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# Maths Challenge... watch

1. (Original post by rieuwa)
Well done "RDOH"! One can see how intelligent people born in June are.... hint hint.....

That was one of the most challenging questions on the ISMTF Junior Maths competition in Paris last year. Ever thought of taking part? The Senior Maths competition for Years 12 and 13 is at my school next year (The British School in the Netherlands).. ask your teachers!

Well done!
Thanks but...

(Original post by ISMTF Website)
2. Contestants must be enrolled in a non selective international school.
Doesn't look like it's for me

2. (Original post by kimoni)
Ok.. last one.. do this, with working please! If anyone else can do this, let me know!
∫ 1 / (x^4 +4) dx
I don't think anybody's done this one ?
Attached Images
3. hard_integral.pdf (45.4 KB, 186 views)
4. Fermat, a little off the point but if you don´t mind me asking the layout of your solution looks rather nice, are you using LaTex or mathtype and Acrobat?

I have been using Mathcad to produce .html, but it really doesn´t do the job well and Im looking for an alternative to create .html and/or .pdf.

thanks

MrM.
5. (Original post by summer)
Fionna you r getting a lot of these tricky questions correct!!
Why didnt your school promote u a year, they should have !!
Good question Summer.

Finally, someone with something sensible to say .
6. Solve this:

A 'snooker' table (measuring 8 metres by 4m) with 4 'pockets' (measuring 0.5m and placed at diagonal slants in all 4 corners) contains 10 balls (each with a diameter of 0.25m) placed at the following coords:

2m,7m...(white ball)
...and red balls...
1m,5m... 2m,5m... 3m,5m
1m,6m... 2m,6m... 3m,6m
1m,7m... 2m,7m... 3m,7m

The white ball is then shot at a random angle from 0 to 360 degrees.
Just to make it clear, a ball is 'potted' if at least half of the ball is in area of the 'pocket'

Assuming the balls travel indefinitely (i.e. no loss of energy via friction, air resistance or collisions), answer the following:

a: What exact angle should you choose to ensure that all the balls are potted the quickest?
b: What is the minimum amount of contacts the balls can make with each other before they are all knocked in?
c: Same as b, except that each ball - just before it is knocked in - must not have hit the white ball on its previous contact (must be a red instead of course).
d: What proportion of angles will leave the white ball the last on the table to be potted?

7. (Original post by Bezza)
We asked our teacher about this and I seem to recall it has something to do with the fact that a logarithm of a negative number is periodic.

We weren't actually shown anything but I've had a think and come up with this:
e^i*pi = e^(1 + 2n)i*pi = -1 for all integers n
ln(-1) = (1 + 2n)i*pi
ln[(-1)^2] = 2ln(-1) = 2(1 + 2n)i*pi = (2 + 4n)i*pi

ok, what I was thinking doesn't actually work then. Crap.
I think you've got the right idea. It's becasue the log function isn't a function in the strict sense when extended to complex (or even negative) numbers. This is because it is multi-valued. A similar example is arcsin, since arcsin(0) = pi, and arcsin(0) = 2pi = mpi
8. (Original post by Fiona_smimms)
Good question Summer.

Finally, someone with something sensible to say .

......erm not really.
Just wanted to see how you would respond.
You see the problem is that your maths has let you down.
however if you were really in year 10 then it should have been April 1989.
9. (Original post by summer)
......erm not really.
Just wanted to see how you would respond.
You see the problem is that your maths has let you down.
however if you were really in year 10 then it should have been April 1989.
LOL. clinical genius
10. (Original post by Bezza)
We asked our teacher about this and I seem to recall it has something to do with the fact that a logarithm of a negative number is periodic.

We weren't actually shown anything but I've had a think and come up with this:
e^i*pi = e^(1 + 2n)i*pi = -1 for all integers n
ln(-1) = (1 + 2n)i*pi
ln[(-1)^2] = 2ln(-1) = 2(1 + 2n)i*pi = (2 + 4n)i*pi

ok, what I was thinking doesn't actually work then. Crap.
I think I've worked out an explanation for this. Not sure if it's mathematically sound though.

ln[(-1)^2] = 2ln(-1) = ln(-1) + ln(-1)
If you then take 2 different values for ln(-1), ie use 2 distinct integers n and m
ln[(-1)^2] = (1+2n)i*pi + (1+2m)i*pi = 2(1+n+m)i*pi
If you then let n=0, m=-1, you get ln[(-1)^2] = 2(1-1)i*pi = 0
11. yup fiona you sound very up yourself and very narrow minded. Wait till you're a bit older before you start ****ging other ppl off, you're not nearly as clever as you think...

any whos...i saw this one somewhere...

1, 11, 21 ,121, 111211, 311221, ...

what are the next 2 terms in this sequence
12. Birthday:
April 29, 1990
Sex:
Not male- eughh
Full Name:
dduughhh
Exams You are studing For:
I got the best gcse results in my college's history
Biography:
i mess about in college, have a larf etc.
Location:
Milton
Interests:
drinkin', chillin, the usual etc...

lol nice profile fiona...are you trying to be funny?!?! I never thought you could be in college in year 10...hmmm the holes in my own knowledge are limitless. so you've already done your gcse's and your in year 10??? AND you got the best results ever...yup of course...'GET A LIFE'
13. (Original post by Mrm.)
Fermat, a little off the point but if you don´t mind me asking the layout of your solution looks rather nice, are you using LaTex or mathtype and Acrobat?

I have been using Mathcad to produce .html, but it really doesn´t do the job well and Im looking for an alternative to create .html and/or .pdf.

thanks

MrM.
It's latex . My first one!
I normally use mathtype for (some) math submissions on this forum. I find it's a lot faster than latex plus it's more intuitive to use. But it does have its limitaions.
Documents can only be so large, otherwise mathtype objects! I've also had to cut down douments I submitted to this forum into two parts because as a single file it just couldn't be handled properly. Also, layout of eqns on mathtype can be a bit tricky.
I originally felt latex was too much trouble - especially for forum submissons - too much to learn - but I'm getting better at it now. And as you said, it's a nice layout, certainly better than mathtype. There're just some of the formatting bits I've gotta get used to yet
14. (Original post by Fermat)
It's latex . My first one!
I normally use mathtype for (some) math submissions on this forum. I find it's a lot faster than latex plus it's more intuitive to use. But it does have its limitaions.
Documents can only be so large, otherwise mathtype objects! I've also had to cut down douments I submitted to this forum into two parts because as a single file it just couldn't be handled properly. Also, layout of eqns on mathtype can be a bit tricky.
I originally felt latex was too much trouble - especially for forum submissons - too much to learn - but I'm getting better at it now. And as you said, it's a nice layout, certainly better than mathtype. There're just some of the formatting bits I've gotta get used to yet
Go to http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/ for a good introduction to latex.
15. (Original post by IntegralAnomaly)
Go to http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/ for a good introduction to latex.
Already got that in my favouirites, ta
I found that intro a bit on the brief side though! For example, in the page on Blank Spaces and Carriage Returns in the Input File, there is mention of how carriage returns in the input file will be ignored for the output file, but no mention of how to create carriage returns, i.e line feeds/line spaces for your document, at least not in that page. I use double-backslash, and it was by chance that I came across that.
That site certainly got me started off well, but now I've got a couple more sites in my favourites and I've got a book on Latex on order from the library
16. (Original post by Whywhywhy)
What a waste of time...if she was actually a 'genius' she would be taking AO Maths or A-Level Maths early. ROFL 'GCSE.' Maths 'geniuses,' the types that go to Cambridge, are spotted at all the Olympiads etc. from an early age. Stop wasting people's time with your GCSE rubbish.
(Original post by Whywhywhy)
What a waste of time...if she was actually a 'genius' she would be taking AO Maths or A-Level Maths early. ROFL 'GCSE.' Maths 'geniuses,' the types that go to Cambridge, are spotted at all the Olympiads etc. from an early age. Stop wasting people's time with your GCSE rubbish.
(Original post by Whywhywhy)
What a waste of time...if she was actually a 'genius' she would be taking AO Maths or A-Level Maths early. ROFL 'GCSE.' Maths 'geniuses,' the types that go to Cambridge, are spotted at all the Olympiads etc. from an early age. Stop wasting people's time with your GCSE rubbish.
(Original post by Whywhywhy)
What a waste of time...if she was actually a 'genius' she would be taking AO Maths or A-Level Maths early. ROFL 'GCSE.' Maths 'geniuses,' the types that go to Cambridge, are spotted at all the Olympiads etc. from an early age. Stop wasting people's time with your GCSE rubbish.
(Original post by Whywhywhy)
What a waste of time...if she was actually a 'genius' she would be taking AO Maths or A-Level Maths early. ROFL 'GCSE.' Maths 'geniuses,' the types that go to Cambridge, are spotted at all the Olympiads etc. from an early age. Stop wasting people's time with your GCSE rubbish.
17. (Original post by Bezza)
We asked our teacher about this and I seem to recall it has something to do with the fact that a logarithm of a negative number is periodic.

We weren't actually shown anything but I've had a think and come up with this:
e^i*pi = e^(1 + 2n)i*pi = -1 for all integers n
ln(-1) = (1 + 2n)i*pi
ln[(-1)^2] = 2ln(-1) = 2(1 + 2n)i*pi = (2 + 4n)i*pi

ok, what I was thinking doesn't actually work then. Crap.
(Original post by Bezza)
I think I've worked out an explanation for this. Not sure if it's mathematically sound though.

ln[(-1)^2] = 2ln(-1) = ln(-1) + ln(-1)
If you then take 2 different values for ln(-1), ie use 2 distinct integers n and m
ln[(-1)^2] = (1+2n)i*pi + (1+2m)i*pi = 2(1+n+m)i*pi
If you then let n=0, m=-1, you get ln[(-1)^2] = 2(1-1)i*pi = 0
I've found another way of thinking about it now which I think's probably better.

If you think about e^ix = 1 then the trivial solution is x = 0, but since e^ix = cosx + isinx, e^ix = e^(ix+2k*i*pi) where k is any integer so e^2k*i*pi = 1
Then ln1 = 2k*i*pi Taking k=0 gives the principal solution ie ln1 = 0
Since ln(-1) = (1+2n)i*pi and ln[(-1)^2] = 2ln(-1) = 2(1+2n)i*pi
As (1+2n) is an integer there will be a corresponding value of k so for every value of ln[(-1)^2] there will be an equal value of ln1.

As I said before, I'm not sure how good this is mathematically but to me it looks ok!
18. You still havent answered my question!

∫ 1 / (x^4 +4) dx

can you do it or not?
19. ∫ 1 / (x^4 +4)

Int ((x^4 + 4)^-1)dx = ln|x^4+4|?
20. (Original post by Neutrinohunter)
∫ 1 / (x^4 +4)

Int ((x^4 + 4)^-1)dx = ln|x^4+4|?
Try differentiating ln |x^4 + 4| and see what you get..
21. (Original post by kimoni)
You still havent answered my question!

∫ 1 / (x^4 +4) dx

can you do it or not?
fiona hadn't done by now, so I guessed she wouldn't be doing it.
And you did want an answer

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