# Question 8

• Nov 27th 2006, 08:54 AM
ThePerfectHacker
Question 8
In Economics there is a rule that if you place money in a bank and each year earn a percentage $R$. Then the amount of years to double your money is, $\frac{72}{R}$

Say you have 100 dollars, and you earn 50% each year. So after 1 year you have 150 dollars and after 2 years you have >200. So it takes 2 years to double the money. And if we use the formula we get,
$\frac{72}{50}=1.44$ but this is only an approximation and the real answer can only involve integer answers. So this says that 1 year is too little and that 2 is enough. So the formula works.

Now, here is the problem. Explain where this number comes from.

NOTE: This is an applied math problem you do not need to be formal, can use graphs, can use calculators, can approximate and do whatever you want. As long as you can nicely explain this phenomena.
• Nov 27th 2006, 10:08 AM
OReilly
Quote:

Originally Posted by ThePerfectHacker
In Economics there is a rule that if you place money in a bank and each year earn a percentage $R$. Then the amount of years to double your money is, $\frac{72}{R}$

Say you have 100 dollars, and you earn 50% each year. So after 1 year you have 150 dollars and after 2 years you have >200. So it takes 2 years to double the money. And if we use the formula we get,
$\frac{72}{50}=1.44$ but this is only an approximation and the real answer can only involve integer answers. So this says that 1 year is too little and that 2 is enough. So the formula works.

Now, here is the problem. Explain where this number comes from.

NOTE: This is an applied math problem you do not need to be formal, can use graphs, can use calculators, can approximate and do whatever you want. As long as you can nicely explain this phenomena.

I don't see how this formula $\frac{{72}}{R}$ works.

Lets say you have 100 dollars and you earn 40% each year.

According to formula $\frac{{72}}{{R}}$ you need $\frac{{72}}{{40}} = 1.8$ years to double the money. Approximately 2 years according to formula.

But that is not correct. You need 3 years to double that money.
After first year you have 140 dollars, after second 196 and after third 274 dollars. So, you need 3 years.

Formula doesn't work.
• Nov 27th 2006, 10:17 AM
ThePerfectHacker
Quote:

Originally Posted by OReilly

Formula doesn't work.

I think I should have said "for small precentage values" because in reality no place gives so much precent.
• Nov 27th 2006, 10:40 AM
OReilly
Quote:

Originally Posted by ThePerfectHacker
I think I should have said "for small precentage values" because in reality no place gives so much precent.

OK.

I don't understand what question is.

Do we have to find exact formula for integer solution?
• Nov 27th 2006, 11:18 AM
TriKri
If we say $\begin{array}{cc}\text{lim}\\R \rightarrow 0\end{array}$ (I don't really knoh how to use limens, but I do so anyway), then the formula for the number of years until doubling the amount:

$N\ =\ \frac{log(2)}{\displaystyle{log\left(1+ \frac{R}{100\ \%}\right)}}\ =\ \frac{ln(2)}{\displaystyle{ln\left(1+ \frac{R}{100\ \%}\right)}}$
Can be as well written as
$N = \frac{ln(2)}{R/100\ \%} = \frac{100\cdot ln(2)\ \%}{R}$
since
$ln(x) = x - 1$ when $\begin{array}{cc}\text{lim}\\x \rightarrow 1\end{array}$.

Now $100 \cdot ln(2)$ isn't 72, it is actually closer to 69.3. So we would get the formula
$N = \frac{69.3\ \%}{R}$
assuming that $\begin{array}{cc}\text{lim}\\R \rightarrow 0\end{array}$. But that can't true, since the amount would not increase at all and the number of years until doubling would be infinite. Thus it's unfair to use the constant 69 or 69.3. I believe 72 is more of an average value for "different common R-values". To see what R-value they have used to get the value 72, we'd have to solve the equation

$\left(1 + \frac{R}{100}\right)^{\frac{72}{R}} = 2\ \Rightarrow\ ...\ \Rightarrow\ \frac{\displaystyle{ln\left(1+\frac{R}{100}\right) }}{R/100} = \frac{ln(2)}{0.72} \approx 0.963$
• Nov 27th 2006, 11:30 AM
TriKri
Does anyone have a tool for solving this equation:
$\frac{\displaystyle{ln\left(1+\frac{R}{100\ \%}\right)}}{R/100\ \%} = \frac{ln(2)}{0.72}$

(If it isn't solvable by hand :o)
• Nov 28th 2006, 07:49 PM
CaptainBlack
Quote:

Originally Posted by TriKri
Does anyone have a tool for solving this equation:
$\frac{\displaystyle{ln\left(1+\frac{R}{100\ \%}\right)}}{R/100\ \%} = \frac{ln(2)}{0.72}$

(If it isn't solvable by hand :o)

You need Lambert's W function to solve this in closed form.

RonL
• Nov 29th 2006, 07:19 AM
galactus
I looked into this topic some years back. It is an interesting little formula.

It's derivation is quite clever.

From what I can gather, the '72 rule' is most accurate with rates ranging from 6%-10%. For very 3% pts away from 8%, the value of 72 is adjusted by 1 point.

For instance, if we use a rate of 11%, then we will see the denominator is 73.

The higher your rates goes, it's better to use a numerator larger than 72.

If we start with the well-known formula $A=P(1+\frac{r}{n})^{nt}$ and let n=1, we get:

$A=P(1+r)^{t}$

Since we want the money to double, we have:

$2P=P(1+r)^{t}\Rightarrow{2=(1+r)^{t}}$

Solve for n and we get:

$t=\frac{ln(2)}{ln(1+r)}$

$t=\frac{ln(2)}{r}\cdot\frac{r}{ln(1+r)}$....[1]

Now, the trick is to use the Taylor series ln(1+r) expanded around 1.

$ln(1+r)=r-\frac{r^{2}}{2}+\frac{r^{3}}{3}-\frac{r^{4}}{4}+\frac{r^{5}}{5}-\frac{r^{6}}{6}+......$

So, $\frac{r}{ln(1+r)}$ is:

$\frac{1}{1-\frac{r}{2}+\frac{r^{2}}{3}-\frac{r^{3}}{4}+\frac{r^{4}}{5}-\frac{r^{5}}{6}+.....}$

In this particular series, the farther away we go the smaller the terms become:

interest rates are usually small decimals whose powers will be small and the denominator gets bigger as we go, thereby, making them even smaller.

So, we'll only use 6 terms, let's say. We probably don't need that many, but what the heck.

If we use 8% has a starting point, we get a fairly accurate representation of the '72 rule'.

$\frac{0.08}{ln(1+0.08)}\approx{1.039487}$

From [1]:

$t\approx{\frac{ln(2)}{r}\cdot(1.039487)}\approx{\f rac{0.72}{r}}$

We can use some other interest rate, like our previously mentioned 11% and see what we get.

$\frac{0.11}{ln(1.11)}\approx{1.05404}$

$ln(2)\cdot{{1.05404}}=0.7306$

See?. the rate was adjusted by approx 1 point because of the higher rate of 11%. Which is 3% higher than 8%.

72 is a nice average to use, though.

That's my 2 cents worth.
• Dec 4th 2006, 09:34 AM
ThePerfectHacker
I learned this rule in 12th Grade when my Economics teacher taught the class the rule. After he mentioned it I was curious to explain why this number comes into the equation.

This is mostly a statistics question. The explaination will be based on an analouge of the method of least squares.
The method of least squares says an approximating curve to a set of finite points is best when the sum of its squares is minimal (if anybody wants me to lecture on this I would. This is one of the things in math I love talking about). The analoague here is that we want to approximate a curve with another curve. And since we have to find the sum of the squares of the points he need to use an integral. Here is another way to look at it. One curve is very closely approximated to the given curve means when we compute the area between the actual curve and approximated curve we get a very small number. For simplicity reasons we compute the square of the area and try to minimize it.

Okay returning to the problem.
Say the bank offers an increase of $r$ in the money. And $P$ is the initial amount of money. Then,
1 year $P+Pr=P(1+r)$
2 years $P(1+r)+Pr(1+r)=P(1+r)^2$
3 years $P(1+r)^2+Pr(1+r)^2=P(1+r)^3$
...
In general,
$P(1+r)^n$ after $n$ years.

We need to find the value of $n$ such that.
$P(1+r)^n=2P$. Note, this value of $n$ might not be an integer. Say for example we get $n=1.5$ it does not mean 1.5 years because the bank only pays once a year not half a year. But what this number tells us is that $n=1$ is too little and $n=2$ exceeds $2P$ but this is the answer. Anyway, you understand what I am saying.

So we need to solve,
$P(1+r)^n=2P$ who places zero dollars in a bank?
$(1+r)^n=2$
This tells us that the doubling time does not depend the initial amount.
Take the logarithms (any base),
$n\ln (1+r)=\ln 2$
Thus,
$\boxed{ n=\frac{\ln 2}{\ln (1+r)}}$
Now let us graph this curve (shown below).

Wow! This curve looks familar.
This looks like a hyperbola.
So it has the equation,
$y=\frac{k}{x}$
Where $k$ is the ideal constant that will minimized the error.

Look at the picture.
The red curve is the exact value of $n$.
And the black curve is the ideal approximating hyperbola.
$y=.72/x$
Note they are almost identical when the rate of interest is small.
And that happens to be true, no bank gives for $r=5$:) .
So the reasonable interval is,
$.02\leq r\leq .1$

By what I said before we need to minizime,
$f(k)=\int_{.02}^{.1} \left( \frac{\ln 2}{\ln (1+x)}-\frac{k}{x} \right)^2 dx$
Open Parantheses,
$f(k)=\int_{.02}^{.1} \frac{\ln^2 2}{\ln^2 (1+x)}dx-k\int_{.02}^{.1}\frac{2\ln 2}{x\ln (1+x)} dx+k^2\int_{.02}^{.1} \frac{1}{x^2}dx$
Simpson's rule,
$f(k)=20-56.55k+40k^2$
Ideal value is when,
$f'(k)=0$
Thus,
$-56.55+80k=0$
Thus,
$k=.706$
Of course it depends what the reasonable interval is.
When I did this problem my first time I got something like,
$k=.7223$