more! Product rule differntiaion C3

why? which module you doing at the moment?

I'm first year uni maths......but somehow having to look through many books to read up on certain parts.....sometimes the notes arent that understandable....

oh, i was thinking of doing maths at uni, but a joint module as i wouldn't surive doing a full degree on just maths and yet i couldn't not do maths at all... are you doing an all maths degree? hows it going?

Aslong as you don't do analysis you'll be fine. Everything else seems rather straight forward in comparison (after doing Math/FMath). I'm sure it will get easier though.

argh, i'm not doing further maths! is that a bad thing???

It will be harder yes. But if the uni doesn't require FMath (they do here), then they should teach you it once you arrive, so not all bad. Just means you'll be learning it in a few weeks rather than a few years

i should/ve started doing it in the first place, but i didn't know i would want to do a degree in maths at all! i'll just have to make sure now that all my uni' choices don't want Fmaths otherwise i'm dead..

Oh cool........have u dne Q1 part b then? I keep getting other limits like nx etc......
Nope...Q5 I had a go at.....but really unsure whether its correct....oh...btw, (i'm assuming u've dne 1401), how do u differentiate n times .....?

And btw, Analysis is in all Maths with *** courses....so u cannot avoid it :P

(at least in UCL anyway)

i should/ve started doing it in the first place, but i didn't know i would want to do a degree in maths at all! i'll just have to make sure now that all my uni' choices don't want Fmaths otherwise i'm dead..

i should/ve started doing it in the first place, but i didn't know i would want to do a degree in maths at all! i'll just have to make sure now that all my uni' choices don't want Fmaths otherwise i'm dead..

Differentiate the equation a few times (3/4 times).. off top of my head think it was

(4-x)y'' + 2y' + Piy
(4-x)y''' + 4y'' + 2y'' + Piy'
(4-x)y'''' + 4y''' + 6y''' + Piy''

Repeat, find a relationship between different differentials... prove the relationship by induction. Done. No I'm just about to start Q1 (b). Q5, prove its a Cauchy sequence like he did in class, thus proving the sequence is convergent.

Kool! Was doing it late at night yesterday lol (1401) so i was a bit lazy wi differentiating it so many times haha.....I'll try it out now actually....

Q5: this is what i have:

|Xn+1-Xn|<1/(2^n)
then 1/(2^n) < 1/n

so we want 1/n < e => n>1/e

Then choose N = 1/e

so |Xn+1-Xn| < 1/(2^n)< 1/n < e
i.e. |Xn+1-Xn| < e for all n>N

Would this also work? or is it necessary to introduce an 'm'

You need to introduce an m, with that working you aren't proving that x_n is convergent.

given E>0 you need to find an N, such that m>N and n>N implies |x_n - x_m| < E

Have you done 3 (c)?

EDIT: Look at the notes from fridays lecture, after the first part its practically identical minus the |x_2 - x_1| which are in the notes.

aaah I c.....3(c):

Well subsequence: X_2n contains all even terms
and X_2n+1 contains all odd terms

both subsequences converge to l
so together, the two subsequences contain all terms in X_n
i.e. all terms converge to this l


U think this is a vallid arguement?

I assume that -1 is an exponent?

Should be $(x-1)[-2(2x+1)^{-2}]+(2x+1)^{-1}$

yes i did get that, so do you just leave it like that and it doesn't need simplifying? cos i think thats where i go wrong, i try to simplify it and then it goes wrong. lol.

i simplyfied it to get:

$(2x + 2) (2x + 1) ^-2 + (2x + 1)^-1$

is this wrong?

That should be (-2x+2), and you can further simplify it.