Work and solution we are given f x e x and p 4 x a bx

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= 0? Work and Solution: We are given f ( x ) = e x and P 4 ( x ) = A + Bx + Cx 2 + Dx 3 + Ex 4 , and we need their derivatives. f ( x ) = e x f 0 ( x ) = e x f 00 ( x ) = e x f 000 ( x ) = e x f (4) ( x ) = e x P 4 ( x ) = A + Bx + Cx 2 + Dx 3 + Ex 4 P 0 4 ( x ) = B + 2 Cx + 3 Dx 2 + 4 Ex 3 P 00 4 ( x ) = 2 C + 6 Dx + 12 Ex 2 P 000 4 ( x ) = 6 D + 24 Ex P (4) 4 ( x ) = 24 E Now, using a = 0 we can determine the constants A , B , C , D and E . In the functions first. f (0) = e 0 = 1 = P 4 (0) = A A = 1 Now in the first derivative. f 0 (1) = e 0 = 1 = P 0 4 (0) = B B = 1 Now in the second derivative. f 00 (0) = e 0 = 1 = P 00 4 (0) = 2 C C = 1 2 Now in the third derivative. f 000 (0) = e 0 = 1 = P 000 4 (0) = 6 D D = 1 6 Now in the forth derivative. f (4) = e 0 = 1 = P (4) 4 (0) = 24 E E = 1 24 3
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So, the function P 4 ( x ) = 1 + x + x 2 / 2 + x 3 / 6 + x 4 / 24. 2 Here’s the graph. -2 -1 0 1 2 1 2 3 Figure 2: Partial graphs of y = x + 1 [red], P 2 ( x ) [blue], P 4 ( x ) [green] and f ( x ) [black]. Yes, the forth degree polynomial is a better fit than is the second degree polynomial. This, of course can be repeated ad infinitum . And the pattern for the n th degree polynomial is as follows: P n ( x ) = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! + · · · + x n n ! . And, in fact e x can be written as, e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! + · · · + x n n ! + · · · = X i =0 x n n ! , Which is a power series. You should also be able to verify that this power series is convergent for all x . So here we can state (unlike the opening example) that e x = X i =0 x n n ! is true for all x . That is, a transcendental function can be written as if it were a polynomial. Okay, now let’s do something a little dangerous here. We know that d d x [ e x ] = e x , so let’s differentiate the power series term-by-term to see what we get. d d x 1 + x + x 2 2! + x 3 3! + · · · + x n n ! · · · = 1 + x + x 2 2! + x 3 3! + · · · + x n n ! · · · Yes, that’s incredible, but we nonetheless expected this. 2 You should recognize the denominators. Yes, they’re factorials. 4
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3. Let’s try to find the power series representation of f ( x ) = 1 x + 2 . I’d like to suggest long division (we’ll do this in class) to see if there’s a pattern. Here’s what we’ll get 1 x + 2 = 1 2 - 1 4 x + 1 8 x 2 - 1 16 x 3 + · · · = X n =0 ( - 1) n 2 n +1 x n . However, we could have also taking the approach that we used to find the power series for e x , that is by taking derivatives. Furthermore, this example actually looks like our introductory problem, 1 1 - x = 1 + x + x 2 + x 3 + x 4 + · · · = X n =0 x n . So let’s rewrite this example to look like 1 1 - x . It doesn’t even look possible at first site, but let’s try! 1 x + 2 = 1 2 · 1 1 + x/ 2 = 1 2 · 1 1 - ( - x/ 2) Now let’s use what we know to see if we get the same result. 1 1 - x = 1 + x + x 2 + x 3 + x 4 + · · · = X n =0 x n . 1 2 · 1 1 - ( - x/ 2) = 1 + ( - x/ 2) + ( - x/ 2) 2 + ( - x/ 2) 3 + ( - x/ 2) 4 + · · · = 1 2 · X n =0 ( - x/ 2) n . = X n =0 ( - 1) n 2 n +1 x n . All as expected. Oh, the interval of convergence for this new power series is ( - 2 , 2). So unlike the power series for e x which is true for all x , equating the power series to this rational function 1 x + 2 = X n =0 ( - 1) n 2 n +1 x n only holds for ( - 2 , 2). 5
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2 Differentiation and Integration of Power Series Theorem If the power series X n =0 c n ( x - a ) n has radius of convergnec R > 0, then the function f defined by f ( x ) = X n =0 c n ( x - a ) n is differentiable, and continuous on the interval ( a - R, a + R ).
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