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Hi CurlyBracket,
Welcome to the forum.
Hello,
Welcome to the forum.
Hi nahid,
Welcome to the forum.
Are you sure you've posted the problem correctly? As it stands, I can't see how the latter statement is true. We have:
and sosince obviously But then the problem statement suggests that we should then have whereas the polynomial doesn't have any real roots. (You can show this by differentiating it once to find the stationary point, then again to show that it's a minimum -- then showing that at this minimum, is strictly positive, i.e. doesn't ever cut the x-axis.)Hi Bob,
I think 'continuous' in this context is referring to a growth rate which is 'convertible continuously'.
For example, suppose you have £100 in a savings account which grows at an effective rate of interest of 5% pa (not likely these days but who knows, given current market conditions!). The bank isn't likely to give you the 5% at the end of each year, though -- so they might offer a rate convertible monthly.
So if the savings account offered you a return on your £100 deposit at an effective interest rate of 5% pa offering monthly interest payments, then we want to find the interest rate such that:The bank can then claim to offer you an interest rate of convertible monthly. So every month, you get about on top of your savings. That's the same as getting an effective interest rate of over a year, except you get the equivalent every month.In this case we subdivided our annual effective interest rate into 12 intervals of equal length (months). But there's no reason why we can't continue subdividing into even smaller intervals, which is where the phrase 'convertible continuously' comes from (where the intervals become infinitesimally small). So suppose we have an interest rate convertible pthly and let p go to infinity. We have:where we've used the limit definition of and writing Here, is the nominal rate of interest (growth) per unit time convertible continuously (sometimes called the 'force of interest'). That's what I think is meant by 'continuous' here. In this case we see that and in particular we can accumulate a payment of 100 from time 0 to time t like this:In Hannibal lecter's example the emphasis should probably be on the fact that the rate is continuous, i.e. a continuous growth rate of 5% (i.e. ) which is slightly different to a growth rate of pa (i.e. ).In other words, if someone told you to accumulate 300 at a rate of 5% pa for 10 years you would rightly do
but at a continuous rate of 5% per unit time you would do
No, because continuous growth is modeled with the equation
, where is the ending value, is the initial value, is Euler's constant, is the continuous growth rate, and is the time that has passed.
What's stopping you from taking k = log(1.07)?
6. a^4 + 1
This can't be factored, because sum of squares cannot be factored using real numbers.
As a final thought (because I think I've flooded your thread enough with my ramblings!), here's one more parametrisation you could think about -- but it's pretty volatile, so use with caution.
Let's consider the standard parametric equations for a hyperbola: let satisfy such that for some we have:Then after some algebra analogous to that which was detailed in post #12, we find that
which produces this curve: https://www.desmos.com/calculator/aggjlrrwlk
This also generates solutions to For example, taking yields:from which we find that one solution pair is:
So what does this mean graphically? It turns out that, if we plot the graph of this function (in terms of x and y):
then we get this graph, which should be identical to Bob's graph except without the trivial line y = x: https://www.desmos.com/calculator/xsgqovhxul
For we trace out everything to the right of the line which you can see here: https://www.desmos.com/calculator/ntzw4opu6x...and similarly for we trace out everything to the left of the line which you can see here: https://www.desmos.com/calculator/ehl1zmc5nqYou'll notice that if you try to plot values between, say, and then it doesn't work -- that's because you get complex solutions.For example, taking gives us the following valid solution to :but both of these are complex numbers which won't show up in the (real) Cartesian plane.
Pulling 'like' terms across the equality:
This gives us another tool for generating solutions in and since we can now just choose a particular which will determine and hence and For example, you can show that if we setthen it follows that and hencewhich is the classical solution to Setting gives us the interesting solution:By the way, the proof of the first line in this limit is as follows. Let's split the function up into two parts, i.e. let
so that We'll try to calculate the limits of these functions as separately. If both exist, then the product of these limits exists (and is equal to the limit of the product). We clearly have:For the first function, we have:
The function inside the limit satisfies the conditions of L'Hopital's rule, so this is just equal to:
and so
You also mentioned the product-log function (also called the Lambert W-function). I'll try to explain where this comes from (and why WolframAlpha tends to spit up that kind of result!). First: what is it?
The definition of Lambert's W-function is that it's the inverse of this function below:
So in other words, if then the W-function tells us which values of we need, i.e. OK great, but how does this relate to our equation? Let's go back to this line in post #8:Now instead of putting let's try a different substitution, say... This gives us:But from the definition of the W-function, the value of which satisfies the above must be given by:Re-expressing all of this in terms of and only, we get:So there's another way we can generate some solutions. The problem is that the W-function isn't expressible (generally) in terms of elementary functions -- so given an arbitrary value of your corresponding which satisfies might end up being (irreducibly) some infinite series!The other issue is that I haven't defined the W-function particularly well. Does even have an explicit inverse? Well, no, not really -- as it stands, our W-function would be multi-valued, so that's no good. We instead need to do what's called a 'branch cut' in complex analysis. (This basically means you define the possible inputs for W such that it only spits out unique values.) So the full solution set would look something like:where, for each the function represents a different branch of the W-function for each of the ranges of possible values ofI haven't done a full video on branch cuts but I do motivate them to some extent in this video if you want to have a look. I've also got a few videos on contour integration which -- while not relevant here -- do involve choosing branch cuts, so I can link to those if you want them too.
By the way, one way you can visualise the connection of the W-function to the graphs Bob and ganesh have shown is by thinking of the straight line as the trivial solution (i.e. y = x) and the curved component as being the equation involving the W-function. Indeed, if you plug in x = e into the equation above then we find that x = y = e also.
Hi Bob,
I'm afraid I don't understand this result, where it comes from or how it is significant.
Bob's result can also be seen in another way: as can be seen from the post above, one such 'solution set' is given by the following:
Hi Relentless,
I'm not sure what maths background you have but I'll try to explain what's going on as best I can -- please let me know if anything isn't clear (happy to talk through it in more detail if needed). You'll find the formulae you need to generate the non-trivial solutions in this post and in the first half of post #12.
The standard method for generating such solutions is as follows. Consider the equation Taking logs of both sides gives us:i.e.
Now suppose that for some value This gives us:i.e.
and since then we also haveThis is called a parametrisation (also called a 'change of variables'). Now to generate some solutions to your equation we just need to vary Clearly setting gives us so let's try a different value, say, Simplifying, we get:Note that the cube root of -2 has three solutions: one is real, the other are complex (and conjugate) roots -- so you could generate complex solutions this way, too.
Hi ziabing,
Welcome to the forum. I will assume here the natural numbers include 0 -- otherwise there is no for which To show that this function is surjective, you need to show that every element can be 'hit' by some under your function . So first consider the case where . Can you always find an such that ?Yes, because we can just take in which case -- because must be even -- And since then it's clear that and so it must be that That takes care of the positive integers (and 0) in -- now you just need to do the same for the negative integers in and you're done. So let's consider the case where So, given can you find an such that we'll always have ?I will leave you to finish this off -- but please do post back if you need more help.
Yes, most of these new members have unfortunately joined solely to advertise and we tend to remove them as soon as they post something. In the vast majority of cases these are fairly easy to spot as their e-mail addresses often show up on widely shared spam lists -- however, there are occasionally some that slip through the net particularly if they masquerade as genuine posters to begin with. We do try to deal with these at the earliest possible instance but naturally we cannot patrol the site every hour of the day (though we do our best!).
The forum's current infrastructure thankfully blocks most of this automatically -- for example, without the requirement to verify your e-mail address before posting we'd easily get over a hundred members flooding the forum within a day.
Your conclusion is correct, but the same comments I made on your other piecewise function thread also apply here so I suggest you read that first (it's the same kind of set-up, just with two pieces rather than three).
The question did ask for you to use a graph. So what do you think the graph looks like and in particular what do you think it looks like near x = -1?
Your conclusion is correct -- however, there are a few technical points to note.
Find the limit of x^2 as x tends to 1 from the left side.
What you really want to be saying is that you are finding the limit of f(x) as x tends to 1 from the left, rather than the limit of x^2.
(1)^2 = 1
Strictly speaking, this is how you evaluate the limit of f(x) as x approaches 1 from the left, but this isn't how you evaluate the left-hand limit of x^2, this is just substituting in x = 1 -- although you can get away with it here. (This is because x^2, 2 and -3x + 2 are all continuous functions, so the left and right-hand limits for each function exist, are equal, and are equivalent to their usual limits.) The way you ought to be thinking about it is:
It'll lead to the same answer, but conceptually this is along the sort of lines you'll be required to think through when you look at the epsilon-delta type questions.However, the question did ask for you to use a graph. So what do you think f(x) might look like on a graph? What kind of shape does it have as you move from left to right? And in particular what happens near x = 1?
Welcome!