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[ed boy]

If you have f(n) = 2^n-1, what is lim[n->0, f(n)^n] then?

f(n) = 2^n-1 for all real numbers n.
Therefore, for all n, f(n)^n=(2^n-1)ⁿ=2¹=2.
So lim[n->0, f(n)^n] = lim[n->0, 2] = 2.

However, note that although lim[n->0, x^n] exists for all x, lim[n->0, f(n)] does not exist.

[MagmaMcFry]

If you have f(n) = 2^n-1, what is lim[n->0, f(n)^n] then?

f(n) = 2^n-1 for all real numbers n.
Therefore, for all n, f(n)^n=(2^n-1)ⁿ=2¹=2.
So lim[n->0, f(n)^n] = lim[n->0, 2] = 2.

However, note that although lim[n->0, x^n] exists for all x, lim[n->0, f(n)] does not exist.

Er, I kinda meant this as a counterexample to palsch's wall of text there.

[ed boy]

Can anyone help?

A basic algorithm for word prediction:
1-Assign each word in your library an integer, starting from 1. Suppose there are n such words in your library
2-Create M, a nxn matrix of zeros. If you want computational efficiency, you'll want a language/module that supports sparse matrices.
3-Calibrate the model with some sample text. Whenever word i is followed by word j, increase the value in row i and column j of M by 1.
And your model is calibrated!

Here are two ways to predict:
1-If we have word i and want to predict the next on, then let S be the sum of all the values in row i of M.
2-Generate a uniform random integer R between 1 and S.
3-Start summing the values in row i again, and as stop as soon as you reach R. The column that you stop in is the predicted word.
Alternatively:
1-Go through row i, keeping track of the highest value so far, H, and how many times it has occured, O.
2-Generate a uniform random integer between 1 and O, R.
3-Go through row i again, counting whenever you find a column with value H. When you reach the Rth such term, that column is the one you want.

Er, I kinda meant this as a counterexample to palsch's wall of text there.

You need to be more careful with your algebra of limits, there. The limit of a product of two functions is only the product of the limits if you can show that all the limits exist and are finite.

[MagmaMcFry]

Er, I kinda meant this as a counterexample to palsch's wall of text there.

You need to be more careful with your algebra of limits, there. The limit of a product of two functions is only the product of the limits if you can show that all the limits exist and are finite.

I used neither a product of limits nor a limit of products. I don't know what you're talking about.

[Skyrunner]

I helped spawn a debate on the properties of 0^0 but I still don't have a conclusive way to show that lim[n->0, (1+1/n)^n]=1...

[MagmaMcFry]

If you replace n by 1/x, you get lim[x->infinity, (x+1)^(1/x)], which obviously converges against 1, since x+1 grows smaller than a^x for all a > 1.

[Jim Groovester]

You need L'Hopital's rule, but you need to fuss around with logarithms and exponents to get it into a form where you can apply L'Hopital's rule.

lim_n->0 (1 + 1/n)ⁿ = lim_n->0 e ^{ln (1 + 1/n)n} = e ^{lim_n->0 ln (1 + 1/n)n}

We can push the limit through the exponent since it's continuous. We'll just look at the expression in the exponent.

lim_n->0 ln (1 + 1/n)ⁿ) = lim_n->0 n ln (1 + 1/n)

We'll let f = ln (1 + 1/n) and g = n^-1, and then use L'Hopital's rule.

f' = 1 / (1 + 1/n) * -1/n²
g' = -1/n²

lim_n->0 f'/g' = lim_n->0 ( 1 / (1 + 1/n) * -1/n² ) / (-1/n²) = lim_n->0 1 / (1 + 1/n) = lim_n->0 n / (n + 1) = 0

Substituting this back in:

e⁰ = 1

[Skyrunner]

If you replace n by 1/x, you get lim[x->infinity, (x+1)^(1/x)], which obviously converges against 1, since x+1 grows smaller than a^x for all a > 1.

Can you expand on the concept? I can't quite see how obvious that is.

@Jim: I haven't learned L'Hopital yet, but my maths teacher thinks he can explain it to me with L'Hoptial's rule. I'll follow up later.

[da_nang]

If you replace n by 1/x, you get lim[x->infinity, (x+1)^(1/x)], which obviously converges against 1, since x+1 grows smaller than a^x for all a > 1.

Can you expand on the concept? I can't quite see how obvious that is.

@Jim: I haven't learned L'Hopital yet, but my maths teacher thinks he can explain it to me with L'Hoptial's rule. I'll follow up later.

If you rewrite it as an exponential expression with base e, the convergence can be seen due to the logarithm growing slower than any power function. Granted, 1/x approaches to zero as x approaches infinity but it approaches it much faster than the logarithm would approach infinity. A comparison would be lim_x->∞ P(x)/Q(x) where P(x) and Q(x) are polynomials and Q(x) is of a higher polynomial degree. That limit will always be zero.

Or with L'Hopital:

Spoiler (click to show/hide)

Differentiable f(x) and g(x) for x in the open interval P\{a}, lim_x->a f(x) = lim_x->a g(x) = 0 v lim_x->a |f(x)| = lim_x->a |g(x)| = ∞

lim_x->a f(x)^g(x) =

lim_x->a e^{g(x)*ln(f(x))} =

e^{lim_x->a g(x)*ln(f(x))} =

e^{lim_x->a ln(f(x))/[1/g(x)]} =

e^{lim_x->a[f'(x)/f(x)]/[-g'(x)/g(x)2]} =

e^{lim_x->a -[f'(x)/g'(x)]*[g(x)2/f(x)]}

Thus if lim_x->a -[f'(x)/g'(x)]*[g(x)²/f(x)] = L and exists, then lim_x->a f(x)^g(x) = e^L

[lordnincompoop]

Can anyone help?

Programming help or representing a stochastic matrix for your second part?

If I'm understanding the logic of your algorithm as you intend it to run, for a given row in your stochastic matrix, you're ignoring all non-maximal entries, so just replace the stochastic matrix with a new one with all non-maximal entries in a row set to zero, and the remaining entries scaled by the appropriate factor.

That's a very good simplification, but... would it really just be sufficient to describe it like that? I was under the impression that such descriptions wouldn't be... rigorous enough. Or that any such operations should be able to be condensed into a function.

Maybe I'm just being stupid?

A basic algorithm for word prediction: <snip>

Thanks, but I already made an algorithm similar to this and implemented it - just need help explaining the last leg of the process. Or is that what you were trying to show?

[Another]

After some thinking and looking around I am now rather convinced that 0^0=1 directly follows from the most common strict definition of exponentiation and can be used without problems.

But. As long as you consider your 0 power an integer number. For real valued powers the definition of exponentiation is different. You may of course say that 0^0.0 is exactly the same as 0^0 and in some cases that would be justified. However if you get 0.0^0.0 as a result of calculus operations (e.g. your integral evaluates to f(x)^g(x) and there is no problem just computing values of f(x) and g(x) at x=0, both being strictly zero) then the value would depend on the nature of functions f(x) and g(x) that were there (those can be innocuous, perfectly defined, continuous at x=0 functions). Here people argued for 980 posts about this problem.

[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).

[Jim Groovester]

That's a very good simplification, but... would it really just be sufficient to describe it like that? I was under the impression that such descriptions wouldn't be... rigorous enough. Or that any such operations should be able to be condensed into a function.

Yeah totally.

I did an internship this summer with a bunch of applied math professors. Nobody would bat an eye if you took an analytical step like this.

[lordnincompoop]

Cool, thanks.

[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.