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Thanks for your reply Ricky.

I've just noticed I screwed up the last couple of paragraphs in this post. They've been updated now, and basically, what I'm looking for is a function which produces a curve which passes through (x1, 0), (x2, 100), (x3, 0), the maximum point of the curve is 100, and x1 < x2 < x3.

Is this possible?

**Tredici**- Replies: 2

Hi guys,

I'm writing an algorithm for a property matching website.

Originally, I needed a function which produces a y value of 0 when x is 350, 100 when x is 500, and 0 when x is 550. So:

By polynomial interpolation I have:

Then I realised that I needed

to be the maximum point of the curve.Thankfully, Cushydom came to the rescue with the following:

Which works tremendously. However, I now realise I need a general function which can work with any y values, provided that:

, , , and are positive, is the maximum point of the curve, andIs this possible? Any help at all would be hugely appreciated.

Cushydom, thanks for everything. You've been really helpful.

How would you go about abstracting this so that it's not dependent on y values?

, where is the maximum, , , and are positive and ?Thank you so much Cushydom.

This works incredibly well. Can I ask how you worked it out?

**Tredici**- Replies: 6

Hi guys,

I'm writing an algorithm for a property matching website, and I've ran into a little problem. Please excuse any incorrect notation or nomenclature, as I'm fairly unpractised.

I need a function which produces a y value of 0 when x is 350, 100 when x is 500, and 0 when x is 550. So:

After a bit of quite simple polynomial interpolation I have:

However, I've just realised, I need

to be the maximum point of the curve. Is it possible to find a function which satisfies this?Thanks in advance for any assistance.

**Tredici**- Replies: 5

Hi All,

This is probably going to be a very simple problem for you masterminds. Please excuse my ignorance here guys but I'm completely out of practice and haven't done this sort of stuff for about five years.

I need a simple function which will produce a value of 0 when 0 is passed into it, a value of 20 when 50 is passed into it, and a value of 0 when 100 is passed into it.

Sounds so simple, but I've literally spent hours getting something even vaguely close. I've gotten closest with one function, a depiction of which is attached to this post.

Thanks in advance for any help at all.

Cheers, David

**Tredici**- Replies: 6

Please help me find dy/dx. I've tried everything. I thought it'd be a case of simple quotient rule application, then chain rule by dividing dy/dt by dx/dt. Maybe it is, I may just be continuously slipping somewhere. Any help at all would be really appreciated. Thanks guys.

**Tredici**- Replies: 3

This one looks as though it'd be easy to me, but when I get stuck in, I end up with silly simultaneous equations, which I can't seem to work with. Any help with this would be really appreciated. Thanks again lads, Tred.

luca-deltodesco, it appears you do have the correct answer, but I really don't understand your method. I fear you may have used techniques I haven't yet been taught, and hence, I doubt they'll need to be implemented in the exam. However, thanks for your input and well done for getting the correct answer.

Stardust, yep, that's it! I understand that, it's one of those answers you think, how the hell didn't I get that!? Haha. Fantastic! Thank you so much. And no, my M1 exam is on Tuesday! Do you get an extra day!?

**Tredici**- Replies: 7

*"A, B and C are three points on a straight road such that AB = 80m and BC = 60m. A car travelling with uniform acceleration passes A, B and C at times t = 0 seconds, t = 4 seconds and t = 6 seconds respectively. Modelling the car as a particle find its acceleration and its velocity at A."*

This the only question in a Mechanics 1 book that I can't answer, or atleast got relatively close to answering. No matter what I try I'm always ending up with 3 unknowns and hence, am unable to apply the UVAST uniform acceleration equations. Someone, please help me with this! I beg of you! I would be very grateful for a worked answer so that I can understand it. Thanks a lot lads, you've helped me loads.

I have the answer, if it's of help:

Although it's smack bang in front of me, I still can't understand it, so could anyone work it through for me? Cheers, Tredici.

**Tredici**- Replies: 5

I'll probably be here a lot recently, considering the core and mechanics revision, so I hope I don't end up flooding this place with questions left, right and centre. If you happen to know of C4 papers worked through, i.e. not just the answers, it would be of great help. Here's the question:

It's probably very simple, as I've said, and I'd like to add, I'd really appreciate it if it were worked through thoroughly, as I'm not looking for answers; I'm looking to understand how you've answered it, which can then be applied to similar questions. Thanks chaps, you've always been of great help to me.

Oops, sorry for spam.

MathsIsFun wrote:

However, I don't think the "none" arrangement is counted

So, P(0,0) = 0!/(0-0)! = 1/1 ... what???

I've had many a discussion with Mathematicians, and they've all agreed that's the logical reasoning behind it, albeit not 'proof'. Admittedly, from my point of view, it seems most reasonable. I'll continue to disregard plugging 1 into the permutations formula and stating it wouldn't work unless 0! = 1 as 'proof', but hey, I'll happilly admit I'm the least knowledgable in Math amongst the members of this community, and hence, I'd understand if you'll disagree.

Just for a bit of fun, my Mechanics teacher noticed somewhat of a sequence within the factorials:

4! = 24

÷4

3! = 6

÷3

2! = 2

÷2

1! = 1

÷1

0! = 1

The following division would be by 0, and hence, 0! is the lowest of the factorials. Also, MathsIsFun, you've put 0 into the permutations formula and ended with 1/1, and then questioned it. Isn't that what I was saying, that 0 could be arranged only 1 way?

Woah, a fair few responses to this thread. I've been thinking about this factorial business, and I'm wondering if a different way of thinking may make understanding why 0! is considered to equal 1 that little bit easier.

I've considered all of the above mentioned formulae during pondering this, but I fail to see how they can be used as proof. Formulae like the permutations and factorial formula where derived with knowledge that any real number less than zero cannot be applied, even fractions/decimals >0,<1 in the Gamma function. In this instance, the likes of Leonhard Euler forever disregarding the application and resultation of negative numbers as 'acceptable' seem sensible mathematicians.

Using n!/(n-1)!=n does seem sensible at first thought, but by accepting the continuation of the rule such that 1= (1)!/(0)!, you must consider the resultant of further applications, like 0= (0)!/(-1)!, and -10= (-10)!/(-11)!, and that's when it becomes slightly less reasonable.

I'm sure we're all in agreeance in respect to a factorial being the product of all positive

integers from 1 to a given number. With this definition strictly in mind, 0 can **not** be applied as it's distance from 1 in a 1,2,3.. sequence is restricted by the realms of negative infinity, using the term 'negative infinity' very loosely. This is all obviously ignoring a vector like existance, where 0 is -1 from 1, which cannot be used here.

If we think more application as opposed to the confinements of laws, rules and constraining equalities maybe it begins to make sense. For example, what exactly is factorial? Well, factorial is just a simpler version of the permutations forumla, where n = r. Therefore, we're given all the different arrangements of n objects which are taken n at a time.

With this in mind, Let's imagine I've just spent a beautiful day by the lake side and have caught myself two wonderfully sized fish, they're so large in fact I'm going to name them Lance and Roy. Now, how should they sit on my plate when I have them for dinner? Well, I can have Roy, Lance or Lance, Roy? (2!=2). Hmmm, decisions, decisions.

Now, the day after, the clouds swell and stain black, and a powerful rainstorm hits the villiage. Being such a poor man, I rely entirely on fish, and must go out, once again, and try to catch some. But nope, not today. I get home, sling my rod to the floor, reach for the cupboard, grab a plate and stare, green with envy, nostalgic with the thought of yesterday's plentiful supper in mind. With the painful sound of my stomach rumbling and the food depriving rain attacking my straw roof, I think, how many times can 0 fish sit on my plate, and that's only 1 way, with no fish whatsoever. .:. 0! = 1

I realise I've just wasted about 5 minutes of your lives, but oh well, atleast I'm now entirely sure you've all gotten my point .

**Tredici**- Replies: 22

More a question.

This may have been asked before, but I've struggled finding anything vaguely similar. Anyway, to the question.

Why does 0! = 1?

**Tredici**- Replies: 0

I'm unsure of exactly where such a question belongs within this forum, but I'd desperately like to know what are the best revision guides available in certain areas of the Edexcel syllabus at A-Level. I plan to retake C2, C3, C4, M1 and S1 in June (No, I'm not terribly poor at Maths, just have high expectations of myself haha). I've been searching all over the web for various revision guides and I see myself purchasing the AS range of revision guides from CGP books,

I was wondering if any of you had personal recommendations for revision material, bearing in mind they'd ideally be for Edexcel, and whether you know of any particular revision guides renowned for aiding success at AS and A2 level, and those to be avoided. I'd also like to mention, if I were to buy the CGP books, I'd still be in search of C3 and C4 revision guides.

It may be odd, but I feel strangely uncomfortable when revising from guides that cover the whole year as oppose to separate modules, I prefer a different book for a different topic, and hence, I'd prefer if any recommendations were module specific.

Thank you very much for your time, in advance. Tredici.

As Ricky said it's always fun to just experiment with these things and see what comes up. Generally you will notice a reoccuring pattern in behaviours.

However, I'm going to jump at this opportunity to actually help someone, thought I'd be the one absorbing all the help around here, not giving it out. .

1. vertical translation of +c

2. vertical stretch of scale factor b

3. horizontal compression of scale factor k, also referred to as horizontal stretch of scale factor 1/k.

4. if (x) = a ^ x and g(x) = a ^ -x, g(x) is a reflection of (x) in the y axis.

Someone correct me if I'm wrong, don't want to be giving out dummy information!

3? Out of 3? Correct? .

Lmao, that's crazy, wouldn't of done it without you guys.

Let's suppose the last question was:

f(x) = 3e^x - **½ ln (x - 2)** as oppose to f(x) = 3e^x - **½ ln x - 2**

Would it then be..

(3e^x)' = 3e^x

(½ ln x - 2)' = is that just, ½ 1/(x-2)? or simplified 1/(2x-4)

∴ '(x) = 3e^x - 1/(2x-4). Not quite so sure on that one, to be honest.

Ok, and I'm back again.

(x) = 3 sin ² x + sec 2x

Ok, I'm not sure if this is correct, but I made that out to be:

(3 sin ² x)' = using chain rule, 3 [ 2 (sin x) (cos x) ] = 6 sin x cos x

(sec 2x)' = 2 sec (2x) tan (2x)

∴ '(x) = 6 sin x cos x + 2 sec (2x) tan (2x)

Please correct me if I'm wrong .

Another question.

[x + ln(2x)] ³

I'm very unsure of whether this is correct, but I ended up with:

[x + ln(2x)]' = (1 + 1/x)

using chainrule I ended up with:

'(x) = 3 [x + ln(2x)] ² (1 + 1/x)

Ok, looking back on that, it looks very wrong. So corrections would never go a miss.

Final One:

f(x) = 3e^x - ½ ln x - 2

Hmm... let's begin

(3e^x)' = 3e^x, right?

(½ ln x)' = ½ 1/x = 1/2x, right?

(2)' = 0

∴ '(x) = 3e^x - 1/2x.

Hmmm.. please let me know how wrong I am on these, lol. Only way to improve, eh?

Thanks again guys, as I've said before, really appreciate what you guys do, I'm surprised you don't charge!

Oh guys I've just noticed, I haven't made myself too clear; I've written a x sin(x) when I meant more along the lines of (a)times[sin(x)]. How would that be differenciated?

Edit: Don't worry, I got it . Silly me.

Wow, this is incredible. I like this community, such drive to offer a helping hand . This has actually been truly helpful. I understand almost all that you've said, apart from the use of 't' in the chain rule, that's confused me, I don't think we use the letter 't'. Anyway, I've got some fantastic questions involving differenciating that I'm hopefully about to power through, with your help that is . I shall report back to this thread should my limited knowledge fail me .

Thanks guys!