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number of terms, the only difference in the absolute equality of the two sides of the equation is that which arises from being an undetermined fraction, mean between 0 and 1. The last term of (81) is called the limit of Taylor's Series.

73.] Examples of Taylor's Series.

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By the method of Ex. 5, Art. 54, if x = cot p, the nth derivedfunction of tan-1 x is found to be

(-)-11.2.3... (n-1) (sin ()" sin no,

by means of which the limit of (83) may be determined.

Ex. 4. Given f(x+y)= f(x)+f(y), where x and y are independent of each other; it is required to find the form of the function.

Expanding f(x+y) by Taylor's Series, we have

f(x)+f(y) = f(x+y)

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now as x and y are independent of each other, the form of the function of y does not depend on ; therefore x and all func

tions of x are constant with respect to y.

f'(x) = a constant =

a;

Hence we may put

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whence also

f(x) = ax, and f(x+y)= a (x + y) ;

a result which palpably satisfies the required conditions.

Ex. 5. f(x + y) = f (x) × ƒ(y), where x and y are independent; to determine the form of the function.

Expanding as before, we have

y2

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f(x) ×f(y) = f(x)+ƒ'(x) { +ƒ” (x) { .2 +ƒ'(x) +

f(y) = 1 +

У
1

1.2.3

f'(x) y f" (x) y2 ƒ""'(x) y3

f(x) I

+

+

f(x) 1.2 f(x) 1.2.3

for the same reason as in the last example, let

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f'(x) = a;
f(x)

.. ƒ'(x)= a f(x);

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...

also f(x) = eax, and f(x+y)= ea(x+y);

:. f(x+y) = eax × eay = f(x) ×f(y),

which satisfies the required condition.

By a similar process let the student prove that, if

f(x+y)+f(x− y) = 2 f(x) × ƒ(y),

And if f(xy) = f(x)+f(y),

f(x) = cos ax.

f(x) = loga;

in the last example he will find it convenient to assume y = 1 + where his undetermined.

For h let a-x be

74.] Also (81) may take another form. substituted; then we have

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and let ɑ and æ be mutually interchanged; then

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whereby any function of x, which is capable of expansion in the form, is expanded in ascending powers of x — a.

As an example, let f(x) = log x; then

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SECTION 4.-Change of the equicrescent variable, and transformation of differential expressions.

75.] From the supposition which we are at liberty to make that one of the variables involved in an equation should increase by equal increments, and therefore that the several differentials of it, after the first, should vanish, problems such as the following arise:

(1) Suppose y =ƒ(x), and that an expression is given involving x, y, and some of the derived-functions or differential coefficients which have been calculated on the supposition that one of the variables is equicrescent; to change the equation into its equivalent, when neither of the variables is equicrescent. Or

(2) To transform it into its equivalent, when the other variable is equicrescent. Or

(3) An expression being given involving a variable, which is either equicrescent or not, and its differentials, and also an equation being given connecting this variable with some other new variable; to eliminate the old variable and its differentials by means of these two equations, and to replace them in the original equation by their equivalents in terms of the new vari

able the new variable being equicrescent or not, as the case may be. Or

(4) It may be required to replace the variables and their differentials in a given differential expression, by their equivalents in terms of new variables, which are connected with the old variables by means of a sufficient number of given equations: the old and the new variables being equicrescent or not, as the case may be.

All these several processes involve transformations of differential expressions, and because such expressions commonly involve second and higher differentials, and one of the variables has been assumed to be equicrescent, they are called changes of the equicrescent variable. The method of effecting them is the same in all cases, viz.

To replace the expression, which has been simplified by the condition of a variable being equicrescent, by its complete value when no such modification has been made, and then to introduce the other conditions which the problem requires.

76.] Thus to solve the first two of the four cases above.

dy day

Let the given expression involve x, y, d' dx2; the

differential coefficients, as their form indicates, having been calculated on the supposition that x is equicrescent; it is required to replace these several differential coefficients by their equivalents, when a is not equicrescent.

Since by equations (2) and (3), Art. 54,

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=

dx

dx5

(85)

(d3y dx — d3x dy) dx — 3 (d2y dx — d2x dy) d2x

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; (86)

and similarly the equivalents of the other differential coefficients may be determined; the second members of the equations (85) and (86) being the complete values when neither nor y is cquicrescent.

If x is equicrescent, the equations are identical.

If y is equicrescent, then d2y = d3y = ... = 0; and the original quantities must be replaced by their equivalents, viz.

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The form however of the nth equivalent is too complicated to be of any use in the solution of a given example.

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valents, (a) when neither a nor y is equicrescent, (3) when y is equicrescent.

d2y
dx2

(a) Replace by its value as given in equation (85), and

multiply by (da)3;

x (d2y dx - d2x dy) + dy3 — dy dx2 = 0.

(B) Let d2y = 0, and we have

d2x dx 2

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+

dy2

dy

(da)2 – 1 =

0.

Ex. 2. To transform (dy2+dx2) + a dx d2y = 0, when æ is equicrescent, into its equivalents, (a) when neither x nor y is equicrescent, (B) when y is equicrescent.

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(dy2+ dx2) + a (d2y dx - d2x dy) = 0.

(B) And if y is equicrescent, d3y = 0, and we have

(dy2+ dx2) — a d2x dy = 0,

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dx2) d2x
= 0.

77.] In (3) and (4) of the cases of Art. 75, suppose the given

dy d2y

a

expression to involve y, x, da' dx2'...... wherein ≈ is the equi

crescent variable, and suppose the new equation to be of the

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