How to find an approximation to $1 - \left( \frac{13999}{14000}\right )^{14000}$?

Question

I want to find an approximation to the expression $$ 1 - \left( \frac{13999}{14000}\right )^{14000} $$

I tried by taking logarithm $$ \ln P = \ln\left(1 - \left(\frac{13999}{14000}\right)^{14000}\right) \approx - \left(\frac{13999}{14000}\right)^{14000} $$

What should I do next? Is my procedure correct?

Answer 1

Note that the expression of interest is of the form

$$1-\left(\frac{n-1}{n}\right)^n=1-\left(1-\frac{1}{n}\right)^n$$

with $n=14000$. Recalling that

$$\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}$$

we have

$$1-\left(\frac{13999}{14000}\right)^{14000}\approx 1-e^{-1}$$

Answer 2

For a solution that gives you control over the precision of the result, you can do the following. First, write the expression as such:

$$1−\left(\frac{13999}{14000}\right)^{14000} = 1−\left(\frac{14000 - 1}{14000}\right)^{14000} = 1−\left(1 - \frac{1}{14000}\right)^{14000}$$

Then use the Binomial theorem

$$(1 + x)^n = \sum_{k\,=\,0}^{n} {n \choose k}\, x^k$$

with $x = -\frac{1}{14000}$ and $n = 14000$. Now, you definitely don't want to compute all those binomial coefficients. The first few terms should be enough to give you a useful approximation.

For example, stopping at $k=1$:

$$(1+x)^n = 1 + n\,x + \frac{n\,(n-1)}{2}\,x^2$$

Then,

$$1−\left(\frac{13999}{14000}\right)^{14000} = 1−\left(1 - \frac{1}{14000}\right)^{14000} \approx 1 - \left(1 + n\,x + \frac{n\,(n-1)}{2}\,x^2\right)$$

or

$$1−\left(\frac{13999}{14000}\right)^{14000} \approx -n\,x - \frac{n\,(n-1)}{2}\,x^2$$

Note that $x$ was chosen negative so the leading term above is actually a positive number (it's actually just 1). The result, then, is

$$1−\left(\frac{13999}{14000}\right)^{14000} \approx 1 - \frac{13999}{28000}$$

You probably don't want to stop at $k=1$, though, for a better approximation. The point is, the more terms you use, the better your approximation.

Answer 3

Another way of doing it : consider $$A=\Big({13999 \over 14000}\Big )^{14000}$$ $$\log(A)=14000 \log\Big({13999 \over 14000}\Big )$$ Now, consider the very fast converging series expansion $$\log\Big({{1+x} \over {1-x}}\Big )=2 \Big(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\Big)$$ and make ${{1+x} \over {1-x}}={13999 \over 14000}$ which gives $x=-\frac{1}{27999}$. The terms in $x^3$ and above do not play almost any role; so $$\log(A)\approx -14000 \times 2\times\frac{1}{27999}=-\frac{28000}{27999}=-1-\frac{1}{27999}$$ So $$A=e^{-1-\epsilon}=e^{-1}e^{-\epsilon}$$ $$1-A=1-e^{-1}e^{-\epsilon}$$ Now, since $\epsilon$ is very small $e^{-\epsilon}\approx 1-\epsilon$ which makes $$1-A=1-e^{-1}(1-\epsilon)=1-\frac 1e +\frac \epsilon e$$ with $\epsilon =\frac{1}{27999}$.

Using the numbers, the last approximation gives $$\color{red}{0.632133697}849$$ while the "exact" value would be $$\color{red}{0.632133697}771$$

Edit

Making the problem more general, considering $$1-\left(\frac{n-1}{n}\right)^n$$ using the same approach but including a few higher order terms in the expansions, you would get $$1-\left(\frac{n-1}{n}\right)^n=1-\frac 1e+\frac 1e \,\Big(\frac 1 {2n}+\frac 5 {24n^2}+\cdots\Big)$$ Using this formula would lead to an approximated value equal to $$\color{red}{0.6321336977710}56$$ while the "exact" value would be $$\color{red}{0.6321336977710}70$$