I wrote the code but still the answer is not correct using delta function

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the two recursive formula are as follows:
C(k,h+1)=D1*(k+2)*(k+1)*C1(k+2,h)+lambda*omega*dirac delta(k,h)-lambda C1(k,h)-k2*((Sum(r=0 to k) sum(s-0 to h)C1(k-r,s)*C2(r,h-s)
2nd equation
C2(k,h+1)=D2*(k+2)*(k+1)C2(k+2,h)-K2*((sum(r=0 to k) sum(s=0 to h) C1(k-r,h)*C2(r,h-s)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
D_1=0.0001
D_2=0.0001
lambda=3;
omega=1
k_2=1
syms x
syms t
C1=zeros(200,200);
C2=zeros(200,200);
series(x,t)=sym(zeros(1,1));
for k=2:1:150
C1(1,1)=1;
C1(k,1)=0;
end
for k=2:1:18
C2(1,1)=6;
C2(k,1)=0;
end
for h=1:1:150
for k=1:1:10
S=0;
for r=1:1:k
for s=1:1:h
if r==1&& h+1==1 for delta condition
D=1;
else
D=0;
end
S=S+C1(k-r+1,s)*C2(r,h-s+1);
end
end
C1(k,h+1)=(D_1*k*(k+1)*C1(k+2,h))+(lambda*omega*D)-(lambda*C1(k,h))-(k_2*S)/h;
C2(k,h+1)=((D_2*k*(k+1)*C2(k+2,h))-(k_2*S))/h;
end
end

Answers (1)

Walter Roberson
Walter Roberson on 16 Jun 2017
Your formula has dirac delta(k,h) which would typically indicate the k'th derivative of the dirac delta with respect to h.
The 0'th derivative or normal dirac delta, dirac(h), is a distribution (not a function) that is nominally defined as +infinity at h == 0 and as 0 otherwise, but is used more for its property that constant * dirac(h) integrates to the given constant at the point where h is 0 (this is something that cannot happen with a true mathematical function; it takes a distribution.) The k'th derivative of the dirac delta is typically (-1)^k * infinity when evaluated in isolation.
Basically a dirac delta only make sense to use if you are going to do symbolic integration (not numeric integration either): a dirac delta that is not being integrated is going to contribute either 0 or an infinity to the function and that infinity is going to mess up your calculations.
Your code has
if r==1&& h+1==1 for delta condition
D=1;
else
D=0;
end
which attempts to define D as 0 except where r == 1 and h == 0, and r is (for reasons that are not immediately clear) looping to k. That is not correct outside of an integral: outside of an integral, D should become +/- infinity (the sign depending on r).
  2 Comments
YOGESHWARI PATEL
YOGESHWARI PATEL on 17 Jun 2017
r depends on k.so depending upon value of k the loop will run for summation.If this is not correct then can u suggest that with term i should replace it so my answer is correct
Walter Roberson
Walter Roberson on 17 Jun 2017
Change
if r==1&& h+1==1 for delta condition
D=1;
else
D=0;
end
to
if r==1&& h+1==1 for delta condition
D=-inf;
else
D=0;
end
Better yet, rewrite your summations so that each of the possibilities is summed separately. That would include not using D at all, replacing
(lambda*omega*D)
with
(-1)^.r * inf
in the cases where h is 0, and dropping the (lambda*omega*D) term completely for cases where h is non-zero.

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