Solution to exercise 4.1.1

Firstly we should note that our expression can be transformed to an equivalent form: \begin{equation*} \sum_{j=1}^{j=N}b_j\left(\sum_{i=1}^{i=j}a_i\right) \end{equation*}

Then we should also note that there is no need to calculate the inner sum from scratch for each new “j” – we can update it from the previous j-iteration. Thus the expression value can be calculated with a single loop.

For a convenience in the solution presented below we implement the sequences “a” and “b” with separate functions.

<sxh c> N = input(“Give N : ”);

function av = a(i)

av = sin(2*pi*i+1);

endfunction

function bv = b(i)

bv = 1/(1+i);

endfunction

totalsum = 0; asum = 0; for j=1:N

asum = asum + a(j);
totalsum = totalsum + b(j) * asum;

endfor

printf(“Expression value = %g\n”,totalsum); </sxh>

<texit> \begin{lstlisting} N = input(“Give N : ”);

function av = a(i)

av = sin(2*pi*i+1);

endfunction

function bv = b(i)

bv = 1/(1+i);

endfunction

totalsum = 0; asum = 0; for j=1:N

asum = asum + a(j);
totalsum = totalsum + b(j) * asum;

endfor

printf(“Expression value = %g\n”,totalsum); \end{lstlisting} </texit>