also its first differential coefficient p'(x) between those values, then will f'(x) vanish for at least one intermediate value.

For if '(x) were always positive or always negative between x=α and x=b, p(x) would be continually increasing or continually decreasing between those values (Art. 44), and therefore could not vanish for both xa and x=b, =b2 which would be contrary to the hypothesis. Hence p′(x) must change sign and therefore vanish for some value of x intermediate between xa and x=b.

118. The same thing is obvious at once from a figure. For, suppose the curve y=p(x) cuts the axis at A

(x=a, y=0) and B (x=b, y=0), then it is obvious that if the curve y=4(x) and the inclination of its tangent be

continuous between A and B, the tangent line must be parallel to the axis of x at some intermediate point P.

It is also clear that the tangent may be parallel to the axis of x at other points between A and B besides P as in Fig. 18, so that it does not follow that '(x) vanishes only once between two contiguous roots of p(x)=0.

119. The same proposition is thus enunciated in books on Theory of Equations: "A real root of the equation '(x)=0 lies between every adjacent two of the real roots of the equation p(x)=0"; and is known as Rolle's Theorem.

120. Remainder after the first n terms have been taken from Taylor's Series.

There is much difficulty in giving a rigorous direct proof of Taylor's Series, as might be expected from the highly general character of the result to be established. It is therefore found easier to consider what is left after n terms of Taylor's Series have been taken from 4(x+h). If the form of this remainder be such that it can be made smaller than any assignable quantity when sufficient terms of the series are taken, the difference between p(x+h) and Taylor's Series for p(x+h) will be indefinitely small, and under these circumstances we shall be able to assert the truth of the theorem. The following investigations of an expression for the remainder are taken, with few changes, from Bertrand's "Traité de Calcul Différentiel et Intégral."

Let R denote the remainder after n terms of Taylor's Series have been taken from p(x+h); so that

h2

hn-1

p(x+h)= p(x)+hp′(x)+2¡p" (x) + ... + (n-1)!?" - 1 (x)+R .....(1)

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Put R- (X-x)"P, a form suggested by the remaining

n!

terms of Taylor's Series. Consider the function formed by writing instead of x throughout the left-hand member of equation (2) except in P, which is therefore independent of z. Call the function thus obtained F(z). Hence

We shall assume that

(2) and all its differential coefficients up to the nth inclusive are finite and continuous between the values x and X of the variable z.

It is clear from equation (2) that F(x)=0, also by putting *= X in (3) we have F(X)=0; also F(z) and F'(z) are finite and continuous between these values of the variable z. Hence F'(z) vanishes for some value of z intermediate between x and X, say for z=x+0(X−x), where ◊ is a proper fraction. Differentiating equation (3) with respect to z, the terms alternately destroy each other except at the end of the series, and we have left

that value of z being taken which makes F'(z) vanish. Hence, remembering that X-x=h, the true value of R

where is a proper fraction.

If then the form of the function (x) be such that by

h n

inaking n sufficiently great the expression "(x+0h)

n!

can be made less than any assignable quantity however small, we can make the true series for p(x+h) differ by as little as we please from Taylor's form

h2

p(x)+hp'(x)+1⁄2¿′′(x)+.....to ∞.

The above form of the remainder is due to Lagrange, and the investigation is spoken of as Lagrange's Theorem on the Limits of Taylor's Theorem.

121. A different form of the remainder is due to Cauchy. In equation (2) put R= (X-x)P and proceed as before, then, instead of equation (4), we shall have

which vanishes as before for some value of z between 2=x and 2=X=x+h, say for z=x+0h; whence

122. Another form is obtained by Schlömilch and Roche by assuming a slightly different form for R, viz.,

(1-0)n-p-1hn

( n − 1 ) ! ( p + 1) 8" ( x +0h).

The last form includes the two former as particular cases; for putting p+1=n it reduces to Lagrange's result, and putting p=0 it reduces to Cauchy's.

123. The corresponding forms of remainder for Maclaurin's Theorem are obtained by writing 0 for x and x for h, when the three expressions investigated above become respectively

124. The student should notice the special cases of equation (6), Art. 120, when n=1, 2, 3, etc., viz., p(x+h)=4(x)+hp'(x+0,h),

h2

p(x+h)=4(x)+hp'(x)+2!P (x+0h),

etc.

All that is known with respect to the in each case being that it is a proper fraction.

125. Geometrical Illustration.

It is easy to give a geometrical illustration of the equation p(x+h)= p(x)+hp'(x+Oh).