Suppose the sequence $(U)_{n\ge 1}$ defined by $U_n = \int_0^1(1-x)^n\exp(2x)\,dx$.
How can I prove that:
$\forall n\in \mathbb{N}\setminus{0} \quad \frac{1}{n+1}\le U_n\le \frac{\exp(2)}{n+1}$.
$\forall n\in \mathbb{N}\setminus{0} \quad2U_{n+1}=(n+1)U_n-1$ and conclude the limit.
I tried to...
@pourjour I don't know if it is useful, but this integral equals $$n!\lim_{t_{n+1}\rightarrow1}\int_0^{t_{n+1}}\cdots\int_0^{t_1} e^{2t_0}\,dt_0\cdots\,dt_{n}$$
@amWhy i recommend getting MSE on your phone. it's like a nicotine patch. having to type MathJAX on a tiny keyboard will really make you ask yourself which posts are worth it.
@amWhy every time I think I should quit MSE, it's always immediately followed by "but I'm not sure how- hey, maybe that would make a good meta post!- ugh, damn it."
Hey, if there are any teacherly types around with enough experience to figure out how to point the asker of this question in the right direction, that could be a good thing: math.stackexchange.com/questions/451457/…
They seem a bit more confused about the basic concepts than I'm able to deal with at the moment.
there is a certain amount of logic you need to grasp on an intuitive level before you learn any kind of math. beyond that you shouldn't need anything special to start learning abstract algebra.
Hello, I hate to bother you guys, but I'm having a lot of trouble understanding the answer to a question I asked, but I can seem to put it in words very well. Could someone please offer me some clarification?
@TobiasKildetoft math.stackexchange.com/questions/450967/… - the main point I'm having trouble with understanding in the whole epsilon delta thing is why delta does not have to approach zero as epsilon approaches zero - doesn't that mean that x doesn't converge to c as f(x) converges to L?
The answerer mentions something about a monotonic delta(-) function, but to be honest I'm not in an analysis course of any sort, and I'm finding his mathematics hard to understand
In my quite non-rigorous calculus course I was given the common "intuitive" approaching values definition, but I can't link it to the epsilon-delta definition
I mean, if δ(ϵ) is an increasing function, how can the limit converge if delta gets bigger and bigger?
@inkyvoyd because delta is just something we need to make sure exists with certain properties
if we pick $\delta(\varepsilon)$ in the "best" way, then this measures how close around the given point we need to look for the function values to all be within $\varepsilon$ of the fixed value (the limit)
but for some functions, we can pick any $\delta$ we wish, no matter which $\varepsilon$ is given to us
@robjohn huh, perhaps Mathematica looks at that series as if it were a continuously differentiable function in $n$ and then perform the differentiation with respect to $n$.
Well, with the "approaching-value" intuition that most calculus courses give, to me it seems that delta must approach zero as epsilon approaches zero - am I correct, or I am getting something completely wrong?
@Chris'ssis Note that if we write the sum as $$\frac1n\sum_{k=1}^n\frac12\left(\frac{\log(n-k+2)}{\log(k+1)}+\frac{\log(k+1)}{\log(n-k+2)}\right)$$ each term is greater than or equal to $1$, but perhaps it is close to $1$ most of the time and the ends of the range are insignificant.
@inkyvoyd you are missing what the meaning of the $\delta$ is
it is how close we "need" to go to the point in order to have aproached the limit sufficiently (sufficiently meaning compated to the given $\varepsilon$)
but from some point, it might be that we don't need to get any closer to have approached sufficiently
@Chris'ssis to add that many terms something like Euler-Maclaurin is probably used, and that might be where the error is creeping in. I have to look at my reasoning more closely.
@Chris'ssis I need to convince myself that that is true, and that the same summation method is not fooling both Mathematica and maple
@TobiasKildetoft That statement about delta helped make things a bit more clear to me - do you know of any examples where delta actually increases as epsilon decreases?
I think the squeeze theorem is the best way. Combining Cauchy-Buniakovski-Schwarz inequality + Stirling formula + Stolz theorem, led me to the conclusion that in the right side I get $1$.
@TobiasKildetoft the example that the answerer to my question mentioned a monotonically-nondecreasing function δ(-). Is that something "close" to what I'm searching for?
@inkyvoyd I guess one thing that might be part of the confusion is that if the function does approach some value, then we can actually pick $\delta(\varepsilon)$ to be strictly decreasing and having it converge to $0$
the reason is that if some $\delta$ works for a given $\varepsilon$, the any smaller number will also work.
@TobiasKildetoft In the case of the constant function though, any delta can be picked for any epsilon - but if the deltas chosen are larger and larger, then how can x possible approach c? Or is it rather that as x gets closer and closer to c, the requirements are still always met?
@TobiasKildetoft I think you've given me all the information I need to know at least twiceover. I'll give it a night and read over it again; hopefully something will click. Thank you very much for your time and patience!
When was the background text in the comment field changed? Now it reads "Use comments to ask for more information or suggest improvements. Avoid comments like"
Seems like someone changed the text to something longer than allowed
@JayeshBadwaik Hello! Just an earliesh (in my day) hello. I'm off to my first day at the hospital (I'll be occupied with lots of days at the hospital for tests, treatment, for awhile, so won't be around so much!) Good time for some extra-curricula reading on my part.
@JayeshBadwaik A little (embarrassing) secret: I've got a "kick-ass" laptop still in the box it was shipped in, yet to be tapped (6 months now) - just because of my attachment to my current (weaker) laptop and netbook. Plus, sheer procrastination on the task of transferring data, and installing the loads of software I have!
I have a question. What book were you reading when you were asking about the problems with Riemann Integration? We are doing Riemann Integration now and I would like to have some problems the kind of which you were solving then.
@Batgirl What do you do? Math student? (I saw your answers so you are not trolling.)
I have a question. What book were you reading when you were asking about the problems with Riemann Integration? We are doing Riemann Integration now and I would like to have some problems the kind of which you were solving then. Where the upper limits and lower limits might not converge and similar.
@robjohn did you see Aryabhata's proof here? math.stackexchange.com/questions/307028/…. Every time I see it I feel creeps ... I'm totally amazed by the power of AM-GM. Far, far too beautiful to be true.
@robjohn glad you like it. Actually, Stolz theorem came to my there, but nothing more. When I saw that proof ... wow ... it's like then one sees God for a second. :-)
Suppose the sequence $(U)_{n\ge 1}$ defined by $U_n = \int_0^1(1-x)^n\exp(2x)\,dx$.
How can I prove that:
$\forall n\in \mathbb{N}\setminus{0} \quad \frac{1}{n+1}\le U_n\le \frac{\exp(2)}{n+1}$.
$\forall n\in \mathbb{N}\setminus{0} \quad2U_{n+1}=(n+1)U_n-1$ and conclude the limit.
I tried to...
