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Euler's number for the calculation of compound interest

  • 04-05-2012 1:18am
    #1
    krd
    Banned (with Prison Access) Posts: 3,455 ✭✭✭


    I've been trying this. But it hasn't worked. I'm not sure how it's meant to

    A = Principal (1+R/n)^nt is the standard formula. r interest rate, n number of times per year calculated, t is number of years.

    The formula I've seen for Euler's number is e^Rt....But I'm not getting answers that are anywhere near correct.


Comments

  • #2
    Biruni
    Registered Users Posts: 8 ✭Biruni


    Probably because the formula is (1+1/n)^n...
    at least where compound interest is concerned.

    The higher n is, the closer the answer to e.


  • #3
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    This might clarify things:

    http://en.wikipedia.org/wiki/E_%28mathematical_constant%29#Compound_interest

    The formula is (1+r/n)^(t*n), not quite what Biruni said. But as was said, the larger n is, the closer to e^rt we get.

    R= 0.05
    (t)1 2 3 4
    Freq
    1 1.05 1.1025 1.157625 1.215506
    2 1.050625 1.103813 1.159693 1.218403
    4 1.050945 1.104486 1.160755 1.21989
    12 1.051162 1.104941 1.161472 1.220895
    52 1.051246 1.105118 1.161751 1.221285
    365 1.051267 1.105163 1.161822 1.221386
    e^(Rt) 1.051271 1.105171 1.161834 1.221403

    I hope that makes sense, see the spreadsheet for calcs


  • #4
    krd
    Banned (with Prison Access) Posts: 3,455 ✭✭✭krd


    Yakuza wrote: »
    The formula is (1+r/n)^(t*n), not quite what Biruni said. But as was said, the larger n is, the closer to e^rt we get.

    It converges.

    So, it's pretty useless if there are not that many terms.

    Does this problem crop up much with the e^x function?


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