Difference Equations and Differential Equations Convergence Exercise

[Metronome]

I am interested in Exercise 12 from Section 3.3 of Differential Equations With Boundary Value Problems by Polking, Boggess, and Arnold...


Exercise 12 references Exercise 11, to which the official solution is...


Exercise 12 part (a) is straightforward; solve [imath]\frac{dQ}{dt} = .06Q,\ Q(0) = 2000, Q(10) =[/imath] the solution, which turns out as [imath]\$3644.24[/imath].

I think the point of this problem requested by part (c) is that the solutions to the difference equations at the [imath]10[/imath] year point are supposed to monotonically increase as we increase [imath]m[/imath], and converge to the differential equation solution as [imath]m[/imath] approaches [imath]\infty[/imath]. Since there's nothing special about the [imath]10[/imath] year point, we can be slightly more ambitious and expect the entire particular solution curve to converge as such.

I'm just a bit stuck on how to get there by following part (b). The formula generated in Exercise 11, [imath]P(n) = 2000(1 + \frac{.06}{m})^n[/imath], doesn't converge to the differential equation solution [imath]Q(t) = 2000e^{.06t}[/imath] for large [imath]m[/imath], but [imath]P[n] = 2000(1 + \frac{.06}{m})^{mn}[/imath] does, so I think I need to get an extra [imath]m[/imath] into the exponent.

I could apply the substitution [imath]t = \frac{n}{m}[/imath], [imath]Q(t) = P(n)[/imath] to the difference equation solution to get it into the right form, but this feels superficial; the substitution would never be undone, and I could just as easily make any substitution to get almost any desired answer.

What else can I do to introduce this missing [imath]m[/imath] in a careful fashion?

 

[mario99]

You are calculating the limit [imath]m \rightarrow \infty[/imath] of [imath]\displaystyle P(n) = 2000\left(1 + \frac{0.06}{m}\right)^n[/imath] without defining what is [imath]n[/imath]. In fact, [imath]n[/imath] depends on [imath]m[/imath]. In other words, [imath]n = tm[/imath]. For example, when you soved question 12 for daily (365 times per year), did you use [imath]n = 10[/imath] or [imath]n = 365[/imath] or [imath]n = 10 \times 365[/imath]?

 

[Metronome]

You are calculating the limit [imath]m \rightarrow \infty[/imath] of [imath]\displaystyle P(n) = 2000\left(1 + \frac{0.06}{m}\right)^n[/imath] without defining what is [imath]n[/imath]. In fact, [imath]n[/imath] depends on [imath]m[/imath]. In other words, [imath]n = tm[/imath]. For example, when you soved question 12 for daily (365 times per year), did you use [imath]n = 10[/imath] or [imath]n = 365[/imath] or [imath]n = 10 \times 365[/imath]?

I indeed think this was my problem! I was naturally inclined toward [imath]n = 10[/imath], but I should use [imath]10 * 365[/imath].

I guess the right way to think about [imath]n = tm[/imath] is as a decomposition of [imath]n[/imath] rather than a substitution. A difference equation solution in [imath]n[/imath] doesn't care what calendar we use; we could just count the compounding periods as the money enters the account without any socially constructed measure of time (we are free to name the counted compounding periods "years," "months," etc., but the math doesn't ask for it). But such indifference is absurd for continuous compounding, as the paradigm would involve enumerating over an uncountable infinity of compounding periods, so we must decompose [imath]n[/imath] into some socially constructed measure of time [imath]t[/imath] in units of "years," "months," etc., which will remain constant as [imath]n[/imath] grows large, and some [imath]m[/imath] which effectively converts between the two measures of time and acts as a dial through which [imath]n[/imath] can be indirectly pushed to [imath]\infty[/imath] without affecting [imath]t[/imath], and through magic or something [imath]m[/imath] also happens to appear in the coefficient of the difference equation in such a way that makes [imath]e[/imath] appear in the limit!

 

[mario99]

I indeed think this was my problem! I was naturally inclined toward [imath]n = 10[/imath], but I should use [imath]10 * 365[/imath].

I guess the right way to think about [imath]n = tm[/imath] is as a decomposition of [imath]n[/imath] rather than a substitution. A difference equation solution in [imath]n[/imath] doesn't care what calendar we use; we could just count the compounding periods as the money enters the account without any socially constructed measure of time (we are free to name the counted compounding periods "years," "months," etc., but the math doesn't ask for it). But such indifference is absurd for continuous compounding, as the paradigm would involve enumerating over an uncountable infinity of compounding periods, so we must decompose [imath]n[/imath] into some socially constructed measure of time [imath]t[/imath] in units of "years," "months," etc., which will remain constant as [imath]n[/imath] grows large, and some [imath]m[/imath] which effectively converts between the two measures of time and acts as a dial through which [imath]n[/imath] can be indirectly pushed to [imath]\infty[/imath] without affecting [imath]t[/imath], and through magic or something [imath]m[/imath] also happens to appear in the coefficient of the difference equation in such a way that makes [imath]e[/imath] appear in the limit!

I think the point of this problem requested by part (c) is that the solutions to the difference equations at the [imath]10[/imath] year point are supposed to monotonically increase as we increase [imath]m[/imath], and converge to the differential equation solution as [imath]m[/imath] approaches [imath]\infty[/imath]. Since there's nothing special about the [imath]10[/imath] year point, we can be slightly more ambitious and expect the entire particular solution curve to converge as such.

When I read your sentence above I instantly knew [imath]n[/imath] is related to [imath]m[/imath] and the only way [imath]P[/imath] converges to [imath]Q[/imath] if and only if [imath]n = tm[/imath]. Also, I am very familiar with the limit of this structure [imath]\displaystyle \left(1 + \frac{0.06}{m}\right)^m \rightarrow e^{0.06}[/imath] as [imath]m \rightarrow \infty[/imath].

Part (b) of problem 12 demonstrates this idea: as you increase the compounding [imath]m[/imath], you increase your money. This is another way to see that [imath]P[/imath] is eventually converges to [imath]Q[/imath].

I understand that in Finance there are a lot of similar formulas that define [imath]n[/imath] in different ways which can even trick the professionals in this field.

 

[Metronome]

The next thing I'm inclined to try is using the discovered relation [imath]n = tm[/imath] to go from the difference equation [imath]P(n + 1) = (1 + \frac{.06}{m})P(n), P(0) = 2000[/imath] to the differential equation [imath]\frac{dQ(t)}{dt} = .06Q(t), Q(0) = 2000[/imath] without solving either of them.

The problem I'm confronting is that the solution converges in the limit of [imath]m \rightarrow \infty[/imath] as was the point of the exercise, but the differential equation exists in the limit of [imath]\Delta t \rightarrow 0[/imath]. I guess I definitely still need [imath]m \rightarrow \infty[/imath], but maybe I should switch back to the interpretation that [imath]n[/imath] is fixed (i.e., at [imath]10[/imath] years) and [imath]t \rightarrow 0[/imath] to compensate for [imath]m \rightarrow \infty[/imath]?

Is there a way to proceed here?

 

[mario99]

The next thing I'm inclined to try is using the discovered relation [imath]n = tm[/imath] to go from the difference equation [imath]P(n + 1) = (1 + \frac{.06}{m})P(n), P(0) = 2000[/imath] to the differential equation [imath]\frac{dQ(t)}{dt} = .06Q(t), Q(0) = 2000[/imath] without solving either of them.

The problem I'm confronting is that the solution converges in the limit of [imath]m \rightarrow \infty[/imath] as was the point of the exercise, but the differential equation exists in the limit of [imath]\Delta t \rightarrow 0[/imath]. I guess I definitely still need [imath]m \rightarrow \infty[/imath], but maybe I should switch back to the interpretation that [imath]n[/imath] is fixed (i.e., at [imath]10[/imath] years) and [imath]t \rightarrow 0[/imath] to compensate for [imath]m \rightarrow \infty[/imath]?

Is there a way to proceed here?

I don't understand what you are trying to do. If you don't solve the two equations, you cannot use both [imath]m \rightarrow 0[/imath] and [imath]\Delta t \rightarrow 0[/imath]. To use [imath]m \rightarrow 0[/imath], you need know the solution [imath]P(n)[/imath] and you can express the derivative of the solution [imath]Q(t)[/imath] by [imath]\Delta t\rightarrow 0[/imath].

 

[mario99]

I don't understand what you are trying to do. If you don't solve the two equations, you cannot use both [imath]m \rightarrow 0[/imath] and [imath]\Delta t \rightarrow 0[/imath]. To use [imath]m \rightarrow 0[/imath], you need know the solution [imath]P(n)[/imath] and you can express the derivative of the solution [imath]Q(t)[/imath] by [imath]\Delta t\rightarrow 0[/imath].

I meant [imath]m \rightarrow \infty[/imath]

 

[Metronome]

I don't understand what you are trying to do. If you don't solve the two equations, you cannot use both [imath]m \rightarrow 0[/imath] and [imath]\Delta t \rightarrow 0[/imath]. To use [imath]m \rightarrow 0[/imath], you need know the solution [imath]P(n)[/imath] and you can express the derivative of the solution [imath]Q(t)[/imath] by [imath]\Delta t\rightarrow 0[/imath].

Yeah, so my target would be something like [imath]\lim_{\Delta t \rightarrow 0} \frac{Q(t + \Delta t) - Q(t)}{\Delta t} = .06Q(t), Q(0) = 2000[/imath].

I might be able to bypass the use of [imath]m \rightarrow \infty[/imath] entirely. I noticed that there are at least two methods that give the same solution to the original problem of getting just the solutions to converge. Using the relation [imath]n = mt[/imath], it is possible to substitute [imath]n[/imath] out of the solution and push [imath]m[/imath] to [imath]\infty[/imath], or to substitute [imath]m[/imath] out of the solution and push [imath]n[/imath] to [imath]\infty[/imath]. That the second method exists seems to follow from your initial insight that I was mistakenly holding [imath]n[/imath] fixed at [imath]10[/imath] as I pushed [imath]m[/imath] to [imath]\infty[/imath], whereas they should both approach [imath]\infty[/imath] together; that is...[math]n_{approching\ \infty}\ =\ m_{approaching\ \infty}\ t_{not\ changing}[/math]One new idea I might be able to use is [imath]\Delta n = \frac{1}{m}[/imath]. This might be incorrect, because if [imath]\Delta n[/imath] should be measured in compounding periods, then [imath]\Delta n = 1[/imath], trivially. I'm not super educated in discrete calculus, but I believe in at least some contexts [imath]\Delta n[/imath] is indeed either assumed or proven to be [imath]1[/imath] (such as in the discrete derivative which is [often?] a simple difference rather than a difference quotient).