Are we talking about your real world, or a mechanical engineer’s? You probably don’t ever have a system of ODEs that you’d like to solve, but the engineer does almost nothing else.
At the core of mathematical statistics is the notion of maximum likelihood estimation, which involves calculus - in particular, the maximization of likelihood functions.
As for solving systems of differential equations, think in terms of multivariable fluid dynamics.
I don’t know why I opened this thread, I was lost after the first posting! However, I find all this stuff quite fascinating even though I don’t understand a word of it.
If you put one dollar in the bank, what is the interest rate needed for that dollar to have doubled in 10 years. Or:
I’m pretty sure that this is no challenge to most of the people in this thread but it is to me (note: it is not homework, though!). I was trying to figure this out the other day, and I was pretty sure that there was a time that I did know how to find the answer. Those days, however, are long gone, but I sure do hope someone can enlighten me.
If you’re looking for the interest rate, you should write the equation as (1 + i)[sup]10[/sup] = 2. Then raise both sides to the .1 power to get 1 + i = 2[sup].1[/sup], and i = 2[sup].1[/sup] - 1, which comes out to about 7.2%.
The relevant property of exponentiation is that (x[sup]a[/sup])b = x[sup]ab[/sup]. Since 10 * .1 = 1, that’s how you undo the 10th power in the original equation.
Just for reference in case your bank compounds interest at a different rate than annually. From math.com: "
P = C (1 + r/n) [sup]nt[/sup]
where
P = future value
C = initial deposit
r = annual interest rate (expressed as a fraction: eg. 0.06)
n = # of times per year interest is compounded
t = number of years invested
"
so after doing some algebra; r = ((P/C)[sup]1/nt[/sup] -1)/n.
So for the initial problem: if your bank compounds interest quarterly then
r = (2[sup]1/40[/sup] -1)/4
r ~ 6.99%
If interest is compounded continuously, you want the rate r that makes
e[sup]10r[/sup] = 2.
10r = ln(2)
r = (ln 2)/10 = about 6.93%.
Note that the amount of money you start with ($1 or $1,000,000) doesn’t affect how long it takes to double.
To see where these formulas come from:
At an interest rate of r, if you never compound [i.e., never earn interest on any new money in addition to your original deposit], then you earn money at the rate r * your original deposit. [This is the defining property of interest: you earn money at a rate proportional to the amount you have]. So after time t, you have original money + r * original money * t. That is to say, your money multiplies by a factor of 1 + rt.
Now, let’s start accounting for compounding. That is to say, let’s start earning interest on not only our original deposit, but on any new money which comes into it via interest. (Nice feedback loop there, eh?). That is to say, instead of always earning money at the rate r * original money, we keep updating this to r * current amount of money.
How do we calculate compounded interest? Well, one way is by not actually compounding all the time; instead of constantly updating the earning rate to r * current amount of money, we’ll only do this after a “compounding period” passes. For example, maybe we only update the earning rate at the beginning of each month. In this case, what happens after time t? Well, suppose we go through N many compounding periods in this time. Then, each compounding period has length t/N, so over each compounding period, our money multiplies by 1 + rt/N, by the reasoning of the first paragraph above. Since there are N of these multiplications in total, the cumulative effect is that our money multiplies by (1 + rt/N)[sup]N[/sup]. [We can reword this in different ways; for instance, in terms of the frequency n = N/t of compounding periods, this becomes (1 + r/n)[sup]nt[/sup]. Or, in terms of the ratio k = n/r between the frequency of compounding and the interest rate, this becomes (1 + 1/k)[sup]krt[/sup] = ((1 + 1/k)[sup]k[/sup])[sup]rt[/sup].]
Ah, but that’s all rather ugly, isn’t it? Why not do away with the staggered updates of our earning rate, and just continuously compound (i.e., make sure that our earnings rate is always r * our current amount of money)? Well, we can! In this case, after time t, our money will have multiplied by a certain factor. What factor? Well, we can just make up a name for this factor; we’ll call it exp(r, t).
What do we know about exp(r, t)? Well, continuous compounding is like discrete compounding with an extremely high frequency of compounding. Thus, we can approach exp(r, t) by taking any of the above formulas for discrete compounding, and making the frequency of compounding very, very large. In particular, using the last formula above, we have that exp(r, t) is approached by ((1 + 1/k)[sup]k[/sup])[sup]rt[/sup], as k gets very, very large. I.e., whatever ((1 + 1/k)[sup]k[/sup]) approaches, raise that to the power rt.
Alright, now what? Now, let’s just make up a name for whatever ((1 + 1/k)[sup]k[/sup]) approaches as k gets very large; call this special number “e”. So we have that exp(r, t) = e[sup]rt[/sup]; remembering what this was all about, this means e[sup]rt[/sup] is, by definition, the amount our money multiplies by at interest rate r after time t, using continuous compounding (as is nicest).
Of course, none of the above was restricted to describing money alone; we can use interest rates to describe any situation where something’s rate of increase is proportional to how much of it there is.
In particular, now’s the time to let my secret agenda comes into play: we’ll pull Euler’s theorem out of the end of the above.
The above tells us that, by definition, e[sup]rt[/sup] is the amount something multiplies by at continuous interest rate r after time t; i.e., if you have some quantity whose rate of change is always r * its current value, then after time t, it will have changed by a total factor of e[sup]rt[/sup].
Suppose someone is walking around a circle of radius 1 (centered at the origin) at speed 1. Then the rate of change of his position (i.e., his velocity) is always equal in magnitude to his distance from the center of the circle (as both are 1); however, it’s turned 90 degrees from this (he’s not walking towards or away from the center of the circle; if he were facing the center of the circle, he’d be moving to his side). Thus, his position is essentially growing with interest rate “90 degree rotation”. So, if we let “i” denote “90 degree rotation” (as makes sense, since doing this twice is turning all the way around to face the other way; that is to say, i^2 = -1), we have that e[sup]it[/sup] denotes the effect on his position after time t.
But, by definition, π is the amount of time it will take for him to get to the opposite point on the circle (i.e., for his position to multiply by -1). Thus, we have that e[sup]iπ[/sup] = -1. More generally, by definition, a “radian” is how much this guy rotates by in 1 unit of time, so have that e[sup]it[/sup] denotes rotation by t radians. And that’s the general form of Euler’s theorem. It’s that simple; you all now know exactly how it’s derived.
To put it another way, all the theorem is stating is that, if you stored a vector of length 1 in an eccentric bank account at a continuous interest rate of 90 degree rotation per year, then your balance will just keep rotating around a circle at a speed of 1 (in particular, after π many years, your balance will have gone halfway around the circle and negated itself, and after 2π many years, your balance will have gone all the way around the circle and come back to its original value). Which makes sense, since when one moves around a circle, one’s velocity is always perpendicular to one’s radius (thus, one is moving with an interest rate of 90 degree rotation per some unit of time).
I like what tim-n-va and ultrafilter said, but here’s my… uh - reply:
Your basic Differential Calculus is an extension of algebra to include rates of change. (Integral Calculus reverses DC and projects aggregate changes over a span built up from a rate of change, or most often, varying rates of change.)
(The span can and often is time, but it can be other things such as a physical line, temperature, etc.)
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Here you may want to skip over to the part after the triple asterix.
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Now, if you have a chart of World Population over time, the yearly growth rates (births minus deaths) that become apparent are like differentials. Make a separate chart with those little jumps stretched out vertically and you have a second function that is the derivative of the first. Because it was derived.
The only problem is that all those yearly growths make only a bumpy, finite picture of what is going on. Monthly or weekly would be better. And so on. If we reverse this (IC) and happen to have a function to start with that instantaneously represents the pattern of varying growth rate, we can have the aggregate population growth over the period of time involved, which would be added to the original population count at the beginning of the projection.
This is why, BTW, it has been said that algebraic solutions would be incomplete. Pre-calculus is often called finite mathematics, because it would involve the lumpy, grainy approach of adding change over finite increments of the span.
(At one time the overall World Growth Rate was projected at a steady approximate 2%. This would result in a doubling every 35 years. An octupling in a little over a century. You could add a zero to any round figure over a period of less than 120 years. And so on. This was viewed by most analysts as a serious problem, especially in combination with a – presumed – steady per capita pollution output. Since those days many nations have reduced their growth rates without draconian measures – in most cases. And several nations now have the opposite problem: depoplulation. This has lead to a somewhat smaller estimated population in 2000, for example.)
While projections over fairly long periods of time are problematic, they work rather well with, say, 5 year spans. We can safely presume that the rate of change of the rate of change (growth rate) will be zero, that is to say, the rate of change is constant. A more refined approach would be to study recent changes in the rate of change and project those over the next 5 years. This, BTW, is an example of double integration.
Y’know I think a better illustration is with distance, speed and acceleration over time.
I’ll start with an old joke, minus the stereotyping. A cop stops a dumbass for speeding. He says that the dumbass was going over 90 miles an hour. How can that be, asks the dumbass, since I have only been driving half an hour? (This could also be put as: I’ve only covered 45 miles.)
With your most “basic” function, distannce over time, speed over time is the derived function of the former. We can use that to project that, if unchanged, the driver would indeed have covered 90 miles in the full hour. Interested in the rate of change of the speed? That’s your acceration over time, the second derivative, which actually may not be a zero function, since the speed may not be constant. If we can pin down the acceleration pattern (maybe the dumbass speeds up a certain level when the sign for the town and distance is first spotted?) we can integrate and get the speed function. Then the second integration would point out where along the road the driver actually is, at any particular time.
I hope that helps. (I was originally going to end with budgeting/non-budgeting patterns, and how saver versus spender makes more difference than how much money you start out with. But I think that speed and acceleration make a better illustration.)
- Jack***
Hey, Jack, Blue, whatever, why de’ heck is 1 not a prime number?! Huh?!
There are no whole numbers that go into it a whole number of times. So it has no factors, hence is not what dey call a composite number. So it hasta’ be prime.
Whadya say? Huh? Huh? Betcha’ can’t answer that one, canya’?
- Johnny Boy
The short answer is that a natural number is prime if it has exactly two distinct natural divisors. The precise statement of that definition tends to get mangled somewhere along the way from textbook author to textbook to elementary math teacher to elementary student.
The reason behind that is that if you disallow one as a prime, the fundamental theorem of arithmetic–that every number can be written as the product of powers of primes, and that that factorization is unique–is short and sweet. If you allowed 1, then it would become much less so, especially in more general rings.
Well, they don’t allow sock puppets here for any member, for any reason, so I have to answer this one in a way that makes it appear that I’m talking to myself. (Imagine!)
There is absolutely no reason that there has to be a “binary choice” here. There probably are other examples where a single or a handful of exceptions go beyond such a dichotomy, although I’m sure I don’t know of any, offhand. Anyway, 1 among the positive integers is a special case.
And specifically, most modern definitions of prime start off with “a number greater than one…”
But that, of course, doesn’t really answer anything. It still raises the question of WHY mathematicians (modern ones, anyway) would want to leave off this non-composite number from the primes.
So here it is, in a nutshell:
Mathers find a certain theorem very useful. It’s called the Fundamental Theorem of Arithmetic, and it states that all positive integers can be represented as the product of primes in (essentially) just one way. In the case of primes it is the trivial “product” of just that single prime itself.
For instance, 6 is 3X2 or, if you prefer, 2X3, but these are two ways of saying the same thing. We can say that 48 = 3 X 2^4 or 3 X 2 X 2 X 2 X 2. Again, we are saying the same thing.
But if one is considered a prime, then all bets are off!
6 could be 3 X 2 X 1 or … X 1 X 1. We can put one in any number of times, generating an endless supply of representations.
One is specifically called a unit. How many units are there, and what does it mean, exactly?
When we are talking about the system of integers, there are only 2, 1 and -1. There are no doubt more if we are talking about the system of complex numbers, but the important thing is that if we are talking about positive integers, there is only one! <Groan!>
The definition is that a unit divides one. -1 goes into it -1 times, being that negative numbers squared have positive results.
In the complex number plane, there are also + i, -i, and I believe, all the complex numbers going around zero at a distance of 1. These all have a “magnitude” of 1, magnitude being an extension of absolute value.
There are actually more abstract, less practical number systems defined in recent decades. Many of these have multiple units, and the sensible thing to do is to first identify all “numbers” that “evenly” divide whatever multiplicate identity element is discovered. Then you can go about identifying the actual prime numbers.
BTW, the Greeks had no such modern approaches to positive integers. They used “proto” to descibe primes when they first focused on them, and 1 was included as a proto number.
- "Jack"
ETA: Obviously I’m not a fast enough writer or typist to avoid being interrupted when talking to myself. Anyway, thank you,
ultrafilter.
That’d be an odd way to put it. There are whole numbers that go into 1 a whole number of times; namely, 1 goes into it once.
The definition of “prime” as “It has exactly two factors: 1 and itself” isn’t actually the directly useful one (let’s call these “schoolyard primes” to avoid ambiguity). Instead, the more directly useful one is this: a number is prime if, whenever it divides a product of finitely many factors, it divides one of those factors (but, again, to avoid ambiguity, let’s call these “Euclidean primes”).
In particular, 1 isn’t Euclidean prime because 1 divides the empty product (i.e., the product of zero many factors, which is 1) without dividing any of its factors (since the empty product doesn’t actually have any factors).
The punchline, of course, is that, if one looks at the positive integers, the Euclidean primes and the schoolyard primes are the same (a simple but perhaps worthwhile exercise for any student to prove for themself).
“But wait”, you object. “Suppose I look not just at the positive integers, but at all the integers. Then -2, -3, and so on are Euclidean primes too. And so is 0.”
You’re right. And if you broaden your scope to all real numbers, then 2 isn’t prime anymore, in either the schoolyard or the Euclidean sense, since, e.g., it’s divisible by 2/3 and it multiplies by 0.5 to give the empty product and so on. So, much is context-sensitive. Convention is that “prime number” means “among the positive integers”, unless otherwise specified.
Still, the natural thing to do now is to say: “You know what? All that really matters about Euclidean primes is what they divide; their identity is unimportant beyond that. So we’ll say that two numbers are associated if they both divide each other, and consider them to be the same, for such purposes as talking about what the Euclidean primes are”. Thus, in the integers, 2 and -2 are the same Euclidean prime, 3 and -3 are the same Euclidean prime, and so on.
But 0 is still the odd man out.
Which is fine. For many purposes, it is natural to consider 0 a prime. But if one is really concerned about this, one can finally define “prime element” as “non-zero Euclidean prime (up to associatedness)”.
ETA: Goddamn, this thread has passed me by. This was in response to TBJ’s original post (which I guess I might as well quote at the top now).
Well, I sort of get that 0 would count as an Euclidean prime, along with the “nega-primes.” But I’m used to thinking of 0 as "ultra-composite" because any number will divide it, and the insertion of 0 into any non-zero product changes it into zero product.
((( BTW, have you ever heard of the math teaser: What is (a-x) * (b-x) * (c-x) * (d-x) * (e-x) * (f-x) * … all the way up to the twenty-sixth factor? )))
- Jack