Estimating $(\frac{37}{38})^{36}$

Question

I'm working through a bank of probability questions from 1965, so pre-calculator era (I think?). The solution to one of the questions confidently states that $\left(\frac{37}{38}\right)^{36}\approx 0.383$. Is this obvious/is there a trick here?

Answer 1

During the pre-calculator days, calculations such as what you're requesting (i.e., those involving using exponents, roots, products or quotients) were quite often done approximately using a logarithm table and/or a slide rule (although in fairly special cases such as here where $\frac{37}{38} \approx 1$, plus the exponent $36$ is relatively large (when considering convergence of a limit expression for $e^x$) and close to the $37$/$38$ fraction values, then certain short cuts using these details, such as done in several other answers here, may be used instead).

Using the properties of logarithms, for some base $b$ (most likely $10$ since that is what most tables were in although, in theory, you could use any base $\gt 1$, e.g., $e$), results in

$$x = \left(\frac{37}{38}\right)^{36} \implies \log_{b}(x) = 36(\log_{b}(37) - \log_{b}(38)) \tag{1}\label{eq1A}$$

A log table for base $b$ would then be used to estimate the values of $\log_{b}(37)$ and $\log_{b}(38)$ (e.g., see Log Table: How To Use Logarithm Table, With Examples & Practice Questions and 4 Ways to Use Logarithmic Tables - wikiHow for more details), usually to $4$, $5$ or $6$ significant digits each, then determine the difference, likely by hand as it's not too difficult.

Next, either by hand or by using a slide rule (e.g., see How to Use a Slide Rule (with Pictures) - wikiHow for how to use one), determine the product of this with $36$, thus giving an approximate value of $\log_{b}(x)$ in \eqref{eq1A}.

The next step is to take the fractional part of the result (also called the mantissa) and use the log table again to find the anti-log, i.e., determine approximately what value the log corresponds to, in particular a value between $1$ and $10$, and then possibly apply a decimal place adjustment. In this case, the value will be negative, so you will will need to use the value when added to $-1$ (which is the "characteristic") to get the appropriate mantissa to use, so after determining its anti-log, move the decimal point to the left by one.

Finally, due to the various cumulative errors inherent with using a log table (and a slide rule if it was also used), you may wish to determine an appropriate number of significant digits to keep, with this apparently being done in your example, to come up with a final approximate result of $0.383$.

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For your specific example, using the log table base $10$ given in Log Table: How To Use Logarithm Table, With Examples & Practice Questions, looking at the row with $37$, and column $0$ (since the value we're using is, to $4$ significant digits, actually $37.00$, there's no need to use the "Mean Difference" section here), gives after determining its characteristic is $1$, a result of

$$\log_{10}(37) \approx 1.5682 \tag{2}\label{eq2A}$$

and, similarly for $38$, gives

$$\log_{10}(38) \approx 1.5798 \tag{3}\label{eq3A}$$

Calculating by hand next gives that

$$\log_{10}(37) - \log_{10}(38) \approx -0.0116 \tag{4}\label{eq4A}$$

Next, after multiplying by hand, \eqref{eq1A} gives

$$\log_{10}(x) \approx -0.4176 = -1 + 0.5824 \tag{5}\label{eq5A}$$

Using the log table now to find where $5824$ is located gives it's in row $38$ and between columns $2$ and $3$. Since column $2$ has $5821$, which is $3$ less, the mean difference column indicates a value of $3$, giving an approximate result of $3823$. Due to the characteristic of $-1$ in \eqref{eq5A}, this then gives

$$x \approx 0.3823 \tag{6}\label{eq6A}$$

Here, rounding to $3$ decimal places gives $0.382$, which is a bit less than what your estimate is. There are several possible reasons for this difference, with the most likely one being that a more accurate logarithm table (e.g., with $5$ or $6$ significant digits instead of $4$) was used in its calculation.

Answer 2

Recall: $$\lim\limits_{n\to\infty}\left(1+\frac{x}{n}\right)^n = e^x.$$

So for a large $n:$ $$\left(1-\frac{1}{n}\right)^{n-2} = \left(\dfrac{1}{1+\dfrac{1}{n-1}}\right)^{n-2}=\dfrac{n}{n-1}\cdot \left(1+\dfrac{1}{n-1}\right)^{1-n}\approx\dfrac{e^{-1}n}{n-1}$$

Answer 3

The Binomial expansion is never a good approach to this type of problems because the binomial coefficient rise very sharply, and you cannot take just the first terms.

In this case you may put $$ \eqalign{ & \left( {{{37} \over {38}}} \right)^{36} = \left( {1 - {1 \over {38}}} \right)^{36} = {1 \over {\left( {1 - {1 \over {38}}} \right)^2 }}\left( {1 - {1 \over {38}}} \right)^{38} \approx \cr & \approx {{e^{\, - 1} } \over {\left( {1 - {1 \over {38}}} \right)^2 }} \approx {{\left( {1 + {2 \over {38}}} \right)} \over e} \cr} $$

Answer 4

One solution is to set $x = 1/36$ and expand around $x=0$.

By this approach we have $$ \left(\frac{37}{38}\right)^{36} = \left( \frac{1+x}{1+2 x}\right)^{\frac{1}{x}} = \exp \left( \frac{1}{x} \log\left( \frac{1+x}{1+2x} \right) \right) = \exp\left( -1 + \frac{3 x}{2} + O(x^2) \right) = \frac{2+ 3 x}{2 e} + O(x^2) $$ plugging in the value of $x$ yields $$ \left(\frac{37}{38}\right)^{36} \approx \frac{25}{24 e} = 0.383\ldots $$

Answer 5

hint

$$\frac{37}{38}=1-\frac{1}{38}$$

$$(1-x)^n\approx 1-nx+\frac{n(n-1)}{2}x^2$$

Answer 6

Hint. Use

$$ \left ( 1- \frac{1}{38}\right ) ^{36} \sim 1- \frac{36}{38}+ \binom{36}{2} \frac{1}{38^2} $$