# How to estimate numbers like $(19/20)^{30}$

Is it possible to estimate by hand what is the value of expresion like $(19/20)^{30}$?

$$19/20 = 0.95$$ but $$(19/20)^{30} \approx 0.2146$$

So it is totally different number.

• Estimate in one's head, sure. $(1-1/20)^{20}$ is about $e^{-1}$, so our number is about $e^{-1.5}$. Dec 1, 2015 at 22:06
• A better estimate for $1/e$ is $(1-1/20)^{19.5}$ Dec 1, 2015 at 23:00

Use $(1-\frac 1N)^M=e^{M\log(1-1/N)}\approx e^{-\frac M N}$ to the first order or:

$$(1-\frac 1 N)^M\approx e^{-\frac {M}{N-1/2}}\approx e^{-\frac M N-\frac M {2N^2}}\approx e^{-\frac M N}(1-\frac M{2N^2})$$ to the second order together with $e^3\approx 20$ ($20.1$ if you want 3 significant digits) to get $$(19/20)^{30}\approx e^{-30/20}={e^{-3/2}}\approx 20^{-1/2}=\frac {\sqrt 5}{10}\approx 0.22$$ or more precisely $$(19/20)^{30}\approx \frac {1}{\sqrt{20.1}}(1-\frac 3 {80})\approx 0.215$$ which agrees with the real value ($\approx 0.2146$) in all significant digits.

We can use the binomial theorem: $$(1-x)^{30}=1-{30\choose 1}x+{30\choose 2}x^2+O(x^3)$$ where $O(x^3)$ is small when $|x|$ is small. In our case, $x=\frac{1}{20}$, which is indeed small. Hence a first approximation is $$(1-x)^{30}\approx 1-30x+435x^2$$ so $$\left(\frac{19}{20}\right)^{30}\approx 1-30\left(\frac{1}{20}\right)+435\left(\frac{1}{20}\right)^2=0.5875$$ As you can see, it's not that good an approximation, as $\frac{1}{20}$ isn't small enough in comparison to the rather large exponent of $30$. We can do better by taking more terms: $$(1-x)^{30}\approx 1-30x+435x^2-4060x^3$$ so $$\left(\frac{19}{20}\right)^{30}\approx 1-30\left(\frac{1}{20}\right)+435\left(\frac{1}{20}\right)^2-4060\left(\frac{1}{20}\right)^3=0.08$$

We're getting closer, but we really need a few more terms to get a decent estimate. However, if we were trying to find $\left(\frac{199}{200}\right)^{30}$ instead, we would have $$\left(\frac{199}{200}\right)^{30}\approx 1-30\left(\frac{1}{200}\right)+435\left(\frac{1}{200}\right)^2=0.860875$$ while the exact answer is $0.860384\ldots$, i.e. very close.

• So it's not a good way for those numbers... Dec 1, 2015 at 20:54
• It's a decent way, but you need eight or nine terms before you get a good estimate. Dec 1, 2015 at 20:55
• @rop, there is no absolute answer for whether it is a "good way" or not when it comes to approximation methods. It's just a question of how close you need to be.
– Paul
Dec 1, 2015 at 22:09

You could use the Taylor series expansion of $\left({19 \over 20}\right)^x$ if you know an approximation of $\log {19 \over 20}$.

Suppose the Taylor approximation is $$T (b,x) = 1 +\sum_{n=1}^b { (x \cdot \log{19 \over 20} )^n \over n!}$$

Now, $\lim_{b \to \infty} T(b, 30) = \left({19 \over 20}\right)^{30}$, but you only need $7$ terms to get three significant digits of your approximation; $T(7, 30) \approx 0.21397 \dots$

Even if you approximate $\log {19 \over 20} \approx -0.05$, then your approximation will be $\frac{3191}{14336} \approx 0.222 \dots$

If you have access to base 10 logarithm tables, there would be a good and ancient way to find your answer: $\log_{10} (19/20)^{30}=30\log_{10} (19/20)=30(\log_{10} 19-\log_{10} 20)$. Once you have found the logarithms in your table and done the one subtraction and one multiplication, you can find the "anti-logarithm" ($10^{\log_{10} (19/20)^{30}}$) again using a logarithm table to get your answer.