§2. Series expansions
387
which he did by the method expounded in Chap. II, n° 22: he calculated the
successive powers in the series y, substituted the results in (6) and wrote that
the total coefficient of each monomial x 3 , x 5 , etc. is zero, whence obtaining
relations between the coefficients of (5) enabling him to calculate them oneby-one. For example
x = (x + a3x3 + ... ) + (x 3 + ... )/3.2 + ...
where the ... contain no further x or x 3 , whence a3 = -1/2.3. It follows that
x = (x - x 3 /6 + a5x5 + ... ) + (x - x 3 /6 + ... )3/6 +
+ 3(x + ... )5/40 + ... = (x - x 3 /6 + a5x5 + ... ) +
+ (x 3 - 3x 2 .x 3 /6 + ... )/6 + 3(x 5 + ... )/40 + ... =
x + (a5 - 1/12 + 3/40)x 5 + ...
where the ... contain no further x 5 , whence
a5 = 1/12 - 3/40 = 1/120 = 1/2.3.4.5,
etc.
Further, integration provided him the series for 10g(1 +x) which, inverted,
produced the exponential series. We have seen that
y = 10g(1 + x) = x - x 2 /2 + x 3 /3 - ... ;
let us try to extract x in the form
first we will have
where the unwritten terms are of degree ~ 3 in y. So we must have a2 = 1/2
to eliminate the terms in y2. This done,
the terms in y and y2 cancel, as they must, and to eliminate the terms in y3
we must have a3 - 1/2 + 1/3 = 0, whence a3 = 1/6 = 1/3!, etc.
This rudimentary method does not provide general formulae, which Newton confines himself to extrapolating from the first coefficients - if you believe
in the perfection of Creation, you can go on to the end -, nor did it prove that
the series obtained converged; but of course Weierstrass and his successors
traversed the same path, including for power series in several variables. Given
a converyent power series of the form
387
which he did by the method expounded in Chap. II, n° 22: he calculated the
successive powers in the series y, substituted the results in (6) and wrote that
the total coefficient of each monomial x 3 , x 5 , etc. is zero, whence obtaining
relations between the coefficients of (5) enabling him to calculate them oneby-one. For example
x = (x + a3x3 + ... ) + (x 3 + ... )/3.2 + ...
where the ... contain no further x or x 3 , whence a3 = -1/2.3. It follows that
x = (x - x 3 /6 + a5x5 + ... ) + (x - x 3 /6 + ... )3/6 +
+ 3(x + ... )5/40 + ... = (x - x 3 /6 + a5x5 + ... ) +
+ (x 3 - 3x 2 .x 3 /6 + ... )/6 + 3(x 5 + ... )/40 + ... =
x + (a5 - 1/12 + 3/40)x 5 + ...
where the ... contain no further x 5 , whence
a5 = 1/12 - 3/40 = 1/120 = 1/2.3.4.5,
etc.
Further, integration provided him the series for 10g(1 +x) which, inverted,
produced the exponential series. We have seen that
y = 10g(1 + x) = x - x 2 /2 + x 3 /3 - ... ;
let us try to extract x in the form
first we will have
where the unwritten terms are of degree ~ 3 in y. So we must have a2 = 1/2
to eliminate the terms in y2. This done,
the terms in y and y2 cancel, as they must, and to eliminate the terms in y3
we must have a3 - 1/2 + 1/3 = 0, whence a3 = 1/6 = 1/3!, etc.
This rudimentary method does not provide general formulae, which Newton confines himself to extrapolating from the first coefficients - if you believe
in the perfection of Creation, you can go on to the end -, nor did it prove that
the series obtained converged; but of course Weierstrass and his successors
traversed the same path, including for power series in several variables. Given
a converyent power series of the form