New Concept (2 Viewers)

[dawso]

nah ac, thats my (and i think u mite be suprised, alot of 4unit students' problem) we go hey, look, this is 2unit only, we dont do 2unit, no need 2 learn any of teh formulas and never revise is, it will be a big prob come hsc time but this is where i struggle in 3 + 4unit, harder 2unit (except in topics like integration) where the topics cross over...

 

[~ ReNcH ~]

Series and sequences still appear in MX2 exams, often in Q6-8...sometimes you have to establish a proof by induction or find the sum of a complex series among other things. It's part of Harder MX1 and such questions quite often actually prove to be the most challenging in the entire paper.

EDIT: By "complex", I don't mean that they involve complex numbers, but rather that it's difficult to observe any obvious pattern.

 

[Slidey]

I'm fine with sequences and series (beyond 4unit level even).

Just don't mention approximate methods of integration. Heh heh.

 

[~ ReNcH ~]

I'm fine with sequences and series (beyond 4unit level even).

Just don't mention approximate methods of integration. Heh heh.

Yea...that damn Simpson's Rule screwed me over in last year's 2U CSSA trial. I was shocked to see it appear in the 1998 CSSA MX2 paper - they'd better not do it again

 

[...]

eh

dunno if this should be 3u or 2u

<b> prove </b> that:

lim _{(x --> 0)} (sin x) / x = 1

 

[dawso]

yeah but rench and steele just dont count.....lol, but thats wat i mean, the simpsons rule and trapezoidal rule, i know how 2 do em, but dont ask me the formula please...

 

[dawso]

Last edited by bubz : Tommorow at 08:42 PM. lol, thats funny, nice work, very speedy edit i suppose....

anyway, thats 3unit, um, cant remember how u do it but, any 3unit text 2 shud hav it or steele will prob post a solution by the time my reply comes through...

 

[...]

yeah, the statement is used in 3unit
but u can still test it using a calculator by pluggin in numbers to see if it works

 

[dawso]

yeah, the statement is used in 3unit
but u can still test it using a calculator by pluggin in numbers to see if it works

u hav been easily fooled man, the entire point of this topic is because u cant just plug things in2 the calculator,

exibit a) sub x = 0 in, omg, it is 0/0 which is undefined, not 1, hm......

round 1 - dawso

soz bout the coccyness m8, lol, i dunno...

 

[dawso]

i think the coccyness started wen i proved steele wrong on 2 seperate occasions....

 

[...]

hahahah err..

thats the point of this limit??
what does the graph look like, when x <b> approaches </b> 0, not when it is at 0

 

[dawso]

lol, i realised that and steele warned me, but pffft, i will still maintain wat i said, lol

 

[~ ReNcH ~]

eh

dunno if this should be 3u or 2u

<b> prove </b> that:

lim _{(x --> 0)} (sin x) / x = 1

CM_Tutor (don't know where he's gone, but he was the resident maths genius when he was on BoS...for those of you who didn't know of him) proposed the use of L'Hopital's Rule (not in the HSC) for evaluating limits...

Suppose you have:
lim (x-->a) f(x)/g(x)

According to L'Hopital's Rule:
lim (x-->a) f(x)/g(x) = f'(a)/g'(a)

There are conditions for this rule to work, but I can't remember what exactly they are...I know it works for lim (x-->0) sinx/x and it makes the working much easier...

* I wish I used it to check my answer in last year's MX1 limit question...coz I made a careless mistake there

 

[acmilan]

Ok, by popular demand, i am posting the first challenge question.

 

[Slidey]

CM_Tutor (don't know where he's gone, but he was the resident maths genius when he was on BoS...for those of you who didn't know of him) proposed the use of L'Hopital's Rule (not in the HSC) for evaluating limits...

Suppose you have:
lim (x-->a) f(x)/g(x)

According to L'Hopital's Rule:
lim (x-->a) f(x)/g(x) = f'(a)/g'(a)

There are conditions for this rule to work, but I can't remember what exactly they are...I know it works for lim (x-->0) sinx/x and it makes the working much easier...

* I wish I used it to check my answer in last year's MX1 limit question...coz I made a careless mistake there

The conditions are: The denominator and numerator must be differentiable at x=a and they must BOTH be zero at x=a. I think you can also use it with limit to infinity, but I don't know much about that and it's not really in the HSC.

The proof I use is like a squeeze: you basically get it so that 1<lim sinx/x<1 for x->0, so is 1. I can post it if you want.

 

[...]

WD slide rule
squeeze is one method

theres another method that i think u might like to see (check your PM for it )

 

[acmilan]

Hehe we are currently in the process of learning all this stuff for differential calculus

 

[Slidey]

WD slide rule
squeeze is one method

theres another method that i think u might like to see (check your PM for it )

Thanks. That's one of Chen's PDFs I haven't read.

 

[withoutaface]

I won't participate on areas which are my strong point, but things like geometry, locus, et cetera, I'm quite weak at, so I'd still like to try such problems.

I agree with shaf. I would like this to be extended to the ME2 forum aswell. I have a number of interesting problems for that forum.

Olympiad papers are packed with 4u challenge questions