{} = 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...
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.
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
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.
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...
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...
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...
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...
@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...
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...