Approximating Compound Interest with the Taylor Series for Logarithms

# taylor expansion in compounded interest


# taylor expansion in compounded interest












You invest $1, 000$ dollars for $10$ years at a $5$% yearly interest rate. After each year the interest paid is reinvested at the same rate.

(a) Represent the total amount A after ten years in the form $A = S(1 + x)^n$. What are $S$, $x$, and $n$ here?

$ S = 1000, x = .05, n =10$

(b) Compute A (up to cents) using a calculator.

Plugging in the corresponding numbers

$A=1628.89$ dollars

This is where I begin to struggle with the problem

(c) If one writes $A$ in the form $A = S(1+x)^n = S\exp(n\log(1+x))$ and uses the approximation $log(1 + x) ≈ x$ for small $x$, then one obtains $A ≈ Se^{nx}$. Which value for A is obtained with this approximation method?

(d) A better approximation is $log(1 + x) ≈ x − 1/2 x^2$. Explain how this is related to finding the taylor approximation for $\log(1+x)$ which is

$log(1+x)=x-x^2/2+x^3/3-x^4/4+x^5/5+...=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$

(e) Based on the approximation in (d) develop an approximation formula for A in terms of S, x, and n. Use it to compute A in our specific example.

## Answer by caverac (score 2, accepted)

https://quant.stackexchange.com/a/36269

> (c) If one writes $A$ in the form $A=S(1+x)^n=S e^{n\ln(1+x)}$ and uses the approximation $\ln(1+x)\approx x$ for small $x$, then one obtains $A\approx S e^{nx}$. Which value for $A$ is obtained with this approximation method?

This is is based on the properties

\begin{eqnarray} \ln (a^b) &=& b\ln a \tag{1a}\\ e^{\ln a} &=&a \tag{1b} \end{eqnarray}

Therefore

$$ A = S(1+x)^n \stackrel{(1b)}{=} Se^{\ln(1 + x)^n} \stackrel{(1a)}{=} Se^{n\ln(1 + x)} \tag{2} $$

Now, you can use the fact that when $|x|< 1$ then

$$ \frac{1}{1+x} = 1 - x + x^2 - \cdots = \sum_{k=0}^{+\infty}(-1)^k x^k $$

if you integrate both sides, you get

$$ \ln(1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3}- \cdots = \sum_{k=0}^{\infty}(-1)^k \frac{x^{k+1}}{k+1} \tag{3} $$

such that for small $x$ it is enough to take just the first term in this sum

$$ \ln(1 + x)\approx x \tag{4} $$

Eqn (2) then becomes

$$ A_1\approx Se^{nx} = 1648.72 $$

I will use the subscript $1$ to emphasize we are taking just the first term in the series

> (d) A better approximation is $\ln(1 + x) \approx x - x^2/2$. Explain how this is related to finding the Taylor approximation of $\ln(1 + x)$ which is $$ \ln(1 + x) = x -\frac{x^2}{2} + \frac{x^3}{3}- \cdots = \sum_{k=0}^{\infty}(-1)^k \frac{x^{k+1}}{k+1} $$

You just need to take the first two terms in Eq. (3). Here is the thing, if $x$ is small $x^2/2$ is smaller, and $x^3/3$ is even smaller, and $\dots$ So the more terms you include the better, but the smaller the correction in the overall result. Up to second order

$$ A_2\approx Se^{n(x - x^2/2)} = 1628.24 $$

which looks almost like the exact result you found before. If you're to include up to third order you'd see how it approaches even better the exact solution.

> e) Based on the approximation in (d) develop an approximation formula for $A$ in terms of $S$, $x$, and $n$. Use it to compute $A$ in our specific example.

In general if you take up to $N$ terms in the sum this is what you get

$$ A_N \approx Se^{n(x - x^2 - \cdots + (-1)^{N+1}x^{N}/N)} $$

Below there's some results

\begin{eqnarray} A_{\rm exact} &=& 1628.89 \\ A_1 &=& 1648.72 \\ A_2 &=& 1628.24 \\ A_3 &=& 1628.92 \\ A_4 &=& 1628.89 \end{eqnarray}