Mathematics - 2016-12-29 (page 3 of 5)

GFauxPas

16:00

they're both "correct"

no?

DHMO

well

not having a_n x^n seems strange

GFauxPas

it jumps from starting from 0 to starting from 1 to starting from zero, I'm confused

DHMO

what?

z/(1-z)^2 is good

our natural numbers start from 0 also

completely consistent

GFauxPas

would the proof need to change then

Mithlesh Upadhyay

-2

Q: Solving recurrence relation $ T(n/s)+T(7n/10+6)+O(n) $

The solution of the recurrence relation: $$T(n) = \begin{cases} \Theta(1) & \text{if } n \leq 80 \\ T(n/s)+T(7n/10+6)+O(n) & \text{if } x > 80 \end{cases}$$ is: $O(\lg n)$ $O(n)$ $O(n \lg n)$ None of the above Please, note that I posted this question as it is $T(n)=\c...

DHMO

16:05

@GFauxPas i can change the proof

Mithlesh Upadhyay

Why they voted down :(

Secret

Is noting that $\forall x\in S, ax=b \in S$ sufficient to show that the left semigroup action $a$ is surjective in $S$, even if $S$ is infinite?

GFauxPas

DHMO if we're generating $1, 2, 3, \ldots$ then it would "make sense" that the index of summation be $1, 2, 3, \ldots$, no?

DHMO

@GFauxPas and it would also make sense to have a_0 = 1

GFauxPas

yes

maybe

why not start at 1

why not have $a_i = i$

DHMO

16:18

exactly

so we would need z/(1-z)^2

and start at 0

Mary Star

@AlessandroCodenotti We have that $\begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}=\begin{pmatrix}-1 & 0 \\0 & -1\end{pmatrix}=-I$, right?

Alessandro Codenotti

Right

Semiclassical

A power series typically includes a constant term, so it's (usually) natural to include an index of zero.

Alessandro Codenotti

Hm, this set of 2x2 matrices can be thougth of as a vector space over $\mathbb R$, what's its dimension? What's a basis for it? @MaryStar

GFauxPas

commented

DHMO

16:25

@GFauxPas whether we start with a_0 = 0 or a_1 = 1, we are still going to get z/(1-z)^2

GFauxPas

so wheres the $1/$ coming from, is it a mistake?

DHMO

unless you do funny shit like G(z) = sum a_{n+1} x^n

instead of sum a_n x^n

Semiclassical

You'd need to have 1+2z+3z^2+... in order to have 1/(1-z)^2.

DHMO

@GFauxPas can you see it is 1z^0 + 2z^1 + 3z^2 + ... as @Semiclassical said

Semiclassical

two ways to see that: either anti-differentiate term-by-term to get 1+z+z^2+...=1/(1-z) (taking the constant of integration to be 1)

Mary Star

16:27

$\begin{pmatrix}a & b \\ -b &a\end{pmatrix}=a\begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix}+b\begin{pmatrix}0 & 1 \\ -1 &0\end{pmatrix}$

So, a basis is $\left \{\begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix}, \begin{pmatrix}0 & 1 \\ -1 &0\end{pmatrix}\right \}$.

Therefore, the dimension is $2$.

Is this correct?

Alessandro Codenotti

Yup

Semiclassical

alternatively, (1-z)*(1+2z+3z^2+...) = (1+2z+3z^2+...) - (z+2z^2+3z^3+...) = 1+z +z^2+z^3+... = 1/(1-z).

DHMO

or just use (general) binomial theorem

Alessandro Codenotti

So, you have a 2 dimensional vector space over $\mathbb R$ with a basis made by 1 (the identity) and a square root of -1. What does that remind you of?

GFauxPas

commented

$iz^i$ is more natural than $(i+1)z^i$

Semiclassical

16:29

true. i like the second way best, since it emphasizes that the differences between coefficients are constant.

GFauxPas

oh then you cant have a constant term

Semiclassical

eh, you do have a constant term. it's just zero :P

DHMO

@GFauxPas t'as fait quoi?

GFauxPas

I dont speak french

DHMO

what did you do

never mind you reverted your revert lol

GFauxPas

16:33

$Cx^{-1}$

that won't cause any problems

DHMO

@GFauxPas ??

GFauxPas

trying to find a compormise

start at $-1$

DHMO

what the hell are you trying to do

0x^-1 + 1x^0 + 2x^1 ???

GFauxPas

$Cx^{-1} + 0x^0 + 1x^1 + 2x^2 + \ldots$

Semiclassical

...

that's garbage.

DHMO

16:34

how is that a compromise

do you have any idea what you are doing

GFauxPas

my latex is broken

Mike Miller

what's going on here

GFauxPas

I'm just being silly

Semiclassical

your latex doesnt really matter here.

There's no reason to do anything in the realm of Laurent polynomials.

GFauxPas

It was a joke. Went over like a lead balloon :(

Semiclassical

16:35

Generatingfunctionological practice @MikeMiller

Mike Miller

lol

Balarka Sen

I think I might finally be able to fix my sleep pattern today

GFauxPas

Hey DHMO I've been working on your $|W|$ let me show you

Mike Miller

if you say so

Semiclassical

I'd say the solution is simply that the generating function is $z(1-z)^{-2}$.

DHMO

16:36

I concur

Semiclassical

with the reasoning that $z\frac{d}{dz}$ induces $z^n\mapsto nz^n$.

DHMO

and with the reasoning that N starts with 0

and a_i = i+1 seems weird

a_(i+1) z^i also seems weird

Semiclassical

Well, in the context of term-by-term differentiation, the point is that doing so does two things: it gives $n$, but it also lowers the degree by 1. to avoid the second part, one does z d/dz not just d/dz.

that's even more important if one wants to do the GF of the natural squares, since if you do d/dz twice you'd get z^n -> n(n-1)z^(n-2) not n^2 z^n.

in that respect, z*d/dz is the more natural operator than d/dz.

(that's also why it's harder to compute sums of natural powers than sums of consecutive integer products: it's easier to differentiate n times than to operate n times with z*d/dz)

Alessandro Codenotti

@Marystar did you see my message above? I forgot to ping you

Aresloom

has every linear differential equation the generell solution e^λx, if so is it the only one? I.e. is e^{ix},e^{-ix},e^{x} and xe^{x} the basis of the solutions of a linear differential equation?

GFauxPas

16:42

Plenty of work to do with the branch cut

but that's $|W_0(z)|$

with level curves

Semiclassical

Lambert W?

GFauxPas

yes, DHMO asked me to make him a visualization of $|W|$

so I'm trying level curves at first

DHMO

@GFauxPas so that's a 3d graph compressed?

GFauxPas

no, a level curve graph is a graph of $f(v) = c_1, c_2, \ldots, c_n$ constants

Semiclassical

WolframAlpha agrees with that in the upper half-plane: wolframalpha.com/input/…

But not in the lower.

GFauxPas

16:46

uh oh I dont want to disagree with WA lets see what went wrong

Semiclassical

And I'd be suspicious of it too, on the grounds that $W_0(\overline{z})=\overline{W_0(z)}$ and therefore the absolute value should be symmetric across the real line.

Possibly a branch cut issue.

DHMO

@Semiclassical but the contour is continuous

Semiclassical

Contour?

DHMO

I thought the blue lines are the contours

Perturbative

Hey everyone, has anyone read through Munkres' Topology: A First Course fully? If so how long did it take you to complete it? I'm just looking for some sort of benchmark

Semiclassical

16:50

Oh, thought you were saying something about contour integration.

Balarka Sen

Level curves, not contours. And they don't look continuous to me.

DHMO

@BalarkaSen generates a contour plot of $f$ as a function of $x$ and $y$.

Balarka Sen

Yeah, that's not standard terminology.

Semiclassical

Eh, semi-standard.

Balarka Sen

Better choice of word, yes.

In mathematics you mostly call it a level curve. In geography, contours, I suppose.

Semiclassical

16:52

the better phrase isn't contour but contour line

GFauxPas

imaginary parts discarded in coercion

Semiclassical

^ Hrm.

That seems suspect.

GFauxPas

well there's the problem

> absW(4,1)
[1] 1.222854
> absW(4,-1)
[1] 1.215419

Semiclassical

Yeah. The thing I'd noticed is that the bottom half looks like what you'd get from the real part.

ew.

s.harp

Is the spectrum of an unbounded operator "scary"?

I ask because I'm interested in "positive unbounded operators", but I'm a bit afraid of using a definition via the spectrum

Mike Miller

16:56

yes

GFauxPas

ahah, its not my code, it's THEIR code

> lambertW(4+i)
[1] 1.215419+0.134647i
> lambertW(4-1i)
[1] 1.215419
Warning message:
In lambertW_base(z, ...) : imaginary parts discarded in coercion

Semiclassical

yikes.

Secret

Hey guys, I have a function $f: \mathbb{R}\rightarrow \mathbb{R}$ on the reals $\mathbb{R}$ such that $\forall x\in \mathbb{R}, f(x)$ either fixes $x$ or maps to $y \neq x$. This function also have a left inverse $g$ thus $f$ is injective. Is there enough information here to show that it is surjective?

DHMO

@GFauxPas try lambertW(4)

Semiclassical

Here's what I get if I have Mathematica do a contour plot of |W_0| in the upper half-plane and Re(W_0) in the lower half-plane: i.sstatic.net/ehie9.png

DHMO

16:58

@GFauxPas can you make sure that your comment make sense?

GFauxPas

that's my code, you ignored the imaginary part on the bottom half plane?

Mary Star

@AlessandroCodenotti It reminds me of $\mathbb{C}$, where the basis over $mathbb{R}$ is $\{1, i\}$ and it holds that $i^2=-1$, right?

GFauxPas

yup, thats it

Alessandro Codenotti

It is isomorphic to C indeed

Semiclassical

It joins up on the positive real line because the imaginary parts are zero there.

But that's all kinds of silly. Why have a function for the Lambert W function that doesn't handle imaginary parts correctly?

DHMO

17:00

@GFauxPas I mean your comment on proofwiki.org/wiki/Talk:Generating_Function_for_Natural_Numbers

@GFauxPas that's easy to get around; just make a new function that checks if the imaginary part of the argument is negative

and use the W(z') = W(z)'

Semiclassical

@DHMO Or plot |W(x+i|y|)|

DHMO

right

Mary Star

@AlessandroCodenotti Ah ok... And since C is a field, it follows that K is also a field, right?

DHMO

@MaryStar was ist K?

GFauxPas

Oh $W(\bar{z}) = \bar{W}(z)$ ?

Mary Star

17:04

@DHMO $K=\{\begin{pmatrix}a & b \\ -b &a\end{pmatrix} \in \mathbb{R}^{2\times 2} \mid a,b\in \mathbb{R}\}$

GFauxPas

I can't guarantee that it makes sense DHMO because I dont really understand generating functions very well

DHMO

@GFauxPas then understand it

GFauxPas

what's wrong with my comment

DHMO

@MaryStar denn wenn $A \in K$ haben wir $A^2 = (a^2-b^2)I$?

@GFauxPas i have absolutely no idea what you are talking about

GFauxPas

how about now

Bellmondo

17:08

No problem

Bitrex

Hi again MSE :)

GFauxPas

Okay I'm trying another implementation of LambertW

Bitrex

I have been away too long!

GFauxPas

this is open source so there's no quality control

I could submit a bug report to the creator but I probably wont

DHMO

@GFauxPas I don't understand the second sentence

GFauxPas

17:09

there is no second sentence

check again

DHMO

@GFauxPas i meant the second sentence

you deleted the third sentence

GFauxPas

I delted the second one lso

"This implementation should return values within 2.5*eps of its counterpart in Maple V, release 3 or
later. Please report any discrepancies to the author or translator"

I guess I will do that

Secret

0

Q: Necessary and sufficient condition to show the surjectivity of an action if the existence of right inverse is not known

Given a semigroup $S$ with possibly infinite elements , it is known that there is an element $a$ such that its left semigroup action on all other elements $x \in S$ either fixes it or map to a distinct element i.e. $ax=x$ or $ax=b$ where $b \in S$ and $b \neq x$. It is also known that a left inve...

semigroups

Mary Star

@DHMO Wir haben dass $A^2=\begin{pmatrix}a^2-b^2 & 2ab \\ -2ab &a^2-b^2\end{pmatrix}$, oder nicht?

DHMO

@GFauxPas I guess you should just write one yourself

@MaryStar ah, du bist richtig

Astyx

17:14

Hi chat

Mary Star

@Astyx Hello!!

Secret

Astyx: A bit wrong time to be on cause everyone is too busy discussing about the Lambert W function plot

Astyx

@MaryStar How are you ?

Mary Star

I am fine and you?

Astyx

@Secret I'll just watch then :)

I'm fine, thanks

GFauxPas

17:17

what's the workaround you suggested?

W(z*) =(W(z))*?

DHMO

@GFauxPas yes

and even better:

16 mins ago, by Semiclassical

@DHMO Or plot |W(x+i|y|)|

@MaryStar du bist von Deutschland?

GFauxPas

Well I'd want a working function even outside of the plot

DHMO

@GFauxPas I see

if you want a working function

I suggest you write one yourself

wikipedia

wolfram also has asymptotic expansion

Mary Star

@DHMO Halbwegs. Meine Mutter ist von Deutschland. Woher bist du?

DHMO

@MaryStar ich bin von Hong Kong

denn woher ist deine vater?

GFauxPas

17:22

this one automatically switches to asymptotic for large values, and it allows branch choice, id rather not start from scracth

DHMO

@GFauxPas alright

Astyx

I thought that was information you didn't wish to communicate @DHMO :p

Mary Star

@DHMO Grieche

GFauxPas

LW<-function(z,b = 0)
{
  return(ifelse(Im(z)>0,
    lambertW_base(z,b),
    Conj(lambertW_base(Conj(z),b))
    ))
}

magic

Sophie

@GFauxPas what's this language? I don't like it

Astyx

17:24

^

GFauxPas

R. whaat don't you like? I don't have to use ifelse

I can use standard if and else

if you'd prefer

it's like ? in Java

actually I can probably not use a return statement since there's only one line

Sophie

yes that's what's trigggering me I like
if(x) {
y;
else {
z;
}

DHMO

@Astyx I dont care lol

Sophie

thanks for breaking the indent

DHMO

people change all the time

Astyx

17:26

I'm not blaming you, I just find it amusing

DHMO

:p

GFauxPas

LW<-function(z,b = 0){
  if(Im(z)>0){
    lambertW_base(z,b)
  }
  else{
    Conj(lambertW_base(Conj(z),b))
  }
}

better?

Sophie

yes

Astyx

Much better

GFauxPas

R has algorithms to determine when you don't need a return() statement but I don't know exactly what they are

en.wikipedia.org/wiki/%3F: here's the ifelse shortcut which you don't like

Astyx

17:29

What are you trying to do ? @GFauxPas

DHMO

@GFauxPas programiz.com/r-programming/return-function

GFauxPas

except it doesnt include the ":" when I like to wikipedia

oh DHMO so there you go, if the end of the logic is an object it assumes you're returning the object

thanks

TIL

@Astyx this package I downloaded online implements the Lambert W function, but it has a nasty bug where it discards the imaginary partwhen $\operatorname{Im}(z) < 0$

Astyx

Ah right

robjohn

@Secret these are the terms for the series of $e^{2x}$ and $e^{-2x}$...

Astyx

What is a good method used for computing the Lambert W function ? Do we usually use Newton's method ?

DHMO

17:34

14 mins ago, by DHMO

wikipedia

14 mins ago, by DHMO

wolfram also has asymptotic expansion

Secret

robjohn: I see

Astyx

Yeah but that's not very practical is it ?

DHMO

@Astyx why not?

user379685

Could someone help me find a formula for this sum? $\sum_{i=1}^{n}\frac{1}{2n+i}

Astyx

The convergence radius is quite small and the rest seems unnecessarily complicated

DHMO

17:38

@Astyx there are two formulas

one when |z| < 1/e

the other otherwise

Astyx

@user379685 Dont forget $

DHMO

@user379685 e.g. when n=5, it would give 1/11 + 1/12 + 1/13 + 1/14 + 1/15

which I don't think have a closed form

GFauxPas

better

robjohn

5

A: Lambert function approximation $W_0$ branch

Here is a post that I made on sci.math a while ago, regarding a method I have used. Analysis of $we^w$ For $w\gt0$, $we^w$ increases monotonically from $0$ to $\infty$. When $w\lt0$, $we^w$ is negative. Thus, for $x\gt0$, $\mathrm{W}(x)$ is positive and well-defined and increases mo...

Astyx

Thanks

DHMO

17:39

@robjohn @Astyx good luck using the Newton method on $\Bbb C$

robjohn

@user379685 do you want the limit of that sum?

user379685

no

i know how to calculate the limit

i'm interested in the partial sum

Astyx

$\sum_{i=1}^{n}\frac{1}{2n+i}$

$H_{3n} - H_{2n}$

DHMO

$\displaystyle \sum_{i=1}^{n}\frac{1}{2n+i}$

Astyx

Don't think there's much more to it

robjohn

17:41

@user379685 I can give you an asymptotic formula, a function in terms of special functions, what is it you want?

DHMO

@robjohn you replied the wrong dude

and i'm interested in the formula

Astyx

The simplest way to compute it, efficiently

user379685

@Astyx hmm i see thanks

Astyx

But I think your link answers my question, thanks again @robjohn

user379685

Is there a formula for the sum of consecutive square roots?

Astyx

17:47

I doubt it

But I might be wrong

You can get an asymptotic equivalent easily enough

DHMO

@Astyx do you know the asymptotic formula for H_3n - H_2n?

(except 1.5, that is)

@user379685 look, we don't have closed forms for everything

Astyx

I don't know it off by heart no

But it's easily derived from the one of $H_n$

DHMO

12

A: Is there a partial sum formula for the Harmonic Series?

No, there is no nice closed form for the harmonic numbers. There are some very accurate approximations that are easily computed; $$H_n\approx\ln n+\gamma+\frac1{2n}-\frac1{12n^2}$$ is quite good, where $\gamma\approx 0.5772156649$ is the Euler-Mascheroni constant.

user379685

@DHMO it's because there are none or noone has found any?

DHMO

alright

Semiclassical

17:49

$H_n \sim \gamma + \ln n$, so $H_{3n}-H_{2n} \to \ln(3/2)$

DHMO

@user379685 I've no idea

Astyx

Well obviously there is one

But it's not getting simpler than a sum of $n$ terms

But that's a closed form

robjohn

$$\log\left(\frac32\right)-\frac1{12n}+\frac5{432n^2}-\frac{13}{31104n^4}+\frac{‌95}{1679616n^6}-\frac{1261}{80621568n^8}+\frac{5275}{725594112n^{10}} +O\!\left(\frac1{n^{12}}\right)$$

Semiclassical

I've actually been on a harmonic sum asymptotics / Euler-Maclaurin formula kick lately

Due to the following problem in this month's American Mathematical Monthly:

GFauxPas

whats AMM?

DHMO

17:52

@robjohn thanks

American Mathematical Monthly

The American Mathematical Monthly is a mathematical journal founded by Benjamin Finkel in 1894. It is published ten times each year by the Mathematical Association of America. The American Mathematical Monthly is an expository journal intended for a wide audience of mathematicians, from undergraduate students to research professionals. Articles are chosen on the basis of their broad interest and reviewed and edited for quality of exposition as well as content. In this the American Mathematical Monthly fulfills a different role from that of typical mathematical research journals. The America...

Semiclassical

Compute $$\sum_{k=1}^\infty\left(1+\frac12+\frac13+\cdots +\frac1k -\ln k-\gamma -\frac{1}{2k}+\frac{1}{12k^2}\right)$$

GFauxPas

qh

Balarka Sen

@Semiclassical is that by our resident superstar

DHMO

@Semiclassical heh, easily solved with the formula above

Semiclassical

Actually, no @DHMO

DHMO

17:54

@Semiclassical why not?

Semiclassical

The above series is asymptotic, and if you try to resum it in the above manner you'll get something that doesn't converge

Sophie

@user379685 $H_n=\gamma+\ln(n)+\frac{1}{2n}+\sum_{h=1}^\infty \frac{1}{2hB_hn^{2h}}$ I think

DHMO

@Semiclassical oh

Semiclassical

I think I've managed to compute it to my satisfaction, though, using the Poisson form of the Euler-Maclaurin formula.

GFauxPas

do they have crosswords?

Semiclassical

17:55

Not that I know of.

user379685

@Sophie what's the last term?

GFauxPas

was only half series

Semiclassical

heh.

GFauxPas

FREUDIAN SLIP

Semiclassical

They do have problems you can submit solutions to, though.

And this is one of them.

Sophie

17:56

@user379685 the infinite summation? I'm not sure that is correct but it involves Bernoulli numbers

robjohn

@Semiclassical One could leave the $\frac1{12k^2}$ term out of the sum and it would still converge.

Semiclassical

Would it? Hm.

Astyx

Yeah it would

$$\pi^2\over 72$$

Semiclassical

It'd better, come to think of it, since that portion of the sum is certainly finite.

Sophie

@robjohn how is this possible? Because $\sum_k \frac{1}{12k^2}$ converges

Semiclassical

17:59

If $\sum_k (a_k+b_k)$ converges and $\sum_k a_k$ converges, then $\sum_k b_k$ converges as well.

DHMO

@Semiclassical really?

what about conditional convergence?

Semiclassical

...hrm hrm hrm.

point. in the present case I think it doesn't matter---all the terms are positive, so convergence must be absolute---but you're right to call me on that.

Sophie

@DHMO do you have a counterexample?

DHMO

I don't

Astyx

No that's not relevant

The difference of two congering series is a convering series

Or rather a series whiches general term is the difference of the general terms of two converging series converges

Semiclassical

18:03

Yeah. So I was right originally.

Thought there might be a subtlety I was forgetting.

Astyx

There isn't :)

Semiclassical

(the sum of two divergent series, on the other hand, can certainly be convergent. but that's a different matter entirely)

s.harp

that depends obviously on how you would define such a sum

Sophie

$\infty-\infty=\pi$

Semiclassical

I had in mind the rather trivial example 1/n+(-1/n)=0.

Astyx

18:06

$1-1 = 0$ is even more trivial

Semiclassical

point.

Balarka Sen

It depends on how you define addition for two divergent series

Astyx

I think we mean addition of the general term

Semiclassical

the proper statement: the difference in partial sums of two sequences can be a convergent sequence even if the partial sums themselves don't converge.

Balarka Sen

Sure

Kaj Hansen

18:12

bacon

Balarka Sen

bacons are nice

robjohn

@Semiclassical $$\frac12\left(1+\gamma-\log(2\pi)+\frac{\pi^2}{36}\right)$$

Semiclassical

You got that way faster than my ego would prefer :/

Kaj Hansen

Hell yeah @BalarkaSen. Crispy or chewy? I like slightly crispy myself.

robjohn

@Semiclassical It helps to realize that $$\sum_{k=1}^nH_k=(n+1)H_n-n$$

Kaj Hansen

18:14

Good God @Semiclassical, you pretty much perfectly articulated a common feeling of mine

Semiclassical

The way I did it was to write down the Euler-Maclaurin formula with integral remainder term, show that said remainder term essentially is the individual term to be summed, and then write said integral remainder as a sum of terms.

That lead to a series which mathematica was perfectly happy to resum.

robjohn

@Semiclassical Eww... that sounds messy.

Semiclassical

Lol.

Hence why it took me a while.

robjohn

Let
$$
S_n=\sum_{k=1}^n\left(1+\frac12+\frac13+\cdots +\frac1k -\ln k-\gamma -\frac{1}{2k}+\frac{1}{12k^2}\right)
$$
Since
$$
\begin{align}
\sum_{k=1}^nH_k
&=\sum_{k=1}^n\sum_{j=1}^k\frac1j\\
&=\sum_{j=1}^n\sum_{k=j}^n\frac1j\\
&=\sum_{j=1}^n\frac{n-j+1}j\\
&=(n+1)H_n-n
\end{align}
$$
we have
$$
\begin{align}
S_n
&=(n+1)H_n-n-\log(n!)-n\gamma-\frac12H_n+\frac{\pi^2}{72}+O\!\left(\frac1n\right)\\
&\sim\left(n+\frac12\right)\left(\log(n)+\gamma+\frac1{2n}\right)-n-\frac12\log(2\pi n)-n\log(n)+n-n\gamma+\frac{\pi^2}{72}+O\!\left(\frac1n\right)\\

Semiclassical

hnnngghh.

that's good.

Astyx

18:16

Haha

Semiclassical

The only advantage for the Euler-Maclaurin approach is that it might generalize more easily.

Say, to partial sums of 1/k^j.

robjohn

@Semiclassical I used Euler Maclaurin to get the expansions of $H_n$ and $\log(n!)$.

Semiclassical

True.

Just using Stirling's approximation for $\log n!$ probably doesn't quite work.

robjohn

Well, I used Stirling to get the $\frac12\log(2\pi)$ as the constant for EMSF

Semiclassical

Right.

Oh well.

I'm not sure I'd have been able to get myself to write up a solution for it, so it's just as well.

user379685

18:25

Do you think this wolframalpha.com/input/… has a finite formula?

Semiclassical

There's still a geometry one in this month's problems that I'm interested in; I've got a brute-force analytic geometry solution which works, but it's probably not the most elegant.

robjohn

@user379685 The limit as $n\to\infty$ does, I think

Semiclassical

Agreed. I think you might be able to interpret it as a Riemann sum in that limit?

user379685

It should be interpreted as a Riemann sum but i'm looking for a formula where n is finite

robjohn

@user379685 $\frac\pi6$

Semiclassical

18:30

I think you'll be looking a long time.

robjohn

@user379685 I doubt there is one for the finite sum, at least without using some esoteric special function

user379685

:c

Semiclassical

Most I could expect is that EMSF yields some interesting approximations.

Not sure how well that deals with sums that depend on the upper limit of summation, though.

Probably it can be taken care of, but I don't remember.

mathtron999

Hi, does anybody know a good website with plenty of problem solving practice prblems?

Null

math stack exchange

mathtron999

18:37

Yes, but any other ones

Sophie

18:57

how do you prove that $\sum_{k=0}^{\infty}\left(\sum_{i=nk+1}^{n(k+1)}\frac{1}{i}-\frac{1}{k+1}\right)‌=\ln(n)$?

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