I don't understand why f(u) has to trend towards 0 as u approaches infinity. Since lim ( ln u / u) is zero, I think f is not required to approach 0. What am I missing?
@@TheEternalVortex42 I mean, at the end of the day you just need to show that all the leftover pieces trend to zero, which is a bit more involved than Michael showed, but the main structure still stands. Thanks!
A more usual approach is use big-O notation instead of this abstract f. Goes to show that this technique is really much more important and usable that it usually is thought of.
okay so i did some math on my own I did some approximations: 2^(1/2x) with 1+ln(2)/2x x^(1/2x) with 1+ln(x)/2x if you do these two the limit becomes: ln(2) + ln(x)/2x * (ln(2)-ln(sqrt(x))) so at infinity the result should be ln(2) (i have not seen the video yet)
@jeremy.N --You are missing needed grouping symbols: 2^[1/(2x)] with 1 + ln(2)/(2x) x^[1/(2x)] with 1 + ln(x)/(2x) If you do these two, the limit becomes: ln(2) + ln(x)/(2x) * [ln(2) - ln(sqrt(x))]
or you just replace u=1/x and expand both exponential functions to first order as exp(-u*ln(u)) = 1-u*ln(u)+O(u²*ln²u). the term u*ln(u) behaves just like u and tends to 0 as u does, so the higher orders will drop. putting that in the limit and cancelling, we immediately get the result
math task: find all minimums and maximum values of an nth-order polynomial equation, ie quintic equation or higher. just start with the 0th order equation, then 1st order linear, 2nd order quadratic. etc.
Slightly confused about what Michael was up to here. He writes u^1/u = e^(ln u / u) - fair enough, but let's call term on the right, e^v where v = ln u / u (= ln ax / ax, where a might be 1 or 2). The point being here that v -> 0 as x -> inf He then, in effect, does e^v = 1 + v + e^v - (1 + v) = 1 + v + O(v^2). He actually pulls out a v^2 from the O(v^2) terms Why have we done this? Is it because we have a previously factored "x" lurking outside the bracketed-off x^1/x terms and v^2 ~ O(1/x^2), so that when we take that x back into the bracket, we get terms of the form xv that are going to contribute meaningfully to the limit plus terms ~ O(1/x) which don't Is that it?
I sometimes wonder about these ever more arcane limits & integrals. Do they have any grounding in reality, or are they just conjured up for the sake of it, so that the chaps can practice their mathy chops.? As the goodly, late Prof James Clerk Maxwell once opined "Hoots Mon, yer aff yer Heid!"
I can answer your question; I'm programming some random walk simulations for physics, and I routinely come across such limits. For example, it might be that I want to take the limiting case where an exponent goes to zero, and at the same time a threshold controlling the step size of the random walk goes to infinity. Plugging in extremely small or large numbers in the simulation would make it too unstable, so I have to calculate the limiting value and use that.
Perhaps some would take that form. One that I saw as an upper bound in a computer science lecture was lim n->∞ n(2^(1/n) - 1) which I recognised to just be ln(2) from the (a^h -1)/h term you get when differentiating a^x
10:51 It seems like Virgina is a good place!
I stopped there once! Truly beautiful
I don't understand why f(u) has to trend towards 0 as u approaches infinity. Since lim ( ln u / u) is zero, I think f is not required to approach 0. What am I missing?
Yes, you're right. It would have been better to call f(u) the whole thing multiplied by (ln u / u)^2, since that entire term does have to go to 0
@@TheEternalVortex42 I mean, at the end of the day you just need to show that all the leftover pieces trend to zero, which is a bit more involved than Michael showed, but the main structure still stands. Thanks!
Yes it didn't have to, all needed was that it was bounded and had a finite limit.
@ davidetassinari -- The quantity of which you're taking the limit needs grouping symbols around the argument of the logarithm: ln(u)/u.
In fact f(u) tends to 1/2! =1/2 as u tends to infinity
A more usual approach is use big-O notation instead of this abstract f. Goes to show that this technique is really much more important and usable that it usually is thought of.
Exactly what I did, you just have a O(ln2(x)/x2) or a o(lnx / x) which is much easier to handle I agree.
4:24 The first term of that sum is (constant) 1/2, so the sum does not approach zero. Arguing that it's bounded is enough for the proof, though.
At 3:45, u^(1/u) is infinity^0 not 1^0... But, ln(u)/u does go to zero as L'Hospital's rule easily shows. Thus, the series does add to 1.
"Almost heaven, west VIRGINIA..."
okay so i did some math on my own
I did some approximations:
2^(1/2x) with 1+ln(2)/2x
x^(1/2x) with 1+ln(x)/2x
if you do these two the limit becomes:
ln(2) + ln(x)/2x * (ln(2)-ln(sqrt(x)))
so at infinity the result should be ln(2)
(i have not seen the video yet)
@jeremy.N --You are missing needed grouping symbols:
2^[1/(2x)] with 1 + ln(2)/(2x)
x^[1/(2x)] with 1 + ln(x)/(2x)
If you do these two, the limit becomes:
ln(2) + ln(x)/(2x) * [ln(2) - ln(sqrt(x))]
or you just replace u=1/x and expand both exponential functions to first order as exp(-u*ln(u)) = 1-u*ln(u)+O(u²*ln²u). the term u*ln(u) behaves just like u and tends to 0 as u does, so the higher orders will drop. putting that in the limit and cancelling, we immediately get the result
It's in fact necessary to first show that f(x) is bounded as x grows, and then that (ln x)^2/x -> 0.
math task: find all minimums and maximum values of an nth-order polynomial equation, ie quintic equation or higher. just start with the 0th order equation, then 1st order linear, 2nd order quadratic. etc.
No time for the word "stop"? Why cut so early?
Slightly confused about what Michael was up to here.
He writes u^1/u = e^(ln u / u) - fair enough, but let's call term on the right, e^v where v = ln u / u (= ln ax / ax, where a might be 1 or 2). The point being here that v -> 0 as x -> inf
He then, in effect, does e^v = 1 + v + e^v - (1 + v) = 1 + v + O(v^2).
He actually pulls out a v^2 from the O(v^2) terms
Why have we done this?
Is it because we have a previously factored "x" lurking outside the bracketed-off x^1/x terms and v^2 ~ O(1/x^2), so that when we take that x back into the bracket, we get terms of the form xv that are going to contribute meaningfully to the limit plus terms ~ O(1/x) which don't
Is that it?
Hi,
10:52 : missing "top".
This one was all over the place and not well explained. Sorry Michael.
that's a good place to what?
0
it took me 12"
I sometimes wonder about these ever more arcane limits & integrals. Do they have any grounding in reality, or are they just conjured up for the sake of it, so that the chaps can practice their mathy chops.? As the goodly, late Prof James Clerk Maxwell once opined "Hoots Mon, yer aff yer Heid!"
Maybe not in real life, but in other sciences, even finance, they do pop up
I can answer your question; I'm programming some random walk simulations for physics, and I routinely come across such limits. For example, it might be that I want to take the limiting case where an exponent goes to zero, and at the same time a threshold controlling the step size of the random walk goes to infinity. Plugging in extremely small or large numbers in the simulation would make it too unstable, so I have to calculate the limiting value and use that.
@@Nikolas_Davis The good Prof would have approved of your Maths being at the service of Physics!
Perhaps some would take that form. One that I saw as an upper bound in a computer science lecture was
lim n->∞ n(2^(1/n) - 1)
which I recognised to just be ln(2) from the (a^h -1)/h term you get when differentiating a^x
Could you have used Generalised Puiseux Series?
1 min ago lets go!