bloodhound

the problem is there are too many forums and sub forums in here. so it takes ages to browse everything

me and my mate have a forum at

http://www.marrouche.net/forum we also discuss mathematics there.

lol . thats a lot.

I have about 11 hrs a week of lectures and about 3 hours of problem class and 1 hour group tutorial a week. but i never go. i got told off for not attending problem classes at all.

yeah. newton raphsons method is much much faster. but the function must be nice and several conditions must be satisfied for convergence to the solution ure looking for .

I think, if u descrive the events properly, and then bang in the numbers in the bayes therom, or theorem of total probability, u get the answer. but i am to lazy to do anything like that.

Maths is Nottingham is also one of the best if u look at the tables. The timetable is very light though. so thats good. i have a fren who studies maths at imperial. also another frens roommate also does maths. and when i look at the work they get and their timetable, i thank god that i went to nottingham.

but what is the condition of e so that it loses mga of energy each bounce

i often dont go to bed at all so that i wont get up late and miss the exams. i did that two days in a row. ... and i needed to revise badly as well. i once went to lecture about 2 mins before it finished. just because i had to hand in a probability assignment. that module is over now. have statistics and mathematical structurs now. along with the core modules.

VECTOR spaces is probably the most interesting. Its absolutely mind boggling when u think about it.

i used to love stats at A-level. but now its just boring. maybe the lecturer is boring. so many tests to learn. we get huge blocks of lecture notes. and i often miss the lectures cos i am too lazy to wake up at 11 am

Because i am doing a project on bouncing balls, i am also looking at a situation where the ball bounces down a stair.

What i need, is to find out the required value of e for this question

ok , u have a ball bouncing down a stair. forget about the horizontal stuff, u can either think of it as looking head on to the stairs. or imagine a ball bouncing on a surface, but after each bounce the surface moves down a fixed height. basically a stair. u get me rite.

if u look at the graph. we DROP a ball at rest from height H, and the ball as coeff of restitution e, all i want is the condition for e so that the blue heights are equal. or equivalently the brown heights are equal. i would prefer the blue heights.

Take the height of the steps to be "a" or any letter u like

anyway . cheers

glad to be of some help. hows maths in warwick. i heard its one of the best. was having a look at some of u problems in the competition sections. quite good questions.

Since pple have asked this question before in other forums. ill post my solution to it anyway

take a function f where f (x)=ln(x)/x

therefore a^b=b^a if and only if f(a)=f(b). now differentiating f(x) u will find that there is a maximum at x=e. therefore the function f is increasing for x<e and decreasing for x>e. therefore its not possible for both a,b <e or a,b >e cos then f(a) will never be equal to f(b) for a<b

so the only solutions possible is a<e and b>e . now we have 0<a<e. and a is an integer. therefore a=1 or a=2. putting in a=1 we get b=1 which is not a solution as we need a<b. the only other soultion is a=2. puting in a=2 and solving f(b)=f(2) by similar method or however u want we get b =4 . therefore the only solution to a^b=b^a for 0<a<b is a=2,b=4

anyway. i did this question for integer solution of a^b=b^a given 0<a<b

here it is

take a function f where f (x)=ln(x)/x

therefore a^b=b^a if and only if f(a)=f(b). now differentiating f(x) u will find that there is a maximum at x=e. therefore the function f is increasing for x<e and decreasing for x>e. therefore its not possible for both a,b <e or a,b >e cos then f(a) will never be equal to f(b) for a<b

so the only solutions possible is a<e and b>e . now we have 0<a<e. and a is an integer. therefore a=1 or a=2. putting in a=1 we get b=1 which is not a solution as we need a<b. the only other soultion is a=2. puting in a=2 and solving f(b)=f(2) by similar method or however u want we get b =4 . therefore the only solution to a^b=b^a for 0<a<b is a=2,b=4

i dont think its possible to find a general solution for x^y=y^x. something do with trascendental numbers and fucntions

yeap; the previous post has said everything i was about to say in this one.

oh i love maths so much

have a look at some sequences where the denominators tend to 0 . u will find that the limit of the sequence doesnt have to be infinity.

Anyway. I am a Maths Student at University of Nottingham. Just about to finish my first year. Have exams in may/june. Ok then. introductions over, ill give u a started question:

Find all integer solutions for a^b=b^a given that 0<a<b.

Its a really nice question I found in a book. It has a nice solution as well.

ooh dont bother with that question. i see that its been asked in other sub forums

HEHE

anyway later