#$&* course Mth 174 7/27 6 course Mth 174
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Given Solution:
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Given Solution: ** The maximum possible error of the degree-3 Taylor polynomial is based on the fourth derivative and is equal to the maximum possible magnitude of the n = 4 term of the Taylor series. The present function is x^(1/3). Its derivatives are f ' (x) = 1/3 x^(-2/3), f '' (x) = -2/9 x^(-5/3), f ''' (x) = 10/27 x^(-8/3), f '''' (x) = -80/81 x^(-11/3). All these derivatives are undefined at x = 0. Since all the derivatives are easily evaluated at x = 1, we expand about x = 1. The maximum possible magnitude of the fourth derivative f''''(x) = -80/81 x^(-11/3) between x = .5 and x = 1 occurs at x = .5, where we obtain | f '''' (x) | = 80/81 * (.5^-(11/3) ) = 12.54 approx. We will still have a valid limit on the error if we use the slight overestimate M = 13. So the maximum possible error is | 13 / 4! * (.5 - 1)^4 |, or about .034.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: ********************************************* Question: Query problem 10.3.10 (3d edition 10.3.12) (was 9.3.12) series for `sqrt(1+sin(`theta)) what are the first four nonzero terms of the series? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1+x)^(1/2) = 1 + x/2 + [(.5)(.5-1)x^2]/2! + [(.5)(.5-1)(.5-2)x^3]/3! ... = 1 + x/2 + x^2/8 + x^3/16 ... sin theta = theta - theta^3/3! + theta^5/5! - theta^7/7! ... Substitute sin theta for x from series (1+x)^(1/2): sqrt (1 + sin theta) = 1 + (1/2)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...) - (1/8)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^2 + (1/16)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^3 ... About x = 0: sqrt (1 + sin theta) = 1 + theta/2 - (theta^2)/8 - (theta^3)/16 ... confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: We substitute the Taylor expansion of sin(x) into the Taylor expansion of 1 / sqrt(x) and ignore terms with powers exceeding 4: Expanding y = sqrt(x) about x = 1 we get derivatives y ' = 1/2 x^(-1/2) y '' = -1/4 x^(-3/2) y ''' = 3/8 x^(-5/2) y '''' = -5/16 x^(-7/2). Substituting 1 we get values 1, 1/2, -1/4, 3/8, -5/16. The degree-4 Taylor polynomial is therefore sqrt(x) = 1 + 1/2 (x-1) - 1/4 (x-1)^2 / 2 + 3/8 (x-1)^3/3! - 5/16 (x-1)^4 / 4!. It follows that the polynomial for sqrt(1 + x) is sqrt( 1 + x ) = 1 + 1/2 ( 1 + x - 1) - 1/4 (1 + x-1)^2 / 2 + 3/8 (1 + x-1)^3/3! - 5/16 (1 + x-1)^4 / 4! = 1 + 1/2 (x) - 1/4 (x)^2 / 2 + 3/8 (x)^3/3! - 5/16 (x)^4 / 4!. Now the first four nonzeo terms of the expansion of sin(theta) give us the polynomial sin(theta) = theta-theta^3/3!+theta^5/5!-theta^7/7!. We note that since sin(theta) contains theta as a term, the first four powers of sin(theta) will contain theta, theta^2, theta^3 and theta^4, so our expansion will ultimately contain these powers of theta. We will expand the expressions for sin^2(theta), sin^3(theta) and sin^4(theta), but since we are looking for only the first four nonzero terms of the expansion we will not include powers of theta which exceed 4: sin^2(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^2 = theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4 sin^3(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^3 = theta^3 + powers of theta exceeding 4 sin^4(theta) = (theta-theta^3/3!+theta^5/5!-theta^7/7!)^4 = theta^4 + powers of theta exceeding 4 Substituting sin(theta) for x in the expansion of sqrt(1 + x) we obtain sqrt(1 + sin(theta)) = 1 + 1/2 (sin(theta)) - 1/4 (sin(theta))^2 / 2 + 3/8 (sin(theta))^3/3! - 5/16 (sin(theta))^4 / 4! = 1 + 1/2 (theta-theta^3/3! + powers of theta exceeding 4) - 1/4 (theta^2 - 2 theta^4 / 3! + powers of theta exceeding 4)/2 + 3/8 ( theta^3 + powers of theta exceeding 4) / 3! - 5/16 (theta^4 + powers of theta exceeding 4)/4! = 1 + 1/2 theta - 1/4 theta^2/2 - 1/2 theta^3 / 3! + 3/8 theta^3 / 3! + 2 / ( 4 * 3! * 2) theta ^4 - 5/16 theta^4 / 4! = 1 + 1/2 theta - 1/8 theta^2 + 1/16 theta^3 - 5/128 theta^4. STUDENT SOLUTION: Rewrite as sqrt (1+sin theta) = (1+sin theta)^(1/2) Produce Taylor series for (1+x)^(1/2) using binomial expansion with p = 1/2: (1+x)^(1/2) = 1 + x/2 + [(.5)(.5-1)x^2]/2! + [(.5)(.5-1)(.5-2)x^3]/3! ... = 1 + x/2 + x^2/8 + x^3/16 ... Produce Taylor series for sin theta sin theta = theta - theta^3/3! + theta^5/5! - theta^7/7! ... Substitute sin theta for x in series for (1+x)^(1/2): sqrt (1+ sin theta) = 1 + (1/2)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...) - (1/8)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^2 + (1/16)(theta - theta^3/3! + theta^5/5! - theta^7/7! ...)^3 ... sqrt (1+ sin theta) = 1 + theta/2 - (theta^2)/8 - (theta^3)/48 ... INSTRUCTOR RESPONSE: Your approach should work; you got -1/48 for the theta^3 term, whereas I believe the correct coefficient is -1/16. Otherwise your expansion agrees with that obtained below. However a different approach, which doesn't involve the binomial expansion, was used above. What is the Taylor series for `sqrt(z)? RESPONSE --> Using f(x) = sqrt (z) will not work because f(0) and all derivatives of f(x) evaluated at 0 = 0. Let f(x) = sqrt (1+x) Using binomial expansion: sqrt (1+x) = 1 + x/2 - x^2/8 + x^3/16 + ... Substituting z-1 for x (since 1 + (z-1) = z) sqrt (z) = 1 + (z-1)/2 - (z-1)^2/8 + (z-1)^3/16 + ... What is the Taylor series for 1+sin(`theta)?
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RESPONSE --> Since 1 is a constant, its Taylor series is itself (if f(x) =1, f(0) =1 and all derivatives of f(x) = 0 Taylor series for sin (theta) = theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,, Adding the two series gives 1 + sin (theta) = 1 + (theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,, = 1 + theta - theta^3/3! + theta^5/5! - theta^7/7! + ,,, How are the two series combined to obtain the desired series?
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RESPONSE --> See previous response. Since a Taylor series is a power series it follows the rules of a power series, including the rule that power series can be added term by term.
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STUDENT COMMENT: The Taylor series for sqrt(z) provides an interesting way to manually calculate approximations of square roots. Along with a table containing some reference values, your calculator uses Taylor expansions to evaluate functions.
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