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Original post by Insight314
That's pretty much how it is at the start, but make sure you follow through it since the rigor is important. Have you tried the example sheets? Questions 1 to 5 are accessible after finishing Chapter 1: Complex numbers.


No, but they look easily doable (so I will attempt them now) (I wonder if Siklos steals adapts questions from these example sheets occasionally)
Original post by Mathemagicien
No, but they look easily doable (so I will attempt them now) (I wonder if Siklos steals adapts questions from these example sheets occasionally)


They are pretty easy compared to STEP questions, so wouldn't think so. :biggrin:
Original post by Insight314
They are pretty easy compared to STEP questions, so wouldn't think so. :biggrin:


These particular questions, sure, but I'd imagine that there exist some example problems which can be expanded upon to make a nice STEP problem, even if its only STEP I
(edited 7 years ago)
Original post by Mathemagicien
These particular questions, sure, but I'd imagine that there exist some example problems which can be expanded upon to make a nice STEP problem, even if its only STEP I


Oh yes, example sheets get very difficult and interesting IIRC. They try to make example sheets more difficult than Tripos, so I could imagine doing some insanely difficult questions (more difficult than STEP) in the future. For now, the first questions are pretty straightforward, although they are still interesting.

Btw, did you get to the nice "Lemma-Proof-Theorem-Proof" bit in Dexter's lecture notes? The one with proving the theorem exp(z1)exp(z2)=exp(z1+z2)\exp(z_1)\exp(z_2) = \exp(z_1 + z_2) using the lemma n=0m=0amn=r=0m=0rarm,m\displaystyle\sum_{n=0}^\infty \displaystyle\sum_{m=0}^\infty a_{mn} = \displaystyle\sum_{r=0}^\infty \displaystyle\sum_{m=0}^r a_{r-m, m}.

I wonder if we are meant to know lemmas like these, I would never have thought of doing that.
(edited 7 years ago)
Original post by Insight314
n=0m=0amn=n=0m=0rarm,m\displaystyle\sum_{n=0}^\infty \displaystyle\sum_{m=0}^\infty a_{mn} = \displaystyle\sum_{n=0}^\infty \displaystyle\sum_{m=0}^r a_{r-m, m}.



That ain't look right. Maybe r for n on the RHS?
Original post by EricPiphany
That ain't look right. Maybe r for n on the RHS?


Whoops, thanks, it was a typo.
Original post by Insight314
Whoops, thanks, it was a typo.


That looks like a very fun identity :wink: ngl

and thanks for the rep, I got prsom.
Original post by EricPiphany
That looks like a very fun identity :wink: ngl


What comes out of it is even more fun!
Original post by Insight314
What comes out of it is even more fun!


Oh I'm sure... lol
Original post by EricPiphany
Oh I'm sure... lol


Unparseable latex formula:

\exp(z_1)\exp(z_2) &= \displaystyle \sum_{n = 0}^\infty\displaystyle\sum_{m = 0}^\infty \frac{z_1^m}{m!}\frac{z_2^n}{n!}\\ &= \displaystyle\sum_{r = 0}^\infty\displaystyle\sum_{m = 0}^r \frac{z_1^{r - m}}{(r - m)!}\frac{z_2^m}{m!}\\ &= \displaystyle\sum_{r = 0}^\infty\frac{1}{r!} \displaystyle \sum_{m = 0}^r \frac{r!}{(r - m)!m!}z_1^{r - m}z_2^m\\ &= \displaystyle\sum_{r = 0}^\infty\frac{(z_1 + z_2)^r}{r!} = \exp(z_1 + z_2)



Look how they took out the 1r!\frac{1}{r!} to get the binomial coefficient, and then reverse binomial expansion to get to (z1+z2)r(z_1 + z_2)^r. Damn!!
Original post by Insight314
Oh yes, example sheets get very difficult and interesting IIRC. They try to make example sheets more difficult than Tripos, so I could imagine doing some insanely difficult questions (more difficult than STEP) in the future. For now, the first questions are pretty straightforward, although they are still interesting.

Btw, did you get to the nice "Lemma-Proof-Theorem-Proof" bit in Dexter's lecture notes? The one with proving the theorem exp(z1)exp(z2)=exp(z1+z2)\exp(z_1)\exp(z_2) = \exp(z_1 + z_2) using the lemma n=0m=0amn=r=0m=0rarm,m\displaystyle\sum_{n=0}^\infty \displaystyle\sum_{m=0}^\infty a_{mn} = \displaystyle\sum_{r=0}^\infty \displaystyle\sum_{m=0}^r a_{r-m, m}.

I wonder if we are meant to know lemmas like these, I would never have thought of doing that.


These types of sums are common in stuff like combinatorial proofs and stats in summation algebra. Im starting probability btw :smile:.


Posted from TSR Mobile
Original post by Insight314
Unparseable latex formula:

\exp(z_1)\exp(z_2) &= \displaystyle \sum_{n = 0}^\infty\displaystyle\sum_{m = 0}^\infty \frac{z_1^m}{m!}\frac{z_2^n}{n!}\\ &= \displaystyle\sum_{r = 0}^\infty\displaystyle\sum_{m = 0}^r \frac{z_1^{r - m}}{(r - m)!}\frac{z_2^m}{m!}\\ &= \displaystyle\sum_{r = 0}^\infty\frac{1}{r!} \displaystyle \sum_{m = 0}^r \frac{r!}{(r - m)!m!}z_1^{r - m}z_2^m\\ &= \displaystyle\sum_{r = 0}^\infty\frac{(z_1 + z_2)^r}{r!} = \exp(z_1 + z_2)



Look how they took out the 1r!\frac{1}{r!} to get the binomial coefficient, and then reverse binomial expansion to get to (z1+z2)r(z_1 + z_2)^r. Damn!!


Ye, that doesn't look like something I would've ever thought of. tbf, mathematicians knew it was true before they went looking for a proof, so one had to exist somewhere.
Original post by physicsmaths
These types of sums are common in stuff like combinatorial proofs and stats in summation algebra. Im starting probability btw :smile:.


Posted from TSR Mobile


Wow. So this is like a pretty usual lemma? I got so excited when I saw how this simple identity is used to prove such a general result, and then how they took out the 1r!\frac{1}{r!} to get to binomial coefficient and binomial expansion just made me rock solid (not literally ofc).

I haven't done stats since Year 9, but I talked to that experienced teacher I was telling you about before and he adviced me to do V&M then Groups and then Analysis, and not to touch D&R nor Probability since D&R requires VC and Probability should be taught sensibly to me since it is first time I do big boy stats.
(edited 7 years ago)
Original post by EricPiphany
Ye, that doesn't look like something I would've ever thought of. tbf, mathematicians knew it was true before they went looking for a proof, so one had to exist somewhere.


Yeah, but it looks so beautiful and that binomial expansion just amazed me when I first saw it.
Original post by Insight314
Yeah, but it looks so beautiful and that binomial expansion just amazed me when I first saw it.


true.
it's possible that whoever wrote that proof started from the end and worked his/her way up.
started from the bottom now we're here.
Original post by Insight314
What is the difference between a lemma and a corollary? :confused:


corollary comes after the proof, things that follow without much more work.
lemma is before the proof, something you want to use in the proof.
Original post by EricPiphany
started from the bottom now we're here.

LOL :rofl:

But yeah, I would think so too. However, physicsmaths did say that this Lemma is a pretty usual one in combinatorial proofs and summation algebra so it may be that they started from the top and got to the bottom.
Original post by EricPiphany
corollary comes after the proof, things that follow without much more work.
lemma is before the proof, something you want to use in the proof.


Yeah, I see. Thanks!

Btw I deleted my post because I thought we are invading this thread with maths a bit too much and I was going to ask this question in the ASoM thread. Wanna migrate there?
Original post by Insight314
Yeah, I see. Thanks!

Btw I deleted my post because I thought we are invading this thread with maths a bit too much and I was going to ask this question in the ASoM thread. Wanna migrate there?


I feel intimidated by all the mathmos over there. I suppose I shouldn't do that to the lovely people of this thread, so you have a point lol
Original post by gasfxekl
I know a guy with erdos number 2


So do I; and since I have published with him, my erdos number is 3. :u:

There are quite a lot of us...:colonhash:
(edited 7 years ago)

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