When c_3=0 in (3) , this means c is infinity which implies for (2) y=-2 when x=0, and for (1) the expression is undefined because y=infinity/infinity.

Note that y=-2 does not only apply to x=0. It applies to any x.

Also, the equation 1 is not undefined. Why would you say that? Generally, infinity/infinity is indeterminate, which simply means that you don't know what the value is. It can be any value depending on the details of what the expression actually is. Once you create a specific expression and take a limit, that expression can take on a meaningful value. In this case the limit is -2, meaning that the ratio itself is well defined, in the limit.

Equations 1, 2 and 3 are basically the same equation in different forms. Also, note that equation 3 has a defined meaning when c3 goes to infinity. Then you get y=2, just as you get y=2 for equation 1 when c=0.

If you are not ready for these concepts, that's fine, but I feel obligated to point out what I believe is the better interpretation of these confusing equations.

Tesla has a good point above. All this confusion can be avoided by taking his point of view, which seems reasonable to me. If you do want to go down the more difficult road, then you need to know the destination and work towards it, and not allow your own preconceived notions to divert you to some unknown land.

 

Just a side question. I was wondering if there also exist integral equations. They do exist. But why aren't they as much popular as differential equations? Perhaps, an integral equation can be converted into a differential equation and as I believe I have read somewhere that many times differentiation could be easier to do than integration so perhaps this is the reason integral equations are dealt so often. Thanks.

Regards
PG

 

Just a side question. I was wondering if there also exist integral equations. They do exist. But why aren't they as much popular as differential equations? Perhaps, an integral equation can be converted into a differential equation and as I believe I have read somewhere that many times differentiation could be easier to do than integration so perhaps this is the reason integral equations are dealt so often.

Yes, integral equations do exist, as you see in the Wikipedia article. The same physics is displayed by either the integral form of the equation or the differential form. Some problems are better solved in one form, but it really depends on the example. Definitely you can convert between the two forms of the equations. Actually, you may have noticed that when you solve differential equations, you sometimes end up converting to integral form and then solving by direct integration.

Here is a link to an example of this that comes up often in electrical engineering. Maxwell's equations, which describe the behavior of electromagnetic fields are used in either differential or integral form, as suited for a given problem. This is a general trend in physics and engineering.

http://en.wikipedia.org/wiki/Maxwell's_equations

 

Note that y=-2 does not only apply to x=0. It applies to any x.

Also, the equation 1 is not undefined. Why would you say that? Generally, infinity/infinity is indeterminate, which simply means that you don't know what the value is. It can be any value depending on the details of what the expression actually is. Once you create a specific expression and take a limit, that expression can take on a meaningful value. In this case the limit is -2, meaning that the ratio itself is well defined, in the limit.

Seriously, I don't know how to thank you for all this help and that too with all the kindness. I hope I'm not exploiting this kindness!

My new query is in blue below at the end.

\( y=2{{1+c e^{4x}}\over{1-ce^{4x}}\) (1)

and divide the numerator and denominator by c to get the following

\( y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}\) (2)

Now, redefine the constant, and let c_3=1/c, to get the following.

\( y=2{{c_3+e^{4x}}\over{c_3-e^{4x}}\) (3)

Now when c_3 is zero, you get y=-2.

c_3 = 1/c

When c_3=0 in (3) , this means c is infinity which implies for (2) y=-2 when for all x, and for (1) the expression is undefined because y=infinity/infinity.

Okay. When c3 = 0, the c would be ∞. Then, don't you the following expression would be indeterminate, as you say? If all three expression are the same equation in three different forms, then it shouldn't be inderterminate, and y=-2 should be the result for all x. Please help me with it.

\(y=\frac{1+\infty e^{4x}}{1-\infty e^{4x}}\)

Best wishes
PG

 

Okay. When c3 = 0, the c would be ∞. Then, don't you the following expression would be indeterminate, as you say? If all three expression are the same equation in three different forms, then it shouldn't be inderterminate, and y=-2 should be the result for all x. Please help me with it.

\(y=\frac{1+\infty e^{4x}}{1-\infty e^{4x}}\)

Best wishes
PG

This gets into an issue of terminology and precision of language, or basically semantics. The form of the equation is indeterminate, but that doesn't mean that you can't determine an answer in certain cases. Mathematicians identify certain conditions as indeterminate. In this case infinity/infinity is the situation of interest.

If I walk up to you on the street and say, "what is infinity/infinity", your answer should be, "It all depends. It could be infinity, zero, one or another number. Please tell me the expressions that are giving you infinity in the numerator and infinity in the denominator, and perhaps I can help you by taking a limit."

If I then say, "my numerator is x and my denominator is x and I'm interested in what happens when x=infinity", then your answer will be "Oh, that's easy. In the limit as x goes to infinity x/x is 1."

If I gave you x/x^2, then your answer would be 0, and if I gave you x^2/x your answer would be infinity. If you doubt these answers, then plot out the functions x/x, x/x^2 and x^2/x and see if you can visualize what the value will be at x=infinity. Is the fact that these are indeterminate forms preventing you from seeing the pattern? I doubt it, and I doubt you would even need to do the plotting.

Now, these are very simple examples and you probably would have simplified the equations as 1, 1/x and x before you even tried to take a limit, but there are more complicated situations that show up.

What if the problem is as follows \({\rm lim}_{x\to\infty} \Bigl({{x+1}\over{x^2+2}}\Bigr)\)? Here you would not be able to simplify the equation before taking the limit. The answer here is zero, but this might not be obvious to someone who has not studied limits before.

I think the basic way to think about this is that the term "indeterminate" is like a road sign that is telling you to slow down because there is a sharp corner coming up. It doesn't mean stop, turn around or jump out of the car, but just to be careful and drive through cautiously.

 

The material is from post #20 of this thread.

\( y=2{{1+c e^{4x}}\over{1-ce^{4x}}\) (1)

and divide the numerator and denominator by c to get the following

\( y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}\) (2)

Now, redefine the constant, and let c_3=1/c, to get the following.

\( y=2{{c_3+e^{4x}}\over{c_3-e^{4x}}\) (3)

Now when c_3 is zero, you get y=-2.

The entries in blue in the quoted post are my additions.

c_3 = 1/c

Suppose that c_3=0, what does it mean for "c". c_3 has taken a specific value, i.e. "0". Infinity is not a number and therefore you can't say c=∞. But you can solve this conundrum (at least to me it looks that way) by using limit. c = lim[c_3 -> 0] 1/c_3=∞.

But as I was saying in this thread that there is a difference between saying that limit exists at a certain point and function is defined at some point and you agreed with me. So, in my opinion, we can't evaluate the expression (1), (2), and (3) by direct substitutions of the values. e.g. You can't say in case of (2) that y=-2 when c=∞. Rather we should do this:

y = lim[c->∞] \( y=2{{1/c+e^{4x}}\over{1/c-e^{4x}}\)

Do you agree with me? The limits of (3) and (2) are equal, i.e. y=-2. Does this mean the expressions are equivalent? In one of the previous posts, you said (1), (2), and (3) are equivalent.

So, you see I'm confused! Please help me with it. And I can understand it would be quite hard for you to track what we were discussing here. Please let me know I can start the discussion anew so that you don't have to refer to previously said material. Thank you very much.

Best wishes
PG

 

Do you agree with me? The limits of (3) and (2) are equal, i.e. y=-2. Does this mean the expressions are equivalent? In one of the previous posts, you said (1), (2), and (3) are equivalent.

So, you see I'm confused! Please help me with it.

Walt Whitman once said, "Do I contradict myself? Very well, then I contradict myself. I am large, I contain multitudes."

Perhaps this is an interesting quote to mention when talking about infinity, and the confusion it can create. The quote might just as well have ended with "I am small, I contain almost nothing", because infinity and zero are very closely tied concepts, and they give rise to very similar conundrums.

Basically,I agree that you are always correct to say that these tricky cases should be expressed as limits. However, some cases are more tricky than others and once you have experience, a simple thing like 1/c going to zero as c goes to infinity, or 1/c going to +infinity as c goes to zero from the right, or 1/c going to -infinity as c goes to zero from the left, will often be shown (and even thought about) as 1/0=infinity and 1/infinity=0 in a given context. Or, you might even see 1/0- or 1/0+ to make it more clear. But, any mathematician is going to side with you, i expect.

When your calulator gave you -inf when you pressed ln(0) you have a case of a shorthand notation. You can be sure that the guy who programmed the calculator is now on the mathematician's all time most wanted list, but that guy is also the hero of the engineering students stuck in the middle of an exam trying to remember how ln(x) behaves near zero, when coming from the right. (by the way, my old HP-28s says "Log Errror: Infinite Result" when I do ln(0))