Hi guys, help me, please!

How can I generate this recurrence formula a
(0)=1;a(1)=0;a(2)=3;a(3)=1/6;for n=2:28;a(n+2)=((6/((n+1)*(n+2)))*sum(a(i)*a(n-i);i=0..n));end

9 Comments

Add one to each index, because you cannot use 0 as an index.
Thank you
Could you help me and write full code for my problem, please!
What is the text of your assignment?
a(0+1)=1;a(1+1)=0;a(2+1)=3;a(3+1)=1/6;for n=2:28;a(n+2+1)=((6/((n+1)*(n+2)))*sum(a(i+1)*a(n-i+1);i=0..n));end
Except that you need to figure out how to rewrite the sum(a(i+1)*a(n-i+1);i=0..n)) part . Hint: use .* instead of * and vectorize
Still I have a problem
This is called differential transform method for solving initial value problem
As I wrote, you need to figure out how to rewrite the sum(a(i+1)*a(n-i+1);i=0..n)) part. MATLAB does not offer any syntax for summation that uses variable=lower..upper notation (not unless you get fairly far into how the Symbolic Toolbox actually works.) MATLAB offers a symbolic summation of the form
symsum(SYMBOLIC_EXPRESSION, SYMBOLIC_VARIABLE, LOWER_BOUND, UPPER_BOUND)
however, the SYMBOLIC_VARIABLE cannot be used to index any array. symsum() is not intended for summation of a small finite number of terms: it is intended for creating formulas, such as
symsum(m^2, m, 1, n)
to get out the formula for the sum of squares of the first n numbers.
You need to find a different way to do the summation of indexed variables.
Thank you Mr Walter

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Answers (1)

MATLAB indexing is 1-based, not 0-based. You will need to adjust your indexing:
a(1) = 1;
a(2) = 0;
:
etc

4 Comments

Mr James, this is a part of solution of an initial value problem using differential transform method
That is irrelevant. You cannot index a variable at index 0.
Mr John, could you help me and write full MATLAB code for my problem, please?
You need to find a different way to do the summation of indexed variables.
Hint: use .* instead of * and vectorize
Hint:
>> A = [1 3 5 7]; A(1:3) .* A(2:4)
ans =
3 15 35

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Asked:

on 2 Oct 2019

Commented:

on 2 Oct 2019

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