Probability in a small interval is ##P. dx##

[Kashmir]

Reif says
" ... variable $u$ which can assume any value in the continuous range $a_{1}<u<a_{2}$. To give a probability description of such a situation, one can focus attention on any infinitesimal range of the variable between $u$ and $u+d u$ and ask for the probability that the variable assumes a value in this range. One expects that this probability is proportional to the magnitude of $d u$ if this interval is sufficiently small"

" Indeed, the probability must be expressible as a Taylor's series in powers of du and must vanish as $d u \rightarrow 0$. Hence the leading term must be of the form $P d u$, while terms involving higher powers of $d u$ are negligible if $d u$ is sufficiently small"So expanding probability function as Taylor series I've

$P(x+d x)=P(x)+\frac{P^{\prime}(x)}{1 !} d x+\frac{P^{\prime \prime}(x)}{2 !} d x^{2}+\cdots$

in limit $dx$ is small we've

$P(x+d x)=P(x)+{P'(x)} d x$

Now how do I make the connection that "probability is proportional to the magnitude of $d x$ if this interval is sufficiently small"?

 

[vanhees71]

The meaning of a probability distribution is that $P(x) \mathrm{d} x$ is the probability to find the random variable $X$ to take a value in an "infinitesimal interval" of length $\mathrm{d} x$ around $x$. I has nothing to do with a Taylor expansion of $P$ around $x$.

 

[PeroK]

Reif says
> ... variable $u$ which can assume any value in the continuous range $a_{1}<u<a_{2}$. To give a probability description of such a situation, one can focus attention on any infinitesimal range of the variable between $u$ and $u+d u$ and ask for the probability that the variable assumes a value in this range. One expects that this probability is proportional to the magnitude of $d u$ if this interval is sufficiently small;

>- Indeed, the probability must be expressible as a Taylor's series in powers of du and must vanish as $d u \rightarrow 0$. Hence the leading term must be of the form $P d u$, while terms involving higher powers of $d u$ are negligible if $d u$ is sufficiently small.So expanding probability function as Taylor series I've

$P(x+d x)=P(x)+\frac{P^{\prime}(x)}{1 !} d x+\frac{P^{\prime \prime}(x)}{2 !} d x^{2}+\cdots$

in limit $dx$ is small we've

$P(x+d x)=P(x)+{P'(x)} d x$

Now how do I make the connection that "probability is proportional to the magnitude of $d x$ if this interval is sufficiently small"?

You've calculated $P(x + dx)$. What you want to calculate is $$\int_x^{x + dx}P(x')dx'$$Which, for small enough $dx$ is approximately $P(x)dx$. You don't need a Taylor expansion to see this.

 

[Kashmir]

Indeed, the probability must be expressible as a Taylor's series in powers of du and must vanish as du→0. Hence the leading term must be of the form Pdu, while terms involving higher powers of du are negligible if du is sufficiently smal

Why does the author then discuss Taylor expansion @PeroK, @vanhees71

 

[PeroK]

Why does the author then discuss Taylor expansion @PeroK, @vanhees71

No idea unless you tell us the context.

 

[Kashmir]

This is the whole passage.

Please see the footnote marked with aestrik* . Also the figure are missing, I've only this xeroxed photocopy.

 

[PeroK]

First, it's clear that an integral over a small interval is approximately the function value at (any) point in the interval times the width of the interval. It's the area under the curve, right?

You can prove that rigorously for a continuous function using an epsilon-delta proof. There is no need to express the function as a Taylor series. Not least because the function need not even be differentiable.

Given that the author hasn't actually done the proof, it's safe to say he mentioned Taylor series without realising that's not necessary.

Move on. It takes long enough to learn QM without worrying about things like this.