Official TSR Mathematical Society

ok this is the game theory:
two people have stolen something, police can not arrest any of them if they dont confess.if one of them confesses the other one is arrested and jailed for one year, and the one that has confessed is freed. its the same argument for the other person. but if they both confess each will be prisoned for 6 moths. what do you think they will do?

Cooperatively they will shut the something up

guys you all possibly know this. but i have a problem on game theory. the thing is that it assumes none of the people involved in the game will work collaboratively with the other people involved. what if they do. are we sure that they will get the highest profit ( none goes to prison)?

Your original question was



if they work collaboratively they both 'win'
unless I'm not understanding something about the question

yes if they do work collaboratively. but the whole point is that people are selfish and they go for personal profit.
and end up with something less than what they can get

ok. if you want my opinion dont go for a new calculator. find your answer in radians and divide them by 3.1416( eg if you get 1.5708 radians divide it by 3.1416 and what youll get is 1/2) then times it by pi (e.g. 1(pi)/2) but if you want the method of working with a fancy calculator ill get one of my friends to post it tonight on this thread. hope it helps
by the way tell me what exam you have. itll be better.

oh wow i just tried it and that's much better, thank you so much!!!!!!!!!!!!!!!!!!!!!!!! i have aqa maths core 3 and core 4 exams by the way

honestly thank you really i am so grateful i was freaking out before about this but really thanks!!!!

BMO1 1966

A regular polygon ABCD.... is such that $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$ .

How many sides does the polygon have?

I'm hoping to see a better solution than mine. I ended up using De Moivre's theorem which seems inappropriate for BMO1.

There is no need for anything other than basic geometry here...

Any chance you could post this basic geometry solution. I confess I'm not too good at geometry.

...

Are you interested how to work it out, or just in the actual number?
If these are rough percentages, anything more than $70.33\%$ should do.

$\displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!$
How can we prove this?

$\displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!$
How can we prove this?

Can someone help me with this maths problem or know anyone that can. thank you :-)
In steady state heat transfer the temperature T satisfies the Laplace equation. In two dimensions with Cartesian coordinates x and y this equation is d2T/dx2 +d2T/dY2=0
This is a partial differential equation, and general methods of solving it are beyond the scope of this module - wait for the second and final year! However there are ways of getting an approximate solution to the equation by realising that it is a mathematical statement of the law of conservation of energy - in this case heat energy. One such method is the finite difference method.

One way to see how such methods work, is to consider a square area ABCD of a heat-conducting plate:

Figure 1
The values T0, T1, T2, T3, and T4 represent respectively the temperatures at the central point (or node) and the four surrounding equally distant nodes. T0 is at the centre of the sqaure, T1 goes through the middle of the points AD and conects to T0 on the sqaure. T3 goes through the middle ofthe points CB and connects to T0. T2 goes through the middle of the points DC and connects to T0 and T4 goes through the middle of the points AB and connects to T0. For the sake of simplicity suppose the length of side of the square is h, and that each of the temperature nodes are a distance h apart.

The law of conservation of energy in this case says that the net heat flowing into ABCD must equal the net heat flowing out. The heat flowing across a boundary of ABCD is proportional to the temperature gradient and the length of that boundary. Thus conservation of energy says
Heat in + Heat in = Heat out + Heat out
across AC across CD across AB across BD

Approximately this gives
(T0-T1/h)h +(T0-T4/h)h = (T2-T0/h)h+(T3-T0/h)h
or rearranging T0=1/4(T1+T2+T3+T4) .
ie temperature at one node = average of temperatures at four surrounding nodes.
Why is this approximate, and how can it be made more accurate? Write some words on this, and to see if you’ve really understood, do the following:
Work out the equation when ABCD is a rectangle but not necessarily a square, with the lengths
AB = CD = distance from T1 to T0 = distance from T0 to T3 = h
and the lengths
AC = BD = distance from T4 to T0 = distance from T0 to T2 = k

A simple case
A flat plate as shown in Figure 2 has the temperature on three edges held at T = 0; the left hand side and the top and bottom the fourth edge is held at a temperature of T = 100, which is on the right hand side. T1 and T2 are on the central nodes of the grid.
Figure 2


An approximate value of the temperature at certain points can be obtained as follows:
• Place a regular square grid over the plate as shown.
• Name the temperatures at the two central nodes of the grid T1 and T2.
• Then according to the finite difference model above, the values of T1 and T2, which are approximate values of the temperature at those points, can be found by noting that the temperature at each node is the average of the temperatures at the four surrounding nodes.

This results in two simultaneous equations which can be solved for T1 and T2. The equations are:
T1=(0+0+0+T2)/4and T2=(0+0+100+T1)/4

Solve these equations in two ways. Verify your answers.

$\displaystyle\int_0^1 (-\mathrm{log}\ x)^n\ \mathrm{dx} =n!$
How can we prove this?