How to explain light reduction in Helmert demonstration (Chi square distribution)? I search to understand how Helmert find his second equation (2) in his article that explain Chi Square distribution function.
You can find Helmert article from 1876 on Goettingen University Site (PDF page = 208) (Book page = 193)

In this article Helmert write a very important equation that is THE basis of Chi Square function's formula.
I have translated in english what he written in german.

If $n = 1$, the probability that $ε_1^m$ is between $σ_m-dε/2$ and $σ_m+dε/2$ is equal to
$$
 φ(σ_m)_1.δ_m=\int_{\sqrt[m]{σ_m-δ_m/2}}^{\sqrt[m]{σ_m+δ_m/2}}ψ(ε)dε
$$
where index 1 has been added to $φ(σ_m)$ expression because $n = 1$.
Since $δ_m$ is supposed to to be a differential, this expression can be written as following form
$$
 φ(σ_m)_1 δ_m=\big(\sqrt[m]{σ_m+δ_m/2}-\sqrt[m]{σ_m-δ_m/2}\big).ψ(\sqrt[m]{σ_m})
$$
Now, supposing that $σ_m$ can never be equal to zero, we obtain, AFTER LIGHT REDUCTION, following equation
$$
 φ(σ_m)_1 δ_m=\frac{ψ(\sqrt[m]{σ_m})}{mσ_m^{(1-\frac{1}{m})}}δ_m
$$

QUESTION: What are these LIGHT reduction ?
How is it possible from
$$
 φ(x) δ_m=\big(\sqrt[m]{x+δ_m/2}-\sqrt[m]{x-δ_m/2}\big).ψ(\sqrt[m]{x})
$$
to deduce that
$$
 φ(x)=\frac{ψ(\sqrt[m]{x})}{mx^{(1-\frac{1}{m})}}
$$
Or how it is possible that
$$
 \big(\sqrt[m]{x+δ_m/2}-\sqrt[m]{x-δ_m/2}\big)=\frac{\sqrt[m]{x}}{mx}δ_m
$$
 A: This equality can be explained using Taylor serie for $\sqrt[m]{1+x}$
if -1 < x < 1 then
$$
\sqrt[m]{1 + x} = 1 + \frac{1}{m}x - \frac{(1+m)}{2!\space m^2}x^2 + \frac{(1+m)(1+2m)}{3!\space m^3}x^3\space\space\space\space(1)
$$
$$
\sqrt[m]{1 - x} = 1 - \frac{1}{m}x - \frac{(1+m)}{2!\space m^2}x^2 - \frac{(1+m)(1+2m)}{3!\space m^3}x^3\space\space\space\space(2)
$$

The first expression
$$
\sqrt[m]{x+δ_m/2}-\sqrt[m]{x-δ_m/2}
$$
can also be written as
$$
\sqrt[m]{x+x.δ_m/2x}-\sqrt[m]{x-x.δ_m/2x}
$$
putting $x$ in evidence we obtain
$$
\sqrt[m]{x(1+δ_m/2x)}-\sqrt[m]{x(1-δ_m/2x)}
$$
extracting x from $\sqrt[m]{...}$ we obtain
$$
\sqrt[m]{x}.\big(\sqrt[m]{1+δ_m/2x}-\sqrt[m]{1-δ_m/2x}\big)
$$
now considering Taylor serie (1) and (2) and neglecting term greater than $x^2$  because they are insignicant, we obtain
$$
\sqrt[m]{x}\space\big((1+\frac{δ_m}{2mx}-\frac{(1+m)}{2!\space m^2}x^2)-(1-\frac{δ_m}{2mx}-\frac{(1+m)}{2!\space m^2}x^2)\big)
$$
in resolving expression we obtain
$$
\sqrt[m]{x}\space\big(2.\frac{δ_m}{2mx}\big)
$$
that is equal to
$$
\frac{\sqrt[m]{x}}{mx}δ_m
$$
The tip used by Helmert is to use Taylor series for $\sqrt[m]{1+x}$ and $\sqrt[m]{1-x}$ and to cut terms greater that $x^2$ because they are insignificant.
