recurrence relation for any given 'n'.

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Venkata 2017-6-20

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评论： Venkata 2017-7-7

采纳的回答： Walter Roberson

How to compute A_j^(2n) for any 'n' using A_j^(2).

Here, n=2,3,4,...; and j=1:n-1.

Any kind of help is highly appreciated.

Thank you.

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Walter Roberson 2017-6-21

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https://ww2.mathworks.cn/matlabcentral/answers/345464-recurrence-relation-for-any-given-n#answer_271385

If you have the Symbolic toolbox, you could use evalin(symengine) or feval(symengine) to call into MuPAD in order to take advantage of solve() of a rec() recurrence structure

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Walter Roberson 2017-6-27

编辑：Walter Roberson 2017-6-27

在 MATLAB Online 中打开

Guessing that there is a vector of initial constants, A(J)^1, J = 1..n-1, and calling that vector A_1:

function result = Aj_2(j, A_1)
  result = A_1(1:j) * A_1(j:-1:1).';
end
function result = Aj_2n(j, two_n, A_1)
  if two_n == 2
    result = Aj_2(j, A_1);
  else
    result = 0;
    for mu = 1 : j
      result = result + Aj_2n(mu, two_n - 2, A_1) .* Aj_2(j - mu + 1, A_1);
    end
  end
end

You can invoke this by creating a specific vector of values:

function result = A_driver
  N = 7;                  %for example
  A_1 = randi(5, 1, N);   %for example
    result = zeros(1, N-1);
    for j = 1 : N-1
       result(j) = Aj_2n(j, 2*N, A_1);
    end
  end

With large enough values of N, the cost of recursion might add up significantly (though you would probably risk exceeding 2^53, the point at which you stop being able to distinguish adjacent integers.) I was going to show a way in R2017a and later to improve that speed, but the code above is fast enough for the values I tried that it is not worth going to the trouble.

Walter Roberson 2017-7-6

I tossed together the attached. I did not test it with functions that return multiple outputs.

Aj_2n_better and Aj_2 are the same as last time. memoizefun is new (I just wrote it) and A_driver_cache is almost the same as the previous A_driver_better except for calling memoizefun instead of R2017a's memoize()

Note: you should test that the results are the same as the previous... they should be, but I did not test that.

Venkata 2017-7-7

Dear Roberson,

Sure, I'll check them.

I owe you big time, Mr. Roberson.

Thanks a ton.

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