How can I get a value not NaN when calculating ''inf/inf''?

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]

[Chanhp Jeong]

In mathmatics, not using matlab.

I believe that when n=∞, (2*n)/(n+1)=2 .

I want a get this kind of value not 'NaN'

[madhan ravi]

syms n
limit((2.*n)/(n+1),n,inf)

[Chanhp Jeong]

Thank you for your comment. Seems that's how i can solve inf/inf.

However, i still have difficulty adjust your comment into my for loop.

Any further help for this? thanks

[madhan ravi]

you need to explain what you are trying to obtain from the calculation ? from your code I don't understand r's calculation

[Chanhp Jeong]

I'd like to get r(1) , r(2), r(3) , ....., r(11).

when I solved it with my hand r(1)=inf , r(2) = 90+50*2^(1/2)

[madhan ravi]

can you upload the photo of your solution by hand ?

[John D'Errico]

Please don't encourage people to upload pictures of solutions. They can as easily type in the results they computed.

[madhan ravi]

My apologies @John D‘Errico

[Chanhp Jeong]

Thank you for your help. This is how i solved r(2)

[John D'Errico]

Note that inf*90/(inf+90) is NOT 90. It is anything you want it to be, once inf is in there. Instead, you need to use limits, as I show. Yes, the limit is still 90 in the end, but once you have inf in there, you cannot solve it.

[John D'Errico]

BTW, The reason why I suggest not encouraging people to use a picture, is because pictures are often unreadable. The writing is poor, or the contrast low, or the text is simply too small, etc.

Answer 1

[John D'Errico]

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.

[Chanhp Jeong]

Thanks you for the detailed explanation.

I learnt a lot from your comments. Seems i should reestablish the concept of inf.

And also 'eps' you shown here will be quite useful for my future matlab works.

Thank you