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