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where and (These are called polar co-ordinates.) Then: -- i.e. raising both sides to the power of Expanding the brackets on both sides (because we want on one side and on the other), we get:

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 In particular, we again see Bob's result showing up here too. Note that our expression for in terms of wasn't defined when i.e. when However, if we consider the limit instead, we get:So as then and so we have:i.e. and both converge to as approachesBy 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 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.

Relentless wrote:

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 my previous post, one such 'solution set' is given by the following:

Clearly this doesn't make much sense when We can, however, consider the limit as so let's do that instead. Setting gives us:One way of interpreting this result is that is exactly the point at which the trivial solutions (that's everything on the line y = x) and the non-trivial solutions (that's the curved bit of Bob's graph) meet.

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).

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. 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:and similarly you can generate (infinitely) many similar solutions by plugging in some additional values of (By the way -- these types of solutions are called 'algebraic numbers'.) OK, so how about negative solutions? Well, let's try setting This gives us:

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.

nycmathguy wrote:

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.

nycmathguy wrote:

(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 -3*x* + 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:

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!