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Thanks both -- I've edited the original post back to what it was (and it should stay that way).

Hi David,

Is your goal to sit the A-level exams privately or are you looking to self-study the courses for general interest?

In addition, **Disfruta las Matemáticas** is the Spanish version of the website.

Hi jadewest,

Yes, that's right.

One way you can check your answer is just to consider some different fence post heights. Suppose you had a fence post that was 49 inches tall -- this is clearly within the 2 inch margin of error. According to the inequality in option A, we have:

|x - 48| = |49 - 48| = 1

and this is less than or equal to 2, so the inequality works in this case.

Similarly if you choose a fence post height of 47 inches, your inequality will still hold.

If you had a fence post of 51 inches then your inequality shouldn't hold -- and indeed in this case we have |x - 48| = |51 - 48| = 3, which is not less than or equal to 2. So your inequality correctly distinguishes between fence posts within a 2 inch margin of error and those that aren't.

Hi CurlyBracket,

Yes, that's right. In general:

These are sometimes also called the binomial coefficients, because they represent the coefficients (i.e. the 'leading terms') in the expansion of A shorthand way of denoting the above is:which you might see on your scientific calculator.

Note that an important point is whether or not the order matters. For example, suppose you have four cards labelled (A,B,C,D) and you want to choose 2 out of those 4. How many different ways are there to do it? If we don't care about the order, the combinations are just:

(AB, AC, AD, BC, BD, CD)

which represents different ways. On the other hand, if the order does matter then we'd have:(AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, DC)

which represents different permutations. If the order of the objects we've selected is of interest then we use a slightly different formula:Hi CurlyBracket,

First, you're being asked to consider placing 2 pebbles in an 8 x 8 grid with 64 squares. So the total number of possibilities we need to consider is '64 choose 2' - which just answers the question "how many different ways are there of choosing 2 squares out of 64"? This gives us:

Now we need to work out how many different ways two pebbles can be arranged in the same **column**. Well, how is this choice made? First you (a) choose one of the 8 columns that you're going to put your pebble in, and then (b) choose a square in the column you've selected to put the second pebble. There are therefore

ways to do this (you first pick one of the 8 columns, and then once you've put the pebble down you've got 7 squares left to choose from in that column).

Next, how many different ways can two pebbles be arranged in the same **row**? This is again 56 (just rotate the board 90 degrees and you're back to dealing with columns again).

So the probability is:

Let me know if that makes sense - happy to clarify anything if needed.

An interesting extension of this problem to check your understanding is to consider the case where you've got an *n* x *n* chessboard. What would the probability be in that case? How about the case where you have more than two pebbles? What happens if the pebbles are allowed to share a diagonal?

Hi MaddSci3ntisT,

Welcome to the forum.

I say yes -- a google search of their e-mail address (include the quotation marks around it so you get exact matches) reveals the same poster also joined freemathhelp under the username 'e**aycr*tics' and 'Keira' claims to be a brand manager for the site. Moreover there are also a few blog sites you can find containing (IMO) obviously bogus reviews of the service.

What do I think? I think 'keiraec' has signed up to a bunch of forums like this one pretending to need help with something by copying and pasting someone else's question, when 'keiraec' is likely only interested in using our forum for the free ad space (note that they joined nearly two years ago and this was the only post they made). This strategy is a hallmark of people who sell e**ay-writing services online: they come to our forum and post something which looks genuine to try to 'fit in' but it almost always comes packaged with a third party link, either affixed to their profile, their signature, hidden somewhere in their post, etc. But that's just my opinion.

By the way, I think the same applies to 'Muneera123' in post #8 and 'stivemorgan' [sic] in post #4 who has also used their signature to advertise their e**ay services. I also previously deleted a post which came just before Mathegocart's which was doing the same thing (this is what bumped the thread up again so it looks as though there is a 2-year gap between posts #11 and #12), so clearly this thread attracts a lot of the same types of people. I imagine just using the word 'e**ay' attracts bots to this thread.

It is unfortunately not just a recent phenomenon: it has been happening daily since I became a mod here and probably before then too (based on the lengthy ban list). In most cases the decision to delete/ban is straightforward since many of the posters you mention have often already been reported multiple times on places like 'Stop Forum Spam' or otherwise have an e-mail address which a simple google search reveals that they've done the same thing on half a dozen other forums. In other cases it can be much more difficult because some new members will deliberately post something harmless and then edit in a link a few days later, or they might pose as an ordinary member only to start eventually posting ads. These (particularly the former) are generally harder to spot and are less easy to deal with: I generally avoid banning posters unless I have good reason to and prefer to have some evidence of wrongdoing if at all possible but these types are admittedly (somewhat) rarer.

New users have to click on an activation link sent to an e-mail address to register which cuts down the number of spam registrations significantly: without this we would easily have a hundred different spammers a day (this is no exaggeration). As I understand there are also restrictions which prevent most new members (with < x posts) from posting links, which is why we have lots of new daily registrations from spam accounts but of those only a small proportion end up posting ads, and there are also word censors which filter out a few too. Consequently the spam that does make it through is, IMO, annoying but not unreasonably so, given (a) the nature and number of spam posts, (b) the overall forum activity susceptible to clicking on said posts and (c) the administrative resource available. At the moment (a), (b) and (c) are reasonably balanced and therefore the level of spam is, in my view, manageable. Regrettably we have in the past had the occasional new member post some particularly awful content on here -- and I can't be certain that some of the people who visit this forum (members or not) didn't fall victim to it where we may not have been able to deal with the issue immediately -- but fundamentally the most determined spammers/advertisers will find a way around any precautionary measure and so we tend to operate on a best endeavours basis given the tools we have.

In the meantime please continue to report any posts you see like this as it helps us deal with them more quickly: we often check the forum a few times a day but there will always be the odd one or two that get through when none of the mods/admins are active.

Bob wrote:

I'm still stuck with

. Prove thatEasy to show it's a possible value for x+y and the graph suggests it's unique, but I still cannot prove it properly. Hint please.

Hi Bob,

CurlyBracket wrote:

(4 - 9/2) ² = (5 - 9/2) ²

4 = 5

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 theHi 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 convertibleIn 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

Mathegocart wrote:

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:

Similarly we can obtain the classical solution by setting

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.

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 Setting gives us the interesting solution: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 approaches

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.

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 the post above, 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). **You'll find the formulae you need to generate the non-trivial solutions in this post and in the first half of post #12.**

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: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.