i don't know this site's homework-tagging rule, but this asker says "i have thought of the following formula", when a picture of a theorem (not a formula) is displayed...and that theorem comes from some notes online https://www.google.com/#q="theorem 2.1 suppose a starts"
@robjohn as regards $1-\frac1{4n^2\log\left(1+\dfrac{k^2}{n^2}\right)}$, I only used the fact that $\log(1+x)\le x$ for all $1\le k \le n$. I showed this what the DCT for product, but I didn't consider $k>n$. Am I correct?
@robjohn the point I don't get is why I should consider the cases $k>n$ since it's about $\prod_{k=2}^{n}$.
@robjohn Since $\lim_{n\to\infty} \sum_{k=2}^{n }\frac1{4k^2} <\infty$, the theorem may be applied and we immediately get the desired answer. I don't manage to see where I could possibly be wrong.
@robjohn I former teacher told me: "you considered $1\le k \le n$, but we should also take into account $k>n$." I think he is wrong, but I cannot get in contact with him now and clarify these things.
@Chris'ssis to use either monotone convergence or dominated convergence the function must be monotone on the whole domain, not just the part where it is not $0$. This is why I noted that each term is greater than $1-\dfrac1{k^2}$
@Chris'ssis you made the estimate there that $a_{n,k}\le\frac2{k^2}$, and although you said it was for $1\le k\le n$, it is true for all $n$ if we set the terms to $0$ when $k\gt n$.
I read in my calculus book that "Partial sums of Taylor and Maclaurin series can be used as polynomial approximations to more complicated functions." I have tried this but they fail, in particular for the LambertW function.
@robjohn But If I only have these power series, there is not much I can do to evaluate them for higher values $n$ in $x=\frac{n-\frac{11}{8}}{\exp (1)}$
@robjohn I think I got the whole point. So, both Tannery and DCT tell us that we need to somehow "break" the independence between $n$ and $m_n$ in order to place the limit inside the summation. This is possible only if we consider a domainated functions, let's say, from $k=1$ to $\infty$ and NOT from $k=1$ to $\lim_{n\to\infty} n$ or $\lim_{n\to\infty} n^2$ or any other example.
@robjohn All these statements were a little bit confusing to me, but now things become clear.
@robjohn it took me a bit to clarify these things since I thought I know the point but I was on the wrong track. Well, sometimes it happens (especially when you learn things on your own).
what is the most generalized algebraic structure that the space of real sequences is an example of? We have addition and multiplication in the natural sense.
@robjohn this is also a bit troublesome and I'll tell you why imagebin.org/272596
@robjohn in the last line we have $$\sum_{k=1}^{\infty} b_k =\sum_{k=1}^{\infty} \lim_{n\to\infty} a_{n,k}=\lim_{n\to\infty}\sum_{k=1}^{\infty} a_{n,k}$$
@robjohn in exercises we usually meet the form $\lim_{n\to\infty} \sum_{k=1}^{m_n}$
@robjohn I would have written $$\sum_{k=1}^{\infty} b_k =\sum_{k=1}^{\infty} \lim_{n\to\infty} a_{n,k}=\lim_{n\to\infty}\sum_{k=1}^{m_n} a_{n,k}$$
or as in Tannery's theorem order $$\lim_{n\to\infty}\sum_{k=1}^{m_n} a_{n,k}=\sum_{k=1}^{\infty} \lim_{n\to\infty} a_{n,k}=\sum_{k=1}^{\infty} b_k$$
@robjohn after the first equal sign there is no more dependence between $n$ and $m_n$.
Thanks, it's a CAD problem but it boils down to just maths I think. I have a sphere that I need to rotate, and the program I use needs me to give the rotation about 3 fixed axes. This'd be easy if I happened to want to rotate about one of those axes, but I don't.
so I need a function to transform the smooth rotation about my axis into rotation about the 3 fixed axes
(if I've used any technical terms, assume I didn't mean them technically, I don't know much math)
@robjohn This seems to be the key sentence relevant to my problem: " In these applications it is very difficult, if not impossible, to find the function itself" and: "However, there are methods of determining the series representation for the unknown function." (copy pasted from the page you sent as a link)
$f(x) = \sum _{n=1}^{\infty } \frac{x n^{k n} (-x)^n (k!)^n}{(n+1) (k n)!}+x+1$ should be the function if I integrated correctly
OK, easy question: what's a good, free computer algebra system for doing symbolic calculations? In particular, I want to be able to look at quotients of free algebras by specifying relations. Something easy to use would be highly appreciated: I don't have much experience using CASs.
@Chris'ssis Sorry, I slept then went to the park. If we simply define $a_{n,k}=0$ for $k\gt m_n$ and extend the summation to $\infty$, I think this is simply DCT.
@JackM I would have tried to program something in python, but implementing symbolic multiplication seemed hard. There is probably a module out there though...
@Shahab The most general? For sure it's a real vector space, but more generally it's an abelian group. More generally it's a semigroup. Are you sure you don't mean "most specific"?
k is the variable: Clear[n, k, x, y, yy] k = 1; a1 = 1 + x + Sum[(n^(k n) (-x)^n x (k!)^n)/((1 + n) (k n)!), {n, 1, k*80}];
x = (1 - 11/8)/Exp[1]; N[2*Pi*Exp[1]*a1, 20]
x = (2 - 11/8)/Exp[1]; N[2*Pi*Exp[1]*a1, 20]
But I don't know what this function: $$f(x) = \sum _{n=1}^{\infty } \frac{x n^{k n} (-x)^n (k!)^n}{(n+1) (k n)!}+x+1$$ is, other than that it is exponentiated Zeta function limits of truncated Dirichlet series for logarithms of n.
*Integrated.
k determines where to truncate the Dirichlet series. k goes to infinity is a non truncated Dirichlet series.
@robjohn the main idea is this: if we can dominate the $a_{n,k}$ by $b_{k}$ and our summation is dominated by $\sum_{k=1}^{\infty} b_{k}<\infty$, then we can go your way and write that $a_{n,k}=0$, for $k>m_{n}$ because when $n$ goes to $\infty$ , $m_{n}$ goes to $\infty$, and thus you can choose any $m_{n}$ because things remain the same. Therefore, one can safley choose $\sum_{k=1}^{\infty} \lim_{n\to\infty} a_{n,k}$.
@MatsGranvik Most of the plots are outside of the radius of convergence, according to my calculations, so they are simply blowing up too slowly to show in the number of terms you are using.
Anyone here good with calculus and finding limits? I am only just beginning calculus, and I am having a hard time finding the limits of trigonometric functions. For more details, you can see my question here. math.stackexchange.com/questions/513562/…
This is an example that doesn't entirely fit to our theorem but we can use it to see that there are some case when $m_n$ cannot be neglected. So $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n+k}=\log(2)$ and $\lim_{n\to\infty} \sum_{k=1}^{2n} \frac{1}{n+k}=\log(3)$ and so on. Our theorem guarantees that we avoid such cases by the dominated summation.
@Chris'ssis I didn't say that we can simply ignore the limits of summation, I said that in the cases where we want to apply DCT, we should (or need to) be able to ignore them.
@Chris'ssis dominated summation is dominated integration. They are identical
@robjohn I accidentally stared a message. But knowing the radius of convergence that you gave above makes it even more interesting. Keeping Andre LeClaires 11/8 in the formula, the approximation tends to get ever closer to the first zeta zero. Not very close though. But closer.
That is as long as the expression is within the radius of convergence.
No that is not entirely right said of me. Getting too close to the radius of convergence makes it fail.
Clear[n, k, x, y, yy] k = 8; a1 = 1 + x + Sum[(n^(k n) (-x)^n x (k!)^n)/((1 + n) (k n)!), {n, 1, k*80}];
x = (1 - 11/8)/Exp[1]; N[x, 12] N[2*Pi*Exp[1]*Exp[Log[a1]], 20]
x=-0.137954790439 Zeta zero approximation: 14.223359028041127982
If I post a question at math.stackexchange.com, then close the window, will I be notified on the StackOverflow page I have open if that question gets a reply, right?
I don't really know how the stack exchange sites work together.
@JackM Not too late, thanks to notifications :) OK, I thought sage might be recommended. I've already found some documentation that seems to suggest it does the sort of thing I want... thanks!
@PedroTamaroff after some manipulation, like taking log of the product, then it should be written as $$ e^{\displaystyle \int_0^{\ln(2)} (1-e^x) \ dx}$$
@PedroTamaroff because it can be viewed as a Riemann sum.
At any rate, I'm too tired now to think further. Preparing here to go to sleep.
I'll create these days a pack with questions like this one.
@DanielR I didn't leave yet. I wanna tell you the answer is $3/e$.
@DanielR and if we want to prepare the product for a contest this year, all we have to do is to rearrange the expression under the product sign and get $2013/e$.
We can write things as $$\lim_{x\to1^{-}}\sum_{n=0}^{\infty}x^n(\log(1+x^{n+1})-\log(1+x^{n}))$$ and then note $u_n(x)=\log(1+x^{n})$ that leads us to $$\int_0^{\log(2)} (1-e^u) \ du$$
@robjohn then, the other version yields $$\lim_{x\to 1^{-}} \prod_{n=0}^{\infty}\left(\frac{1+x^n+x^{n+1}}{1+x^{n-1}+x^n}\right)^{\large x^n+x^{n-1}}= \int_0^{\log(3)} (1-e^u) \ du$$
@robjohn @Chris'ssis Is there some method you use to determine that these limits should have simple analytical results? I FIGHT and FIGHT with my limits and integrals to find closed-form answers and frequently there aren't any.
@KevinDriscoll There is no telling if a complicated limit is going to have a closed form. The ones that Chris's Sis comes up with are not your random limits. They are developed with the answer in mind, usually.
@KevinDriscoll Or at least an idea of why the limit should exist
Also I doubt anyone works for Wolfram here, but they just need to remove the MeijerG 'function' from the software altogether. It is completely useless.
