answer? Also, what number should you use for $x_0$? – Frank Vel Dec 12 '14 at 15:09
As far as I remember any real number would work, but $x_0=x$ (or $\frac{x}{2}$ which is a better guess for $x>2$) does a least. – Henrik supports the community Dec 12 '14 at 15:16
It might be that only positive numbers work, but if your first guess for the square root of a real number is negative... – Henrik supports the community Dec 12 '14 at 15:20

John Hughes · Answer 2 · 2014-12-12T15:16:46.417

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 $$ \log_2(x) = \frac{\ln(x)}{\ln (2)} \approx \frac{10}{7} \ln(x) $$

score 1 · Answer 3 · answered Apr 01 '16 at 07:36

$x_{n+1}=x_n+2 x_n (A-x_n^2)/(3 x_n^2+A)$ 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.

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

3 Answers3