$$\lim_{x \to 0} \left[\tan \left(\dfrac{\pi}4 +x \right)\right]^{\frac1x}$$

I have simplified it to:

ew

$e^{\lim_{x \to 0} \dfrac{1}{x} \left(\tan\left(\dfrac{\pi}{4} +x\right)-1 \right)}$

yeah, that's a good call

@Semiclassical what do i do next?

It's in $0.\infty$ form

It helps to write $f(x)=\tan(x+\pi/4)$

Okay.

Then?

note that f(0)=tan(pi/4)=1

then the limit is of the form $\displaystyle \lim_{x\to 0} \frac{f(x)-f(0)}{x}$

Which should look familiar.

Heya @Eulb

Hiya @BalarkaSen

Unable to follow @Semiclassical

Unable to follow why it's of that form, or what to do next?

(@Eulb is a compatriot from hbar who has plans to hang out in this chat now)

@Semiclassical Familiar to?

familiar from differential calculus.

@Semiclassical limits is taught before differential calculus ...

Any other method to evaluate it?

So they're having you do this limit before doing derivatives?

Yes

Well, that's irritating

Just to be clear where this would go: by definition, $\displaystyle f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}$

So you'd evaluate $\frac{d}{dx}\tan (x+\pi/4)=\sec(x+\pi/4)^2$ at $x=0$, i.e. $\sec(\pi/4)^2 = 2$

I dont want to do it that way...

Fair enough. I just wanted to make the destination clear

namely, to show that $\frac{1}{x}(\tan(x+\pi/4)-1)\to 2$ as $x\to 0$

Not sure about the best way tbh. One starting point would be to do the angle-sum formula for tangent

which I never remember tbh

i know it, let me see.

doesnt help

eh, it does. but it takes a bit of doing

I get;

$e^ {\lim1/x (2tan x/(1-tanx))}$

Do you want me to use taylor expansion of tan x then

Maybe focus on just what $\tan(x+\pi/4)-1$ simplifies to

You're allowed to use Taylor series, but not derivatives?

@Semiclassical Yeah

...huh

thats wierd

Strange, i know :P ...

normally thats a few course later

from when u learn limits

the problem with doing the taylor series is that you need the taylor series around x=0

Is the answer e

should be e^2 I think

But, here's how to use the sum formula

you've got $$\tan(x+\pi/4) = \frac{\tan x+\tan (\pi/4)}{1-\tan x\tan (\pi/4)}=\frac{\tan x+1}{1-\tan x}$$

i did that only ,... then subtracted 1

yeah. That gives you $\tan(x+\pi/4)-1 = \frac{2\tan x}{1-\tan x}$

at this point, you can note that as $x\to 0$ then $\tan x\to 0$

Which means the denominator just goes to 1

@Semiclassical Then taylor series for numerator only?

to cancel the xs

Yeah. You end up with $\frac{2\tan x}{x}$

alternatively, you have $\tan x=\sin x/\cos x$, and the denominator goes to 1 as x->0

$e^{\lim_{x\to 0}{2+2x^3/3 }/(1-\tan x)}$

so you've basically just go $\lim_{x\to 0} 2\dfrac{\sin x}{x}$

= e^2

Yeah

The thing I'd stress is that, ultimately, the limit can be boiled down to $\lim_{x\to 0}\frac{\sin x}{x}=1$. How one computes that limit boils down to how sin(x) is defined.

@Semiclassical is my method correct?

looks fine

If you define sin(x) through its Taylor series, then sin(x)/x -> 1 as x->0 is automatic

$\lim_{x\to \infty}\dfrac{f(x)}{g(x)}$, f(x) and g(x) are polynomial functions

Can you tell me why: if degree of f(x)= degree of g(x), then the limit is the ratio of the leading coefficients of f(x) and g(x)

try factoring out x^degree from both top and bottom, and considering the resulting functions

These results are intuitive^

@Semiclassical What to consider about them, they are still in $\infty/ \infty$ form

no, they're not

take (x^2+2x)/(x^2-1) for instance

then?

then do what I suggested.

right, got it!

mmkay

that's obviously a specific example, but the same principle holds

$$\lim_{x\to 0}\left\{\dfrac 2{x^3}(\tan x- \sin x )\right\}^{2/x^2 }$$

ew

But my guess is that the Taylor series approach is the smart one here

@Semiclassical Not really

@Semiclassical you dont like these questions :p ?

can't say I do

they seem tedious without being interesting

^

define "interesting" @mercio @semi

why should they

they don't seem to give insight into new methods to solve difficult problems

these questions would look tedious to any reasonable person who has done mathematics beyond hard/tedious calculus problems

because you can just hit them with taylor series

After writing the taylor series does the limit become distributive or not?

I mean am I allowed to substitute 0 inside after writing the taylor series and do the cancelling of the x^3s

distributive ?

you are only allowed to do legal steps

@mercio Wrong word. I know. Cant think of the proper one.

I mean $\lim_{x \to a}{(f(x)^{g(x)}} =\lim_{x \to a}{f(x)}^{\lim_{x \to a} g(x)} $

Is this valid here or not?

Because in many cases its not valid.

well what do you think

$\tan(x) - \sin(x) = x^3/2 \cdot (1 + x^2/4 + \mathcal{O}(x^3))$ is the relevant Taylor series here.

I'd also like to know when it is valid and when it is not

@mercio I think its valid?

Because when f(x) is kind of independent of g(x) then we can do that ...

well think again

what is the limit of $g(x)$

$\infty$

then it's not even finite

what theorem are you applying to say that the limit and the exponentiation commute

I dont know ...

we werent taught that rigorously ...

have you tried computing a few values of the thing with small $x$

numerator is tending to 0 for small x

$\tan x \approx x, \sin x \approx x $

what about the denominator

its tending to 0 too

(and $\tan x \approx x, \sin x \approx x$ is not enough to deduce that their difference converges to $0$)

how to go about solving the question then?

taylor series

of?

of everything

never mind.

actually it's hard to compute it for small $x$

you need lots of precision

the Diophantine equation $2x+3y=515$ has infinitely many solutions, but what about the positive solutions? are they also infinite?

\o @CooperCape

hola

what brings you to this degenerate opium den

Felt like being in just one of the known to be bad chats wasn't enough...

jk... just came to see some maths

@Twink x can't be bigger than 515/2 and y can't be bigger than 515/3, so there's limits on how big x,y can be

lol

meanwhile x,y have to be at least one

so there are 85 solutions?

so there's only going to be finitely many possibilities for solutions and so only finitely many solutions

that sounds reasonable

but I'm not sure if they're 85 :(

I started with (256,1)

and 256/3=85.3

you also need y to be odd

since if it were even then 2x+3y will be even as well

but I'd start with 515/3=171 +2/3

so any odd y from 1 to 171 should do

each is of the form 2n+1 from 0 to 171, so n=0 to 85

hence, 86

right! :)

and that is the very problem with the way high school teaches calculus

yo

o. .o

anyone here familiar with stochastic processes?

@DrewBrady maybe... but it's usually best to just ask your question or state your point of confusion - if people can help, they will.

^

suppose I take an Ehrenfest urn (which can be a seen as a discrete time markov chain on d + 1 states) and then make it a continuous time Markov chain by having it transition when a Poisson process with rate parameter d \lambda jumps. How do I compute the probability transition matrix P(t) of this continuous time markov chain?

exponential of something

?

let Q = lim t -> 0 of 1/t (P(t) - I), then P(t) is exp(tQ)

@mercio where did that come from

dragon common sense

mercio, why? also, how do I identify the generator matrix Q?

(if I don't know P!)

well you have P(t1) P(t2) = P(t1+t2) for every t1 and t2

okay.

so?

so if such a limit exists, it shouldn't be too difficult to show that P(t) is exp(tQ)

uuuh

alright. I have notes that show this, but how do I know/identify Q?

anyway there should be loads of literature about this ?

well

you consider a very small time step

for continuous time, dt -> 0, thus P(t) -> 0 too?

and assume that only 1 jump will happen

and ponder very hard about the probability that a jump from k to k+1 or k-1 happens in a very short time

that probability should be proportional to dt (in the limit dt -> 0)

and the proportionality coefficient should be directly linked to your parameter dlambda

so you have to remember what that parameter means

okay but to compute with P(t) = e^(Qt), you have to know what Q is. How do you compute Q?

@AaronHall , nah P(0) is the identity

I realize it is P'(0), but you can't actually use that if you don't already know what P' or P is.

you can if you relaize that you only need to know the limit of (P(t) - I)/t and not allf of P precisely

and you have to use your asumption

about a poisson process with rate parameter d lambda

for you, what does this mean

what is this information trying to tell you

Doesn't this just define a counting process N_t with P(N_t = j) = exp(-lambda t) * (lambda t)^j / j!?

"by having it transition when a Poisson process with rate parameter d \lambda jumps"

what is the probability that such a process jumps, in a time interval dt with dt very small

Well the probability that the nth jump J_n = inf { t > 0 : N_t >= n} occurs at or before time t is the same as the probability that by time t, N_t >= n. i.e., it is 1 - P(N_t < n)

o..o'

?

I am reading the wikipedia page

about poisson process

and I'm wondering where they are hiding the one fundamental thing about poisson processes

that the behaviour of N(t1+t2) conditioned on N(t1) = 0 is exactly the same as N(t2)

so that really, at each little time interval you ar rolling a dice and having an small probabiliy that "something happens"

and you just kinda keep rolling that dice

and that as that time interval goes to 0

that probability is equivalent to lambda * dt

hm

yeah thats the memoryless ness

which is what you want

a poisson process of parameter lambda is telling you "every very small interval, there is probability lambda * dt that something happens"

now the actual construction of such a process and showing that it satisfies that and the memoryless property

is a bit technical

but it's not the important thing

what do you think the matrix Q is going to look like ?

what would P(t) look like for t very very small

a completely different (and atrocious) approach is to write P(t) as P0 * probability that 0 jump have happened + P1 * probability that 1 jump has happened + P2 * probability that 2 jumps have happened + ..., where Pn is the transition matrix for n jumps

and then you are halfway through this replacing the probabilities that n jump have happened with their complicated expressions from the wikipedia page

that you realize you are writing down exactly the power series for exp(tQ)

I ranted again

when somebody votes down your post, is there a way to find out who did it and whhy?

not really

that does not seem fair or right to me

if somebody does not like me they can simple vote down all my posts

if somebody did that it would be caught by robomoderator and reverted

I see

by the way, here is a link to my post: money.stackexchange.com/questions/95659/…

It is not math releated

is any1 here

I am

u know some measure theory?

sorry, no

I know some measure theory, but I'm no expert

okay this is a simple question

it is well known that every nonnegative measurable function can be approximated by a sequence of simple functions, yes?

yes

@MatheinBoulomenos hi!

in the proof of it

i can never understand why we need the measurability of f

@LeakyNun hi

where is this ever used?

the idea is to break the image of f into intervals

@MatheinBoulomenos what is so special about R that every symmetric R-matrix is diagonalizable?

and then define simple functions to be those images on those particular sets

so why do we need the measurability of f?

@Hawk so the construction I've seen was something like this

Let $A_{i,n} = f^{-1}([\frac{i-1}{2^n},\frac{i}{2^n}))$

And let $B_n = f^{-1}([n,\infty))$

Then $s_n = \sum_{i=1}^{n2^n} \frac{i-1}{2^n}\chi_{A_{i,n}} + n\chi_{B_n}$

You may have seen something a little bit different, but at the end of the day you take preimages of certain intervals under $f$ and then linear combinations of characteristic functions of those intervals

right...but why do we need to assume f is measurable?

The reason we need measurability of $f$ is that we require that if we write a simple function $s = \sum_{i=1}^n a_i\chi_{A_i}$, that the $A_i$ are measurable

@Hawk because you want your functions to be simple, i.e. the characteristic functions need to be measurable, i.e. the sets need to be measurable

Sniped

sure

(Also hello @Leaky and @Mathein)

So if we drop the measurability from both f and s

your functions wouldn't be simple

we can surmise that there exists increasing (s_n) -> f?

your functions would be linear combination of characteristic functions of non-measurable sets

using the same construction?

*possibly non-measurable

sure

so why do we sometimes say "simple measurable functions"?

if simple functions have to be measurable?

eh... google only gives me simple functions

Yeah, I've only seen them called simple functions

You ask for measurability because you're trying to define integration

So $s = \sum_{i=1}^n a_i\chi_{A_i}$, then you define $\int s = \sum_{i=1}^n a_i\mu(A_i)$

If the $A_i$ are not measurable... rip