Discrete Mathematics with Graph Theory (3rd Edition) 127

Discrete Mathematics with Graph Theory (3rd Edition) 127 -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 5.1 125 32. We use induction on n with no = 2. For n = 2, certainly PI --+ P2 is a valid argument. Now assume PI --+ P2 that k 2: 2 and the result is true for k. Given the premises PI --+ P2 P2 --+ P3 Pk-l --+ Pk Pk --+ Pk+l we note that the first k - 1 yield PI --+ Pk by the induction hypothesis. Together with Pk --+ Pk+l, we obtain the conclusion PI --+ Pk+l by the chain rule. So the result is true by the Principle of Mathematical Induction. 33. lin = 1, xl = x, d~x = 1 = l x o and the formula is correct. Now suppose that k 2: 1 and the formula is correct for n = k; that is, suppose that d~xk = kxk-l. We must prove that the formula is correct for n = k + 1; that is, we must prove that d~xk+l = (k+ I)X(k+ l )-l = (k+ l)xk. Using the product rule, d d d d
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: _xk+l = _(x k . x) = (_xk) . X + xk-x = kx k-l . X + xk . 1 = kxk + xk = (k + l)xk dx dx dx dx as required. By the Principle of Mathematical Induction, we conclude that the formula is correct for all n 2: 1. r 1 -X 2r + 1 1 _ x4 34. When n = 1, rr;=l (1 + x 2 ) = 1 + x 2 and 2 = --2 = 1 + x 2 and so the formula is I-x I-x correct. Now suppose that k 2: 1 and the result is true for n = k; that is, suppose k 2k+1 II( 1 2r)=I-x +x 1 2 -x r=l We must prove that it is correct for n = k + 1; that is, we must prove that k+l Now II (1 + x2r) r=l k = (II (1 + x2r)) (1 + X 2k + 1 ) r=l 2 k + 1 -_ (11-_Xx2 ) (1 + x 2k + 1 ) (by the induction hypothesis) as required. By the Principal of Mathematical Induction, we conclude that the formula is correct for all n 2: 1....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online