Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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Hi jadewest,

The key bit is here:

jadewest wrote:

each number increase on the scale indicates an intensity 10 times stronger than the previous number on the scale.

In other words, if you've got an earthquake A measuring 1 on the Richter scale and earthquake B measuring 2 on the Richter scale, earthquake B has an intensity 10 times stronger than earthquake A. Does that make sense?

What would happen if earthquake A measured 2 on the Richter scale and earthquake B measured 4 on the Richter scale -- what would the difference in intensity be then?

More generally any cubic can be reduced to a depressed cubic by substitution -- and then the method above can be applied.

What have you tried? In what context were you given this problem?

The answers to these questions will help us understand what type of solution you are looking for.

Jeremy Desmond wrote:

Thank you Ganesh for that extra interesting information about pi. But you still haven’t explained how we know pi has an infinite number of decimal places containing non-repeating digits. You stated that computers have calculated pi correct to 1.33 x 10^13 decimal places but how do we know the end is not just round the corner? Is it possible that one day a computer will calculate the exact value of pi to a finite number of decimal places?

Hi Jeremy,

Welcome to the forum.

The reason that pi doesn't have a final digit is because it is an irrational number, and irrational numbers have decimal expansions which continue forever. In other words, to answer your question it suffices to:

(1) Prove that pi is irrational, and then

(2) Prove that all irrational numbers have an infinite, non-recurring decimal expansion.

There are quite a few proofs that pi is irrational, some of which are more complex than others. One of the most common proofs is Lambert's, where he essentially (a) shows that all infinite continued fractions are irrational and then (b) finds an infinite continued fraction for pi, which automatically implies that it must be irrational from part (a). If you're interested in continued fractions I've got some videos about it on my YouTube channel (although it doesn't discuss the continued fraction of pi). This takes care of (1).

Now for (2). Suppose instead that you could find an irrational number which didn't have an infinite, non-recurring decimal expansion. Let's say for example, the number 0.12345123451234512345... This decimal expansion has an infinitely recurrent pattern (the '12345' bit). But we see that if we let x = 0.12345123451234512345... then:

Multiplying both sides by 100000:

Subtracting x from both sides:

And finally, dividing both sides by 99999 gives us:

But hang on -- if we can express it as a fraction of two integers, it must be a rational number! (This is pretty much the definition of what it means for a number to be rational.) Similarly, if pi had a 'final digit' we could also express it as a fraction -- for example, if pi terminated after 5 decimal places, i.e. 3.14159, then we could write that as 314159/100000, which is a rational number.

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

Abbey78336 wrote:

Give factors of 8.81632653061

8.8132653061 is roughly 432/49, which is the same as the coefficient of the middle term that you wrote in your problem here:

Abbey78336 wrote:

2x^2 + (432/49x) = 3

Exactly what have you been asked to do?

Looks good to me.

You can also type the pi symbol in text form by typing the following:

`:pi`

which produces π.

Hi Bob,

Yes, agreed -- the answer will mirror the binomial expansion of (f + g)^3, with the same binomial coefficients, just with number of derivatives rather than powers.

Hi Bob,

No I don't think you are on the wrong track -- your f1 function works (it is three times differentiable on but not four times) because of the cusp at , and you are right that there are a range of possible solutions we could have just by adding constants (or even polynomial terms that vanish after taking successive derivatives).I think the two questions are intended to be separate (the condition that f, g are three-times differentiable is probably just there to ensure that f''' and g''' exist rather than bearing any relation to Q1). I am interpreting (f.g)^(3) as meaning the third derivative of (f times g) but happy to be corrected by the thread starter.

Hi YHWH,

Welcome to the forum.

YHWH wrote:

1.Find a function f: R → R that is three times differentiable on R but not four times.

It might be easiest to start with a function that isn't differentiable at a single point (say, x = 0) and work backwards.

Consider a function which returns 1 for positive values of x and -1 for negative values. What function could you differentiate to get that?

YHWH wrote:

2.Let the functions f and g: R → R be three times differentiable. Calculate (f .g)^(3).

This is a demonstration of Leibniz's rule for differentiation. Use the product rule to determine the first derivative -- then differentiate again, and again, and see if you can spot a pattern.

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.