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Interest rate homework problem

  1. Jan 27, 2005 #1
    Hello all

    For discrete compunding, we have after n years [tex] (1+r)^n [/tex] where r is the interest rate. IF we receive m interest payments at a rate of [tex] \frac {r}{m} [/tex] then our discrete compounding equation becomes [tex] (1+ \frac{r}{m})^m = e^{m\log(1+(\frac{r}{m}))} \doteq e^r [/tex] After time t we will have [tex] e^{rt} [/tex]. My question is, how do they receive the approximation of [tex] e^r [/tex]? Could we look at this as a differential equation such that if we have an amount [tex] M(t) [/tex] in the bank at time t, how much will it increase from one day to another? So [tex] M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ... [/tex] 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?

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

    Finally the equation [tex] e^{-r(T-t)} [/tex] 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. :smile:
     
  2. jcsd
  3. Jan 27, 2005 #2

    dextercioby

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    1.They used an aproximation...That is for very small [itex] \frac{r}{m} [/itex]

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

    using the definition of "e"...


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

    Daniel.
     
  4. Jan 27, 2005 #3
    Thanks. How did they get this: [tex] M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ... [/tex]
     
  5. Jan 27, 2005 #4
  6. Jan 27, 2005 #5
    oh so basically for separation of variables we have [tex] \frac {dy}{dx} = g(x)f(y) [/tex] then the solution is [tex] \int \frac{dy}{f(y)} = \int g(x) dx [/tex]
     
  7. Jan 27, 2005 #6

    dextercioby

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

    Daniel.
     
  8. Feb 5, 2005 #7
    Could someone please show me how they got this: [tex] M(t+dt) - M(t) \doteq \frac{dM}{dt}dt + ... [/tex] (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, [tex] \frac{dM}{dt} = rM(t) [/tex] could someone please show me how they seperate the variables?

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


    Thanks :smile:
     
    Last edited: Feb 5, 2005
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