How to approximate

Question

I was reading a book and saw this approximation $(1 - 10^{-3})^{1023} \approx 2^{-1.476}$

I am wondering how it is calculated.

Answer 1

Making it more general, you wonder what is $x$ if $$(1-\epsilon)^{1023}=2^x$$ Taking logarithms, you find $$x=1023 \frac{ \log(1-\epsilon)}{\log(2)}$$ When $\epsilon$ is small, you have a very good approximation $$\log(1-\epsilon)\approx -\epsilon-\frac 12 \epsilon^2+\cdots$$ So, $$x\approx -1023\frac{\epsilon+\frac 12 \epsilon^2 }{\log(2)}$$ Ignoring the second term and using $\epsilon=10^{-3}$, we then have $$x\approx -\frac{1.023}{\log(2)}\approx -1.47588$$

Answer 2

For large $n$ we know that

$\left(1-\frac{x}{n}\right)^n\approx e^{-x}$

$(1 - 10^{-3})^{1023}=(1 -\frac{1}{1000})^{1023}=(1 -\frac{1,023}{1023})^{1023}\approx e^{-1.023}$

$2^x=e^{- 1.023}$

$x=\log_2(e^{- 1.023})=\frac{-1.023}{\ln(2)}\approx-1.476$

Thus $(1 - 10^{-3})^{1023}\approx 2^{-1.476}$

Answer 3

@Claude Leibovici provided a nice way to solve this using the polynomial approximation for $\log(x)$, where $x$ is small. Let me give an answer that digs deeper than a polynomial approximation:

We recall a commonly accepted definition for $e$, the natural number: $$ e = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n $$ When we try to approximate $e$, we simply find a large number $n$, such as $1000$, and evaluate it in a formula to find $e$. For example, with $n=1000$, we get $$ e \approx \left(1 + \frac{1}{1000}\right)^{1000} \approx 2.7169 $$ a pretty good approximation.

Now, what if my exponent is different than the denominator inside the parentheses? Do I still get $e$? To find out, let's do some algebra. $$ \left(1+\frac{1}{n}\right)^m = \left(1+\frac{1}{n}\right)^{n \cdot \frac{m}{n}} = \left(\left(1+\frac{1}{n}\right)^n\right)^{\frac{m}{n}} \approx e^{\frac{m}{n}} $$ Hmph. Notice that this requires $n$ to be large, but $m$ can be any value, small or large.

Now, it is also the case that $$ e = \lim_{n \to \infty} \left(1-\frac{1}{n}\right)^{-n} $$ (notice the two sign changes). Doing the same sort of algebra gets us $$ \left(1-\frac{1}{n}\right)^{m} = \left(\left(1-\frac{1}{n}\right)^{-n}\right)^{-\frac{m}{n}} \approx e^{-\frac{m}{n}} $$

We can then try to apply this to your question. $$ x := \left(1-10^{-3}\right)^{1023} = \left(1-\frac{1}{1000}\right)^{1023} \approx e^{-\frac{1023}{1000}} = e^{-1.023} $$ We then throw in the change-of-base formula to see that $$ \log_2 x = \frac{\log_e x}{\log_e 2} \approx \frac{-1.023}{0.693} \approx -1.476 $$

Whoa, fr$\large e \Large e \LARGE e\Huge e \LARGE e \Large e \large e$aky!

Answer 4

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$\ds{\pars{1 - 10^{-3}}^{1023} \approx 2^{-1.476}:\,?.}$

\begin{align} \color{#f00}{\pars{1 - 10^{-3}}^{1023}} & = 0.999^{1023} = \bracks{2^{\ln\pars{0.999}/\ln\pars{2}}}^{1023} = 2^{\color{#f00}{1023\ln\pars{0.999}/\ln\pars{2}}} \end{align} $$ \begin{array}{|c|}\hline\\ \ds{\quad\color{#f00}{1023\,{\ln\pars{0.999} \over \ln\pars{2}}} = \color{#f00}{-1.476\color{#00f}{62}}\quad} \\ \mbox{}\\ \hline \end{array} $$