@r9m The Beta function would have been my fallback if I couldn't get a recursion. When I saw the simple form the OP gave, I knew that a recursion was possible.
@Chris'ssis @robjohn I was trying this .. and thinking if we can compute the higher powers $$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{3}$$ and $$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{4}$$ ? :-)
I'm having some doubt in real analysis which i actually think that is a doubt regarding first order logic, free-variables, etc ... Could anyone help ? postimg.org/image/vur5jxf7t
A function and its graph completely determine one another, so theres no such a thing as having a graph but not knowing the function it represents, or vice-versa
Nick, i'm telling you completely know the function ... you don't know an algebraic equation which summarizes the action of the function from the domain to the co-domain
for example, by knowing graph in R^2 whose points are (0,1) and (1,2 ), we completely know its function ... its function is f = { (0,1) , (1,2) } which is a subset of R^2 . But now, we can also think of an elegant and quick way to summarize the action of the function from the domain into the codomain : f(x) = x+1 in that case .
@BalarkaSen Doing that without special functions $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_{n+1}^2}{n+2}-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+2}\left(1+\frac{1}{2^2}+\cdots +\frac{1}{(n+1)^2}\right)=\frac{1}{3}\log^3(2)$$
@BalarkaSen You like these questions as far as I know.
@nerdy No, no, imaginary numbers are complicated. For example, the first thing one would imagine after knowing that $\sqrt{-1}$ is not a "number" (i.e., not real) is whether there are other such phenomenon of having an operator act on a number (including $\sqrt{-1}$, i.e., complex numbers) and not giving out a "number".
^ that might be an elementary question, but there is hardly an elementary answer
I've faced this question before from 5th graders and had to tell them that this is highly nonelementary to prove. It's called the fundumental theorem of algebra : $\Bbb C$ is algebraically closed.
even if doesnt fully understand rigorously wether there are other dimensions or not , you will still have introduced imaginary numbers in a natural fashion, making it seem not really strange
IceBoy, no problem.. also check the other articles
By the **Cauchy product**, we have that $$\underbrace{\left(\sum_{n=0}^{\infty}H_{n+1} x^n\right)\cdot \left(\sum_{n=0}^{\infty}\frac{x^n}{n+1}\right)}_{\displaystyle \frac{\log^2(1-x)}{x^2(1-x)}}=\sum_{n=0}^{\infty}\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1} x^n$$ Rewriting $\displaystyle \frac{\log^2(1-x)}{x^2(1-x)}$ as a **Cauchy product**, we have that $$\underbrace{\frac{1}{x^3}\left(2\sum_{n=0}^{\infty}\frac{H_{n}}{n+1} x^{n+1}\right)\cdot \left(\sum_{n=0}^{\infty}x^{n+1}\right)}_{\displaystyle \frac{\log^2(1-x)}{x^2(1-x)}}=\sum_{n=0}^{\infty} 2\sum_{k=0}^{n}\frac{H_{k}}{k…
Do you know any nice way of expressing
$$\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$$
?
Some simple manipulations involving the integrals lead to an expression that also uses
the hypergeometric series. Is there any way of getting a form that doesn't use the HG function?
Hey everyone, I want to ask you about the Gödel's Incompleteness Theorem. My little research tells me the following is true. Could you confirm this?
The Gödel's Incompleteness theorem applies if and only if the logical system is effective (understandable and followable by a computer) and proves/contains as axioms the axioms shown by the Robinson Arithmetic (RA).
effective is not quite that strong (it just means the axioms are recursively enumerable, i.e., you could write a finite program such that a computer could read it and then output the axioms - but this program doesn't need to terminate)
@r9m did you start working on $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_{n+1}^2}{n+2}-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+2}\left(1+\frac{1}{2^2}+\cdots +\frac{1}{(n+1)^2}\right)=\frac{1}{3}\log^3(2)$$? :D
and it's not quite that it contains the axioms of arithmetic (dunno what RA is; I only know the proof as applied to Peano arithmetic), but rather that one can define in the language the appropriate symbols, and then the appropriate axioms as pertains to these symbols, @mathh
that's what's meant by being capable of expressing PA
@MikeMiller I'm not sure what you mean. Do you have in mind that the logical system simply has to represent, not necessarily directly, the axioms of arithmetic? If so, then well we can take the axioms of arithmetic to thus be our theorems the system has proved, albeit indirectly, but I really don't understand what you have in mind.
@mathh e.g., one needs to be able to define the naturals, a successor function, and 0; these do not need to be in the symbols of your theory, and then prove the axioms of peano (robinson?) arithmetic for these symbols. as an example of defining something not in the structure itself, consider the reals with their field axioms. we still can define $x \geq 0$: define $x \geq 0$ iff $\exists v$ such that $v \cdot v = a$, and by addition can extend this to a $\leq$ relation defined on the reals
I'm probably a bad person to explain things coherently but I learned this all from Enderton's book back when I knew it
for a stupid example, imagine your axioms were precisely the peano axioms, but instead of 0 you had $7$, and instead of the successor function you had the retriever function. it's just the same axioms, but with the objects renamed; but it's not precisely peano arithmetic, and one needs to interpret peano arithmetic (in a trivial way: $0 := 7$, $S := R$, etc)
@MikeMiller This still makes me think your point is that you need not have the axioms as theorems/axioms in our logical systems directly and in the exact same symbols used by the RA, you only need to have them true in some, not necessarily direct (could be indirect), sense. In which case my other comment says everything.
@MikeMiller If the axioms are shown to be true or are taken as axioms, only in different symbols instead, they still are true, and I'm not negating this.
And I am not sure which words in my definition of 'effective' you weren't satisfied with. Please elaborate?
'not quite strong', you said. But understandable and followable doesn't look strong to me at all?
@mathh If it is a question worth putting on main, it will get more exposure there, and it will give the people who help you some reputation for answering.
@mathh Well, those aren't precise. I interpreted what you said to be that it was finitely axiomatized for some reason, but still, "understandable and followable by a computer" is not a mathematical definition. Anyway, as robjohn said, you should post this on main - I'm the only one here who's responded and I'm clearly not the right person to help!
@Chris'ssis well, that is what the rest of the proof is trying to show. I wanted to make the steps small so that they are easy to follow. I could have done it in 4 lines with bigger steps.
@Chris'ssis So that people can buy it easily and cheaply? For example, I buy all my books from amazon.com, lol. It has got the best price, shipping aside.
@robjohn The only place with many math books is the university libraries which I don't have free access to, and even they do not have all the math books on amazon.com!
@Khallil Hello! Preparation is over, I had the exam this morning. It went okay, despite some logistical challenges (got there at 8:50, they'd started at 8:30 despite advertising 9:00).
I hate it when that happens. I was on time to one of my exams two years ago, but the damn fools jumped the gun and started 10 minutes early. They insisted that I wouldn't get any time (that I should have gotten) because of my 'lateness'.
So that kind of threw me. And the function that I had to investigate had a couple of $e^2x$ terms in the numerator, and my notes didn't print properly, lots of blanks including the chain rule! So it's touch and go, and I'm frustrated because I'd have sailed through with more organized preparation, but if I should fail, it just means there are some courses I can't take (until I pass this, next opportunity in January)
@Khallil I wasn't the only one, and we argued for more time. But it wasn't good for my focus, nor was being up all night. But again better preparation, and the excuses would have been irrelevant. Buck stops here
Which reminds me... I'm very thankful to the people here for their help, encouragement and patience the last few days. Is there anything I can donate to to say thanks?
Let $S^N$ be a unit $N$-sphere. Let $f:S^N\to\mathbb{R}$ be a function. Let $\bf{\Sigma}$ be a unit $(N-1)$-cell, consider the function $g:S^N\to\bf{\Sigma}$ such that, for any $\hat{a}\in S^N$, $g(\hat{a}) = \hat{\theta} = [\theta_1,\theta_2,...\theta_i..\theta_{N-1}]$, where $$\theta_i = \frac...
@robjohn The last result is not just a random result, it is also one of the main bricks of my elementary evaluation (no use of special functions) of $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{H_n^2}{n}$.
Do you know any nice way of expressing
$$\sum_{k=0}^{n} \frac{H_{k+1}}{n-k+1}$$
?
Some simple manipulations involving the integrals lead to an expression that also uses
the hypergeometric series. Is there any way of getting a form that doesn't use the HG function?
