Getting NaN greater values in a function
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Hey there, I am trying to plot a function, that works perfect for smaller values but as I input the larger values I get NaN as output. Below is the code for what coded:
clearvars;
iterations = [10, 100, 1000];
count = 1;
p = 0.5;
for n = iterations
figure();
hold on;
for k = 1:n
binomial = (factorial(n)) / (factorial(k) * factorial(n - k)) * (p^k * (1-p)^(n-k));
plot(k, binomial, '*');
end
end
As shown in the above graphs, I get the output for the iterations 10 &100, but there is no output for for 1000. Even if I replace 10 by 1000 I get the same output so I am sure its nothing related to the indexing.
采纳的回答
warning('off')
iterations = [10, 100, 1000];
count = 1;
p = 0.5;
for n = iterations
figure();
hold on;
for k = 1:n
%binomial = nchoosek(n,k) * p^k * (1-p)^(n-k);
binomial = exp(sum(log(n-(0:k-1)))-sum(log(1:k))) * p^k * (1-p)^(n-k);
plot(k, binomial, '*');
end
end
2 个评论
Yousaf Farrukh
2022-10-7
It works!! Thanks. Can you help me understand why is it different from what I wrote? I believe nchoosek function is same as 
so why does it give NaN?
so why does it give NaN?
"I believe nchoosek function is same as ... so why does it give NaN?"
Because calculating using factorials will very quickly get values well beyond those which can be represented using binary floating point numbers. Which is why NCHOOSEK uses another algorithm (hint: TYPE).
更多回答（1 个）
What is the factorial of 1000 in double precision?
factorial(1000)
It overflows. If we computed it symbolically, using arbitrary precision arithmetic:
vpa(factorial(sym(1000)))
That's a lot larger than realmax so it is not surprising it overflows.
realmax
Most likely you're dividing Inf by either Inf (if the denominator overflows), NaN, or 0 (if the denominator underflows.)
Inf ./ [Inf, NaN, 0]
The plot function doesn't display NaN values.
If you use nchoosek as @Torsten suggested it avoids computing such large numbers and then dividing a large numerator by a large denominator.
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