course Sorry about all the bugs in this one. I had to restart the program twice. ?y?s??????????T??assignment #014?????????????
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12:04:36 query problem 10.1.23 (3d edition 10.1.20) (was 9.1.12) g continuous derivatives, g(5) = 3, g'(5) = -2, g''(5) = 1, g'''(5) = -3
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RESPONSE --> a) Find Taylor polynomial of degree 2 for x near 5: To do this I need to take the general form of a Taylor polynomial and insert the values of the derivatives at the right points. This gives me a polynomial that looks like: P_2(x-5) = 3 - 2(x-5) +(1/2)(x-5)^2 P_3(x-5) = 3 - 2(x-5) +(1/2)(x-5)^2 - (1/2)(x-5)^3 b) Approximate g(4.9) To do this, all I did was plug in 4.9 for x into both equations: P_2(4.9) = 3.205 P_3(4.9) = 3.2055
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12:04:41 what are the degree 2 and degree 3 Taylor polynomials?
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RESPONSE --> See above.
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12:04:47 What is each polynomial give you for g(4.9)?
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RESPONSE --> See above.
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13:00:16 What would g(4.9) be based on a straight-line approximation using graph point (5,3) and the slope -2? How do the Taylor polynomials modify this tangent-line estimate?
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RESPONSE --> The approximation that was created was not a straight-line approximation. That would have been the tangent line approximation. The Taylor polynomials take this tangent-line approximation and turn it into an approximation of any power. As the power increases, so does the approximation over the interval of convergence.
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?|????z?S?=???€?? assignment #014 ????????????? Physics II 11-13-2007
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14:30:57 query problem 10.1.23 (3d edition 10.1.20) (was 9.1.12) g continuous derivatives, g(5) = 3, g'(5) = -2, g''(5) = 1, g'''(5) = -3
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RESPONSE --> Had to attend another class. Restarted.
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14:31:01 what are the degree 2 and degree 3 Taylor polynomials?
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RESPONSE --> .
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14:31:04 What is each polynomial give you for g(4.9)?
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RESPONSE --> .
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14:31:07 What would g(4.9) be based on a straight-line approximation using graph point (5,3) and the slope -2? How do the Taylor polynomials modify this tangent-line estimate?
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RESPONSE --> .
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15:24:45 query problem 10.1.35 (3d edition 10.1.33) (was 9.1.36) estimate the integral of sin(t) / t from t=0 to t=1
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RESPONSE --> a) Estimate with a Taylor polynomial, n=3. The first thing I did was find the derivatives of sin t: f(0) = sin 0 = 0 f'(0) = cos 0 = 1 f'' = -sin 0 = 0 f''' = -cos 0 = -1 Putting these together into a Taylor polynomial results in this: P_3 = t - t^3/3! I then inserted this into the integral: int( (t - t^3/3!) / t, t, 0, 1) = int( 1 - t^2/6, t, 0, 1) Then I found the antiderivative: t - t^3/18 Finally, I found the overall value by using the first fundamental theorem (F(b) - F(a)): F(1) - F(0) = 1 - 1/18 -(0) = 17/18 Roughly equivalent to 0.9444 b) Estimate for n=5 I did the same thing for this equation, first creating a Taylor polynomial using derivatives: P_5 = t - t^3/3! + t^5/5! = t - t^3/6 + t^5/120 I found the integral: int(1 - t^2/6 +t^4/120, t, 0, 1) The antiderivative: t - t^3/18 + t^5/600 And finally, the value using the first fundamental theorem: 1 - 1/18 + 1/600 = 0.94611
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15:24:50 what is your degree 3 approximation?
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RESPONSE --> See above.
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15:24:55 what is your degree 5 approximation?
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RESPONSE --> Up.
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15:27:56 What is your Taylor polynomial?
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RESPONSE --> I believe I put it in above.
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15:50:37 Explain in your own words why a trapezoidal approximation will not work here.
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RESPONSE --> Hmm. Not quite sure. Would it be because trapezoidal approximation uses the origin, 0, and therefore fails because it tries to do division by 0?
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15:51:23 Query problem 10.2.25 (3d edition 10.2.21) (was 9.2.12) Taylor series for ln(1+2x)
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RESPONSE --> Ok.
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{???????w????assignment #014 ????????????? Physics II 11-13-2007
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15:57:09 Query problem 10.2.25 (3d edition 10.2.21) (was 9.2.12) Taylor series for ln(1+2x)
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RESPONSE --> I hit the wrong button somewhere...
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16:00:36 show how you obtained the series by taking derivatives
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RESPONSE --> After the first derivative I realized that I could use negative exponents on (1+2x), which would make it easier than dealing with the fraction line: f'(x) = 2(1+2x)^-1 f'' = -4(1+2x)^-2 f''' = 16(1+2x)^-3 f^4 = -96(1+2x)^-4
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16:09:55 how does this series compare to that for ln(1+x) and how could you have obtained the above result from the series for ln(1+x)?
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RESPONSE --> The value of x is doubled for the original function (ln(1+2x). In comparing the Taylor series, each part in ln(1+2x) is multiplied by 2^n.
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16:10:42 What is your expected interval of convergence?
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RESPONSE --> I expected the interval to be half as large because the values of x were doubled, but I had no reason other than that. Any suggestions would be greatly appreciated!
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16:12:21 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Series are a lot different from the previous topics I have worked with, but it is good to see that basic calculus topics such as derivation and integration apply to many different topics.
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