gamma function error in calculation
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The above formula is coded as follows:
syms x a
Y=sym(zeros(1));
Y(1)=0;
a=1/2
for i=1:4
if i==5
A=1
else
A=0
end
if i==4
B=1
else
B=0
end
Y(i+1)=simplify((gamma(a*(i-1)+1)/gamma((a*(i-1)+3/2))*(A-Y(i)+((2*B)/gamma(5/2)))));
end
disp(Y)
But it is showing the calculation error Y(5)=1 but the value is shown in MATLAB is as follows:('2535301200456458897054207582575/2535301200456458802993406410752'). Ecxept Y(5) all values are zeros
Antworten (1)
syms x a
Y=sym(zeros(1));
Y(1)=0;
a=1/2
for i=1:4
if i==5
A=1;
B = 0;
Y(i+1)=simplify((gamma(a*(i-1)+1)/gamma((a*(i-1)+3/2))*(A-Y(i)+((2*B)/gamma(5/2)))));
else
A=0;
B =1 ;
Y(i+1)=simplify((gamma(a*(i-1)+1)/gamma((a*(i-1)+3/2))*(A-Y(i)+((2*B)/gamma(5/2)))));
end
if i==4
B=1;
A = 0;
Y(i+1)=simplify((gamma(a*(i-1)+1)/gamma((a*(i-1)+3/2))*(A-Y(i)+((2*B)/gamma(5/2)))));
else
B=0;
A = 1;
Y(i+1)=simplify((gamma(a*(i-1)+1)/gamma((a*(i-1)+3/2))*(A-Y(i)+((2*B)/gamma(5/2)))));
end
end
disp(vpa(Y,4))
3 Kommentare
VBBV
am 16 Apr. 2022
For each if condition , there are few unknown variables e.g. either A or B in the equation.
So, to compute whole equaton for each iteration, values have to be defined with A and B or computed to get final matrix Y.
yogeshwari patel
am 17 Apr. 2022
Torsten
am 17 Apr. 2022
You mix symbolic and floating point parameters in the calculation of Y(5). This will result in reduced accuracy.
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