Therefore, every time the difference is taken, the exponent of the Dower of which the residual p-1, is the root, will be increased by anity; whence in taking the difference n times, the exponent of the power, of which the residual r—1 is the root, will be increased to n; whence ▲apx=a(p—1)*xp* as was to be demonstrated. (9.) Corollary. Hence if p be a fraction equal to 1 then the n will be (-1)x and this will be ob q* for p, o if r==or=p, then in all Theorem 4. (10.) The difference of v w is equal to either of the factors multiplied into the difference of the other, plus the difference of the former multiplied into the succeeding value of the latter. Let v be increased by any quantity v, and w by any quantity w, and Then (v+v)(w+w) is the same function of v+v and w+w, which vw is of v and w; then the difference between these two similar functions is =vw+v\w+w) which was to be demonstrated, o let w+w=w1, then will Av w=vw+v w ̧. Problem 3. To find the difference of the factorial mix ̧ Rule. Multiply the exponent m of the given factorial into another factorial, of which the first factor is one more, and the exponent one less, than that of the given factorial. K By rule A(x+3)s1x =5(x+4)111. Ex. 3. Find the difference of (x—3—111. By rule ▲(x—3)—n—1)(x ‡ 1—3)”—211 — (n—1)(.2—2)r—1 ̧ Problem 4. To find the nth difference of a factorial, of which the first factor a %, the number of factors m, and the common difference, unity. Rule. (12.) Form a factorial of which its first factor is equal to the exponent of the given factorial, and its exponent equal to number, expressing tl.e number of differences; and the common difference of its factors is minus unity: multiply the factorial thus formed into another, which is the given factorial, having its first factor increased, and its exponent diminished by the number of differences to be taken. of the given factorial thus changed, into the same factorial, having its exponent diminished by unity. (14.) Change 1 Rule. - into : then form a factorial, of which its factors are negative, diminishing by unity, the first factor being the same as the exponent of the given factorial, and the exponent the same as the number which expresses the number of differences, then multiply the factorial, thus formed, into another factorial which is the same as the factorial changed, except that the exponent is diminished by a number equal to the number of differences. By rule ▲(x+2)-211 − (−2) 41ī × (x+2)—2—411 — 120 or= (x+2)(x+3)(x+4)(x+5)(x+6)(x+7)' (x+2)218* 23.4.5 N. B. in these two examples the negative sign is placed above each of the factors, in order to save room, as is sometimes done, with regard to the indices of logarithms. Problem 7. To find the difference of the product of two variable quantities. Rule. (15.) Multiply either of the factors into the difference of the other, and to the product add the difference of the first factor multiplied into the succeeding value of the second. Example. Find the difference of yz. By nile, yz is the one factor multiplied into the difference of the other. And yx, is the difference of the first factor of the forr er product, multiplied into the succeeding value of the second factor I Therefore, Ayz=yz+yz. 1 Problem 8. To find the nth difference of a quantity of the form ar«, whether ▾ De considered as a whole number, or as a fraction. Rule. (16.) Multiply the constant co-efficient a into the nth power of the root of the exponential diminished by unity, and this product again into the exponential itself. Example 1. Find the fourth difference of 3*. By rule, ^3*=(3—1)* × 3*=2* × 3′′. Example 2. Find the third difference of 2*+3, 3 By rule, 2*+3=(2—1)3 × 2*+3—2*+3. Example 3. Find the second difference of 5.(-1) that is, of 2 1 By rule, ▲ 5 (¦)"=5(¦— 1 )* × (};)"=5(—3) ̊ × -—. 3* To find the difference of the logarithm of x, by art. Functions, we have log. 1+x=b(x— x2 x3 1+x=b(x + +, &c.) For 1+x substitute ལ and, consequently, for æ substitute, and we shall have logarithm logarithms, and the method of differences, log. -log.x=(log. ); whence (log. 2)= % or if b = 1, this difference will become the difference of the Neparian logarithm of z. Problem. To find the difference of the exponential a as expressed in a series in terms of z. Whence ▲a==a*(Ax+ 12 +1.2.3+ &c. ·) Theorem 5. (17.) The 7th order of differences of the series of powers, 1", 2", 3", 4", &c. is nalī For the general or xth term of the series is a"; by decompo sition x2 = x2111+Bx”—111 +Cx2¬2+Dx”—311 + . . . . . . L. Now it is evident that the difference of the first term can be taken, as often as n contains units, that of the second term as often as n-1 contains units, that of the third term, as often as n-2 contains units, and so on. But as the series contains n+1 terms, and as the last term is constant, in taking the difference of all the terms, the last term will become zero; and the last but one will become constant. Therefore in taking the second difference of the remaining terms, the last but one will become zero; and the last but two will become constant, and so on; and as the last term of the series that is newly differenced, is destroyed in taking the last difference, it is evident that in differencing n times, that all the terms will be taken away except the first, and that its exponent will be diminished by n, and will, therefore, be equal to zero. |