(1 point) The Taylor series of f, centered at a, is f(x) = f(a) + f'(a)(x – a) + žf"(a)(x – a)2 + . In this problem you

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1 Point The Taylor Series Of F Centered At A Is F X F A F A X A Zf A X A 2 In This Problem You 1 (465.46 KiB) Viewed 16 times

(1 point) The Taylor series of f, centered at a, is f(x) = f(a) + f'(a)(x – a) + žf"(a)(x – a)2 + . In this problem you will use this formula with f(x) = VX and a = 100.1 to approximate V105. This is the best value of a to use since it is close to 105, and you can compute f(a) without a calculator. In fact, you can easily compute f'(a) and f" (a) without a calculator, too. Try to do this whole problem without a calculator. The first approximation is V105 ~Ti(105) = f(a) + f'(a)(x – a) = = You might remember this as the linear approximation from Calculus I. The quadratic approximation to f near x = a gives V105 ⓇT2(105) = f(a) + f'(a)(x – a) +Žf"(a)(x – a)2 = = = Warning! WebWork might have marked those approximations as correct even if they are wrong. This is because WebWork marks an answer correct if it is within 0.1 per cent of the correct answer. A good way to see if you did it right it to check the errors of the approximations. You don't need a calculator for this part either, since WebWork can do the calculations for you. The error (positive or negative) of the first approximation is = Ti(105) – V105 The error (positive or negative) of the second approximation is T2(105) – V105 =

(1 point) (a) Compute 54 (the 4th partial sum) of s = Σ 32 E 5 n3 n=1 S4 = Note: if giving a decimal approximation to this part, you may need to enter more digits than usual. 0 (b) Estimate the error in using S4 as an approximation of the sum of the series (i.e. use I f(x)dx 2 R, Estimate = (c) Use n = 4 and 00 Sn + f(x) dx < s < Sn + +1, f(x) dx to find a better estimate of the sum: < s < Note give each answer accurate to (at least) five decimal places.