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I'm having troubles understanding how to isolate for the interest with either the future value or present value equations. Say we know what $K$ and $n$ are, and I'm trying to find $j$:

$$\frac{(1+j)^n-1}{j}=K$$

or

$$\frac{1-v^n}{j} = K$$

where $v = \frac {1}{1+j}$

lulu
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user281096
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  • second equation should be: (1-v^n)/j = K – user281096 Nov 04 '15 at 18:40
  • Note: I reformatted your question pretty extensively...please check to make sure I didn't accidentally change your meaning. In particular, I couldn't tell whether you wanted the first equation as I wrote it or if you wanted $(1+j)^{n-1}$ instead. – lulu Nov 04 '15 at 18:46
  • No that's what I wanted thank you ! It makes way more sense. I just want to note that the these 2 equations are independent of each other (i.e. the Ks don't equal). I'm looking to find out how I can get the value of j with the first equation. and how I can get j with just the second equation. Thanks ! – user281096 Nov 04 '15 at 18:51
  • Unless I'm missing something (always possible) these equations don't have simple closed solutions. Numerical methods work well. – lulu Nov 04 '15 at 18:54
  • Here's an a question example : ((1+j)^30 -1)/j=41.487451 and the answer is j= 0.0214273. But I have no idea how to get that answer – user281096 Nov 04 '15 at 18:56
  • @user281096 I misinterpretated your question. But agree to lulu. – callculus42 Nov 04 '15 at 18:57
  • Wolfram alpha is your friend! just do a mesh. I just ran it in excel...took only a few seconds. But...it's a 30 degree polynomial! Easy to rule out quick solutions. – lulu Nov 04 '15 at 19:10
  • By a "mesh" I mean: just compute $(1+j)^{30}-41.487451j-1$ for $j={.01,.011,.012,...}$. You'll see that it changes sign between $.021$ and $.022$. So then, start at $.021$ and use a finer mesh. There are better numerical methods available, but this works and it is very fast. – lulu Nov 04 '15 at 19:12
  • If I get this kind of question on a test I need to be able to do it without technology ? – user281096 Nov 04 '15 at 19:12
  • Well, ok. I don't see how to get the answer $.0214273$ without a computer. Doesn't mean there isn't a way to do it! I often miss things. Like I say, I'd do it numerically. Sorry! – lulu Nov 04 '15 at 19:14
  • @user281096 I´ve edited my answer. – callculus42 Nov 04 '15 at 19:24

2 Answers2

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You have $\frac{(1+j)^{30} -1}{j}=41.487451$. Let $j+1=q \Rightarrow q-1=j$

And let $a=41.487451$

We get $\frac{q^{30} -1}{q-1}=a$

Multiplying both sides by $q-1$.

$q^{30} -1=aq-a$

Putting all terms on the RHS.

$q^{30}-aq -1+a=0$

Substituting back

$q^{30}-41.487451\cdot q +40.487451=0$

This equation can not be solved algebraically. You have to use an approximation method, for instance the Newton-Raphson method.

We know that j is greater than 0 and almost surely not bigger than $10\%=0.1$ since $j$ is an interest rate or something similar. And therfore q is between $1$ and $1.1$ A good initial value for q should be $1.05$

callculus42
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Given that $w= j+1$, you have \begin{align} K &= \frac{w^n-1}{w -1}\\ \end{align}

We know that \begin{align} (w^n-1) &= (w-1)(w^{n-1}+w^{n-2}+\cdots+1) \end{align}

Therefore you can obtain the the value for $w$, and $j$ consequently, by finding the roots of the equation \begin{align} (w^{n-1}+w^{n-2}+\cdots+1) -K = 0 \end{align}

Doing it on paper I wouldn't expect $n\geq3$.

Marco Aguiar
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