Unofficial Maths237 thread (2 Viewers)

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yes, inspired by MJ's official stat270 thread, i've created the unoffical maths237 thread

i know flyin' is doing it , so stop hiding
and i occasionally see flamin' there (for no apparent reason *cough*) so he might able to chip in some comments

post away my friends

ps. flyin': did you do the tute questions for wk1?
i can't do the 4cents + 7cents question proerly

 

[flyin']

yes, inspired by MJ's official stat270 thread, i've created the unoffical maths237 thread

i know flyin' is doing it , so stop hiding
and i occasionally see flamin' there (for no apparent reason *cough*) so he might able to chip in some comments

post away my friends

ps. flyin': did you do the tute questions for wk1?
i can't do the 4cents + 7cents question proerly

p(n): 4a + 7b = n, n >= 18, a, b are in N

n = 18 = 4.1 + 7.2
n = 19 = 4.3 + 7.1
n = 20 = 4.5 + 7.0
n = 21 = 4.0 + 7.3

Suppose p(n) holds for all k < n and n >= 22 (*), then
n
= (n - 4) + 4
= 4x + 7y + 4, using (*) and x, y are in N
= 4(x + 1) + 7y

Hence for all n in N, p(n) holds by strong form of mathematical induction.

that's all from me. typing solutions ain't fun.

 

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flyin:

have you done the assignment

 

[flyin']

If done means complete then no.

Negate this proposition.

 

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haha ok

I've *almost* completed it
and by complete, i mean attempted the questions, but have no idea if I am right or not

any ideas on how to tackle q3? tha matrices one

 

[flyin']

hint for q3: calculate what the matrix is for n = 1, then n = 2, then n = 3, and it should be obvious. then the rest of the question is easy.

 

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yeah thats what i was told

but then its more of a "proof by observation"

it might happen for n=1,2,3,4,5 ...
but how to we mathematically prove it is true for all natural numbers of n

edit: oh wait, i think i get what you mean now..let me give it a bash

 

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flyinnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn


assignment time

 

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flyin'


assignment time

have you doing 1) b)
the sequence is 1,1,2,2,3,4,4,8,5,16,6,32,7,64
i was wondering can i split this sequence into two seperate sequence

that is
j(x) = 1,2,3,4,5,6,7,...
g(x) = 1,2,4,8,16,32,...

and then work out the generating function for each sequence and combine it together??
if not; what was your method?


also
for 4)
is that saying

u_1+u_2+u_3+...+u_8 = 30 where u_x (where x = 1,2,3,...,7,8) has no conditions at all?

 

[flyin']

1(b): basically yes, you're on the right track. Except note that the two sequences are:
1, 0, 2, 0, 3, 0, 4, 0, ...
0, 1, 0, 2, 0, 4, 0, 8, ...

4. the conditions are: 1 <= u_1 <= 9, and 0 <= u_i <= 9 for i = 2, 3, ..., 8.

It would not make sense to have, for example, the digit 05555550.

 

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so what exactly is the question asking? (q4)

How many 8-digit numbers are there with digits adding up to 30?


does that mean

55551117 is a 8digit number, because 5 + 5 + 5 + 5 + 1 + 1 + 1 + 7 = 30 ???

and
11111112 is not, because 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2 != 30 ??

 

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flyin'

do you have any idea how to do 1?
thats the only question i'm stuck on

 

[flyin']

Hint:

How do you get a(n), if I gave you a(n-1), a(n-2), a(n-3), etc. ?

How many ways can you a(n), if I just gave you a(n-1), and the numbers in the bracket?

Repeat this for a(n-2) and so on, and it should be obvious.

Another way to look at it. Find the recurrence relation of (1 + 10)* given the same question. Obviously, we have a(n) = a(n-1) + a(n-2), right? [This may already give away the answer, if you understand it or see a pattern.]

Notice: question doesn't ask for what is a(1), a(2), etc, just the a(n). This may be helpful, if you were attempting to find the equation manually by inspection.

 

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this has been a while

but *bump*


flyin' ; how are you planning to study for this exam? and i assume you and your group will do a lot of 'group cramming' together?

 

[flyin']

This is 3 days after the last exam which matters for the group. So I predict by which time, we should all have gone into holiday mode.

How I am planning to study?

I've been reading (note: reading, not doing) tutorial questions and solutions, and assignment questions and solutions. Then when I don't understand something (or more likely, when I cannot remember something which is most of the time), I look at the lecture notes/course notes (online ones). So in short, its read, remember, repeat.

But once I have time after my last important exam, I'll be going through past papers.

I've looked at Rod's strand, and I reckon his section will be very similar to the past papers. They always ask what looks like the same questions. I'm not sure about Gerry's strand.

 

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ah, in terms of knowing the stuff, I am not worried about Rod's part, his part seems easier to understand compared to Gerry's.

You have solutions to tutes? Can i borrow them for a day?

 

[flyin']

tute solutions = solutions from tutes + tute questions i did

I'm surprised other people don't have their own. =p

 

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i don't have the full version thats why


btw, for assignment5, question 1) (a) and (b)

have you read the worked solutions??

for (a) , the solution reads:

a_n^(p) = A₁n + A₂ + A₃2ⁿ + (A₄n² + A₅n)3ⁿ

how did the bold part came to being??

same applies to (b)
it reads:
A particular soltion takes the form a_n^(p) = A₁n³ + A₂n²








edit: for part (a): why was A_2 there?

 

[flyin']

RR: a(n+1) - 3a(n) = 4n + 5(2^n) + 7n(3^n)


* Notice: the 4n on the RHS. A (part of the) particular solution for this is (Jn + K), and not just Jn, for constants J, K. This is why you see A_2. (Details: Table from Chen 16-8)


* Notice the general solution to the LHS is a_n = L(3^n), for constant L. A (part of the) particular solution to the RHS is then [Mn + N(n^2)](3^n), and not just Mn, for constants M, N. This is why you see A_4 and A_5. Read up on "Lifting Trial Functions" (Chen 16-10: 16.7).

 

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ooooooooooooh

how stupid of me, thanks man..

more questions for you tonight after work