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 $\frac{dQ}{dt} = .06Q,\ Q(0) = 2000, Q(10) =$ the solution, which turns out as $\$3644.24$.

I think the point of this problem requested by part (c) is that the solutions to the difference equations at the $10$ year point are supposed to monotonically increase as we increase $m$, and converge to the differential equation solution as $m$ approaches $\infty$. Since there's nothing special about the $10$ 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, $P(n) = 2000(1 + \frac{.06}{m})^n$, doesn't converge to the differential equation solution $Q(t) = 2000e^{.06t}$ for large $m$, but $P[n] = 2000(1 + \frac{.06}{m})^{mn}$ does, so I think I need to get an extra $m$ into the exponent.

I could apply the substitution $t = \frac{n}{m}$, $Q(t) = P(n)$ 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 $m$ in a careful fashion?

 

[mario99]

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

 

[Metronome]

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

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

I guess the right way to think about $n = tm$ is as a decomposition of $n$ rather than a substitution. A difference equation solution in $n$ 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 $n$ into some socially constructed measure of time $t$ in units of "years," "months," etc., which will remain constant as $n$ grows large, and some $m$ which effectively converts between the two measures of time and acts as a dial through which $n$ can be indirectly pushed to $\infty$ without affecting $t$, and through magic or something $m$ also happens to appear in the coefficient of the difference equation in such a way that makes $e$ appear in the limit!

 

[mario99]

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

I guess the right way to think about $n = tm$ is as a decomposition of $n$ rather than a substitution. A difference equation solution in $n$ 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 $n$ into some socially constructed measure of time $t$ in units of "years," "months," etc., which will remain constant as $n$ grows large, and some $m$ which effectively converts between the two measures of time and acts as a dial through which $n$ can be indirectly pushed to $\infty$ without affecting $t$, and through magic or something $m$ also happens to appear in the coefficient of the difference equation in such a way that makes $e$ 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 $10$ year point are supposed to monotonically increase as we increase $m$, and converge to the differential equation solution as $m$ approaches $\infty$. Since there's nothing special about the $10$ 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 $n$ is related to $m$ and the only way $P$ converges to $Q$ if and only if $n = tm$. Also, I am very familiar with the limit of this structure $\displaystyle \left(1 + \frac{0.06}{m}\right)^m \rightarrow e^{0.06}$ as $m \rightarrow \infty$.

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

I understand that in Finance there are a lot of similar formulas that define $n$ 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 $n = tm$ to go from the difference equation $P(n + 1) = (1 + \frac{.06}{m})P(n), P(0) = 2000$ to the differential equation $\frac{dQ(t)}{dt} = .06Q(t), Q(0) = 2000$ without solving either of them.

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

Is there a way to proceed here?

 

[mario99]

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

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

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 $m \rightarrow 0$ and $\Delta t \rightarrow 0$. To use $m \rightarrow 0$, you need know the solution $P(n)$ and you can express the derivative of the solution $Q(t)$ by $\Delta t\rightarrow 0$.

 

[mario99]

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

I meant $m \rightarrow \infty$

 

[Metronome]

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

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

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

 

[mario99]

Using $\displaystyle \Delta t \rightarrow 0$ in a differential equation like that is meaningless. It will not take you anywhere. I am still not sure what you are trying to achieve! If this is what you are trying to do: make $P(n)$ converges to $Q(t)$ without solving the two equations, it is impossible. Also, if you want to find $P(n)$ and $Q(t)$ without solving the two equations, it is also impossible.

If you are trying to do something else, I'm sorry, but I could not understand you.

 

[Metronome]

Using $\displaystyle \Delta t \rightarrow 0$ in a differential equation like that is meaningless. It will not take you anywhere. I am still not sure what you are trying to achieve! If this is what you are trying to do: make $P(n)$ converges to $Q(t)$ without solving the two equations, it is impossible. Also, if you want to find $P(n)$ and $Q(t)$ without solving the two equations, it is also impossible.

If you are trying to do something else, I'm sorry, but I could not understand you.

The exercises in the book were the red trajectory from the top-left square to the bottom-right square. I am now trying to figure out some way to traverse the blue trajectory. If the rightward facing blue arrow is impossible, then is there some intuition of why this is the case? It seems to me that an IVP and its solution are two descriptions of the same real-world system; if we can move to the right via the red arrow, then surely there is some method to move to the right via the blue arrow?

 

[mario99]

The exercises in the book were the red trajectory from the top-left square to the bottom-right square. I am now trying to figure out some way to traverse the blue trajectory. If the rightward facing blue arrow is impossible, then is there some intuition of why this is the case? It seems to me that an IVP and its solution are two descriptions of the same real-world system; if we can move to the right via the red arrow, then surely there is some method to move to the right via the blue arrow?

View attachment 38301

All I know is that if we have a second order boundary value problem which consists of a second order differential equation, we can approximate its solution by a finite Difference Equation.

For these other scenarios:

1. From First order differential equation to Difference Equation.
2. From Difference Equation to First order differential equation.
3. From Difference Equation to Second order differential equation.
3. From Difference Equation to Higher order differential equation.

I have never heard or seen a scenario like these before (although I am one of the rare people who is very interested in DE's and approximations). Now you want to know the reason why mathematicians don't care about the scenarios above! Well I don't know the exact answer, but I can tell you my opinion.

Approximations in general are derived from the Taylor series. Practice has proved that the first 3 terms of the Taylor series give great approximation to a given function. The other terms are garbage because they don't really affect the approximation (in general). It happens that the third term contains the second derivative of that given function which makes the Taylor series works pretty good in second order differential equations. Therefore, a first order differential equation will behave badly by the Taylor series because it only will take the first two terms while third order differential equation will also behave badly as it depends on the fourth term when it does not really affect the approximation.

In other words, mathematicians created discrete mathematics to help them approximate continuous applications. This means they need to convert the differential equation to the difference equation (which is easier to deal with) so that this difference equation helps them understand the differential equation. Not the reverse!

 

[Metronome]

All I know is that if we have a second order boundary value problem which consists of a second order differential equation, we can approximate its solution by a finite Difference Equation.

For these other scenarios:

1. From First order differential equation to Difference Equation.
2. From Difference Equation to First order differential equation.
3. From Difference Equation to Second order differential equation.
3. From Difference Equation to Higher order differential equation.

I have never heard or seen a scenario like these before (although I am one of the rare people who is very interested in DE's and approximations). Now you want to know the reason why mathematicians don't care about the scenarios above! Well I don't know the exact answer, but I can tell you my opinion.

Approximations in general are derived from the Taylor series. Practice has proved that the first 3 terms of the Taylor series give great approximation to a given function. The other terms are garbage because they don't really affect the approximation (in general). It happens that the third term contains the second derivative of that given function which makes the Taylor series works pretty good in second order differential equations. Therefore, a first order differential equation will behave badly by the Taylor series because it only will take the first two terms while third order differential equation will also behave badly as it depends on the fourth term when it does not really affect the approximation.

In other words, mathematicians created discrete mathematics to help them approximate continuous applications. This means they need to convert the differential equation to the difference equation (which is easier to deal with) so that this difference equation helps them understand the differential equation. Not the reverse!

Okie, thanks for the explanation and all the help. I'll maybe try playing around with the same process for second order or slightly more interesting equations, but I'll expect to not find anything. One interesting wrinkle I noticed in all of this is that when we observe/introduce the relation $n = tm$, it makes the solution to the difference equation invalid, because $(1 + \frac{.06}{m})$ is not a constant coefficient (it is not independent of $n$). Crazy stuff!

 

[Metronome]

Note: Application of similar transformation going from first-order difference equation to first-order differential equation in the derivation of replicator dynamics...

Evolutionary Game Theory (Stanford Encyclopedia of Philosophy)

May be useful if I look at this finance problem again in the future.