MATLAB: How to get a value not NaN when calculating ”inf/inf”

for loopinfinf/infnan

I got 'NaN'(which is not 'NaN') for my for loop calculation and I found that it was because of 'inf/inf'
Here is my coding. I kinda simplified it for better readability.
Could anyone help me getting r(1), r(2), …..,r(11) for this problem?
When I solved it by hand, r(2) was 160.71
Thank you
 >> syms x r
>> a=1; b=0; dx=5; t=40;K=10;
>> f(x) = a*x+b;
>> df(x)=diff(f(x));
>> r(1)=inf;
>> for i=1:10
      r(i+1)=K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t));
  end
>> r
r=
[ Inf , NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN]

Best Answer

There is no consistent numeric value you can obtain when taking the ration inf/inf.
Should it be 1? Clearly x/x is always 1, for any non-zero NUMBER x. But inf is not really a number. And since 2*inf is also inf, then logically, we could conclude that if inf/inf was 1, then
2 = 2*1 = 2*(inf/inf) = (2*inf)/inf = inf/inf = 1
Just as easily, we could construct any number of other nonsense results, that is, if inf/inf were defined to take on any value.
The point is, once your computations have reduced to inf/inf, you are already in a deep hole, one from which you will not escape.
Instead, you need to do some thinking. See if you can avoid computing those infs in the first place.
syms x r
a=1; b=0; dx=5; t=40;K=10;
f(x) = a*x+b;
df(x)=diff(f(x));
r(1)=inf;
Now, lets see what happens when you try to start that loop, trying to create r(2).
i = 1;
K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))
ans =
NaN
That expression has a few terms in it.
K*int((1+df(x)).^(1/2),x,i,i+dx)
ans =
50*2^(1/2)
(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))
ans =
NaN
(r(i)*(K*f(i*dx)+t))
ans =
Inf
(r(i)+(K*f(i*dx)+t))
ans =
Inf
So, you have inf/inf, which as I showed before, is an expression that has no reasonable value that can be assigned.
You may have "done it by hand" but if you did so, you made up a result that you decided you like. So, lets go more deeply into that expression for r(2).
You have what is essentially
50*2^(1/2) + inf/inf
and now, you want the inf/inf to magically be 90?
Hmm. I think you are now making up a result. As we saw before, inf/inf is anything you wish it to be. Personally, I think 17 or 42 are better choices for results there. But what have you done? And why do you think the result of
(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t))
ans =
NaN
should be 90?
(K*f(i*dx)+t)
ans =
90
Hmm, so now you have
(inf*90)/(inf+90)
And now you think that
inf + 90 = inf
so 90*inf/inf should be 90, since inf/inf should be 1.
SHEESH!
It seems that you are playing fast and loose with infs here.
Lets see what happens if r(1) is just some large number.
r(1) = 1/eps;
for i=1:10
      r(i+1)=K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t));
  end
double(r)
ans =
  Columns 1 through 7
       4.5035996273705e+15          160.710678118653          145.531748914597          153.120263450657          164.190629789399          175.546135998982          186.482443382689
  Columns 8 through 11
          196.869165257471          206.723594803462          216.097689853589          225.046227726715
Thus, by making r(1) a finite value, while still large in context of double precision arithmetic, it is not infinite. Now we get a value that is at least somewhat more reasonable.
So, can we do this using mathematics? A limit seems like what you really wanted to do.
syms u real
r(1) = u;
for i=1:10
      r(i+1)=K*int((1+df(x)).^(1/2),x,i,i+dx)+(r(i)*(K*f(i*dx)+t))./(r(i)+(K*f(i*dx)+t));
end
r = limit(r,u,inf);
r(1)
ans =
Inf
r(2)
ans =
50*2^(1/2) + 90
r(3)
ans =
50*2^(1/2) + (7000*2^(1/2) + 12600)/(50*2^(1/2) + 230)
double(r(11))
ans =
          225.046227726715
I imagine that is what you wanted to see.
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