Question

Show that the "doubling time" (time for your money to double) is n periods at interest rate$\mathbf{\text{i}}\mathbf{\%}$ per period with $\mathbf{\text{ni}}\mathbf{=}\mathbf{69}$, approximately. Show that the error in the approximation is less than$\mathbf{10}\mathbf{\%}$ if, $\mathbf{\text{i}}\mathbf{\%}\mathbf{<}\mathbf{20}\mathbf{\%}$. (Note that does not have to be the number of years; it can be the number of months with $\mathbf{\text{i}}$=interest rate per month, etc.)

Short answer

It proved that the doubling time is $n$ periods at interest rate $\text{i}\%$ per period with $\text{ni}=69$, approximately; and the error in the approximation is less than$10\%$ . If $\text{i}\%$ is less than or equal to $20\%$.

Step-by-step solution

10%Determine the doubling time is   periods at interest rate   per period with ni=69, approximately; and the error in the approximation is less than 10%. If  i% is less than or equal to20%

As per the given in the question,

${\left(1+\frac{i}{100}\right)}^n=2$

Taking natural logarithm on both sides,

$$\begin{gathered} ln{\left(1+\frac{i}{100}\right)}^n=\ln 2 \\ \therefore n\ln \left(1+\frac{i}{100}\right)=\ln 2 \\ \therefore \ln \left(1+\frac{i}{100}\right)=\frac{\ln 2}{n} \end{gathered}$$

…… (1)

Now, according to equation $\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{4}}}{\mathbf{4}}\mathbf{+}\mathbf{\cdots }\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{<}\mathit{x}\mathbf{⩽}\mathbf{1}$.

In our case,$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{i}\mathbf{/}\mathbf{100}$and it is indeed less than 1 as$\mathit{i}$is less than 100. So, it can apply equation. If it approximates to the first term only

$\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}$ …… (2)

Now in this approximation, the absolute error will be given by the second term. If it denotes the absolute error by epsilon,

$$\begin{gathered} \mathbf{\in }\mathbf{=}\left|\frac{-x^2}{2}\right| \\ \mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}} \end{gathered}$$ …… (3)

Replacing $x$ by $\text{i}/100$ in step (2) and using step (1),

$$\begin{gathered} \ln \left(1+\frac{i}{100}\right)=\frac{i}{100} \\ =\frac{\ln 2}{n} \\ \therefore ni=100\ln 2 \end{gathered}$$

Putting value of $\ln \left(2\right)$ approximated to two decimals,

$$\begin{gathered} ni\approx 100\times 0.69 \\ \therefore ni\approx 69 \end{gathered}$$

Hence, it is proved that $ni\approx 69$.

Determine the relative error

Now, the relative error in $\ln (1+x)$ is:

$\text{relativeError}=\frac{\in }{\ln (1+x)}$

From equations and,

$\text{relativeError}=\frac{x^2/2}{x}$

relativeError $=\frac{x}{2}$

Replacing $\text{x}$by$\text{i}/100$ ,

relativeError $=\frac{i}{200}$

Percentage error is relative error times 100,

percError$=\frac{i}{2}$

percError$⩽\frac{20\%}{2}$ if $i\%⩽20\%$

percError$⩽10\%$ if $i\%⩽20\%\dots \dots \dots .$.

Hence, the required result is proved.

Conclusion:

It has proved that the doubling time is periods at interest rate $\text{i}\%$ per period with $\text{ni}=69$, approximately; and the error in the approximation is less than $10\%$. If is less than or equal to $20\%$.