I want the coding for double summationI wrote the following code but my answer is not correct.

My whole formula is

C(k,h+1)=A*(k+1)*C(k+1,h)-B*C(k,h)+D*(k+2)*(k+1)C(k+2,h)+2*a*D*(summation(r=0 to k)summation (s=0 to h))Kroneckerdelta(r-1,h-s)*(k-r+1)*(k-r+2)C(k-r+2,s))+a*a*D*(summation(r=0 to k)summation (s=0 to h))Kroneckerdelta(r-2,h-s)*(k-r+1)*(k-r+2)C(k-r+2,s))

For kroneckerdelta i had separted defined a function which is as follows: function d = kronDel(i,j)

if i == j
    d = 1;
else
    d = 0;
end

Now first part ,i initial the value using for loop

n=10;
for i=1:1:n+5
    C1(1,1)=0.030792+0.009208;
    C1(i,1)=(0.009208*((-0.00000588235)^(i-1))/factorial(i-1));
 end

the initial values are correct Now what i did was i broke my formula into parts and calculate the first three terms of my formula i.e. upto D*(k+2)*(k+1)*C(k+2,h)using the following for loop

 for j=1:1:n+1
  for  i=1:1:n+1
    C1(i,j+1)=((a*i*C1(i+1,j)-b*C1(i,j)+D1*i*(i+1)*C1(i+2,j)))/j;
      end
end

Now i calculate the 4th term i.e.term containing Keonecker delta using the following loop

 for  i=1:1:n+1
   for j=1:1:n+1
    ab(i,j)=(2*a*D1* kronDel(i-1,j)*((n+1)-i+1)*((n+1)-i+2)*C1((n+1)-i+2,j))/j;
  end
 end

now this 4th term value is wrong.so i want the code for double summation. Check the pdf file aslo

in that k=0 to 100 h=0 to 100

you will get more clear idea I Think the logic is wrong in N+1 but i am not getting it.Kindly guide me

Answer 1

[Torsten]

One hint:

Because of the kroneckerDelta's in both double summations, substitute r=h-s+1 in the first double summation and r=h-s+2 in the second double summation. This way, you are left with two single summations over s.

Best wishes

Torsten.

[YOGESHWARI PATEL]

THEN WHAT ABOUT K IN FORMULA.