Is There a Real Solution for a^n+b^n=c^n as n Approaches Infinity?

[Icebreaker]

Why does it "make no sense" to solve an equation at the limit?

x = x

lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a

Maybe you can explain your definition of "solving an equation" better.

 

[Curious3141]

Why does it "make no sense" to solve an equation at the limit?

x = x

lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a

Maybe you can explain your definition of "solving an equation" better.

Your notation is valid for finite a, but it breaks down for infinite a. In the case of a simple identity, the break-down is not obvious.

Try this :

lim_{x \rightarrow \infty}(x - 1) = lim_{x \rightarrow \infty}x

Does that imply that you've solved x -1 = x ?

EDIT : Made it a much simpler example.

 

[Icebreaker]

No, it implies that as x\rightarrow\infty, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

In other words, x - 1 = x is true iff x\rightarrow\infty.

Or, if you prefer:

f(x)=f(x) \Rightarrow lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y)

However

lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y) does not imply that f(x)=f(x).

 

[Curious3141]

No, it implies that as x\rightarrow\infty, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

I would prefer to say the expressions approach one another, and the limits of the expressions are equal. Now think : is that really an equation ? Have you truly "solved" for a real value of x ? Real numbers have to be finite.

By the same token, in your original problem statement, any real number set (a,b,c) that's put into the equation "at the limit" will be either reduced to zero or be increased without limit. From that you think you can say that "there is no real solution" to the equation at the limit (which is what I think you tried to prove).

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

I just saw your edit :

In other words, x - 1 = x is true iff x\rightarrow\infty

I disagree with this statement. I think you should phrase that as "the limit of x-1 is equal to the limit of x as x tends to infinity). Subtle difference, semantics perhaps, but to me, it makes a lot of difference.

 

[Icebreaker]

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

What kind non-triviality are we talking about here? Because, as I pointed out before:

lim_{x\rightarrow\infty}\frac{1}{x}=a=0

And I disagree, it makes perfect sense to ask the question, even if you are right. Redundant, perhaps, but nothing illegal or illogical. After all, if you are right for any general case, then I can prove it for any special case, and I did. So what's the problem here?

 

[matt grime]

No, it wasn't obvious to me what the question was asking for.

well it was to me.

For me the solution to an equation has to satisfy the equation when it is substituted back into it. Agree ?

but there is no explicit equation to solve, so this is not relevant. there is a pointwise defined function.

Now let me explain my problem with the question. If one tries to solve the equation algebraically "at the limit", the solution might just not work on the original equation.

one doesn't solve it at the limit, one shows that no real a,b,c all strictly positive exist such that the limits agree and are not infinite, whether or not there is a solution for any particular n is not important, what is important is finding if there are real strictly positive a,b,c such that lim a^n + b^n =lim c^n with the limits taken as n tends to infinity (when it exists)

Let's say we decide the values of a and b and want to determine what value of c (if any) exists. The obvious way of doing that is to evaluate c = {(a^n + b^n)}^{\frac{1}{n}}. It is clear that as n tends to infinity, the limit of c = max(a,b). Fair enough ?

yes, and if you evaluate that limit what do you get?

The rest of you post isn't important, evaluating "sequentially? what does that even mean?


[/quote]Now let's try putting that back into the original equation in the form a^n + b^n - c^n = 0, where n tends to infinity. Here we are evaluating each term sequentially. It is clear that no solution set other than the trivial (0,0,0) will actually work upon substitution.

I believe the original poster was trying to use this sort of logic to show that the equation can have no real solutions "at the limit". But to me, the question makes no sense. If you solve an equation with valid algebra, the solution *has* to work. Putting the solutions back in and verifying that they do not satisfy the equation does not mean that there are no real solutions to the equation per se, it means that the statement itself is flawed.

The equation (barring the limit) is in a valid algebraic form. The solutions that I found were with valid algebra. The problem was in the way that solution behaved when the limit was taken. That limit could not successfully be substituted back into the original equation.

The conclusion I draw from that is that it *makes no sense* to speak of solving an equation at the limit. It's fine to define the behaviour of a function at the limit, as you have done, but not try to solve an equation.

Do you still disagree, Matt ?[/QUOTE]


we aren't solving at the limit, i don't see the original post using the phrase "at the limit"; only you appear to be using it. solve isn't the word I'd've used, but that is neither here nor there really. translating the question to one where we do not "solve" isn't very demanding

 

[Curious3141]

Sorry to dredge this up again, but I would like this case resolved :

Is there a stipulation against a, b, or c containing a term in n ?

What about a = {(1 + \frac{1}{n})}^{\ln{(e - 1)}}, b = 1 and c = (1 + \frac{1}{n}) ?

Those are all definitely real numbers. Put them into the equation and the limit simply becomes :

(e - 1) + 1 = e

So why isn't that a real solution set ?

EDIT : Although I don't know if a and c really qualify as "standard" real numbers, but that's why I'm asking the question.

 

[matt grime]

Because they aren't solutions for all n (or indeed any n).

Any three numbers of abs value less than 1 satisfy

lim(a^n+b^n)=lim(c^n)

but the question was asking what are the solutions for all n. The first line of the question stated that

a^n+b^=c^n.

And a and c aren't fixed real numbers, either, since they vary with n.

 

[Curious3141]

Because they aren't solutions for all n (or indeed any n).

Any three numbers of abs value less than 1 satisfy

lim(a^n+b^n)=lim(c^n)

but the question was asking what are the solutions for all n. The first line of the question stated that

a^n+b^=c^n.

And a and c aren't fixed real numbers, either, since they vary with n.

How do you mean "for all n" ? The original question didn't say "for all n" ?

Do you mean : does a particular positive real solution set (a,b,c) exist such that a^n + b^n = c^n holds identically for all natural n ? Because that can immediately be shown to be untrue - just square the thing.

I still am not sure what the question is asking. Certainly there're infinitely many real solutions (a,b,c) for each and every finite value of n. I'm still thrown by the use of the limit notation in the question.

But given the limit notation, I don't see why my solutions (with a and c in terms of n) cannot hold. Please explain this better.

 

[matt grime]

i guess this is the difference in the way we read the question.

 

[Curious3141]

I still think it is a bad question, since we can't seem to agree on the proper interpretation. Thank you for your time.

 

[Icebreaker]

I still think it is a bad question

Well, I respectfully disagree.

 

[Curious3141]

Well, I respectfully disagree.

Then please explain exactly what you meant by the question. Does my solution set in terms of n work ? Your question said nothing about independence from n for the solutions.

 

[Icebreaker]

I deemed it unnecessary to state this "independence" because a violation of such "independence" would require the solution set to hold true for all n, otherwise it would not satisfy the equation a^n+b^n=c^n in the first place. Thus, the solution I want are not variables that change with n, but fixed numbers.

 

[Curious3141]

I deemed it unnecessary to state this "independence" because a violation of such "independence" would require the solution set to hold true for all n, otherwise it would not satisfy the equation a^n+b^n=c^n in the first place. Thus, the solution I want are not variables that change with n, but fixed numbers.

I'm very confused now. What sort of solution set are you looking for ?

1) Did you mean to imply that a particular fixed solution set (a,b,c) must satisfy the equation a^n + b^n = c^n (without the limit notation) as the value of n is allowed to vary ? For example, if (a,b,c) are positive reals that satisfy the equation when n = 2 then the same values will satisfy the same equation when n = 4 ? As I said, this is trivially falsifiable, just by squaring the LHS and RHS of the equation.

2) Did you mean that at least one solution set (a,b,c) can be found for the equation a^n + b^n = c^n for each and every natural number value of n ? Then I heartily agree, in fact, there are an infinite number of solutions for each and every such value.

3) Did you mean are there values (a,b,c) that fail to satisfy a^n + b^n = c^n for all natural numbers n, yet will satisfy the same equation when n is treated as infinite ? Because, as I have said c = max(a,b) is a solution set when either one or both of (a,b) is one or greater. If both (a,b) are less than one, then *all* permutations of (a,b,c) with 0 <a,b,c <1 will satisfy. With no further clarification of the question, I don't see why the other solution in terms of n and e cannot be a solution set as well.

4) Did you mean something else that I am completely missing ?

 

[Icebreaker]

There is at least a solution set {a,b,c} of finite reals for every integer n>2. There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n\rightarrow\infty.

 

[Curious3141]

There is at least a solution set {a,b,c} of finite reals for every integer n>2. There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n\rightarrow\infty.

In that case, the problem is equivalent to the case 3 that I mentioned, and solutions exist (meaning what you set about to "prove" is wrong). Here is the solution set :

(a, b, max[a,b]) is a solution set when either one or both of a, b is one or greater. *All* permutations of (a,b,c) with 0 < a,b,c <1 will satisfy.

 

[Icebreaker]

Then I will impose a restriction on the solution, where {a, b, c} must be integers which are greater than zero; thus reverting back to a special case of FLT, which was what I was set out to prove anyway.

 

[Curious3141]

Then I will impose a restriction on the solution, where {a, b, c} must be integers which are greater than zero; thus reverting back to a special case of FLT, which was what I was set out to prove anyway.

Sigh... you keep changing the question. Are you acknowledging that you were on the wrong track to begin with ?

If it's FLT, you don't need the limit or anything. FLT has already been proven.

 

[Icebreaker]

Actually this is the first time that I changed the question.

I know that a general proof of FLT had already been found; however, that doesn't prohibit me from proving special cases of it. In my original proof, I limited a, b, c and n to integers. I wanted to know if I can extend this proof for all reals of a, b and c, which is why I posted it here anyway.

 

[Data]

There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n\rightarrow \infty.

What precisely do you mean by this statement? My reading is that you are trying to say that if \forall n \in \mathbb{N}^*, \ \mbox{we define} \ S_n = \{ (a, \ b, \ c) \in \mathbb{R}^3 | \ a^n+b^n=c^n \}, \ \mbox{then} \ \lim_{n \rightarrow \infty} S_n = \phi, which is silly, since \forall \ (a, \ b \in \mathbb{R}^+, \ n \in \mathbb{N}^*), \ \exists c\in \mathbb{R} \ \mbox{s.t.} \ a^n+b^n = c^n.

 

[Icebreaker]

Yes, but not if a,b,c are integers, as I explained above what I tried to do.

 

[Data]

Of course not. If we require a, \ b, \ c \in \mathbb{N}^* then there are no solutions \forall n \geq 3 by Fermat's Last Theorem. I was replying to

There is at least a solution set {a,b,c} of finite reals for every integer n>2. There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n \rightarrow \infty.

and

I wanted to know if I can extend this proof for all reals of a, b and c, which is why I posted it here anyway.

 

[matt grime]

I still think it is a bad question, since we can't seem to agree on the proper interpretation. Thank you for your time.

And here we do agree (or at least we can agree that the version of the question posted was ambiguously written). But I don't think this justified you jumping all over the OP with spurious comments about solving things "at infinity" or whatever the phrase was, espcially given your example using logs.

I am also more than confused as to what the OP wanted to even solve, especially if he was extending FLT in some way, since there are obvisouly solutions to a^n+b^n=c^n for all n integer with a,b,c real, and I thus may well have been miscrediting him with showing something he wasn't showing. (Yes, squaring both sides would be better, but proofs aren't unique. Oddly no one pointed out the flaw that you can't exchange limits and quotients in the manner he did)

 

[Curious3141]

And here we do agree (or at least we can agree that the version of the question posted was ambiguously written). But I don't think this justified you jumping all over the OP with spurious comments about solving things "at infinity" or whatever the phrase was, espcially given your example using logs.

I am also more than confused as to what the OP wanted to even solve, especially if he was extending FLT in some way, since there are obvisouly solutions to a^n+b^n=c^n for all n integer with a,b,c real, and I thus may well have been miscrediting him with showing something he wasn't showing. (Yes, squaring both sides would be better, but proofs aren't unique. Oddly no one pointed out the flaw that you can't exchange limits and quotients in the manner he did)

For the record, please note that I did not "jump all over the OP" until said OP started to evince disdain and rudeness by using phrases like "save your breath" and "get it through your head". I would like to think I am fairly civil until provoked.

When a person is wrong AND rude, I don't see the need for tact in return.

 

[matt grime]

Fair enough.

But you should realize it is perfectly possible to solve for things that have limits in them if you're going to criticize that.

For instance, I can "solve" lim (1+x/n)^n=2, the answer being, as you know, x=log2

 

[Curious3141]

Fair enough.

But you should realize it is perfectly possible to solve for things that have limits in them if you're going to criticize that.

For instance, I can "solve" lim (1+x/n)^n=2, the answer being, as you know, x=log2

Yes, that is fair enough. Does the RHS necessarily have to be a constant term (for this sort of "limit equation" to work), or can it be dependent on x as well ?

 

[matt grime]

I could put an x in the rhs if I felt like it since that "equaiton" above is exactly the same as

e^x=2,
so I could ask if there is a (real) x satisfying e^x=3x. (there is, but it won't be possible to find algebraically, I suspect)

 

[Curious3141]

I could put an x in the rhs if I felt like it since that "equaiton" above is exactly the same as

e^x=2,
so I could ask if there is a (real) x satisfying e^x=3x. (there is, but it won't be possible to find algebraically, I suspect)

The reason I'm asking is that (assuming the OP meant something like this), the original question is ambiguous because there're variable terms on each side.

The question can be rewritten in two ways :

1) "Solve" (n -> infinity)lim(a^n + b^n - c^n) = 0 for positive real (a,b,c).

In which case, there are no solutions for any a,b or c greater than or equal to unity. Every permutation of real (a,b,c) with each value less than one is a valid solution.

2) "Solve" (n -> infinity)lim(a^n + b^n)^(1/n) - c = 0 for positive real (a,b,c).

Here the solution is always c = max(a,b)

Both forms are algebraically equivalent, yet the solution set is different depending on which form is used. At least when the constant term is on the RHS, an unambiguous problem statement can be constructed.

 

[matt grime]

Both forms are not equivalent though because of that "limit" word in there (n is a dummy vairable, not actually a variable).