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[da_nang]

Eh. Take two variables x,y ∈ ℝ and let the function f(x,y) = x^y.

If you can show that lim_(x,y)->(0,0) f(x,y) = L exists where L ∈ ℝ, then you have a reason to define f(0,0) = L as a "removable singularity".

If it doesn't then 0⁰ will remain indeterminate and whatever you define it as will not have any effect on the general conclusion.

Since the L = 1 when approaching from y = 0 and L = 0 when approaching from x = 0, the limit does not exist.

Thus by defining it as something else in a specific context, you're simply choosing a specific approach direction. In the case of 0⁰ = 1, you're following the y = 0 approach direction (or some other direction where L = 1 such as y = x).

Would you at least read my posts on that before giving such a response¿ I spent already three posts on explaining that the limit is irrelevant.

Why would it be irrelevant? It's the very reason why it is indeterminate. As such, the value of 0⁰ cannot be determined without knowing the context. Formally speaking, anyway.
Granted, in many of the cases where 0⁰ = 1, the context is known either explicitly or implicitly but straight up saying 0⁰ = 1 without context is erroneous especially in analysis and should be discouraged unless there are plans to change the definition of indeterminate forms (which there aren't AFAIK).

The formal (context-free) value of 0⁰ is therefore not known and cannot be known because of the results from the limits.

Just to steal an example from the wiki: lim_t->0+ (e^-1/t)^at = e^-a. This can assume any positive non-zero value depending on the value of a.

[Another]

I think an even better example is that the value of a definite integral can formally come to be 0^0* and not necessary be 1.

*If you are willing to substitute zero values of 2 functions at one of the points into a formula you got.

[Skyrunner]

How do I define, then solve a function in wolfram alpha?

Say, f[x_]:=3sqrt(3)/2*x(x-1)(x-2)
And I want to solve
(f○f)(x)/x==1. I tried a bunch of stuff but it never gets the point.
Solve[Composition[f,f]

• /x==1, x] doesn't work, eg.

[Another]

How do I define, then solve a function in wolfram alpha?

Say, f[x_]:=3sqrt(3)/2*x(x-1)(x-2)
And I want to solve
(f○f)(x)/x==1. I tried a bunch of stuff but it never gets the point.
Solve[Composition[f,f]

• /x==1, x] doesn't work, eg.

Is wolframalpha necessary? In Sage this stuff can be done this way:

Code: [Select]

sage: f(x)=3*(3)^0.5/2*x*(x-1)*(x-2)
sage: (f(f(x))/x-1).find_root(-10,10)
2.2968062370863764


[Skyrunner]

I want to find all roots over the entire real number range though...

[Another]

I want to find all roots over the entire real number range though...

Code: [Select]

sage: plot((f(f(x))/x-1),0,2.5)   

sage: (f(f(x))/x-1).find_root(0,0.5)
0.27467858771942977
sage: (f(f(x))/x-1).find_root(0.5,0.8)
0.7031937629136243
sage: (f(f(x))/x-1).find_root(0.8,1) 
0.8930529366073069
Along with the root above - that is all.
As for roots outside of [0, 2.5] range - you can most likely prove that by examining first 8 derivatives of your 8th degree polynomial.

[Skyrunner]

So there's no easy way to do it like WA does? Input numbers, get ALL the roots?

I mean, I could manually copy and paste the formula for each x, but it seems like there should be a better way t o do it.

[Another]

After thinking some more when f(x)==x -> f(f(x))==x  and 2 of the roots can be found as solutions to f(x)=x this way.

[RedWarrior0]

The thing with Alpha is that, while it is awesome, it can only do so much. For stuff beyond that, you'd need a dedicated program like Mathematica.

[Another]

To my knowledge there does not exist a universal formula that would tell you even how many real roots an arbitrary high degree polynomial has. A computer program can of course check the sign at every small epsilon step in X for some reasonable interval but that does not provide a guarantee in the general case.

[Mego]

To my knowledge there does not exist a universal formula that would tell you even how many real roots an arbitrary high degree polynomial has. A computer program can of course check the sign at every small epsilon step in X for some reasonable interval but that does not provide a guarantee in the general case.

Why wouldn't Sturm's Theorem work?

[Skyrunner]

To my knowledge there does not exist a universal formula that would tell you even how many real roots an arbitrary high degree polynomial has. A computer program can of course check the sign at every small epsilon step in X for some reasonable interval but that does not provide a guarantee in the general case.

But there are ways to figure it out. For example, couldn't you

The thing with Alpha is that, while it is awesome, it can only do so much. For stuff beyond that, you'd need a dedicated program like Mathematica.

find out all the maxima and minima, and if the product of adjacent max/min is less than or equal to 0 there should be a root between them?

@red: I though WA had mathematica in it :<

[RedWarrior0]

No, Mathematica is something different. Mathematica has a rigid syntax and whatnot.

[Another]

To my knowledge there does not exist a universal formula that would tell you even how many real roots an arbitrary high degree polynomial has. A computer program can of course check the sign at every small epsilon step in X for some reasonable interval but that does not provide a guarantee in the general case.

Why wouldn't Sturm's Theorem work?

Well, good to know that there are precise methods for polynomials. I eventually discovered that Sage uses VAS algorithm and how to make it work in case of polynomials with non-rational coefficients.

Spoiler (click to show/hide)

sage: R.<x>=RR[]                       
sage: f=3*(3)^(0.5)/2*x*(x-1)*(x-2)
sage: (f(f).quo_rem(x)[0]-1).roots()
[(0.274678587719429, 1),
 (0.703193762913601, 1),
 (0.893052936607363, 1),
 (2.29680623708592, 1)]

As mentioned earlier 2 of these roots can be expressed in radicals.

[ZetaX]

Eh. Take two variables x,y ∈ ℝ and let the function f(x,y) = x^y.

If you can show that lim_(x,y)->(0,0) f(x,y) = L exists where L ∈ ℝ, then you have a reason to define f(0,0) = L as a "removable singularity".

If it doesn't then 0⁰ will remain indeterminate and whatever you define it as will not have any effect on the general conclusion.

Since the L = 1 when approaching from y = 0 and L = 0 when approaching from x = 0, the limit does not exist.

Thus by defining it as something else in a specific context, you're simply choosing a specific approach direction. In the case of 0⁰ = 1, you're following the y = 0 approach direction (or some other direction where L = 1 such as y = x).

Would you at least read my posts on that before giving such a response¿ I spent already three posts on explaining that the limit is irrelevant.

Why would it be irrelevant? It's the very reason why it is indeterminate. As such, the value of 0⁰ cannot be determined without knowing the context. Formally speaking, anyway.
Granted, in many of the cases where 0⁰ = 1, the context is known either explicitly or implicitly but straight up saying 0⁰ = 1 without context is erroneous especially in analysis and should be discouraged unless there are plans to change the definition of indeterminate forms (which there aren't AFAIK).

The formal (context-free) value of 0⁰ is therefore not known and cannot be known because of the results from the limits.

Just to steal an example from the wiki: lim_t->0+ (e^-1/t)^at = e^-a. This can assume any positive non-zero value depending on the value of a.

Look, you cant just say "you can't define it continuously, therefore it should not be defined". Continuity is a _bonus_, not the ultimate goal everything has to satisfy.