Question:
Let $f$ be differentiable on $[0,+\infty)$, such as$$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$
I think this problem is similar to this link and (the background)
For my problem I guess that if
$$\lim_{x\to+\in...
@chinamath a limit from the highschool textbook. Maybe you like it and want to compute it. $$\lim_{n\to\infty} \frac{\sqrt[n]{(2n+1)(2n+4)\cdots(5n-2)}}{n}$$
@chinamath Yeah, it's the correct version. It has a very nice closed form. $$\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)\cdots (2^{2^n+1})}$$
@chinamath Usually I would have been very upset for this post, but I understand your deep desire for finding the result of this limit. So, it's OK. Though, I hope you won't post any other question of mine on main. :-)
Hello! Help me please, what is the meromorphic function on a Riemann surface $X$ with values in $\mathbb C^2$? Is it the holomorphic map from $X$ to $\mathbb CP^2$?
@r9m Last night I dreamt I was working on a lot of great questions, one nicer than the other ... but I don't remember anyone of them ... (not this time) :-(
@r9m If I were like I see myself in my dreams, I'd pretty famous. I talk, and at the same time I'm amazed by the way I construct the sentences, the words I use. :-)
@NicolasBarbulesco I would say there is a limited number of stars in our universe, but then comes another universe at the border of our universe, so the total number of stars goes closer to infinity again.
@Chris'ssis my dreams are usually timeless (in the sense .. I have never had a dream where I was aware watch or any other standards of measuring time :} )
@Sawarnik time and relative distance in space :P (although the original T.A.R.D.I.S from Doctor Who is an abbreviation of Time and Relative Dimension in Space) :P
@BalarkaSen it has some interesting connections to the hypergeometric function, some values that I possibly met in the past while dealing with some other hard series.
@robjohn There is a reverse Holder Inequality by Diaz, Goldman, Metcalf, with bounds slightly different from the bound you proved .. $$\left(\int_0^1 f^p \right)^{1/p} \left(\int_0^1 g^q \right)^{1/q} \le C_p \int_0^1 fg \,dx$$ .. where $M_1 > f > m_1$ and $M_2 > g > m_2$, and $p,q > 1$ are cojugate indices .. where $$C_p = \dfrac{M_1^p M_2^q - m_1^p m_2^q}{(pM_2m_2(M_1M_2^{q-1} - m_1m_2^{q-1}))^{1/p}(qM_1m_1(M_2M_1^{p-1} - m_2m_1^{p-1}))^{1/q}}$$
@robjohn ^ mentioned in Mitrinovic Analytic Inequalities book ! (pg 64) .. I haven't seen their original paper .. but the book does not provide a proof of the result
@Chris'ssis You mean $$\sum_{n=1}^{\infty} \dfrac{n^2}{\binom{2n-2}{n-1}^2}$$ ? :)
@r9m I found another instance of a reverse Hölder inequality with a different constant. I will have to look it up to see if it is the same as that one.
@robjohn the discrete version is by Manolova and Docev (mentioned in Mitrinovic: Classical and new inequalities in analysis, pg 125 and the integral version on the following page) .. but without proof .. I haven't seen their original papers either ..
@r9m the answer is not deleted, even if I am logged out. If I have to reload the page, I simply copy the answer text before I reload so that I can paste it if needed after the reload.
@ShuklaSannidhya matrix multiplication is associative, from which you can prove that $A^kA=AA^k$
It can be shown that if $A$ and $B$ have the same eigenvectors, then $AB=BA$ (this is just a side remark, we won't use this to prove the assertion above)
Suppose that we define $A^k=A^{k-1}A$. Then we need to prove that $A^k=AA^{k-1}$
We know that $A^1=AA^0$
Suppose we know for some $k$ that $A^k=AA^{k-1}$, then by definition $A^{k+1}=A^kA=(AA^{k-1})A=A(A^{k-1}A)=AA^k$
If you're given an arbitrary function like sine or exp, is there a general method for determining the minimum number of points in the Cartesian plane that we force the curve to pass through that uniquely determines the curve?
@robjohn I starred your message unintentionally, anyways I need to find a recurrence that gives a generalization of powers of a number as row sums of a lower triangular matrix. Then I might get to integrals that give zeta(2), zeta(3) and so on.
So according to the legends, when the souls have been reborn often enough to reach maturity, they get to Nirvana (meaning: Poof, they vanish). The transcendental world.
@DanielFischer I think he left some time after the deletion if I remember correctly. Btw who is the 900 pushups guy? I can not keep track of all the name changes that goes around.
Question:
Let $f$ be differentiable on $[0,+\infty)$, such as$$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$
I think this problem is similar to this link and (the background)
For my problem I guess that if
$$\lim_{x\to+\in...
