Recent content by mathhabibi

Is not a double variable product because for there to be only two terms either ##x=y+1## or ##y=x-1##. Good observation though.

Using the concepts of Summability Calculus but generalized such that the lower bound for sums and products is also variable, we can prove that the solution to the following PDE: $$P^2\frac{\partial^2P}{\partial x\partial y}=(P^2+1)\frac{\partial P}{\partial x}\frac{\partial P}{\partial...

okay I started my tutoring session will be back later.

No problem. I am a bit sketchy with physics but is it true that ##F_n=m\frac{dv}{dt}##?

I'm not sure It will take me some time but from what I am seeing, after integrating you want to solve for ##x_f-x_i##, is that correct?

In the post you said you are solving for v. Be more clear next time. What even is ##\Delta x##?

which is why you should integrate.

solve for dv/dx and integrate with respect to x.

No. It's ##\ln(n)+\gamma## (where ##\gamma## is the Euler-Mascheroni constant). However, ##\ln n## is pretty close to the harmonic numbers so replacing the numbers with that function shouldn't change anything.

Take the natural log of both sides. Also this isn't the original limit as in the question.

The limit is $$\lim_{n\rightarrow\infty}\left(\frac{H_n}{n^2}\right)^n=\lim_{n\rightarrow\infty}\frac{H^n_n}{n^{2n}}$$Where ##H_n## is the ##n##th harmonic number. The numerator grows much slower than the denominator and so the answer is just ##0##.

I think I can still cite his notebooks somehow. You have contributed something, don't downplay what you've done :)

After taking in all your feedback, I submitted a new version of my paper, where instead the main result is called the simple result and is defined as a lemma, not a theorem. I also solved the "conjecture" in my paper and replaced it with a theorem, deriving an asymptotic for the roots of the...

There's a small typo in the OP but I couldn't change it. The reflection formula should be $$\bar{H}_x-\bar{H}_{2-x}=\pi\cot(\pi x)+\left(\frac1{2-x}-\frac1{1-x}-\frac1x\right)\cos(\pi x)$$But this doesn't change the proof of the conjecture.

I solved the conjecture. Take a look at the function ##\bar{H}_x-\bar{H}_{2-x}+\ln2##, which is just the RHS of the reflection formula added by ##\ln2##. As ##x\rightarrow-\infty## we have that ##\bar{H}_x-\bar{H}_{2-x}+\ln2\sim\bar{H}_x## and the RHS approaches ##\ln2+\pi\cot(\pi x)##. So we...