Advanced Higher Maths 2012-2013 : Discussion and Help Thread

Look whos came out of his rock and entered the forum! Ready to replicate your prelim performance?

Yeah?

When you sub in x = +/- infinity to the divided asymptote your finding out what direction the line is approaching the asymptote at? Or are you asking something else?

Take $y = 1 + \frac {1}{x}$ when you divide the fraction by the highest power of x

$y = 1 + \frac {\frac {1}{x}}{1}$ then you sub in x = + infinty
$y = 1 + \frac {+}{1}$ (the sign shows if the full top line is positive or negative) so y approaches 1 from above and subbing in x = - infinty:
$y = 1 + \frac {-}{1}$ again, showing that y approaches 1 from below so the graph is as shown on wolfram alpha

And so will you!! I would say the same but I know that won't happen.. I have no motivation this year!


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Unconditional-ytis
The illness everyone gets when they get an easy unconditional - I got it too!

I do feel a bit like a hermit I must say nope, not quite yet, hopefully by Wednesday I'll be able to better it though I'm more looking forward to doing the higher paper when I'm out!
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Unconditional-ytis
The illness everyone gets when they get an easy unconditional - I got it too!

I really want to do the Higher Paper I don't know if I will though, I still make the mistake of differentiating instead of integrating every now and again... I'd embarrass myself

Bring the paper home then and give it ago someone will put a marking scheme up so you can check your answers first


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I do feel a bit like a hermit I must say nope, not quite yet, hopefully by Wednesday I'll be able to better it though I'm more looking forward to doing the higher paper when I'm out!


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I told someone to take her paper out and lemme see it when we're going home, so I'll get a little look at least

I've told 6 people I want a copy of the paper! So I may end up with a few copies hahaha!
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And so will you!! I would say the same but I know that won't happen.. I have no motivation this year!

Maybe send one my way? I'm sure I'll find a copy somewhere!

Could anyone look at Question 10 on the 2002 paper and explain to me what to do with it?

For part (a) I get that S(1) = $\frac {n(n+1)}{2}$
for (b): (1-x) S(x) = S(x) - x S(x) but after that I just blank and don't see it.
for (c): same thing, just don't see it

Maybe send one my way? I'm sure I'll find a copy somewhere!

Could anyone look at Question 10 on the 2002 paper and explain to me what to do with it?

For part (a) I get that S(1) = $\frac {n(n+1)}{2}$
for (b): (1-x) S(x) = S(x) - x S(x) but after that I just blank and don't see it.
for (c): same thing, just don't see it

I don't want to lead you through part c until you've got b:

What happens if you divide the equation you've got above by (1 - x)?

For (b), multiply (1+2x+3x²+...+nx^n-1) by (1-x) and take it from thee

$S_n (x) = \frac {S_n (x) - xS_n (x)}{1 - x}$??

Or do you mean the equation in part (a)? In which case I'm not seeing the significance either...

For (b), multiply (1+2x+3x²+...+nx^n-1) by (1-x) and take it from thee

He's already done that, hence how he saw that (1 - x)Sx = Sx - xSx.

Edit: Ignore me, I assumed he took the same route as me to finding it. Just noticed that you can much more easily do it by treating Sx as a term and expanding the bracket.



Exactly!

Now what happens if you sub in the sums...?

Yeah I just want to prove my teacher wrong haha. Predicted me an A which got changed to an N/A, whoops. Nah, me neither :/ pretty much the same with a lot of sixth years I guess!

Maybe send one my way? I'm sure I'll find a copy somewhere!

Could anyone look at Question 10 on the 2002 paper and explain to me what to do with it?

For part (a) I get that S(1) = $\frac {n(n+1)}{2}$
for (b): (1-x) S(x) = S(x) - x S(x) but after that I just blank and don't see it.
for (c): same thing, just don't see it

I've told 6 people I want a copy of the paper! So I may end up with a few copies hahaha!


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For (b), multiply (1+2x+3x²+...+nx^n-1) by (1-x) and take it from thee

Oh! I get it! Thanks I freaking hate series, its the only part of the course thats actually causing me bother now tbh!

So for part c: thats gonna just be a Sum to infinity right?

part c is sum to infinity yes

x