Recent content by TalkOrigin

Yessss, finally I have it (after scratching my head and wasting about 30 sheets of paper)! Thanks, I guess I need to brush up on my algebra more than I realized. When I was previewing the message, the size of the bracket was large next to the q and tiny next to the 1, so it just looked a...

This was a great post, thank you.

Thanks for the detailed response, I think I finally have it! I finally feel like I understand why it works. Specifically relating to this question, this is what I'm getting: \prod_{k=0}^{n+1} (1+q^{2^k})=\frac{1 - q^{2^{n+1}}}{1-q}\cdot (1 + q^{2^{n+1}})=\frac{1 - (q^{2^{n+1})^2}}{1-q} Unless...

Hi, thanks for your replies. I feel like I know the method but that perhaps the concepts are confusing me a little, I asked a more general question on the maths forum about this. Am I right to think that I can just multiply the LHS side by 1+q^{2^{k+2}} and if it gives \frac{1-q^{2^{k+2}}}{1-q}...

Yes, exactly! ok so at least I understand this much now, but when you have a series where it's multiplied and not added, for example (1 + q)(1 + q^2)(1 + q4) ... (1 + q^{(2^k)}) = \frac{1-q^{2^{k+1}}}{1-q} does it change (by "it" i mean the method)? I asked about how to prove this in the...

Ahh ok, thanks so much for the thorough explanation! I think I was doing a few things wrong, one of them being that I thought when using n=k+1, if it's a sequence like a+ar+...+ar^k etc, that you just add $$ar^{k+1}$$ on both sides. Actually, you add it on to the left, and say that if the...

Hi, I'm working through "What is mathematics" by Courant, and in the first chapter he covers proof by mathematical induction. I understand the method, and I do understand the general principle, but I think I'm confused somewhere. Basically, you have to prove something on the LHS is equal to...

? I see a pattern but I don't know where I'm going wrong with my inductive proof; have I done something wrong by doing the standard (k+1) substitution and equating both sides?

i, I'm stumped on a proof, one problem may be that either I don't know how to deal with exponents of this type, or my algebra went wrong somewhere. It's from the first chapter of Courant's book "What is mathematics" (p.18 q4) 1. Homework Statement Prove by mathematical induction: (1 + q)(1 +...

Ok thanks, this is my first dip into proof based mathematics, so I am still learning how to formulate proofs and everything. This book is great though! Thanks for all the help

whoops, sorry i replied with a brainfart

It's actually that 2 divides n, and 3 divides n, not that n divides 2 and n divides 3.

Ah ok, so I have to say ""Suppose 2 | n and 3 | n. As 2 and 3 are both prime, then there is an integer k such that (2)(3)k=n. Thus, (6)k=n and therefore 6 | n." EDIT: Also, could this be generalised to "Suppose a | n and b | n. As a and b are both prime, then there is an integer k such that...

Hi, thanks for the reply. I'm not sure why "2 | n and 3 | n" does not assert there is an integer k such that (2)(3)k = n. I forgot to say at the beginning that n is an arbitrary integer. I would think that if 2 and 3 both divide n, then obviously n is not prime, but can be expressed as a product...

Hi, I wasn't sure whether to post this here or in the pre-calc forum, so apologies if this is in the wrong section. I'm working through Vellemans 'How to prove it' and he gives a proof that if 2|n and 3|n then 6|n (note that a|n means a divides n, just in case this is not standard notation). I...