Brodudedoodebrodude

Totally should be a minus, my bad. Glad I could help =)

Edit: Seems I cant edit my original post =/

Not a problem! The "fixed point" refers to the value(s) of \(f_{n-1}\) that when you work out \(f_n\), you get the same value that \(f_{n-1}\) was, this putting you in a never ending loop and hence is a fixed point (or equilibrium).

Here's the method for finding the fixed point (explained pretty loosely):

-> By definition the fixed point means that the next number should be the same as the current one which is equivalent to saying \(...=f_{n-2}=f_{n-1}=f_n=f_{n+1}=...\)  This means we can forget about the subscripts and treat everything as the same unknown \(f\).

-> This means at the fixed point the equation \(f_n=3f_{n-1}-60\)  transforms into \(f=3f-60\).

-> Now you just solve for \(f\) and that will be your fixed point

In this example, \(f = 30\) and substituting it into the original difference equation we get:

\(f_{n-1}=30\\f_n=3f_{n-1}-60\quad=>\quad f_n=3(30)-60=30\quad=>\quad f_n=f_{n-1}\)

so this is a fixed point.

We have \(f_n = 3f_{n-1}-60\)   and  \(f(1)=f_1=20\)

For a difference equation in the form \(f_n = af_{n-1}+c\) , the general solution takes form of: \(f_n=x+\beta a^n\) where x is the fixed point of the difference equation:

1) \(f(0)=f_0 = \frac{f_1+60}{3}=\frac{80}{3}\)

2) Finding the fixed point \(x\) -> Let \(f_n=f_{n-1}=f\), then:

    \(f=3f+60 \quad=>\quad f=30\) so the fixed point \(x\) is 30.

3) To find \(\beta\) in the general solution, work out the case n=0:

   \(f_0=\frac{80}{3}=30 +\beta 3^0 = 30+\beta*1 \quad=>\quad \beta=\frac{80}{3}-30=\frac{-10}{3}\)

4) Now we have the general solution just plug in whatever term you want to find, in this case the 10th term:

\(f_{n}=30-\frac{10}{3}*3^{n}\\f_{10}=30-\frac{10}{3}*3^{10} = -196800\)

\(3x(2x+1)-5=15x\\6x^2+3x-5=15x\\6x^2-12x-5=0\\a=6, b= -12, c=-5\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4*(6)*(-5)}}{2*6}\\x=\frac{12\pm\sqrt{144+120}}{12}\\x=\frac{12\pm\sqrt4\sqrt66}{12}\\x=\frac{2(6\pm\sqrt66)}{12}\\x=\frac{6\pm\sqrt66}{6}\)

Probably many ways, here's one way where I use all numbers only once:

5 x (18 + 2 - 8 + -4 - 3) = 5 x 5 = 25

Step 1: Divide the first term under the long division line by the first term outside it, \(\frac{x^2}{x}\) in this case

Step 2: Multiply the newest term ontop of the division line with each term on the outside -> \(x*x + x*-3\)

Step 3: Subtract the 2nd line from the 1st -> \((x^2-x) - (x^2-3x) = 2x\)

Step 4: Repeat steps 1 to 3 eg: \(\frac{2x}{x}=2\) ...

Whats left up top is the factorised version.

Method: Try to isolate the x from everything else

In this case:

1: Add x to both sides

2: Subtract 43 from both sides

3: Divide both sides by 4

You now have your answer (x= - 4)

If you're wondering how to factorise this, then you need to think of two numbers that:

-> Multiply together and give you -6

-> Add together to give you -1

Eventually you can do it in your head, but for now it's probably easiest to write down all the whole numbers that multiply together to give you -6;

-1x6 = -6    1x-6 = -6

-2x3 = -6    2x-3 = -6

Now from these numbers which ones add together to give you negative one?

-3 + 2 = -1

So the two numbers you needed are 2 and -3.

Therefore it factorises into \(z^2-z-6 = (z+2)(z-3)\)

Remember that when you divide or multiply by a negative then you have to switch the sign of the inequality:

\(2x+4 < 5x + 3 \\ 2x+4 -4 < 5x+3 - 4 \\ 2x < 5x-1 \\ 2x-5x<5x-1-5x\\-3x<-1\\ \frac{-3x}{-3}>\frac{-1}{-3}\\x>\frac{1}{3}\)

Kind of late reply as I havn't been on but that's what I ended up doing so your answer agrees with me -> The M.C consists of 4 states (0,0) , (1,0) , (0,1) , (1,1) and then you find the transition probabilities for each state using what you've given.

At the bottom of the summation it indicates what it starts on, in this case i starts on 1. At the top of the summation it tells you what the maximum number is that i gets up to. Now you need to sum every number from what you start on up till the maximum.

Your 1st sum translates to $$x_1+x_2+x_3+x_4+x_5+x_6=15$$Your 2nd sum translates to $$x_1+x_2+x_3+x_4+x_5=20$$Your 3rd sum translates to $$\frac{x_1+x_2+x_3+x_4+x_5+x_6+x_7}{7}=8$$

Hope you should be able to do the rest of the problem now =)

The 'tricks' for the first one is that you can take a constant outside of an integral and also that you can split up a single integral of many terms into many integrals of single terms:

$$E[ag(X)+bh(X)]= \int_{-\infty}^{\infty} [ag(X)+bh(X)]f_x(x)dx \\\indent= \int_{-\infty}^{\infty}ag(X)f_x(x)dx+\int_{-\infty}^{\infty}bh(X)f_x(x)dx \\\indent =a\int_{-\infty}^{\infty}g(X)f_x(x)dx+b\int_{-\infty}^{\infty}h(X)f_x(x)dx \\\indent = aE[g(X)]+bE[h(x)]$$

Part b is a very similar process to part a:

$$Var(aX) = E[(aX)^2]-(E[aX])^2 \\\indent =\int_{-\infty}^{\infty}a^2X^2f_x(x)dx-(\int_{-\infty}^{\infty}aXf_x(x)dx)^2\\\indent= a^2\int_{-\infty}^{\infty}X^2f_x(x)dx-(a\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2E[X^2]-a^2(\int_{-\infty}^{\infty}Xf_x(x)dx)^2 \\\indent= a^2(E[X^2]-(E[X])^2)\\\indent=a^2Var(X)$$

I'm not too sure with part c so hopefully someone else can help with that =)

I understand it now, thanks a lot Alan =)

Yeah it does thanks Alan. What's still causing some confusion for me is what is the point of mentioning that the initial distribution for the markov chain is (1/3 , 1/3, 1/3)?

These are the answers provided by my lecturer but yes I also think it's a typo and should be -3 rather than 3, forgot to mention that! 

My problem is just the fundamentals, how does the calculation for P(0.1) look like? Once I see it I can make sense of it (I hope)

A verticle asymptote occurs where a value of x makes the denominator equal to zero but not the numerator:

In this case the only value of x that makes the denominator = 0 is x=-1/2 and this value also doesn't make the numerator = 0 therefore the vertical asymptote is at x=-1/2.

What if the function was this however: $$y(x)=\frac{(x^2+7x+10)(2x+1)}{2x+1}$$

If you dont cancel the 2x+1 and substitute in x=-1/2 you will get 0 for both the numerator and the denominator therefore this wouldnt be  a vertical asymptote. The way to go about this situation is to cancel the 2x+1 to get $$y(x) = x^2+7x+10$$ but you Have to mention that x cannot equal -1/2 as this isn't defined for the original function and is a discontinuity and then you can leave it at that.

Saul from Khan Academy Explains this nicely: https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example