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I would like to know if there was a way to approximate i here without a financial calculator, in the following equation:

$\displaystyle -50000 + \frac{12992}{1+i} + \frac{12992}{(1+i)^2} + ⋯ + \frac{12992}{(1+i)^5} = 0 $.

or written in an other way:

$\displaystyle -3848,52 + \dfrac{1- \dfrac{1}{(1+i)^5} }{i}=0 $

I’m pretty sure there is way to do so, but I don’t know how.

I know that $i=0.0941$ here, but I would like to get it without a financial calculator.

Any help appreciated. Thanks in advance.

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Newton-Raphson works fine. –  AD. Sep 2 '13 at 12:54
    
Note that if it's all multiplied out, it becomes an equation of degree 6. In general, such equations can only be solved by numerical methods (such as the suggestion of Newton's Method made above). –  Gerry Myerson Sep 2 '13 at 13:02
    
Thanks to all for your help. –  XCoder Sep 2 '13 at 13:28

2 Answers 2

up vote 4 down vote accepted

If we multiply everything out, we arrive at:

$$\dfrac{-16 (3125 i^5 + 14813 i^4 + 27190 i^3 + 23130 i^2 + 7505 i -935)}{(1 + i)^5}$$

This means that we are left to solve:

$$f(i) = 3125 i^5 + 14813 i^4 + 27190 i^3 + 23130 i^2 + 7505 i -935 = 0$$

Unfortunately, this is a quintic function which typically requires numerical methods.

A plot shows:

enter image description here

As the comments mention, we can use Newton-Raphson (or a bevy of other root finding algorithms) as (changed variables to $w$ so not to confuse iterate names), so we have $w_{n+1} = w_n + f(w)/f'(w)$, thus

$$w_{n+1} = w_n - \dfrac{3125 w_n^5+14813 w_n^4+27190 w_n^3+23130 w_n^2+7505 w_n-935}{15625 w_n^4+59252 w_n^3+81570 w_n^2+46260 w_n+7505}$$

From the plot, and using a starting point of $w_0 = 0.17$, this method converges quickly (5 steps) to $0.09411009434376908$.

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How'd I miss this earlier? +1 –  amWhy Sep 2 '13 at 14:48
    
The thing is that I cannot do that in a simple way with a basic calculator during an exam where I have one minute by question, isn't it ? –  XCoder Sep 4 '13 at 22:12
    
I'm better trying all supplied multiple choices I think. –  XCoder Sep 4 '13 at 22:12
    
Well, most calculators come with root finding built-in, or it can be coded up (that is how I used to do it) and supply the function and derivative and range and it would find the root. Trying the choices works too. Regards –  Amzoti Sep 4 '13 at 22:32

There is another way to get this, which is not exactly Newton-Raphson, but it's much simpler. I assume you have a calculator which has a table function, such as the TI-30XS Multiview, which is allowed on actuarial exams in the United States. And, the answer by Amzoti would require the same thing any way.

Plug in your equation to a Table function, i.e., a function where you can plug in the equation and then plug in various inputs, $i$, to see what the function evaluates to at $i$. Then plug in 0 and 1. You know the answer is between these. In this problem, you would get a negative number when you plug in 0 and a positive number when you plug in 1. Since the function is continuous, this guarantees there is a solution between 0 and 1. So, you plug in something between 0 and 1, possibly 0.5, or since it's interest rates, maybe you go smaller like 0.1 or 0.05 or 0.2. Then slowly iterate until you get to 0. At each step, you will have an $a$ and a $b$ such that $f(a)$ is positive and $f(b)$ is negative. So, you keep choosing numbers $c$ such that $a < c < b$. This won't necessarily converge as fast but it's a lot simpler and probably faster.

The bisection method is this basic idea, except the bisection method requires that you always choose the midpoint $c = \frac{a+b}{2}$.

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