EXAMPLE 8 Find the Maclaurin series for f(x) = (1 + x)*, where k is any real number. SOLUTION Arranging our work in columns, we have f (x) = (1 + x) * f(0) = 1 f'(x) = k(1 + x ) * - 1 f'(0) = k f"(x) = k(k - 1 )(1 + x)4-2 f"(0) = k(k - 1) f"(x) = k(k - 1)(k - 2)(1 + x)4-3 f"(0) = k(k - 1)(k - 2) f() (x) = k(k - 1) .. . (k - n + 1)(1 + x)k-n f((0) = k(k - 1) ... (k - n+ 1) Therefore the Maclaurin series of f(x) = (1 + x)* is f (")(0) k ( k - 1 ) . . . (k - n+ 1) IM : xn n! n! This series is called the binomial series. Notice that if k is a nonnegative integer, then the terms are eventually 0 and so the series is finite. For other values of k none of the terms is 0 and so we can try the Ratio Test. If the nith term is an, then an+1 k ( k - 1 ) . . . ( k - n + 1) (k - n)x+1 n! ar (n + 1)! k ( k - 1 ) . . . ( k - n+ 1)x" [ k - n) n n + 1 as n -> oo + n Thus, by the Ratio Test, the binomial series converges if | x | 1. The traditional notation for the coefficients in the binomial series is ( * ) k(k - 1) (k - 2) . . . (k - n+ 1) n! and these numbers are called the binomial coefficients.The following theorem states that (l + x)" is equal to the sum of its Maclaurin series. It is possible to prove this by showing that the remainder term RHU) approaches 0, but that turns out to be quite difcult. The proof outlined in Exercise 85 is much easier. The Binomial Series Ifk is any real number and |x| k, then the expression for (5) contains a factor (k It), so (f3): 0 for n > k. This means that the series terminates and reduces to the ordinary Binomial Theorem when k is a positive integer. (See Reference Page 1.) Problem 1. Read Example 8 from Chapter 11.10 (page 806 in 8th edition, page 803 in 9th edition) in the textbook. Let me know by email if you need a copy of this edition. Write the rst three terms of the Taylor series around a=0for \\/1+:n