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8. How many shot are contained in 15 layers of an incomplete pile of shot whose base is a rectangle with 24 shot in one side and 35 in the other?

9. Show that the number of shot in a complete pile whose base is a rectangle, which contains m and n shot in its sides, is

1 m (m+1) (3 n m + 1) if n > m. 10. Prove that the sum of the first n terms of the series

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11. Prove that the sum of the first n – 1 terms of the series

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12. Show by means of the theorem of undetermined coefficients that the sum of the r" powers of 1, 2, 3, 4, . . . . n is 1 s = Ho"+, "4 Am"+ . . . . --A, -īn, where the A's are obtained by substituting successively p = 1, 2, 3, etc., in the equation

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where a = first factor in the first term, n = the number of terms, and d = the common difference of factors.

14. Prove by induction or by the theorem of undetermined coefficients that the sum of the first n terms of the series in 13 may be found by multiplying the last term by the next highest factor and subtracting from this product the product of the first term by the next lowest factor and dividing the difference by (m+1)d. Thus if S, is the sum of the first n terms of the given series,

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729. It has been shown, #73, that the numbers which take the form" are indeterminate, i. e., there is no particular number which

0 represents the value of this fraction.

ExAMPLE 1. For example, the fraction **** takes the form

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a” — 5:r-H 6 — (r– 2) (r–3), a” — 4 T (a 2) (a + 2) and the numerator and denominator are divided by a - 2 (which operation is admissible only in case a is not zero), we have

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a” 4 x + 2 2–3 1 2 + 2 T ; when r = 2. But the division of the terms of the fraction by c – 2 is not admis

which is equal to

sible when r = 2, i. e., we are not allowed to divide both terms of the fraction by 0. Hence it is necessary to define what we mean by the value of the indeterminate form }.

730. Definition of the Value } —Let f(r) and F (a) be respectively the numerator and the denominator of a fraction, and suppose that f(a) = 0 and F(a) = 0, i. e., the numerator and denominator are respectively zero in case c is replaced by a ; then we have

F(x) Jr.—a T F(a) T0

We define the value of the fraction #: for ac = a as the
lim f(r).
a'-- a F(r)
Thus in the preceding example we have

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NotE.-The division of the terms of the second fraction by a-2 is admissible since by definition of a limit, $636, a can not take the value 2.

731. Further Examples.—All the algebraic and transcendental operations used in the previous chapters may be employed to give the expressions which are indeterminate for a particular value of ar, another form which is not indeterminate as a approaches this value. To illustrate:

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This fraction =; for a = 1. Put a = y + 1, then for a = 1, As = 0, and the value of the fraction is

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z ** = *-* becomes " for w=a. r — V2.0°– a” 0 Multiply both numerator and denominator respectively by their

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ExAMPLE 3. The fraction

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To find the true value of this fraction, substitute for sines and cosines their expansions, 3705, and the fraction becomes

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Dividing both terms of the fraction by . the value of the fraction is

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732. The Indeterminate Form 3.-Any number which takes the form # is indeterminate. For, by the definition of a quotient, we must have (#) oc = 2; but a oc = co, where r is any number not zero, i. e., any finite number a multiplied by a number larger than any assignable number, or co. Hence # can have any value

that a can, and therefore # is indeterminate.

To evaluate an indeterminate form of this kind let the fraction be

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but we may write

–– 1
so - Lol ===" for z=a.
F'(z) [*] 1 - 0
at:a f(z) ac=a do

That is, fractions which take the indeterminate form # may be reduced to the indeterminate form } Hence, if a given fraction takes the indeterminate form %, reduce the fraction to the indeterminate form . and proceed as in 4731.

ExERCISE CV

Find the limiting values of the following fractions:

1. :=} for a = 1. Result 1; 2 :=. for ac-2. * : 2; 3. * = for ac-1. * : ; 4. #. for ac- a. * { }*. 5 :=#. for ac-—1. * : ; 6. . . . ; for ac-2. * : ; 7. |-}=: for ac-0. * * (#) 8. # =#!. for ar—b. * . #; 9. o: for ac-1. & 4 2. 10. wo-oo: for ac-1. - “. 0. 11. Yi-Ez- V8 for c=2. * : ;Viš.

V3+ c – V5

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