what is the difference between this and this, the later looks infinitely well recieved >:( ...
sometimes I feel like people are too jumpy to mark questions as off topic ... >8( .. I'm pissed off ... (growls) .. I was typing a cute little solution :| dang !!! :'(
This question is in continuation to this and motivated to answer this question. If I have an answer to it, then I can prove a special case of this question.
Consider two smooth plane curves $C \equiv (X_C(s),Y_C(s))$ and $S \equiv (X_S(s),Y_S(s))$ represented in arc length parametrization. Curve...
Hey there, does anyone remember this reference-request post, I think it had a bounty, where the OP asked for a copy the some pages of a book, because his copy was unreadbale?
First time in this chat, wonder what is happening here, given that there is this wonderful awesome M.SE. Are people just chatting about soccer and stuff? :)
I was asking this question recently: math.stackexchange.com/questions/837054/… Even gave bounty, but didnt get any answere. Do you know how to improve the question?
In English language, what do you mean by "adding decimals"?
To the English speaker, what is a "decimal"?
If I understand this correctly, Americans say "adding decimals" when they mean adding the fractional parts of two decimal numbers. Correct?
Like with 7.056 + 5.67. Here "adding decimals" refers only to adding 56 thousandths and 67 hundredths? Right?
But the entire number 7.056 is a decimal number, and so is 5.67. No?
I know that deci means ten, and it's a decimal (or base-10) numeral system. So in your world, why are the 5 and the 7 not considered to be "decimals"? I don't understand this...
Hi people! I've just made a question concerning integration of a Gaussian-like function, over some region. @DanielFischer, you answered a very similar question 8 mothns ago, but I cannot find the way to connect them... You'll find the link above. If you'd like, take a look! Thanks~!
ALSO, please feel free to edit the title of the question!
ALSO, feel free to upvote if you believe it's an interesting question! For me it's not only interesting but also absolutely useful for me to go ahead!
@sammyg Your question was about the phrase "adding decimals". In what context is that phrase used? You have just restated part of your previous question.
@skullpatrol So? For a number to be a "decimal number", it doesn't necessarily have to be of the form "7.056" with a decimal point, whole number parts and fractional parts. The number "8" might as well be considered a decimal number. No?
Apparently "decimal" comes from medieval Latin word "decimalis" which means "of tenths". In Latin, "decimus" means tenth, and "decem" means ten. So maybe that's why by "decimal" (at least in early and informal math) we mean the decimal representation of a number with fractional parts.
So how's that? Do you buy that? :) That's a new one for me at least, so "decimal" really means tenths, not ten. I mean in this sense, "decimal" really means a number that has fractional parts. So the number 7.086 would be a "decimal" and a number like 8 would be just a regular (whole) number, or "decem" perhaps.
@skullpatrol So in this sense "decimal number" really means a decimal representation of a number with decimal fractions? The word "decimal" is used as a synonym for "decimal fraction"?
@skullpatrol ok
@skullpatrol I'm afraid they will send me back here, or back to the school bench. :P
@Nico i read the question as, "assuming that $G,G^\prime$ are conjugate in $\operatorname{GL}_n(K)$, are they conjugate in $\operatorname{GL}_n(L)$?" but then I reread the question and it instead asks, "assuming that $G,G^\prime$ are conjugate in $\operatorname{GL}_n(L)$, are they conjugate in $\operatorname{GL}_n(K)$?"
@skullpatrol Collage dictionary? Hmm... not sure what kind of dictonary that is. But I'm gonna guess it's one of those Oxford dictionaries. They are usually considered to be most authoritative ones on English language.
$$ \begin{align} \frac1{k^2n^2(k+n)^2} &=\left(\frac1k+\frac1n\right)^2\frac1{(k+n)^4}\\ &=\frac1{k^2(k+n)^4}+\frac2{kn(k+n)^4}+\frac1{n^2(k+n)^4}\\ &=\frac1{k^2(k+n)^4}+\left(\frac1k+\frac1n\right)\frac2{(k+n)^5}+\frac1{n^2(k+n)^4}\\ &=\frac1{k^2(k+n)^4}+\frac2{k(k+n)^5}+\frac2{n(k+n)^5}+\frac1{n^2(k+n)^4}\tag{1} \end{align} $$ Therefore, exploiting the symmetry between $k$ and $n$, we get $$ \begin{align} \sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2n^2(k+n)^2} &=\sum_{k=1}^\infty\frac2{k^2(k+n)^4}+\frac4{k(k+n)^5}\tag{2}
This question is in continuation to this and motivated to answer this question. If I have an answer to it, then I can prove a special case of this question.
Consider two smooth plane curves $C \equiv (X_C(s),Y_C(s))$ and $S \equiv (X_S(s),Y_S(s))$ represented in arc length parametrization. Curve...
@Chris'ssis Oh, wait. I think I can get that from my answer, too.
@Chris'ssis Is there a nice formula for that with a general positive integer exponent in the denominator, as there is with the standard Harmonic numbers in the numerator?
So we have the famous secretary problem, for which the optimal strategy involves the number $n/e$. And then if we want to maximize the product of positive numbers whose sum is $n$, we again choose $n/e$ equal numbers [ignoring rounding issues]. Are these two problems somehow related?
Doing well, thanks, @Chris'ssis ... getting ready to disappear for several weeks :)
@user60887 The denseness of $\mathbb{Q}$ in $\mathbb{R}$ is pretty much equivalent to the existence of sequences of rational numbers converging to any given real number. How you prove it depends on which definitions you use (and which theorems are already proved).
Well my book states that by definitions A set $S$ of real numbers is said to be dense in $\mathbb{R}$ provided that every interval $I=(a,b)$ where $a<b$ contains a member of $S$. The theorem states the set of rational and set of irrationals are both dense in $\mathbb{R}$ which is proved.
@TedShifrin It looks like both problems reduce to (finding the maximum of) $x^{1/x}$. I haven't really analysed the secretary problem, however, to see how it appears there.
Let $5\mid a$, $\gcd(a,b)=1$, and $b\equiv 2\bmod 5$. How can one show that $\sum_{k=1}^{a}k\lfloor\frac{kb}{a}\rfloor\equiv 2\bmod 5$?
Similarly, can we show that if instead $b\equiv 3\bmod 5$, then $\sum_{k=1}^{a}k\lfloor\frac{kb}{a}\rfloor\equiv 3\bmod 5$?
EDIT:
These appear to be special c...
It's quite technical and there's a lot of notation. I have no doubt that there are people around who can understand it, but I — being far from an expert in this kind of stuff — can barely make sense of it.
@nullgeppetto I have no idea how one could evaluate the integral in the general case. Of course if you have specific data, you can numerically evaluate it, but an analytic solution, if at all possible is beyond me.
@DanielFischer, I was afraid of that. The truth is that I need an analytic solution, which I believe (intuitively) that is nor possible. I'am thing of solving the first problem (that problem you solved $8$ months ago), and then apply Hilbert Spaces theory...
@TedShifrin, yes, that's true, for sure. However, the Greek civilization is much older than the Latin, despite the fact that nowdays Greek people (in general, not only politicians) are extremely unappropriate to handle this legacy...
@TedShifrin, I couldn't agree more with you... It's a very bad luck we have been born as humans... Human is the worst "thing" the planet ever saw. Fortunately, there are Mathematics!
@TedShifrin, it's a long story.. Because of my surname.. It's Tzelepis (ok, it might not make sense). That happed many years ago, I was an innocent child back then. I liked it, though, and I kept it. As for the null, when I grew up, start reading poetry and living, I was impressed by the "Zero"..
@robjohn I can easily compute that last series if I use your work (or better say, that relation) above. By the way, that relation you used to solve my question it is somehow familiar to me, I saw it ... (in a proof by Spivey?)
Let $5\mid a$, $\gcd(a,b)=1$, and $b\equiv 2\bmod 5$. How can one show that $\sum_{k=1}^{a}k\lfloor\frac{kb}{a}\rfloor\equiv 2\bmod 5$?
