Creating recurrence relation and plotting relative error

I need to create a recurrence relation that looks like this T[n](x)=(2x)*T[n-1](x)-T[n-2](x) for n=1-25. I need to compute T[n](1/2) and am given that T[0](x)=1 and T[1](x)=x. After calculating the recurrence relation, I need to compare the values to the theoretical value of T[n](1/2)=cos(n*pi/3) and plot the relative error. I am new to MATLAB and have no idea how to do any of this.

I have tried

Tn=@(n) (2*1/2+0)*T(n-1)-1*T(n-2) T(0)=1 T(1)=x for n=2:25 recurrence=T(n) relerror2=abs(theoretical-recurrence)/abs(theoretical) end

and keep getting the error:

Attempted to access T(0); index must be a positive integer or logical.

Error in Problem6 (line 13) T(0)=1

Answer 1

[Guillaume]

Matlab indices start at 1. Therefore T(0) is illegal. If you want to keep most of your code, simply offset the n up by 1, so that T(1) = 0 and T(2) = x) and stop your iteration at 26.

This is not however how you write a true recurrent functions. Here's one way you would write the Fibonacci recurrence: f(n) = f(n-1) + f(n-2) with f(0) = 1 and f(1) = 1:

function f = fib(n)
  if n == 0
     f = 1;
  elseif n == 1
     f = 1;
  else
     f = f(n-1) + f(n-2);
  end
end

You can use this as a guide to write your own recurrent function.

[Tiffany Morris]

Thanks, but I am still confused. I am not sure what I put where you had fib(n). I typed this, which clearly does not work:

function Tn =(2 *1/2*x+0)*Tn(n-1)-Tn(n-2)
if n == 0
   Tn = 1;
elseif n == 1
   Tn = x;
else
   Tn = (2*1/2*x+0)*Tn(n-1)-Tn(n-2);
end

[Guillaume]

The syntax of functions in matlab is:

function [output1, ..., outputn] = functionname(input1, ..., inputm)

In your case,

function result = Tn(n, x)
    if n == 0
      result = 1;
    ...
    else
       result = ...*Tn(n-1)...; 
    end
 end

Note that 2*1/2 in your expression does not make much sense.

[Tiffany Morris]

I have now tried this, which does not output an error, but is not outputting anything. Note that the 2*1/2 you mentioned is because we are using x=1/2. Should I be stating somewhere in the code that x=1/2? Please also note that I know it is unnecessary to add 0, but we are told that an=2, bn=0, cn=-1 for the relation T[n](x)=(an*x+bn)T[n-1](x)+cnT[n-2](x). So, I plugged in those values myself.

function recurrence=Tn(n,x)
      if n==0
          recurrence=1
      elseif n==1
          recurrence=x
      else 
          for n=2:25
          recurrence=(2*1/2+0)*Tn(n-1,1/2)-Tn(n-2,1/2)
          end
      end
  end

[Guillaume]

Well, your recurrence expression does not use x, which it should:

         recurrence = (2*x)*Tn(n-1, x)....

You then call your function simply with:

result = Tn(25, 1/2); %to calculate it for n = 25 and x = 1/2

[Tiffany Morris]

But, how do I get it to calculate for each n so that I can compare it to the theoretical T[n](1/2)=cos(n*pi/3) for each n? It still is not outputting anything. I tried putting the result=Tn(25,1/2) both inside and outside of the else statement.

[Guillaume]

Right, I missed that you want Tn for each n. In any case, a truly recursive function does not work as you run out of stack space. So, the simplest way is to store the result of each iteration, as you did, in a vector. But as I said in my initial answer, you need to offset the n by 1 since indexing start at 1. So, it'll be something like:

nmax = 25;  
x = 1/2;

T = zeros(1, nmax+1);
T(1) = 1;
T(2) = x;
for n = 3:nmax+1
   T(n) = 2*x*T(n-1) - T(n-2);
end

plot(0:nmax, T);

[Tiffany Morris]

Okay, so that at least gives me output now and makes a fair amount of sense to me. Thank you. Now, I need to compute the relative error between that recursive function and T[n](1/2)=cos(n*pi*3). How would I do that? Should I put

theoretical=cos((n-1)*pi*3)

inside the "for" loop? I put (n-1) since I had to increase the index of the recursive relation by 1. If that is the case, then where should I put the case for n=1 i.e. cos(1*pi/3)? I then need to find the relative error between the recursive and theoretical functions and plot that error. How would I do that? I truly do appreciate all the help you have given me. Thank you again.

[Guillaume]

It would be

theoretical(n) = cos((n-1)*pi/3);

That is you want to store the result of each iteration. And, yes it goes into the loop. A good practice is also to preallocate the array before the loop, the same way I've done for T, using zeros.

You calculate the error by subtracting the two values, dividing by the true value and storing the result into a third array. You can actually do that after the loop using memberwise operation (i.e. ./). You can then plot all of these arrays any way you want.