## Tuesday, October 9, 2012

### Examples Sheet 1, #7 and Mathematica

In question #7 of Examples Sheet 1 you will explore how the form of $p_{11}^{(n)}$ differs, depending on whether the eigenvalues of the characteristic polynomial are real and distinct, complex, or repeated. Although it is intended that you do this question without the aid of a computer, you might like to check your answers by writing a small program to do the algebra.

I strongly believe that you should not only understand theory and proofs, but also be able to use modern computing tools to quickly work out examples.

I highly recommend that you install Mathematica and use it while you study and work on examples sheets (in many courses). I personally use it on almost a daily basis. You can download a free copy:

http://www.damtp.cam.ac.uk/internal/computing/software/mathematica/

It is very easy install and learn to use. The time you spend learning to use it (and similarly MATLAB) is a very good investment.

Below are some short Mathematica programs for Examples Sheet 1, #7. I think that a well-motivated student might like do this question by hand, and then subsequently check the answer using Mathematica. By the way, I would not expect an examiner to set a question in tripos that is as difficult to do by hand as doing all of (a) (b) and (c) in an example as hard as #7.
Clear[P,p,p11]

P={{0,1,0},{0,2/3,1/3},{p,1−p,0}};

(* Solution to (a) *)
mu=Eigenvalues[P /. p→1/16]
p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;
Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]
p11[n] /. %[[1]] //Expand

Out[1]= {1,-(1/4),-(1/12)}
Out[2]= {{a->1/65,b->-(2/5),c->18/13}}
Out[3]= 1/65-1/5 (-1)^n 2^(1-2 n)+1/13 (-1)^n 2^(1-2 n) 3^(2-n)

(* Solution to (b) *)
mu=Eigenvalues[P /. p->1/6]
p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;
Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]
p11[n] /. %[[1]] //ComplexExpand

Out[4]= {1,-(1/6)+I/6,-(1/6)-I/6}
Out[5]= {{a->1/25,b->12/25-(9 I)/25,c->12/25+(9 I)/25}}
Out[6]= 1/25+1/25 2^(3-n/2) 3^(1-n) Cos[(3 n \[Pi])/4]
+1/25 2^(1-n/2) 3^(2-n) Sin[(3 n \[Pi])/4]

(* Solution to (c) *)
mu=Eigenvalues[P /. p->1/12]
p11[n_]=a +(b+c n)mu[[2]]^n;
Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]
p11[n] /. %[[1]]

Out[7]= {1,-(1/6),-(1/6)}
Out[8]= {{a->1/49,b->48/49,c->-(6/7)}}
Out[9]= 1/49+(-(1/6))^n (48/49-(6 n)/7)


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