course Mth 174 Are there any tips you can give me on finding intervals of convergence and radius of convergence.
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15:51:58 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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15:54:11 what are the degree 2 and degree 3 Taylor polynomials?
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RESPONSE --> Degree 2 polynomial: 3 + (-2)(x - 5) + (x - 5)^2 / 2 Degree 3 polynomial: 3 + (-2)(x - 5) + (x - 5)^2 / 2 + (-1)(x - 5)^3 / 2
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15:57:39 What is each polynomial give you for g(4.9)?
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RESPONSE --> I plugged in 4.9 into the polynomials as x, which I am not sure about. For the second degree polynomial I got 3.205 and for the third degree polynomial I got 3.2055
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16:09:52 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 --> I do not know exactly what to do here. But since we are using the graph point (5, 3) with the slope of -2 I turned it into the line y = 3 - 2(x - 5). Plugging 4.9 in as x, y = g(4.9) = 3.2 Oncea again, not sure what I was aiming for here.
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16:21:51 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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16:32:57 what is your degree 3 approximation?
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RESPONSE --> My polynomial with degree 3 for sin(t) about t = 0 is sin(0) + cos(0)t - sin(0) * t^2 / 2 - cos(0) * t^3 / 6 which equals t - (1/6) * t^3. I do not know where to go from here.
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16:37:09 what is your degree 5 approximation?
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RESPONSE --> Again, I found the polynomial but not a value answer. t - (1/6) * t^3 + (1/120) * t^5 I do not think I have big problems finding Taylor Polynomials, but I just do not seem to know what to do with them.
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16:37:12 What is your Taylor polynomial?
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16:38:23 Explain in your own words why a trapezoidal approximation will not work here.
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RESPONSE --> Because the integral is from 0 to 1 and sin(t) / t is undefined at t = 0
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16:44:46 Query problem 10.2.25 (3d edition 10.2.21) (was 9.2.12) Taylor series for ln(1+2x)
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16:50:21 show how you obtained the series by taking derivatives
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RESPONSE --> f(x) = ln(1 + 2x) f'(x) = 1 / (1 + 2x) = (1 + 2x) ^ -1 f''(x) = (-1)(1 + 2x) ^ -2 = -1 / (1 + 2x) ^2 f'''(x) = (-1)(-2)(1 + 2x) ^ -3 = 2 / (1 + 2x) ^ 3 About x = 0: ln(1 + 0) + x / (1 + 0) - x ^ 2 / (1 + 0) + 2x^3 / 3 = x - (x^2 / 2) + (2x^3 / 3).... This is my series for n = 3.
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16:51:20 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 --> Unless I am mistaking, it would be the same since 2(x) = 0 and x = 0. Therefor at all times, either (1 + 2x) = (1 + x) = 1
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16:58:40 What is your expected interval of convergence?
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RESPONSE --> I need more practice with finding intervals of convergence. i do not fully understand them
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16:58:43 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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