so that's from $2^nm-2^{n}+1=2^{n}(m-1)+1$ to $2^nm$, it seems like we should be able to prove that an $n$ exists for every $m$ so that that's big enough that it has to have a fifth power in it
@AlexanderGruber You can also do it analytically. For $m>1$, $5(2^nm)^{4/5}\le 2^{(4/5)n}5m^{4/5}$. The second half is constant, so let $n$ be large enough that $2^{(1/5)n}>5m^{4/5}$; then, $4(2^nm)^{4/5}\le 2^n-1$. Now, let $x$ satisfy $x^5\le 2^nm$. Then $(x+1)^5-x^5= x^4+x^3+x^2+x+1\le 5x^4\le 5(2^nm)^{4/5}\le 2^n-1$. Then, I think a proof by contradiction suffices.
Damn, after a certain point, $5(2^nm)^{4/5}$ needs to be replaced by $\lfloor 5(2^nm)^{4/5}\rfloor$.
And in the second line $4(2^nm)^{4/5}$ should be $5(2^nm)^{4/5}$.
@AlexanderGruber I would not have gone with precisely $n=\lceil\log_2(m^4)\rceil$ like adriano did. Anything higher than that should work better at capturing fifth powers as the intervals widen exponentially.
@skullpatrol no math at the moment. I've been developing a new approach as regards quality assurance system, that is I improve some specific areas of it (hopefully they'll be major improvements).
@N3buchadnezzar again you need to use the symmetry and it's of no help to go this way. I mean the 1st way you showed it's enough. This 2nd way brings nothing new.
@Chris'ssis This proof is correct it uses the fact that the $R(x) = 1/(1+x^2)$ satisifies the functional equation thus $R(x)/(x^r + 1)$ is independent of $r$.
@robjohn you used a similar approach to the approach you used yesterday to my question.
@robjohn I suspect $$\lim_{n\to\infty} 2^{n+1} \int_0^{\frac{\pi}{2}} \cos^n (x) \cos(n x) \ dx=\pi$$ can be proved by induction. Actually, the value is $\pi$ for all n.
@N3buchadnezzar I was surprised to find out that in our country kids will learn integrals in the 3rd high school year. Usually, they were thought these in the 4th high school year (the last year). Surely, there were always exceptions with the special kids...
@robjohn unfortunately I have less experience with complex analysis, but I like it very much since it's so powerful. Hopefully, I'm going to manage to read all my complex analysis books and do much better. :-) You know, you need a bit of courage at the beginning... (it's a new world -- the complex one) :-)
Once in a while though one i able to combine the previous knowledge into something new, original and beautiful. Like a buttefly springing from a cukoon.
@N3buchadnezzar The last integral I evaluated by complex analysis was $\displaystyle \int_0^{\infty} \frac{1}{1+x^3} \ dx$, but this one was really easy.
@N3buchadnezzar I have the proof at the bottom of the page I mentioned above, and another that I used when I proved $\sum_{k\in\mathbb{Z}}\frac1{k+z}=\pi\cot(\pi z)$
Did no-one look at this one? Suppose $f:\Omega \subseteq \Bbb R^{n}\to\Bbb R$ is $\mathscr C^1$ and there is $p\in\Omega$ such that $f(p)=0,\nabla f(p)\neq 0$. Then $f(x)=0$ for infinitely many $x$.
So it's equivalent to saying that the level set at zero cannot be finite for a sufficiently smooth function if the gradient is nowhere vanishing in that set. That makes sense.
@Arkamis So basically we use the Implicit Function Theorem to obtain $g:T\subseteq \Bbb R^{n-1}\to\Bbb R$ such that $f(g(t);t)=0$, assuming WLOG that the first entry of the gradient is nonzero.
@N3buchadnezzar Translate and scale $[a,b]$ to get $[0,1]$. Fix 0, and map $(0,1]$ to $[1,\infty)$ by $x\mapsto 1/x$. Then translate that to get $[0,\infty)$.
@Arkamis Yes, I really understood vector spaces algebraically when I realized they are just modules over fields. (This is kind of backwards though, since module theory seems to be partly based on the theory of vector spaces.)
Well there is sort of an English version sort of already. Irresistible integrals. My work is by no means a complete study, just parts I think are cool, and nice integrals with nice techniques, and proofs.
This had been annoying me for ages- so similar to making a substitution and having to replace $dx$ with $du$ to 'fit' the substituted function, you have to bend the original $'dx'$ path (usually the real line) to 'fit' the substituted function?
