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€"(x+) cos n (x + h) = 6"} {cos nx + (a’ + n°) cos (nx + ).h ha

h3 +(a*+ n) cos(n x+24) +(a + no)l cos(nx+30) + &c.}

1.2

1.2.3

If a = cos e, n = sin , Elt+h) cose

cos {(x+h) sin 0} = = 6* cos® {cos(v sin 0) +h cos(x sin0+0) h?

h? cos (v sin @ +20) + cos (v sin 0 + 30) + &c.}

+

1.2

1.2.3

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tan-lx = (-)-1. (r – 1) (1 – 2)... 2. I sin ry. (sin y)',

therefore

h tan-'(x + h) = tan-'x + sin y sin y - sin 2y (sin y)

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1

2

h'

+ sin 3 y (sin y)

h3

- sin 4y (sin y)'

+ &c.

From this development Euler* has deduced many remarkable theorems, some of which are subjoined. In the preceding example let h = – X, then

tan-' (x + h) tan -0 = 0;
therefore tan-?x = sin y. siny.X + (sin y)sin 2y

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2

1

; then

п

y + sin y cos y + 4 sin 2y (cos y)? + Įsin 3 y (cos y)' + &c. Again, let h = -x

(+)

sin y cos y tan-'( + 1) = tan- (3)

+ tan-'Q; therefore +

}

+1 cos y

(cos y) (cos y) (cosy)'

1

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tan

2

sin y

sin 2 y

sin sy

T

sin 4y

+ &c.

2

1

Again, let h = -(1 + x')} =

; then

sin' y

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If we differentiate this series we find

0 = } + cos y + cos 2y + cos 3y + &c. In these formulæ y lies between 0 and 7.

Calc. Dif: p. 380.

(7) Let u = cot-'x, then cot-'fax + h) is easily found from the expression for tan-'(x + h). For since

du cot-' x =

tan2

T

1

-1

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1 + x2?

and we have merely to substitute cot-1 x for tan-'x and to change the signs of the terms beginning with the second : and as in this case y = u, we find

cot-'(x + h) = u – sin u sin u

h

h?
+ (sin u) sin 2u
1

&c.

Secr. 2. Maclaurin's or Stirling's Theorem.

p. 32.

This Theorem, which is usually called Maclaurin's, but which ought to bear the name of Stirling, was first given by James Stirling in his Linea Tertii Ordinis Newtoniana,

Maclaurin introduced it into his Treatise of Fluxions, p. 610, and his name has generally been given to the theorem from an erroneous idea that his work was the first in which it appeared.

The following is the enunciation of the Theorem :

If f(x) be a function of x, and if we represent the values which it and its successive differential coefficients acquire when x = 0, by f (o), f'(o), f'(o), f'(o), &c.; then

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This Theorem is evidently a particular case of that of Taylor.

Ex. (1) Let u = f (x) = (1 + x)}; f(0) = 1,

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Then by the formula (B), Chap. 11. Sec. 1,

du

(-)'1.2...r (1 + 3x)' (1 + 2x + 3x*) –++} x

dar

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f(0) = 1, f'(0) = - 1, 9" (0) = 1.2(1 – 1) = 0, f" (0) = -1.2.3(1 – 3) = 1.2.3.2,

3

$" (0) -- 1.2.3.4.5, 1" (0) = 1.2.3.4.5.

2

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2

Then by Chap. 1. Sec. 1, Ex. 23,
1.2...(r - 1).2-1 1 (» – 1) (» – 2) 1

{1+
dxt
(1 – 2")"-

1.2 x?
1.3 (r – 1)(– 2)(x – 3)(» – 4) i
+

+ &c.}
2.4
1.2.3.4

x1 Therefore,

f(0) = 0, f' () = 1,

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Whence

1 2 3

1.3 to 1, 3.5 x 1
sin-lx = a +

+
+

+ &c.
2 3 2.4 5 2.4.67

value of 1.

It was by means of this series that Newton calculated the

Commercium Epistolicum, p. 85, 2nd Edit. (5) Let u = tan-' (2). By means of Chap. 11. Sec. 1, Ex. 24, we find

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This is Gregorie's series. See Commercium Epistolicum, p. 98, 2nd Edit.

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(6) Let u = sec X;

then f(0) = 1, f(0) = 0, f" (0) = 1, f'" (0) = 0, f" (0) = 5, f" (0) = 0,

f"(0) = 61. Therefore,

5.24

61 v06
sec X = 1 + +
1.2 1.2.3.4 1.2.3 4.5.6

James Gregorie, Ib. p. 99.

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+

+ &c.

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