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Exponential functions - which is the better solution?

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    A town's population grows at 6.5% per annum. How many are in town now, if there will be 15 000 in 4.5 years?

    Please explain which of these solutions is best. Or explain a better solution, please.

    2. Relevant equations

    A(t) = Per(t) <- general approach
    A = P(1+i)t <- compound interest formula (is this too simplified?)

    In both cases, A is the number of people at time, t. P is the current population (solving for P). The rate is r, or i. & e is Euler's number.

    3. The attempt at a solution
    Solution 1.
    A(t) = Per(t)
    15000 = Pe(0.065)(4.5)
    P = (15000)/(e(0.065)(4.5))
    P = 11 195.9287

    Solution 2.
    A = P(1+i)t
    15000 = P(1.065)4.5
    P = 15000/(1.0654.5)
    P = 11 298.4280

    Why the difference? Which solution is better? Is the compound interest formula an over-simplification?

    Many thanks!
     
  2. jcsd
  3. Aug 1, 2013 #2
    The compound interest formula assumes that interest is compounded only once (at the end of) every compounding interval. Therefore, putting in 4.5 for t in the second example isn't really valid. The "general approach" assumes that interest is being compounded continually. In the case of your population growth problem, the general approach is the appropriate formula to use, since children are being born continually, and the birth rate is proportional to the currently existing total population.
     
  4. Aug 1, 2013 #3
    I think the equation [itex] A=Pe^{rt} [/itex] comes from [itex] A=P(1+i/n)^{nt} [/itex] by finding the limit of A as n approaches infinity and using the fact that [itex] \lim_{n\rightarrow +\infty} (1+i/n)^n=e^i [/itex]. This is the definition of e.
     
    Last edited: Aug 2, 2013
  5. Aug 1, 2013 #4

    symbolipoint

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    If you want to make better use of the compound interest formula instead of the continuous growth formula, at least pick a large enough compounding period (and rate adjusted to that period).
     
  6. Aug 2, 2013 #5
    No. This is not correct. The compound interest formula should be A = P(1+i/n)^nt, where i is the annual interest rate, and n is the number of compounding periods per year. So,
    [itex]A=Pe^{ntln(1+i/n)}[/itex]. In the limit of large n, this approaches the continuous compounding formula [itex]A=Pe^{it}[/itex].
     
  7. Aug 2, 2013 #6

    Ray Vickson

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    The difference is due to using a different value of 'r' in the two calculations. To match them up you need to use a "continuous" rate r that satisfies ##e^r = 1.065 \longrightarrow r \doteq 0.0629748.##
     
  8. Aug 2, 2013 #7
    Yes, I'll fix my mistake now.
     
  9. Aug 2, 2013 #8

    Ray Vickson

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    That is not the issue. If 6.5% is the TRUE annual rate of increase, then r ≠ 0.065 in the exponential form. You need r ≈ 0.0629748 in order to have er = 1.065.
     
  10. Aug 2, 2013 #9
    I think this is incorrect. The value of r never changes. It is always 6.5%, or 0.065. However, as you start to increase the compounding periods, the total amount at the end begins to increase. This is the formula that I was talking about previously: [itex] A=P(1+i/n)^{nt} [/itex] where n represents the compounding periods. Now, as n gets larger and larger (meaning you compound the money more and more often) the value of this approaches Pe^rt.

    In brief, just because in the compounding formula the base of the exponent is 1.065 does not mean e^r should equal to 1.065. In fact, it should be greater than 1.065, because now you are compounding it more frequently.
     
  11. Aug 2, 2013 #10

    Ray Vickson

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    The problem stated clearly that "A town's population grows at 6.5% per annum", and I am inclined to treat that statement as meaning exactly what it says, viz., that the true annual growth rate is 6.5%. In other words, at the start of next year the population will be 6.5% higher than it was at the start of this year. That requires r = 0.0629478. I suppose in the context of a particular course/notes/textbook a different usage may have been employed, but without such information I will go with a plain English interpretation. The OP may know better---but in that case, the answer to the OP's question is still that the two rates don't match: if the annual growth is e0.065 - 1, then he/she needs to use an "interest rate" of r' = 6.7159% in the discrete-time model in order to get a match. And THAT is the answer to the OP's original question of why there is a difference.

    This type of issue arises all the time in fields like Finance, when one switches from a discrete-time model to a continuous-time model. It is a very common source of student mistakes. On the other hand, sometimes the confusion/imprecision is built in to a system. For example, when dealing with mortgage payment schemes, if the annual rate is, say 12%, that is treated as a monthly rate of 1%, compounded 12 times per annum. The true annual rate is thus actually more than the stated 12%.
     
    Last edited: Aug 2, 2013
  12. Aug 2, 2013 #11
    Thanks for all the replies!

    I'm going to go with the first reply. I'll use A(t) = Pert because babies are made continuously, not at the end of each year. This isn't a compound interest problem, it's a population growth problem.

    In practical terms, I think the compound interest formula fails because a new born baby has not reached sexual maturity, and cannot contribute to the growth, yet. (Growth in the compounded sense, I suppose.)

    I'm not a mathematician, I'm just a guy trying to learn some math. Thanks for the replies though.
     
  13. Aug 2, 2013 #12

    Ray Vickson

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    I agree: it is not a compound interest problem, but the issue still remains: what value of 'r' should you use? My claim is that to match the problem description (in the context of continuous time) you need to use r = 6.29478% = 0.0629478 in the exponential formula. That will give you a true growth rate of 6% per year.
     
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