How is the Approximation of \( e^r \) Derived in Discrete Compounding?

[courtrigrad]

Hello all

For discrete compunding, we have after n years $$(1+r)^n$$ where r is the interest rate. IF we receive m interest payments at a rate of $$\frac {r}{m}$$ then our discrete compounding equation becomes $$(1+ \frac{r}{m})^m = e^{m\log(1+(\frac{r}{m}))} \doteq e^r$$ After time t we will have $$e^{rt}$$. My question is, how do they receive the approximation of $$e^r$$? Could we look at this as a differential equation such that if we have an amount $$M(t)$$ in the bank at time t, how much will it increase from one day to another? So $$M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ...$$ How do we get the right hand side or approximation? I know it has something to do with a Taylor Series, but could someone please show me?

$$\frac{dM}{dt}dt = rM(t)dt$$ so $$\frac{dM}{dt} = rM(t)$$ Why do we multiply by $$dt$$ in the differential equation? How would we solve this equation? I know the answer is $$M(t) = M(0)e^{rt}$$

Finally the equation $$e^{-r(T-t)}$$ relates the value you will get earlier given that you know the dollar value at time T. Is this a result of the differential equation?

Thanks a lot.

 

[dextercioby]

Hello all

For discrete compunding, we have after n years $(1+r)^{n}$ where r is the interest rate. IF we receive m interest payments at a rate of $$\frac {r}{m}$$ then our discrete compounding equation becomes $$(1+ \frac{r}{m})^m = e^{m\log(1+(\frac{r}{m}))} \doteq e^r$$ After time t we will have $$e^{rt}$$. My question is, how do they receive the approximation of $$e^r$$? Could we look at this as a differential equation such that if we have an amount $$M(t)$$ in the bank at time t, how much will it increase from one day to another? So $$M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ...$$ How do we get the right hand side or approximation? I know it has something to do with a Taylor Series, but could someone please show me?

$$\frac{dM}{dt}dt = rM(t)dt$$ so $$\frac{dM}{dt} = rM(t)$$ Why do we multiply by $$dt$$ in the differential equation? How would we solve this equation? I know the answer is $$M(t) = M(0)e^{rt}$$

Finally the equation $$e^{-r(T-t)}$$ relates the value you will get earlier given that you know the dollar value at time T. Is this a result of the differential equation?

Thanks a lot.

1.They used an aproximation...That is for very small $\frac{r}{m}$

$$\lim_{\frac{r}{m}\rightarrow 0} [(1+\frac{r}{m})^{\frac{m}{r}}]^{r}=e^{r}$$

using the definition of "e"...


2.They multiplied by "dt" to SEPARATE VARIABLES IN THE DIFFERENTIAL EQUATION.It's a standard method...

Daniel.

 

[courtrigrad]

Thanks. How did they get this: $$M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ...$$

 

[vincentchan]

tayl0r series...
If you have not learned yet, read the following link
http://mathworld.wolfram.com/MaclaurinSeries.html
it didn't give the reasoning, but it has a lot of different series expansion

 

[courtrigrad]

oh so basically for separation of variables we have $$\frac {dy}{dx} = g(x)f(y)$$ then the solution is $$\int \frac{dy}{f(y)} = \int g(x) dx$$

 

[dextercioby]

That's right... That's the easiest method among all methods to integrate SOME diff.eqns.

Daniel.

 

[courtrigrad]

Could someone please show me how they got this: $$M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ...$$ (I know from the other posts it is a Taylor series expansion) however could you just explain this a little further?

Also with the separation of variables, $$\frac{dM}{dt} = rM(t)$$ could someone please show me how they separate the variables?

Also how do you get $$e^{-r(T-t)}$$ for the value of the money at an earlier time?


Thanks