18.01 Calculus Jason Starr Fall 2005 Lecture 3. September 13, 2005 Homework. Problem Set 1 Part I: (i) and (j). Practice Problems. Course Reader: 1E-1, 1E-3, 1E-5. 1. Another derivative. Use the 3-step method to compute the derivative of f (x) = 1/√3x + 1 is, f �(x − x −3/2/2 .) = 3(3 + 1)Upshot: Computing derivatives by the definition is too much work to be practical. We need general methods to simplify computations.+ bn+1.� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 18.01 Calculus Jason Starr Fall 2005 2. The binomial theorem. For a positive integer n, the factorial, n! = n × (n − 1) × (n − 2) ×··· × 3 ×2 × 1, is the number of ways of arranging n distinct objects in a line. For two positive integers n and k, the binomial coefficient, n n! n(n− 1) ···(n −k + 2)(n − k + 1) ,= = k k!(n − k)! 13 · 2 ·k(k − 1) ···is the number of ways to choose a subset of k elements from a collection of n elements. A funda-mental fact about binomial coefficients is the following, n n n + 1 + = . k kk − 1 This is known as Pascal’s formula. This link is to a webpage produced by MathWorld, part of Wolfram Research. The Binomial Theorem says that for every positive integer n and every pair of numbers a and b, (a + b)n equals, n n n a + na n−1b + ··· + ka n−k bk + ···+ nabn−1 + b . This is proved by mathematical induction. First, the result is very easy when n = 1; it just says that (a + b)1 equals a1 + b1 . Next, make the induction hypothesis that the theorem is true for the integer n. The goal is to deduce the theorem for n + 1, (a + b)n+1 n+1 n + 1 n+1−k bk = a + (n + 1)a nb + ··· + ka + ···+ (n + 1)abn + bn+1 . By the definition of the (n + 1)st power of a number, (a + b)n+1 = (a + b) ×(a + b)n . By the induction hypothesis, the second factor can be replaced, n n(a + b)(a + b)n = (a + b) a + ··· + a n−k bk + ···+ bn . k Multiplying each term in the second factor first by a and then by b gives, n an+1−k bk + n an−k bk+1 nan+1 + nanb + . . . + � k + . . . + abk+1 n an+1−kbk + n an−k bk+1 n + bn+1+ anb + . . . + + . . . + nabkk−1 Summing in columns gives, n n n n nan+1 + (n + 1)anb + . . . + ( k + k−1 )an+1−k bk + ( k+1 + )an−kbk+1 + . . . + (1 + n)abkf(a)]g(a).� � � � � � � � � � � � 18.01 Calculus Jason Starr Fall 2005 Using Pascal’s formula, this simplifies to, n+1 an+1−k bk + n+1 an−k bk+1 bn+1an+1 + (n + 1)anb + . . . + k + . . . + (n + 1)abn + .k+1 This proves the theorem for n + 1, assuming the theorem for n. Since we proved the theorem for n = 1, and since we also proved that for each integer n, the theorem for n implies the theorem for n + 1, the theorem holds for every integer. n3. The derivative of x . Let f (x) = xn where n is a positive integer. For every a and every h, the binomial theorem gives, n nf (a + h) = (a + h)n = a + na n−1h + ··· + ka n−k hk + ···+ hn . Thus, f (a + h) − f (a) equals, n n(a + h)n − a = na + a n−k hk + ···+ hn . n−1h + ···k Thus the difference quotient is, f (a + h) − f (a) n−1 n n = na +2 a n−2h + ··· + ka n−k hk−1 + ···+ hn−1 . h Every summand except the first is divisible by h. The limit of such a term as h 0 is 0. Thus, → f (a + h) − f (a) n−1lim = na n−1 + 0 + ··· + 0 = na . h 0 h→So f �(x) equals nxn−1 . 3. Linearity. For differentiable functions f (x) and g(x) and for constants b and c, bf (x) + cg(x) is differentiable and, (bf (x) + cg(x))� = bf �(x) + cg�(x). This is often called linearity of the derivative. 4. The Leibniz rule/Product rule. For differentiable functions f (x) and g(x), the product f (x)g(x) is differentiable and, (f (x)g(x))� = f �(x)g(x) + f (x)g�(x). The crucial observation in proving this is rewriting the increment of f (x)g(x) from a to a + h as, f (a+h)g(a+h)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+f (a+h)g(a)−f (a)g(a) = f (a+h)[g(a+h)−g(a)]+[f (a+h)− 5. The quotient rule. Let f (x) and g(x) be differentiable functions. If g(a) is nonzero, the quotient function f (x)/g(x) is defined and differentiable at a, and, (f (x)/g(x))� = [f �(x)g(x) −f (x)g�(x)]/g(x)2 .18.01 Calculus Jason Starr Fall 2005 One way to deduce this formula is to set q(x) = f (x)/g(x) so that f (x) = q(x)g(x), and the apply the Leibniz formula to get, f �(x) = q�(x)g(x) + q(x)g�(x) = q�(x)g(x) + f (x)g�(x)/g(x). Solving for q�(x) gives, q�(x) = [f �(x) − f (x)g�(x)/g(x)]/g(x) = [f �(x)g(x) − f (x)g�(x)]/g(x)2 . 6. Another proof that d(xn)/dx equals nxn−1 . This was mentioned only very briefly. The product rule also gives another induction proof that for every positive integer n, d(xn)/dx equals nxn−1 . For n = 1, we proved this by hand. Let n be s ome specific positive integer, and make the induction hypothesis that d(xn)/dx equals nxn−1 . The goal is to deduce the formula for n + 1, d(xn+1) = (n + 1)x n . dx By the Leibniz rule, n+1) d(x × xn)d(x d(x) d(xn) d(xn)n = = x + x = (1)x n + x . dx dx dx dx dx By the induction hypothesis, the second term can be replaced, n n n d(xn+1)= x + x(nx n−1) = x + nx n = (n + 1)x . dx Thus the formula for n implies the formula for n + 1. Therefore, by mathematical induction, the formula holds for every positive integer
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