****Leaving Certificate: Higher Level Maths Discussion****

Emilyjaneoh

Emilyjaneoh wrote: »

Does anyone have the proof for the amortisation formula? Just heard about it there, and would it take long to learn?

Sorry I seen it, thanks daniel!

plmko

FatRat wrote: »

Also there are two proofs ye are leaving out:

Tn=Sn-Sn-1
Sn= a(1-r^t) / 1-r

@FatRat how do you prove Tn = Sn - Sn-1??

plmko

Daniel2590 wrote: »

Prove √2 is irrational
Construct √2 and √3
Proof by induction:
1+2+...+n
1²+2²+...+n²
1³+2³+...+n³
a+ar+ar²+...+ar^(n-1) = a(1-r^n) / 1-r
De Moivre's theorem for n E N, n=0 & n E Z.
Derive sum to infinity
Amortisation formula

Don't think we need N E Z for De Moivres anymore!

XtotheZ

Saskatchewan wrote: »

I think you mean in the actual mock? Well if you do, it was. Makes no sense because they left out the fact they they pay the same amount, present values etc. Also why would someone who is paying in advance need to pay the 4.5% aer. Was ridiculous! I just did a quick amortisation formula in 6\7 lines and git it right.... Even though its wrong...

oh yeah i did mean the actual mock :P

2thousand14

Emilyjaneoh wrote: »

Does anyone have the proof for the amortisation formula? Just heard about it there, and would it take long to learn?

http://www.projectmaths.ie/documents/teachers/geometric_Seq_ser.pdf

page 14

plmko

PaleMoonlight wrote: »

How do you prove the Amortisation formula? I don't remember learning how and can't find it in my book

You use Sum of Natural Numbers to do it

ms.tinx

Ok so what is like 95% guaranteed to be on paper 1?

eefah

It seems like that for the proofs, even if you don't know them all word for word, you can get through a lot of them just through logic and the steps of induction ....?
Am I right in saying that, or is desperation clouding my judgement?

2thousand14

eefah wrote: »

It seems like that for the proofs, even if you don't know them all word for word, you can get through a lot of them just through logic and the steps of induction ....?
Am I right in saying that, or is desperation clouding my judgement?

once you know how to start it off its grand

ms.tinx

eefah wrote: »

It seems like that for the proofs, even if you don't know them all word for word, you can get through a lot of them just through logic and the steps of induction ....?
Am I right in saying that, or is desperation clouding my judgement?

Well if you know induction really nd ur basics for proofs then ur fine

Mario95

Emilyjaneoh wrote: »

Does anyone have the proof for the amortisation formula? Just heard about it there, and would it take long to learn?

Try this: en.wikipedia.org/wiki/Amortization_calculator#Derivation_of_the_formula
I like the way they did it, its really easy to understand.

ms.tinx

Its so funny we're all here discussing the "tipped topics" rather than actually studying them... :-D ... lol

FatRat

plmko wrote: »

@FatRat how do you prove Tn = Sn - Sn-1??

Sn = T1 +T2 +T3 ... + Tn-1 +Tn
Sn-1 = T1 +T2 +T3 ... + Tn-1

You minuse Sn-1 from Sn and the difference between them is Tn

Sn-Sn-1 = Tn

eefah

Fúck it, I might just take this paper for loss and spend a full day over the weekend revising for paper 2 .....

Days 298

I think this thread has to much emphasis on derivations! :L

Trapozoidal rule with integration may come up and be worth multiplies of them!

XtotheZ

pen or pencil? apparently on the pink paper pencil can be hard to read

Daithi MacG

FatRat wrote: »

Also there are two proofs ye are leaving out:

Tn=Sn-Sn-1
Sn= a(1-r^t) / 1-r

@FatRat How do you prove the Sn=a(1-r^t) / 1-r?

aleatorio

Finding asymptotes of an equation, would the vertical asymptote just be whatever it isn't defined for? And the horizontal would be lim(x-infinity) of the eqn?

little sis...

aleatorio wrote: »

Finding asymptotes of an equation, would the vertical asymptote just be whatever it isn't defined for? And the horizontal would be lim(x-infinity) of the eqn?

Vertical asymptote is letting denominator = 0 so yes
and yes for horizontal also.

eefah

aleatorio wrote: »

Finding asymptotes of an equation, would the vertical asymptote just be whatever it isn't defined for? And the horizontal would be lim(x-infinity) of the eqn?

I think you're right for the vertical asymptote, but I've no idea about the horizontal ones.

aleatorio

little sis... wrote: »

Vertical asymptote is letting denominator = 0
and yes for horizontal.

I don't understand?
But then wouldn't you be dividing by 0?

Edit:
Never mind I get you know, so say it was 1/(x-2) you let the x-2 equal zero and solve for x yep?

little sis...

aleatorio wrote: »

I don't understand?
But then wouldn't you be dividing by 0?

Well you are right, whatever is isn't defined for is when denominator equals zero. we said the same thing in different ways.

ms.tinx

Alri everyone, don't be stressing its only paper 1. Means theres no theorems, trig proofs or geom constructions.... so everybody chill sure 'til be GRAAAND!!! Nd even if we do mess this up there's another 2 days for us all to cram 4 paper 2 :-) its all G

FatRat

Daithi MacG wrote: »

@FatRat How do you prove the Sn=a(1-r^t) / 1-r?

Same principle as the other one but its more multiplication!

Sn = A + Ar + Ar^2 + ... Ar^n-2 + Ar^n-1
You then multiply both sides of this equation by r:
rSn = Ar + Ar^2 + Ar^3 + ... Ar^n-1 + Ar^n

You then minus rSn from Sn like in the other derivation to get the difference between them:

Sn - rSn = A - Ar^n
Using algebra:
Sn(1-r) = A(1-r^n)
Sn = A(1-r^n) / (1-r)

little sis...

aleatorio wrote: »

Edit:
Never mind I get you know, so say it was 1/(x-2) you let the x-2 equal zero and solve for x yep?

yeh

Aspiring

For the proving root 2 is irrational what's the process.

Peregrine

Aspiring wrote: »

For the proving root 2 is irrational what's the process.

Proof by contradiction. Assume it's a rational number which can be written as a/b, where a & b are Z, b is not 0 and a and b have no common factor.

Then prove that the assumption is wrong.

I'll do it out if you want or can you go from there?

TooMuchStudy

Aspiring wrote: »

For the proving root 2 is irrational what's the process.

Let's suppose √2 were a rational number. Then we can write it √2 = a/b where a,b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in its simplest terms, both a andb must be not be even. One or both must be odd. Otherwise, you could simplify.

From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 * b2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2.

This means b2 is even, from which follows again that b itself is an even number!!!
WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So √2 cannot be rational.

Copy and pasted but it explains it extremely well

Daithi MacG

Aspiring wrote: »

For the proving root 2 is irrational what's the process.

Here's a good video: /watch?v=mX91_3GQqLY (It's from youtube, just put that at the end...)

Aspiring

Thanks lads