# Recent content by Venomily

1. ### How can B be an element of C?

{} = sausages {{}} = packet of sausages 0 = Orange { 0, {}, {{}} } = Shopping bag of (Orange + sausages + Packet of sausages). { { 0, {}, {{}} } } = car boot of Shopping bag. We want {{}} i.e. the packet of sausages. It is not related to the Car boot because it is neither a SET nor a MEMBER...
2. ### How can B be an element of C?

Great explanation, I understand now, thanks. But how would I go about 2 (d)? we have the empty set, {}: {A} = { { 0, {}, {{}} } } Judging by what you said, I don't see how {} can be related to the above set? it is neither an element nor a set.
3. ### How can B be an element of C?

im quite desperate.
4. ### How can B be an element of C?

Question 2 (a) how is it possible? B is a set (since A is a set), how can a set be an element of another set? Rather than saying: B is an element of C I thought it would be better to say: B is a subset of C. Also, can someone explain question 2 (d) to me? thanks
5. ### What exactly is a 2nd order differential equation?

Thanks, but can you go through an example with me? actually point out a real life application (which you guys did) but also deriving a 2nd order DE to model it step by step.
6. ### What exactly is a 2nd order differential equation?

A first order DE models the rate of change, e.g. when decay is proportional to time we have the DE: dM/dt = -K.M; this is describing that rate of change mathematically. Am I correct in saying that a 2nd order DE describes the rate of rate of change? Also, can anyone explain any application of...
7. ### STEP question, I don't know what it is asking

please show me what you think the derivative should be.
8. ### STEP question, I don't know what it is asking

I really don't see where this is going, both equations are identical: sin(t).A+cos(t).B = ∫(0->pi) [f(x) sin(x+t)]dx This is the same thing as: sin(t).∫(0->pi)[f(x)cos(x)]dx +cos(t).∫(0->pi)[f(x)sinx]dx = ∫(0->pi) [f(x) sin(x+t)]dx If I tried to differentiate I would just get...
9. ### STEP question, I don't know what it is asking

I said it at the top of the OP: question 5 :smile:.
10. ### STEP question, I don't know what it is asking

I don't know what equation I'll get. How do I make the substitution? "Use the expression (**) to find A and B by substituting for f(t) and f(x) in (*) and equating coefficients of sint and cost". @Bolded, just what? :bugeye: how can i possibly 'substitute' f(t) and f(x) into (*)? if it...
11. ### STEP question, I don't know what it is asking

question 5. I am stuck on the last sentence. Please don't tell me what to do, just what they mean.
12. ### Can you help me choose my maths options?

I am studying Discrete Mathematics and if I do a PhD I would like to base the research on or around quantum computation. Unfortunately I can't take any physics modules at all, so please ignore the physics modules on the pages; I can only pick from the maths modules...
13. ### Can you check my proof by induction?

thanks, I forgot about using known results :p Yes, it becomes very easy now.
14. ### Can you check my proof by induction?

@bold, how would you go about proving it then? rearranging the expression would be tedious with a multiplier (is there a better way?). I thought it would be easier to take the two cases since if you prove one case you imply the other. EDIT: nvm, I found a more refined way of doing it using the...
15. ### Can you check my proof by induction?

1 = 1 1 - 2^2 = -(1+2) 1 - 2^2 + 3^2 = (1+2+3) 1^2 - 2^2 + 3^2 - 4^2 = -(1+2+3+4) and so on. I have to prove that this relationship is true for all natural numbers. This is what I did: clearly it is true for 1, 2, 3 and 4. assume true for n odd: 1^2 - 2^2 + 3^2 - 4^2 ... + n^2 = (1 + 2 + +3...