Thought of 2 Difficult Math Problems

[Izzhov]

I was futzing around with some equations and i came across 2 problems that I'm having trouble solving.
In the first one, you must solve for y as a function of x:
y^x = x^y + 1

For the second one, I know that \sum_{ x=1}^\infty \frac{1}{x^2} = \frac{ \pi^2}{6}, but then what's the exact value of \sum_{ x=1}^\infty \frac{1}{x^3}?

 

[matt grime]

I will define FOO(x) as the function such that FOO(X)^x=x^FOO(X)+1. Assuming this even exists.

The second one is hard, I think (unknown?). Somewhat surprising, really, is that even exponents are 'easy' and odd ones are 'hard'.

 

[Izzhov]

I will define FOO(x) as the function such that FOO(X)^x=x^FOO(X)+1. Assuming this even exists.

All you did here was change the representation of a variable. That doesn't make it any easier... You'd still have to solve it algebraically.

 

[arildno]

Why?
If matt, or somebody else takes the the trouble of proving the existence of Foo(x), the case is over and done with.

 

[Izzhov]

Why?
If matt, or somebody else takes the the trouble of proving the existence of Foo(x), the case is over and done with.

But you haven't solved for y, so you don't know what FOO(x) actually is.
For this problem, I was trying to put it in the form of y=blah, where "blah" is some function of x. Let me give you an example:
If I say "solve this equation for y as a function of x: 2x+y=3," I mean that i want you to put it in the form: y=3-2x. Or, if I said "solve for y: y^x=3," I want it in the form: y=3^(1/x).

 

[arildno]

Set
g(x,y)=y^{x}-x^{y}-1
We see that g(1,2)=0, furthermore, we have:
\frac{\partial{g}}{\partial{y}}=xy^{x-1}-x^{y}\ln(x),
whose value at (1,2) equals 1.

Thus, foo(x) exists in a region about (1,2), and we have:
\frac{dFoo(x)}{dx}=\frac{(Foo(x))^{x}\ln(Foo(x))-(Foo(x))x^{Foo(x)-1}}{x(Foo(x))^{x-1}-x^{(Foo(x))}\ln(x)}
This is a differential equation for Foo(x).

 

[Izzhov]

Set
g(x,y)=y^{x}-x^{y}-1
We see that g(1,2)=0, furthermore, we have:
\frac{\partial{g}}{\partial{y}}=xy^{x-1}-x^{y}\ln(x),
whose value at (1,2) equals 1.

Thus, foo(x) exists in a region about (1,2), and we have:
\frac{dFoo(x)}{dx}=\frac{(Foo(x))^{x}\ln(Foo(x))-(Foo(x))x^{Foo(x)-1}}{x(Foo(x))^{x-1}-x^{(Foo(x))}\ln(x)}
This is a differential equation for Foo(x).

Well, that helps, but it still isn't the answer, because I still don't know what FOO(x) is. (I only know how to do the most basic of differential equations.)

 

[arildno]

Well, that helps, but it still isn't the answer, because I still don't know what FOO(x) is. (I only know how to do the most basic of differential equations.)

So?
Foo is perfectly well-defined now, through the initial condition Foo(1)=2.

 

[Izzhov]

So?
Foo is perfectly well-defined now, through the initial condition Foo(1)=2.

But I didn't want to define Foo, i wanted the equation to be put in the form y=blah, with only x's on the "blah" side. That's what i was asking.

 

[arildno]

Note, by the way, that the line x=0 is a different solution to the equation as well..

 

[arildno]

But I didn't want to define Foo, i wanted the equation to be put in the form y=blah, with only x's on the "blah" side. That's what i was asking.

Well, in a neighbourhood of (1,2), Foo exists, and within that neighbourhood, you may determine the function value of Foo at any point to any desirable degree of accuracy using numerical techniques.

What more do you want?

 

[Izzhov]

Yeah, but if you put domain restrictions on it, you can still put it in the form of y=blah.

 

[arildno]

Set
g(x,y)=y^{x}-x^{y}-1
We see that g(1,2)=0, furthermore, we have:
\frac{\partial{g}}{\partial{y}}=xy^{x-1}-x^{y}\ln(x),
whose value at (1,2) equals 1.

Thus, foo(x) exists in a region about (1,2), and we have:
\frac{dFoo(x)}{dx}=\frac{(Foo(x))^{x}\ln(Foo(x))-(Foo(x))x^{Foo(x)-1}}{x(Foo(x))^{x-1}-x^{(Foo(x))}\ln(x)}
This is a differential equation for Foo(x).

With the initial condition Foo(1)=2, this is enough.
To be nice to you, I'll show you how to find the best linear approximation to Foo close to x=1, i.e, compute the 2-term Taylor series about x=1:

Now, we readily see that \frac{dFoo}{dx}\mid_{x=1}=\frac{2\ln(2)-2}{1}=2\ln(\frac{2}{e})

Thus, we have:
y=Foo(x)\approx{2}+2\ln(\frac{2}{e})(x-1)

EDIT:
I made the wrong sign on the derivative of Foo, meaning that our 2-term Taylor expansion of Foo should be:
y=Foo(x)\approx{2}+2\ln(\frac{e}{2})(x-1)

 

[CRGreathouse]

For the second one, I know that \sum_{ x=1}^\infty \frac{1}{x^2} = \frac{ \pi^2}{6}, but then what's the exact value of \sum_{ x=1}^\infty \frac{1}{x^3}?

The irrational number zeta(3) = 1.2020569... is Apéry's constant. No closed form is known, but the first 2 billion (at least) decimal places have been calculated by Howard Cheng, Guillaume Hanrot, Emmanuel Thomé, Eugene Zima, and Paul Zimmermann.

 

[Izzhov]

The irrational number zeta(3) = 1.2020569... is Apéry's constant. No closed form is known, but the first 2 billion (at least) decimal places have been calculated by Howard Cheng, Guillaume Hanrot, Emmanuel Thomé, Eugene Zima, and Paul Zimmermann.

'K, thanks.

 

[matt grime]

But you haven't solved for y, so you don't know what FOO(x) actually is.

And you don't know what sin(3) is, but I suspect you think that that y=sin(x) is a proper type of solution, don't you?

http://www.dpmms.cam.ac.uk/~wtg10/equations.html

 

[CompuChip]

Yeah, but if you put domain restrictions on it, you can still put it in the form of y=blah.

For some functions, this is just not possible. For example, the error function is defined through an integral equation, and we can describe it's properties, but we cannot write it in closed form.
Same goes for the Bessel function, which is defined as a (specific) solution to some differential equation, and countless others (of which some have special names, like J, B_i, A_n or \operatorname{Foo}(x)).

 

[HallsofIvy]

But you haven't solved for y, so you don't know what FOO(x) actually is.

What do you mean by "actually is"? What "actually is" the Lambert W function, W(x), which is defined as the inverse of the function f(x)= xe^x? What "actually is" the Bessel function of the first order, which is defined as one of the two independent solutions to Bessel's differential equation? For that matter, what "actually is" sin(x) or cos(x) which can be defined in a variety of ways, none of whichy give you an easy formula in which you can just "plug in" a value for x?

For this problem, I was trying to put it in the form of y=blah, where "blah" is some function of x.

That's exactly what matt grime did: it is now in the form y= Foo(x), where Foo is some function of x.

Let me give you an example:
If I say "solve this equation for y as a function of x: 2x+y=3," I mean that i want you to put it in the form: y=3-2x. Or, if I said "solve for y: y^x=3," I want it in the form: y=3^(1/x).

Do you understand that the great majority of functions ("almost all" functions) CANNOT be written in such a way? That they are NOT "rational functions" nor "radical functions" nor even "transcendental functions"?

 

[Izzhov]

And you don't know what sin(3) is, but I suspect you think that that y=sin(x) is a proper type of solution, don't you?[/url]

Yes, because it is in the form of y=blah. I just wanted it in a form where I could plug any number in for x and instantly know what y is, and the only way I know how to do that is to put it in the form y=blah, where there are only x's on the blah side. Also, I do kind of know what sin(3) is, because I can approximate it with an infinite sum. I did not know how to approximate FOO(x) when it was in the form originally given to me.

Do you understand that the great majority of functions ("almost all" functions) CANNOT be written in such a way? That they are NOT "rational functions" nor "radical functions" nor even "transcendental functions"?

If that is the case, then it should at least be proved that the equation I gave you can't be put into the form y=blah (where blah is some function I can approximate on a calculator or using some infinite sum) before you resort to defining "blah" as "FOO(x)."

 

[matt grime]

That really makes no sense. Being transcedental, or not expressible in terms of basic functions has nothing to do with being able to approximate f(x) for any given value of x. I can do that with the FOO function too, or with ERF, or the Lambert function. You really have the wrong idea here.

 

[Izzhov]

That really makes no sense. Being transcedental, or not expressible in terms of basic functions has nothing to do with being able to approximate f(x) for any given value of x. I can do that with the FOO function too, or with ERF, or the Lambert function. You really have the wrong idea here.

But I don't know how to approximate FOO(x) for any value of x, do I?
So, I'll change the terms for solving the problem. To solve the problem, you must put the equation in a form such that I know how to approximate FOO(x) for any value of x.

 

[DeadWolfe]

Newton's method can easily do that.

 

[Izzhov]

Newton's method can easily do that.

Pardon my ignorance, but... what is Newton's method? Do you mean the equation for approximating the derivative of a function, because I have no idea how to apply that to the equation I posted to approximate y for any value of x.

 

[CRGreathouse]

Pardon my ignorance, but... what is Newton's method? Do you mean the equation for approximating the derivative of a function, because I have no idea how to apply that to the equation I posted to approximate y for any value of x.

Here, the first ten links should help.
http://google.com/search?q=Newton's+method

 

[arildno]

But I don't know how to approximate FOO(x) for any value of x, do I?
So, I'll change the terms for solving the problem. To solve the problem, you must put the equation in a form such that I know how to approximate FOO(x) for any value of x.

I have already done that for you.

 

[matt grime]

But I don't know how to approximate FOO(x) for any value of x, do I?

Apparently not.

So, I'll change the terms for solving the problem. To solve the problem, you must put the equation in a form such that I know how to approximate FOO(x) for any value of x.

It's not up to me to know what you do or do not know about numerical methods. Approximating roots is very easy, if tedious - just do repeated bisection if you have to.

 

[HallsofIvy]

But I don't know how to approximate FOO(x) for any value of x, do I?
So, I'll change the terms for solving the problem. To solve the problem, you must put the equation in a form such that I know how to approximate FOO(x) for any value of x.

Newton's method can easily do that.

Pardon my ignorance, but... what is Newton's method? Do you mean the equation for approximating the derivative of a function, because I have no idea how to apply that to the equation I posted to approximate y for any value of x.

That's the fundamental problem with this whole thread! You said "such that I know how to approximate" (emphasis added). Mathematics does not revolve around you. Since we do not know what you can or cannot do, we can't possibly satisfy your conditions.

 

[Dragonfall]

Did I wander into bizzaro world, or are you guys particularly harsh on the OP?

 

[JonF]

I think he meant, “Find an elementary solution for y in terms of x” for the first one. I’m pretty sure you can’t is the answer.

 

[Izzhov]

I think he meant, “Find an elementary solution for y in terms of x” for the first one. I’m pretty sure you can’t is the answer.

Thank you very much!

 

[learningphysics]

Did I wander into bizzaro world, or are you guys particularly harsh on the OP?

I agree. As JonF posted, it seems pretty clear he was asking for an elementary solution of y in terms of x...

Does the question want y as a "relation of" x... or does it have to be a function of x? As arildno posted, x = 0 satisfies the equation with any y... So y isn't a function of x and can't be written as such, since a function would need one y value per x value (but you have (x = 0, y = 0) which is a solution or (x=0, y=1) as a solution etc)

 

[matt grime]

Ah, it's the 'you are mean, you're supposed to be psychic and guess what the OP meant' moment.

 

[arildno]

Thank you very much!

Your skill at communication is truly unmatched.
Your eloquent "y=blah" says it all..

 

[arildno]

I agree. As JonF posted, it seems pretty clear he was asking for an elementary solution of y in terms of x...

Does the question want y as a "relation of" x... or does it have to be a function of x? As arildno posted, x = 0 satisfies the equation with any y... So y isn't a function of x and can't be written as such, since a function would need one y value per x value (but you have (x = 0, y = 0) which is a solution or (x=0, y=1) as a solution etc)

Hmm, you can find local regions where y can be regarded as a function foo(x), as I have shown.

However, not even an explicit linear approximation to foo(x) in the vicinity of (1,2) is enough for the blah-blah guys of the world..

 

[Izzhov]

Your skill at communication is truly unmatched.
Your eloquent "y=blah" says it all..

I actually thought it was fairly self-explanatory. I guess I didn't realize that you thought through every problem at the calculus level. I was looking for a more algebraic approach to it, but I'm fine, since I now have the solution.

 

[matt grime]

Is there some heirarchy that says 'the calculus level' is lower than the 'algebraic level'? You had the solution before, but you just weren't happy with it. The solution was perfectly correct.

 

[Izzhov]

Is there some heirarchy that says 'the calculus level' is lower than the 'algebraic level'? You had the solution before, but you just weren't happy with it. The solution was perfectly correct.

Actually, the whole problem in our communication is that "calculus level" is higher than algebraic level; it's a classic case of the answerer thinking more deeply about the question than the asker was expecting, and answering it at a higher level (i.e. using calculus) than the asker wanted (i.e. using algebra).

 

[arildno]

Actually, the whole problem in our communication is that "calculus level" is higher than algebraic level; it's a classic case of the answerer thinking more deeply about the question than the asker was expecting, and answering it at a higher level (i.e. using calculus) than the asker wanted (i.e. using algebra).

Whichever way you want it, dear, you won't find any other best linear approximation to foo(x) about (1,2) than the one I posted..

 

[Izzhov]

Whichever way you want it, dear, you won't find any other best linear approximation to foo(x) about (1,2) than the one I posted..

'K, thanks.