user400188

can someone correct my rather basic misconception please?
I know that integration by parts can be applied to constant functions, for instance, it is often used to integrate In(x) by noting that In(x)=(1)In(x). However, what is stopping us from writing $\int 1 dx = \int 1\cdot 1 dx $ and then using by parts to arrive at a contradiction. What rule of using integration by parts have I broken here?

Is there a wikipedia like page the mathematical properties of best of N games? E.g., best of three (whoever wins 2 out of the 3 wins the match).

Or, since I doubt anyone knows where one is off the top of their head, where can I look for one? (Wikipedia doesn't have one)

I see, thanks for the insight.
I imagine at least one of those hypotheses on f and f' is that they are analytic, since the Fourier series is a sum of sines and cosines, which are analytic functions, however I am not sure what the others might be?

If I am not mistaken, this equation can't be said to be zero in without knowing what $f(\boldsymbol{x})$ is, however the reference I am using seems to think it is zero.
I'll put the reference here for completeness: https://www.theoretical-physics.net/dev/math/transforms.html

Can anyone please confirm (or explain) why $\left[f(\mathbf{x}) e^{-i \boldsymbol{\omega} \cdot \mathbf{x}}\right]_{-\infty}^{\infty}$ is equal to zero?

$\boldsymbol{\omega}$ is a vector of the same dimensions as $\boldsymbol{x}$. This equation was part of an intermediate step involving the Fourier transform of a derivative. $\boldsymbol{\omega}$ is the variable that the Fourier transform is a function of.

Is a proof that [0,1] is path connected, just:

$A \subset X$ is path connected iff for each pair of points $p$ and $q$ in $A$ there is a continuous function $\gamma:[0,1] \rightarrow A$ such that $\gamma(0)=p$ and $\gamma(1)=q$.

Choose the continuous function, y=x. This is a path from [0,1] to [0,1].

Now, if any function with a domain with the discrete topology is continuous, couldn't we pick $\delta$ to be a step function from $p$ to $q$, where $\delta(0)=p$ and $\delta(1)=q$, and have the topology on $[0,1]$ be discrete?
Then this choosen function would be continuous. But since $p$ and $q$ are arbitrary, we could always pick such a function for any $A$ or $X$. So every $A$ would be path connected, which I don't think is correct.

I remember hearing in undergraduate that 'probability holds up the weight of neutron stars', but I'm not sure if this is misremembering now. When I think about it further, it's the Pauli exclusion principal doing the work, for which there isn't probability involved.
Is the statement incorrect then?

The former, thank you for the reply, I'll look into complex Gaußian variables... wait that just Gaussian random variables.

Because of your reply though, I ended up taking that option again and rethinking it. I went with a multivariate Gaußian distribution with the variance on each of the variables set to 1/4, in order to get the variance for the complex number to be 1/2.

For the Schrodinger equation to hold, the wavefunctions second spatial derivative needs to be defined.
But are there any cases where the solution to the Schrodinger equation is not analytic? I.e. are there any cases where its third derivative or higher is not defined? I can't think of any off the top of my head.(Besides the infinite square well, but that doesn't count because it's unphysical, and also the Schrodinger equation isn't satisified everywhere anyway).

I'm not sure i like either of those, since you are multiplying through by $dt$, and to a lesser extent dividing through by $psi$, which might be zero in some places. Also, I don't think the $d's$ work like that and just absorb $\psi$.
There might be a way to repeat it legally though using dummy variables like $d\psi/dx=dy/dx\cdot d\psi/dy$.

Ok, thank you <3
I had trouble for a while comparing $e^{t\frac{d}{dt}}\psi(x,t)|_{x=0}$ with $e^{-(i/\hbar)\hat{H}t}\psi(x,t)|_{x=0}$, but I think I can show that this is ok, so long as I can take the natural log of both sides.

Can anyone please explain to me the meaning of a derivative as the exponent of an exponential, without any terms to differentiate?
e.g. $e^{-it\hat{D}}$, where $\hat{D}=-\hbar^2/(2m)~\nabla^2$