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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.

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# Homework Help: Interest rate homework problem

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