Quick question guys. For a function $y=f(x)$ that is rotated about the $X$-Axis from $x=a$ to $x=b$, is the surface area not given by $$ \mathrm{SA} = 2\pi \int_{a}^{b} f(x) dx ?$$
i'll rewrite the question now that i have a handle on the latex thanks
i have in front of me that $( 2 \sqrt(5i)) \times ( -2 \sqrt(5i)) = +20 $ . i don't see it. i expand that to $ 2 \times -2 \times \sqrt(25 \times i^{2}) = -4 \times 5 \times \sqrt(i^{2}) = -20i $ ? where am i going wrong? thanks!
dude chantry -- you are so right. i'm reading the problem wrong. the i is not under the root. thx again
Gosh, I am having hard time understanding this question: For a surface $S$ in Monge form $z=f(x,y)$, find a formula for the area of the piece of surface that lies over a region $D$ in the $XY$ plane.
Can anyone give me their interpretation?
I mean, judging from what has been given, I can only assume it is asking me to generalise a formula for surface area of a surface through projections.
That is, projecting a rectangular piece of area on the normal XY plane to the 3D surface.
is there anything wrong with the following argument? if $g(n)=f(n-1)-f(n)$, then $$\sum_{n=1}^\infty n g(n)=\sum_{n=1}^\infty n\left[f(n-1)-f(n)\right]=\sum_{n=1}^\infty \left[(n+1)f(n)-nf(n)\right]=\sum_{n=1}^\infty f(n)$$
i know i'm probably supposed to make it a limit of partial sums or something but i really can't see how it could possibly go wrong
In Fourier analysis, while talking about pointwise convergence, we generally start with the class of functions called, BV functions (functions of bounded variation), which have a finite total variation.
I'd like to avoid this type of classification as they are more complex and I don't really nee...
@Chris'ssis I think you misunderstood my question. I'm looking for series representation of $\dfrac{\ln(1-x)}{1+x}$ that involving harmonic number, not its integral
like differentiating both sides of $$\int_0^x \dfrac{\ln(1-t)}{1+t}\ dt =\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{n+1}x^{k+n} H_k \left(\frac{x}{k+n+1}-\frac{1}{k+n}\right)$$ ?
@robjohn this morning I derived some results absolutely crazy
@Anastasiya-Romanova by the way, I'd like to see if Cleo is able to do this one without using special functions $$\int_0^1 \frac{\log(1-x) \log(1+x)}{1+x} \ dx$$ If you ask me, I think Cleo is not able to do it like that.
@Anastasiya-Romanova I computed that one with the high school tools only (of course, I used the definition of $\zeta(3)$, but this wouldn't be a problem).
@Anastasiya-Romanova $$\frac{\ln^32}{3}-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$ But as I told you, computing this integral by special functions is not interesting to me. So, knowing the answer is not enough. :-)
After substitution I get this integral $$\int_0^1\frac{\ln(1-x)}{x}dx$$. We can use series expansion to evaluate this one but if you use double integral, then maybe you use this method: math.stackexchange.com/a/891300/133248
The problem is that method is using multivariable calculus, we also need to use Jacobian
I also get this integral $$\int_0^1\frac{\ln(1-x)\ln x}{x}dx$$
I wonder, what kind of double integral did you use to evaluate those two without using Jacobian
What kind of high school in this world taught its student about Jacobian?
I don't care about beauty. The fact that they resembles well-known fractals are much cooler.
The holes at rationals are consequence of diophantine errors : one of the characteristics of algebraic numbers are that they can't be arbitrarily well approximated by rationals.
I have never looked at the math they did there, but to my understanding appearance of fractals aren't unusual. The fractal-like formations are due to polynomial iterations from an accumulation point.
@Anastasiya-Romanova I am just interested on the math.
@Anastasiya-Romanova Sorry, I cannot tell you more now. You'll find that in my book that I'll publish somewhere during the next year if I work very hard.
@Anastasiya-Romanova I gave you some clarifications above. I initially thought you were only referring to the series of $\log(1\pm x)$. I didn't use the series of $\log(1\pm x)$, that is my point.
@DanielFischer let $a, b$ be perfect powers $n^2 < a < b < (n+1)^2$. as cubes grow much faster than sqaures, it is expected that for large $n$ there are no more than a singe cube inside teh interval. WLOG, $a$ is a cube and $b$ is a fifth power. then $rad(a) = O(n^{2/3})$ and $rad(b) = O(n^{2/5})$. Let $c = b - a$ which is $< (n+1)^2 - n^2 = O(n)$. Thus $rad(abc)^{1+\epsilon} = O(n^{2/3} \cdot n^{2/5} \cdot n)^{1+\epsilon} = O(n^{31/30+\epsilon})$.
hello I have a really quick question about Poisson can i hit it?
anyone?
I hope someone eventually will see it: We have 3 mails arriving per hour (with a poisson distribution), 2 hours pass by without receiving mails, what is the expected time that we'll get the 1st mail? My answer is that be cause it is memoryless it will wait constantly 1/3 of the hour ---> 20 minutes, a friend says since it's memoryless and 2 hours pass by that we can't find the time. is any oh these answers correct?
@robjohn yes it is, and we don't care about how much time passes without getting a message,right? even if 10^321 hours pass by the system will be waiting about 20 minutes (?)
@user2692669 If the distribution is as described, the expected wait should not change based upon history. However, one may question the validity of the model if $10^{321}$ hours pass without a single event.
@robjohn can you check if I made any mistakes here ?! :) also I read the nice answer you linked to me today .. I am taking an introductory ANT course this sem .. we are learning to use Euler-Maclaurin summation ! :)
@Chris'ssis For when you're back :Let $f$ be a continuous, decreasing function, with a $0$ limit in $\infty$. Let $\displaystyle x>0,\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t$. Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x),\lim_{x\rightarrow0} xf(x)=A$. What is $\displaystyle\lim_{x\rightarrow\infty}\Gamma(x)$ ?
@WillHunting okay .. I had an introductory course in first sem and another one in 3rd sem .. but I wan't to learn a bit about fourier analysis for my ANT classes :-)
@Hippalectryon Since that sum runs over the space of very large numbers because of $x\to\infty$, you have a high flexibility, that means you can even try to use Cesaro-Stolz Theorem, the special case $0/0$, some clever inequalities, or even make use of some advanced tools like Euler–Maclaurin formula. I'm confident you can do it on your own (I'm in the meeting).
@Hippa and then prove that if the disc < 0 then two roots appear in complex conjugates
Conclude the first part of your problem. And then use the newton symmetric polynomials and the reccurence relation to finish the next part by induction.
@robjohn ya !!! I had the exact same question in mind yesterday when I saw it =) lol .. I was away for classes and it was asked and answered while I was away =P
@WillHunting but they don't trigger the automatic scripts, so the community team doesn't want to intercede. It annoys the mods (at least some of them).
@robjohn the tags for that were (integration) and (inequality) .. I checked the (integral-inequality) tag but didn't check the inequality tag .. that's how I missed it :P
@WillHunting 'how to upvote without triggering the script' ?! :O what is that ?
@Hippalectryon How about if I let the variable change $t\mapsto t x$ in $$\displaystyle \gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t$$? What's next then?
Let $f$ be a continuous, decreasing function, with $\displaystyle\lim_{x\rightarrow\infty}=0$.
Let $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t,\displaystyle x>0,$.
Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x)$, suppose that $\displaystyle\lim_{x\rightarrow0} xf(x)=A$.
...
@r9m I don't check tags... I just watch the front page unless a link or something brings another to my attention. I fear that I will miss out on some good question that is out of the set of tags I choose. I like problems in a wide variety of topics.
Let $f$ be a continuous, decreasing function, with $\displaystyle\lim_{x\rightarrow\infty}=0$.
Let $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t,\displaystyle x>0,$.
Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x)$, suppose that $\displaystyle\lim_{x\rightarrow0} xf(x)=A$.
...
