# Getting NaN while using the ratio of power and factorial

8 views (last 30 days)
Yash Khandelwal on 13 Jul 2022
Commented: James Tursa on 13 Jul 2022
1
With the following piece of code, I get NaN
l=0;
lp=2;
d=200;
a2= ((-sqrt(d))^lp);
a3 =((-sqrt(d))^l);
sum1=0
for m=0:500
a1= ((2^m)*(d^m))/(factorial(m))
mul=a1*a2*a3;
sum1=sum1+mul;
end
Is there any way to modify the code to get rid of the issue?
James Tursa on 13 Jul 2022
Cross posted on stackoverflow and answered there over 4 hours ago. If you had simply checked the answer there you could have saved yourself the trouble of posting here and getting essentially the same answer.

Torsten on 13 Jul 2022
Edited: Torsten on 13 Jul 2022
exp(400)*200 is quite large ...
format long
l=0;
lp=2;
d=200;
a2= ((-sqrt(d))^lp);
a3 =((-sqrt(d))^l);
a1 = 1.0;
sum1 = a1;
for m=0:500
a1 = a1* 2*d/(m+1);
sum1=sum1+a1;
end
sum1 = sum1*a2*a3
sum1 =
1.044293406962028e+176
exp(400)*200
ans =
1.044293937952829e+176

Steven Lord on 13 Jul 2022
m = 500;
d = 200;
The numerator and denominator of your expression (for sufficiently large m) both overflow to inf. Dividing infinity by infinity results in a NaN.
numerator = 2^m*d^m
numerator = Inf
denominator = factorial(m)
denominator = Inf
x = numerator / denominator
x = NaN
One potential approach to avoid this is to avoid explicitly computing 2^m, d^m, or factorial(m).
numeratorVector = repmat(2*d, 1, m); % prod(numeratorVector) would effectively give numerator
denominatorVector = 1:m; % prod(denominatorVector) would effectively give denominator
xVector = numeratorVector./denominatorVector;
format longg
x2 = prod(xVector)
x2 =
8.78187252741829e+166
Let's check symbolically.
numeratorSymbolic = sym(2*d)^m;
vpa(numeratorSymbolic) % Pretty big
ans =
1.07150860718626732094842504906e+1301
denominatorSymbolic = factorial(sym(m));
vpa(denominatorSymbolic) % Also pretty big
ans =
1.220136825991110068701238785423e+1134
x3 = vpa(numeratorSymbolic/denominatorSymbolic) % Big but not quite as big as above
x3 =
8.7818725274182843596325298923868e+166
You could be a little more sophisticated / clever if you wanted (preemptively cancelling out factors of 2 in numeratorVector by dividing even values in denominatorVector by 2.) Or you could keep track of x for each value of m then figure out what you need to multiply it by to get x for the next value of m.

Benjamin Thompson on 13 Jul 2022
The factorial function output increases very fast as input increases. See "doc factorial" for details. The output is "inf" for input of 171 or larger.
Steven Lord on 13 Jul 2022
But 400^m overflows for m > 118 so this whole expression becomes infinite or undefined beyond that point.
d = 200;
(2*d)^118
ans = 1.1043e+307
(2*d)^119
ans = Inf

Vijeta Singh Yadav on 13 Jul 2022
Data overflow and underflow problem exists in the code
visit this link for avoid data overflow and underflow problems.