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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.

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.

0⁰ is an expression, just like a function. Yes, functions can be discontinuous. Yes, I can set the value of the function for a specific input arbitrarily. Mathematical constructions et al. But what you need to realize is that it's no longer the same function thus no longer the same expression. Thus if you want to find the value of the original expression, you have to work within its realm, its domain. To do that, you need to know its function.

The most general form of that function is h(x) = f(x)^g(x). If 0⁰ = 1 as the context-free value, then for every x₀ ∈ A, where A is an arbitrary set of numbers, such that f(x₀) = g(x₀) = 0 then h(x₀) = 1 no matter the form of f(x) and g(x) and regardless of context. So if you see a limit that evaluates to 0⁰ you'll then simply equate it to 1. But there are many cases where that is not true. What's the value of (e^-1/02)⁰? Yes, 0⁰ would simply formulas. That would be the case in many areas. And yes it might make my life easier, who knows. But I do have a problem with the uncountably infinite number of defined discontinuities as shown by lim_t->0+ (e^-1/t)^at = e^-a that you've now introduced. These are no longer "removable singularities". Unless you want change the definition of that too.

This will have consequences for indeterminate forms. If you can arbitrarily define 0⁰ = 1, why not 0/0 = 1? You're breaking the very foundation of mathematics. A definition must always be valid in its context otherwise it's not a definition. That's what you're doing by arbitrarily defining the general value of 0⁰. Your definition will cease to be a definition. If you're not defining the general value of 0⁰ but instead defining it within a specific context, then fine. That's your prerogative. My point still stands. Don't expect to be able to expand that definition past your context when there are counterexamples ready to shoot it down.

[ZetaX]

I don't see where you gave a counterexample or contradiction caused by 0^0=1. You only gave discontinuities, nothing more; discontinuties you wouldn't even be able to state if not defining 0^0 at all. Please state and elaborate one such contradiction in detail, and do not use anything that is just a discontinuity.

 You last part is wrong as 0/0=1 is simply not consistent with the axioms of a field, but you obviously could define it that way if you consider whatever other structures where it is no problem.

[da_nang]

0⁰ = 1 won't work in real and complex analysis. Your calculated limits would differ entirely from empirical results. Again, lim_t->0+ (e^-1/t)^at is one occurrence. Your calculated limit would be 1 since the expression evaluates to 0⁰ but the graph clearly approaches e^-a. This is entirely due to the context. The entire point I'm making is that the value of 0⁰ is context-sensitive and as such arbitrarily setting a specific value of it for all applications as you suggest might work in some contexts but won't in others.

Or to put it more mathematically:

lim_t->0+ (e^-1/t)^at = (e^-∞)⁰ = 0⁰ = 1 (according to your 0⁰ = 1 definition)
lim_t->0+ (e^-1/t)^at = lim_t->0+ e^-at/t = e^-a

So, are you saying that 1 = e^-a for all a>0?

[ZetaX]

lim_t->0+ (e^-1/t)^at = (e^-∞)⁰

a) That's completely bogus, please do your limits correctly (the second calculation does). You obviously get nonsense if you make a mistake in calculations.
b) Even if it would be correct (which it isn't!), it would still only be a problem of continuity of something.

I have almost ten years of experience working with complex analysis and I never ever encountered a problem with setting 0^0=1.

[Another]

Let f(x), g(x) be  differentiable, non-negative valued functions, f(0)=g(0)=0, f(1)=g(1)=1. What is the value of Intergal[0,1](g(x)^f(x)*(g'(x)*ln(f(x))+g(x)*f'(x)/f(x)) dx)?
The answer is obviously 1-0^0 if you allow 0^0 be universally defined, but it can have any value depending on the functions.

The problem is that while in most practical situations 0^0 could be assumed to be 1 fine, mathematical definitions should be true in all potentially possible situations.

[da_nang]

lim_t->0+ (e^-1/t)^at = (e^-∞)⁰

a) That's completely bogus, please do your limits correctly (the second calculation does). You obviously get nonsense if you make a mistake in calculations.
b) Even if it would be correct (which it isn't!), it would still only be a problem of continuity of something.

I have almost ten years of experience working with complex analysis and I never ever encountered a problem with setting 0^0=1.

So if a limit evaluates to 0⁰ then I should be able to equate to 1 regardless of context? Because that's the gist I'm getting from you.

We can all agree that regardless of branch, ln(0) = -∞ as e^-∞ = 0.
We can all agree that +-1/∞ = 0 on the extended real number line.
We can also all agree that x^1/ln(x) = e for all x > 0 and x != 1.

Since we can arbitrarily define 0⁰ = 1, shouldn't we also be able to arbitrarily define x^1/ln(x) = e for x=0? In which case, 0^1/ln(0) = 0^-1/∞ = 0⁰ = e.

But 0⁰ = 1.

Thus 1 = e. If not, then why should x^1/ln(x) be undefined for x=0 and 0⁰ be defined?

[ZetaX]

So if a limit evaluates to 0⁰ then I should be able to equate to 1 regardless of context? Because that's the gist I'm getting from you.

We can all agree that regardless of branch, ln(0) = -∞ as e^-∞ = 0.
We can all agree that +-1/∞ = 0 on the extended real number line.
We can also all agree that x^1/ln(x) = e for all x > 0 and x != 1.

Since we can arbitrarily define 0⁰ = 1, shouldn't we also be able to arbitrarily define x^1/ln(x) = e for x=0? In which case, 0^1/ln(0) = 0^-1/∞ = 0⁰ = e.

But 0⁰ = 1.

Thus 1 = e. If not, then why should x^1/ln(x) be undefined for x=0 and 0⁰ be defined?

I never said that you should be allowed to do anything with limits. All I said is that you should stop using/applying "rules" of limits you just invented and which are simply wrong (some of them even if we don't define 0^0 at all). You still ignore that I said that any limit argument is not an argument at all as continuity of everything is not a reasonable axiom to begin with.

You again added random rules without justification. These new rules, and _not_ 0^0=1, are what cause a problem. Your argument that "if you can add that rule, I am allowed to add any other" is simply wrong: mine is based on a lot of reasons, compatibility and not random at all, while you suggest defining something that should not really have a definition at all, or similiar things.

By the way another reason for 0^0: m^n is the number of maps of an n-element set to an m-element set. As a 0-element set is simply empty and there is exactly one map between two empty sets, we get 0^0=1.

[da_nang]

By the way another reason for 0^0: m^n is the number of maps of an n-element set to an m-element set. As a 0-element set is simply empty and there is exactly one map between two empty sets, we get 0^0=1.

In that context, sure. But that isn't context-free. It needs to be more abstract. Pure numbers. It's not because we can count the empty function. It's not because it's more convenient since convenience doesn't imply truth. Using counting arguments to define a general value of an expression isn't that helpful. It's too discrete and needs to be proven to hold in continuous contexts as m and n are integers. For example, your standard factorial definition won't work for non-integer values without the Gamma function. This is because permutations, the "counting argument" of the factorial, only really exist in the integer domain. I mean, how many permutations can you form from 6.5 unique objects? Which then begs the question of what 0.5 objects means.

Just to illustrate the importance of context: 1/s != 1/s, where s is second. Why? Because the left-hand side is in Hertz, while the right-hand side is in rad/s. The numerator units are dimensionless. Or another example would be 1 Nm != 1 J. Why? Because one is in the context of torque, the other in the context of energy. So in the mⁿ case, what you're really doing is (m elements)^{n elements} but "elements" is dimensionless thus you don't see the "units".

[Aqizzar]

I feel like this question is going to be both presumptuously lazy and insultingly simple, but I have scoured Google and cannot find a certain equation.  Mainly because searching for any of these terms turns up thousands of high-rated results that have nothing to do with what I'm looking for.  And please understand that I'm not a physicist, I'm just hoping somebody knowledgeable would like to show off, and that I'm not interrupting anything.

What I'm looking for is an equation that can demonstrate the arc of flight of an object tossed into the air at a given speed and angle.  Assuming no wind resistance and only two dimensions, as simple a scenario as I can make it.  I know that the constant of gravity (on Earth) is 9.8 meters per second of downforce, but I've rearranged variables every way I can imagine and have yet to produce a quadratic formula that can show where an arcing object's position from one second to the next.  Although I guess 'time' is actually the third dimension here.

[Karlito]

Forgotten your high school physics, I take it? 

So, if we have a given initial velocity v and an angle to the ground θ, the first thing to do is rewrite it as its vector components. The part of the velocity that's just in the x-direction we'll call v_x and the part of the velocity that's just in the y-direction we'll call v_y. A little bit of trigonometry gives us:
v_x=vcos(θ)
v_y=vsin(θ)

We can then describe the position of the particle with two equations, one for the x direction and one for the y direction.
x=v_x*t
y=v_y*t-4.9t²

If you want more information, the keywords you'll want to search for are "2D Kinematics". I'm sure Khan Academy has some videos that can explain all this stuff.

EDIT:I guess I should note that there's some implicit assumptions in those equations. Namely that the object is fired from the origin (0,0), and probably some other stuff that's so basic I can't bring it to mind.

[Aqizzar]

Thanks a ton man.  I actually never learned any of that before, I took biology in high school.

My attempt to reverse engineer gravitational physics through programmatic arithmetic was obviously doomed to failure.  I guess the only thing I have to worry about here is adjusting the angle and velocity of the particle from second to second so it keeps producing the same curve, but it should be pretty easy to reverse the equation.

[ZetaX]

By the way another reason for 0^0: m^n is the number of maps of an n-element set to an m-element set. As a 0-element set is simply empty and there is exactly one map between two empty sets, we get 0^0=1.

In that context, sure. But that isn't context-free. It needs to be more abstract. Pure numbers. It's not because we can count the empty function. It's not because it's more convenient since convenience doesn't imply truth. Using counting arguments to define a general value of an expression isn't that helpful. It's too discrete and needs to be proven to hold in continuous contexts as m and n are integers. For example, your standard factorial definition won't work for non-integer values without the Gamma function. This is because permutations, the "counting argument" of the factorial, only really exist in the integer domain. I mean, how many permutations can you form from 6.5 unique objects? Which then begs the question of what 0.5 objects means.

Just to illustrate the importance of context: 1/s != 1/s, where s is second. Why? Because the left-hand side is in Hertz, while the right-hand side is in rad/s. The numerator units are dimensionless. Or another example would be 1 Nm != 1 J. Why? Because one is in the context of torque, the other in the context of energy. So in the mⁿ case, what you're really doing is (m elements)^{n elements} but "elements" is dimensionless thus you don't see the "units".

And again you didn't give an argument against 0^0=1... 
You only said that my definition is not sufficient to define it in every context you may think of, which is obviously correct. You said nothing about it being incorrect, only it being incomplete if e.g. considering reals (it is complete on naturals).

And units are irrelevant here. It's more of a fail of some physicists if they do not account for their units consistenly, but mathematics doesn't even have this problem in the first place.

[lordnincompoop]

Forgotten your high school physics, I take it? 

So, if we have a given initial velocity v and an angle to the ground θ, the first thing to do is rewrite it as its vector components. The part of the velocity that's just in the x-direction we'll call v_x and the part of the velocity that's just in the y-direction we'll call v_y. A little bit of trigonometry gives us:
v_x=vcos(θ)
v_y=vsin(θ)

We can then describe the position of the particle with two equations, one for the x direction and one for the y direction.
x=v_x*t
y=v_y*t-4.9t²

If you want more information, the keywords you'll want to search for are "2D Kinematics". I'm sure Khan Academy has some videos that can explain all this stuff.

EDIT:I guess I should note that there's some implicit assumptions in those equations. Namely that the object is fired from the origin (0,0), and probably some other stuff that's so basic I can't bring it to mind.

That's an imperfect solution, since you're presenting separate equations for each coordinate and not an equation for a parabola. You need to go a step further and eliminate the variable t, as follows:

Let v be the initial velocity, t time, a the angle of launch and g the gravitation constant.

We know the following:

x = vt cos(a)
y = vt sin(a) - 0.5gt^2

Two equations for t can be derived from this.

x/(v cos(a)) = t

Then, taking this into account, y can be further expanded into:

y = v {x/[v cos(a)]} sin (a) - 0.5g{x/[v cos(a)]}^2

Simplify:

y = {[x sin(a)]/[v cos(a)]} - {[gx^2]/[2v^2 cos^2(a)]}

Remember that sin()/cos() = tan(). Thus:

y = x tan(a) - {[gx^2]/[2v^2 cos^2(a)]}

Which is the simplified form. This is the final equation for parabolic projectile trajectories taking into account velocity, angle and gravity.

[da_nang]

You only said that my definition is not sufficient to define it in every context you may think of, which is obviously correct. You said nothing about it being incorrect, only it being incomplete if e.g. considering reals (it is complete on naturals).

Isn't that what I've been saying?

Anyways, this debate has raged since the 19th century and I don't think two forumites are going to change anything. Let's settle this amicably and agree to disagree?

snip

Technically you'd also need to consider the special cases v = 0 m/s and a = π/2 +- π*n but that's just nitpicking.

[Mego]

And, if we want to get even more in-depth for Aqizzar's simple-ish question (which we do, since we're mathematicians), let's build from the ground up!

We're going to start by describing the components for the acceleration vector in two different equations:

a_x(t) = 0
a_y(t) = g

We're ignoring wind resistance and all other forces for the horizontal component, so its acceleration is constant. We only care about gravitational acceleration for the vertical component.

The next step is to turn these into velocity equations. Newtonian mechanics teach us that acceleration is the rate of change of the velocity. Slap that a few times with rudimentary calculus, and it turns into acceleration being the first derivative of velocity. So, to get velocity from acceleration, we have to take a derivative.

v_x(t) = ∫ a_x(t) dt = ∫ 0 dt = C_x
v_y(t) = ∫ a_y(t) dt = gt + C_y

C_x and C_y need to be found, now. They are the initial velocities for each component. Since we don't know them, we're going to label them v_0x and v_0y, because that's so much better than big-C notation.

v_x(t) = v_0x
v_y(t) = gt + v_0y

So now, we need position equations. We once more remember that velocity is the first derivative of position, so it's time to apply another integral:

x(t) = ∫ v_x(t) dt = ∫ v_0x dt = v_0xt + C_x
v_y(t) = ∫ v_y(t) dt = ∫ gt + v_0y dt = 0.5gt² + v_0yt + C_y

Our Cs are the initial positions this time, x₀ and y₀:

x(t) = v_0xt + x₀
y(t) = 0.5gt² + v_0yt + y₀

And there we have it, the classic formulas for projectile motion. With lordnincompoop's post, you now have the problem completely worked through, from base principles to a final solution.