Force of Interest

[joelee88]

Hi guys, just have a question about Force of Interest that has been bugging me since yesterday.

If you refer to the Acted CT1 course note, on page 12, Section 2.1:

It gives the formula i^(p) = p[(1 + i)^(1/p) - 1]

It then says when p approaches infinity, the limit will be 0.0488 (as per the graph shown).

Well from my understanding, if p approaches infinity, we "substitute" infinity in the formula i^(p) = p[(1 + i)^(1/p) - 1], which will then give i^(infinity) a value of 0, am I correct? Is that is so, why is 0.0488 the limit and not 0 instead?

Thanks a lot in advance for your help.

 

[John Lee]

Well from my understanding, if p approaches infinity, we "substitute" infinity in the formula i^(p) = p[(1 + i)^(1/p) - 1], which will then give i^(infinity) a value of 0, am I correct? Is that is so, why is 0.0488 the limit and not 0 instead?

Try larger numbers of p on your calculator and you'll see it doesn't tend to zero.

 

[joelee88]

Hi John,

Notice that you're one of the most active administrators here. Thank you very much for taking the time to help all of us here. Appreciate it

I actually did try it with a very big number:

9999999999[(1 + 0.05)^(1/9999999999) - 1] = 0

As you can see it still gives a value of 0. Did I misunderstand anything? Could you please enlighten me?

Thanks a lot!

 

[td290]

\[
\begin{array}{ll}
p[(1+i)^\frac{1}{p}-1]&=p[e^{\frac{\ln (1+i)}{p}}-1]\\
&=p\left[\frac{\ln (1+i)}{p}+\frac{1}{2!}\left[\frac{\ln (1+i)}{p}\right]^2+\frac{1}{3!}\left[\frac{\ln (1+i)}{p}\right]^3+\ldots\right]\\
&=\ln (1+i)+\frac{(\ln (1+i))^2}{2!p}+\frac{(\ln (1+i))^3}{3!p^2}+\ldots
\end{array}
\]
Now let \(p\rightarrow\infty\). All the terms except the first become zero and you're left with \(\ln (1+i)\).

 

[deepakraomore]

Hi John,

Notice that you're one of the most active administrators here. Thank you very much for taking the time to help all of us here. Appreciate it

I actually did try it with a very big number:

9999999999[(1 + 0.05)^(1/9999999999) - 1] = 0

As you can see it still gives a value of 0. Did I misunderstand anything? Could you please enlighten me?

Thanks a lot!

The formula you are using is the relationship between effective and nominal rates. The force of interest is the limiting value for small small intervals. If you use highest value as p means the interest earn in 1/p time interval will be very small. This is explained in CMP.

 

[John Lee]

Hi John,

Notice that you're one of the most active administrators here. Thank you very much for taking the time to help all of us here. Appreciate it

I actually did try it with a very big number:

9999999999[(1 + 0.05)^(1/9999999999) - 1] = 0

As you can see it still gives a value of 0. Did I misunderstand anything? Could you please enlighten me?

Thanks a lot!

Oops! The problem is that your calculator doesn't store enough digits.

Using slightly less figures:

\(99999[1.05^{1/99999}-1] = 0.04879....\)

 

[Calum]

Now let \(p\rightarrow\infty\).

 

[td290]

Glad someone like it!