course Mth 174 ??????wK????????ye·assignment #015
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14:06:58 Query 10.4.8 (was 10.4.6 3d edition) (was p. 487 problem 6) magnitude of error in estimating .5^(1/3) with a third-degree Taylor polynomial about 0.
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RESPONSE --> To find the bound for the third degree Taylor polynomial, first I took derivatives of the original function to the n+1 derivative: f' = (1/3) (.5)^(-2/3) f'' = (-2/9) (.5)^(-5/3) f''' = (10/27) (.5)^(-7/3) f^4 = (-70/81) (.5)^(-3) So the Taylor polynomial must be less than this fourth derivative according to the theorem: |f^4(x)| <= (-70/81) (.5)^(-3) = -6.91 So M = 7. Then we can insert the values into the theorem formula: (7/4!) (0.5)^4 = 0.0183 So the bound for this function is about 0.0183.
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14:07:39 What error did you estimate?
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RESPONSE --> I estimated the 3rd degree Taylor polynomial error. I found that the bound to the error was about 0.0183.
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14:08:20 What function did you compute the Taylor polynomial of?
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RESPONSE --> I computed the Taylor polynomial of (0.5)^(1/3).
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14:11:37 What expression did you use in finding the error limit, and how did you use it?
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RESPONSE --> I used the expression ( M/ (n+1)! ) |x|^(n+1) To find M, I looked at the (n+1) derivative of (0.5)^(1/3) which I found to be (-70/81)(0.5)^(-3). I then picked a value of M which would be greater than the value of the 4th derivative at that point (0.5). From there it was plug and chug.
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15:09:24 Query 10.4.20 (was 10.4.16 3d edition) (was p. 487 problem 12) Taylor series of sin(x) converges to sin(x) for all x (note change from cos(x) in previous edition)
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RESPONSE --> Here I was supposed to show that the Taylor series of sin(x) converges to sin(x) for all x, and was told to use the following limit: lim( x^n / n!, n->infinity) = 0 To prove this I need to prove that En(x) -> 0 as n -> infinity. We know that the largest vaue of sin(x) and its derivatives is 1, so M=1 for this formula. | En(x) | = |sin x - Pn(x) | <= ( |x|^(n+1) ) / (n+1)! Since En(x) is less than or equal to its Lagrange error bound, if the error bound converges, so does En(x). Now we can look at the limit given above, that for any given n, that particular value converges to 0. Therefore, the overall series has a limit of 0 and the series converges.
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15:09:30 explain how you proved the result.
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RESPONSE --> See above.
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15:12:09 What is the error term for the degree n Taylor polynomial?
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RESPONSE --> For this particular problem, M=1 since the maximum value of sin(x) and its derivatives is 1. Therefore the error term turns into: |x|^(n+1) / (n+1)!
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15:54:41 Can you prove that the error term approaches 0 as n -> infinity?
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RESPONSE --> To prove this, we need to look at the values of the error term as n -> infinity. So let's pick a value of x and look at how the value changes for large values of n: x = 20 20^15 / 15! = 25,058,226 20^30 / 30! = 4,047,993 20^40 / 40! = 13,475 20^60 / 60! = 1.38x10^-4 So as the values of n increase, the value of the error term approaches 0. For a value of n greater than 2x, the value of the function is multiplied by less than 1/2. The more the result is divide by a number larger than 2, the closer it gets to 0. This can be generalized that for n > 2|x|, the error term converges.
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15:56:12 What do you know about M in the expression for the error term?
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RESPONSE --> M is the bound for the maximum value of the nth derivative. Since the nth derivative of sin(x) is going to be +-cos(x) or +-sin(x). The maximum value of any of the derivatives is 1, so we know that M will be at most 1.
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16:07:59 How do you know that the error term must be < | x | ^ n / ( n+1)! ?
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RESPONSE --> The value given by the Lagrange error bound is simply that, a maximum value for the error. Basically it this function takes the actual function at a point and subtracts from it the Taylor polynomial at that point. The error term is always smaller than the bound term because it is the bound for the next derivative.
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16:09:48 How you know that the limit of | x | ^ n / ( n+1)! is 0?
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RESPONSE --> I think I explained it before. Basically as soon as n becomes larger than x, the overall term is being multiplied by numbers less than 1. Since the limit deals with infinity, as the value of n approaches infinity the value of denominator grows faster than the value of the numerator, making the overall value become smaller, moving towards 0.
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16:24:48 Query problem 10.3.10 (3d edition 10.3.12) (was 9.3.12) series for `sqrt(1+sin(`theta))
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RESPONSE --> I first converted this function into a binomial style function: (1 + sin('theta) )^(1/2) Next I found the first three terms for (1 + x)^(1/2): 1 + (1/2)x + (1/2)(-1/2)x^2 I then found the first three terms for sin('theta): 'theta - 'theta^3 / 3! + 'theta^5 / 5! And substituted: 1 + (1/2)('theta - 'theta^3 / 3! + 'theta^5 / 5!) + (-1/4)('theta - 'theta^3 / 3! + 'theta^5 / 5!)^2 Then multiplied out: 1 + 'theta/2 - 'theta^3/12 + 'theta^5/240 - 'theta^2/4 + 'theta^4/24 .... Since we only want the first four terms, that leaves us with: 1 + 'theta/4 - 'theta^3/12 + 'theta^4/24
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16:26:14 what are the first four nonzero terms of the series?
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RESPONSE --> I found the first four terms to be: 1 + 'theta/4 - 'theta^3/12 + 'theta^4/24
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16:26:19 Explain how you obtained these terms.
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RESPONSE --> See above.
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16:29:58 What is the Taylor series for `sqrt(z)?
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RESPONSE --> The series would be equal to (z)^(1/2), which we could then use the binomial series pattern on. I think the pattern is as follows: (1/2)z + (-1/4)z^2 / 2! + (3/8)z^3 / 3! ...
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16:34:17 What is the Taylor series for 1+sin(`theta)?
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RESPONSE --> The series would be the series for sin('theta) with a 1 added: 1 + x - x^3/3! + x^5/5! ...
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16:37:43 How are the two series combined to obtain the desired series?
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RESPONSE --> Substitute 1+sin('theta) into 'sqrt(z) for z: (1/2) (1 + x - x^3/3! + x^5/5!) + (-1/4) (1 + x - x^3/3! + x^5/5!)^2 / 2! + (3/8) (1 + x - x^3/3! + x^5/5!)^3 / 3!
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16:38:29 Query Add comments on any surprises or insights you experienced
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RESPONSE --> It is rather neat to be able to create a function that wraps near another function, especially when that function may itself be rather difficult to deal with.
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