# taylor expansion in compounded interest [closed]

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.

## closed as off-topic by LocalVolatility, Helin, Quantuple, JejeBelfort, amdoptOct 4 '17 at 16:55

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(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}