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Similarly,

dy

{1 – «®'() y\'(x)} = 4 (3).

dy

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2

đ%

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Dividing the one by the other,

dx
d.x
φ (8)

= f (x) suppose.
dx

y (x)

dy
Differentiating with respect to X,
dog da dx

f'(x)
dx dy dx dx dy da ldy
Differentiating with respect to y,
dz dx d’x

12

f
dx dy dy dx dy?

dy
dz dr
Multiplying by

and subtracting,
dy

dx
doz
dr dy dy

ď %
2

+ dy) d x2 dy dx dy da dy This is the general equation to surfaces generated by the motion of a line which constantly rests on two given lines while it remains parallel to a fixed plane. (22) Eliminate the arbitrary functions from

z = q (ay + bx). (ay - bx). Taking the logarithm we have

log x = log (ay + bx) + log y (ay ), and as the functions are arbitrary their logarithms are also arbitrary functions, and we may replace them by the general characteristics F and f. Therefore, differentiating with respect to x and y successively,

1 dx

b F' (ay + bx) bf'(ay ),

2

2

(

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= 0.

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2

1

()

=

2

=

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Differentiating again,
1 dz da

b? F" (ay + bx) + b? f" (ay bx),
z dix?

dx, 1 d’z

d.

aF"(ay + bx) + a'f"(ay bx). dy

(dy) Multiplying by a', bo and subtracting, we obtain as the result of the elimination of the functions fd : dz

dx

dz
a?

Idy z dy
(23) Eliminate the arbitrary functions from

(1) a f (a) + y Q (a) +x\, (a) = 1,
where a is a function of x, y, and x given by the equation
(2)

x f'(a) + y Q'(a) + xy' (a) = 0;
f', o, y' being the differential coefficients of f, , y.

Differentiating (1) with respect to X,
{rf'(a) + y D' (a) + xy' (a)}

+f(a) + y (a) Tv

0;

dx which by the condition (2) is reduced to

f(a) + y (a)

dx In the same way, differentiating with respect to y, we have

da

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d :

= 0.

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dx

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Since from these two equations it appears that

and dz

are both functions of a, the one may be supposed to dy be a function of the other, and we may write

di

F

dr Eliminating the function F from this equation there results

dx

dz dx? dy This is the differential equation to developable surfaces.

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dy)

2

= 0.

(24) Eliminate the arbitrary function from the equation

y

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y Since 20

is a homogeneous function of m dimensions, we know that

du

du
+y
+ X

= mu.
dy

d.

du

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d x

(25)

If u = f (x, y) = F (r, z), and
p = 0 (ax +cx) = 4 (ax by),

1 du 1 du 1 du
then

+
+

= 0.
a dx

Cdx du du dr du da

+ dr do

dx dx'

b dy

dc

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1 du +

= 0.

and therefore

+

a dr

b dy

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CHAPTER V.

TIE

APPLICATION OF THE DIFFERENTIAL CALCULI'S TO

DEVELOPMENT OF FUNCTIONS.

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This theorem, the most important in the Differential Calculus, and the foundation of the other theorems for the development of Functions, was first given by Brook Taylor in his Methodus Incrementorum, p. 93. He irtroduces it merely as a corollary to the corresponding theorem in Finite Differences, and makes no application of it, or remark on its importance. The following is the statement of the theorem :

If u = f (2) and x receive an increment n, then

du
du

d'u

n
f (x + 1) = 1 + ht

+
da dr 1.2 dix 1.2.3

+ &c.

If we avail ourselves of the method of the separation of the symbols of operation from those of quantity, this theorem may be expressed in a very convenient form, which is useful in various parts of the Integral Calculus: viz. d h? d?

d f (x + 1) = {1+h

+ &c.} $ (2') 1.2 dix 1.2.3 d 23

+

+

dx

d n

dx =

f (x). It is frequently convenient to use Lagrange's notation, and to represent the successive differential coefficients of .f (r) by accents afised to the characteristic of the function. In this way Taylor's Theorem is written

23 f (x + 1) = f (x) + f'(o)h + f"(r)

+ &c.

+

f''(2')

1.2

1.2.3

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n

If we stop at any term, as the nth, which is f(n-1) (x) h

the error committed by neglecting the l'e1.2 (n - 1) maining terms lies between the greatest and least values which f (x) (r + On)

can receive; where 0 is less than 1.

1.2 This is Lagrange's Theorem of the limits of Taylor's Theorem. See Lagrange, Calcul des Fonctions, p. 88. Also De Morgan's Differential Calculus, p. 70. Ex. (1) Let f(x) = (a + x)". Then

n(n-1) (a +x+h)"=(a+x)'+ n(a+X)"-1h + (a+x)* -? h?+ &c.

). 1.2

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1.2

n

If we stop at the nth term the error lies between the greatest and least values of als +01) (log a)"

The least value is found by making 0 = 0, and the greatest by making 0 = 1, and therefore the error lies between

h"

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1.2

n

(3) Let f () = log 2. Then since by Chap. 11. Sec. 1, Ex. 11,

d'
(log x) = (-)"-1

(r – 1) (r – 2) ... 2.1
dr"

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h log (x + h) = log x +

1 h?

1 13

+ -&c., 2 x 3

2

and the error of stopping at the nth term lies between

and I

nu (.t + h)"

nilin

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