Page:Squaring the circle a history of the problem (IA squaringcirclehi00hobsuoft).djvu/70

are all integers divisible by $p$. This follows from the fact that $\textstyle \sum \limits _{m=1}^{m=n}(C\beta _{m})^{r}$ is expressible in terms of those symmetrical functions which

consist of the sums of products of the numbers $C\beta _{1},C\beta _{2},\ldots$; and these expressions have integral values.

(4) Let $K_{p}$ denote the integer

$(p-1)!c_{p-1}+p!c_{p}+\ldots +(np+p-1)!$,

which may be written in the form

$\phi ^{(p-1)}(0)+\phi ^{(p)}(0)+\ldots +\phi ^{(np+p-1)}(0)$.

In virtue of what has been established in (3) as to the values of

$\phi ^{(p-1)}(0),\phi ^{(p)}(0),\ldots$

we see that $K_{p}A$ is not a multiple of $p$.

"We examine the form to which the equation

$A+e^{\beta _{1}}+e^{\beta _{2}}+\ldots +e^{\beta _{m}}=0$

is reduced by multiplying all the terms by $K_{p}$. We have

r=np+p-I

r=p-l

Q r+i O r+2

"*" T "*" 7 ~ / ^T +

The modulus of the sum of the series

${\frac {\beta _{m}}{r+1}}+{\frac {\beta _{m+1}}{(r+1)(r+2)}}+\ldots$

does not exceed

${\frac {|\beta _{m}|}{r+1}}+{\frac {|\beta _{m}|^{2}}{(r+1)(r+2)}}+\ldots$,

and this is less than $e^{|\beta _{m}|}$; hence we have

$c_{r}{\beta _{m}}^{r}\left\{{\frac {\beta _{m}}{r+1}}+{\frac {{\beta _{m}}^{2}}{(r+1)(r+2)}}+\ldots \right\}=\theta _{r}|c_{r}{\beta _{m}}^{r}|e^{|\beta _{m}|}$,

where $\theta _{r}$ is some number whose modulus is between 0 and 1.

The modulus of $\textstyle \sum \limits _{r=p-1}^{r=np+p-1}\theta _{r}|c_{r}{\beta _{m}}^{r}|e^{|\beta _{m}|}$ is less than $\textstyle e^{|\beta _{m}|}\sum \limits _{r=p-1}^{r=np+p-1}\theta _{r}|c_{r}{\beta _{m}}^{r}|$,