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MathSofiya Group Title

Please help....Please help....Taylor series f(x)=sinx a=1 n=3 \[0.8 \le x \le 1.2\] \[sinx\approx T_3(x)=sin(1)+\frac{cos(1)}{1}(x-1)-\frac{sin(1)}{2}(x-1)^2-\frac{cos(1)}{6}(x-1)^3\] (b) Use Taylor's Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\] lies in the given interval. \[\left|R_3(x)\right|\le \frac{M}{4!} {\left|x-1\right|}^4\] I'm almost there.....

  • 2 years ago
  • 2 years ago

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  1. MathSofiya Group Title
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    here is an example that me and @smoothmath did http://openstudy.com/study#/updates/500af8b7e4b0549a892f4a6d

    • 2 years ago
  2. SmoothMath Group Title
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    Which part is giving you trouble?

    • 2 years ago
  3. SmoothMath Group Title
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    It looks like you made some mistakes in constructing the Taylor Series.

    • 2 years ago
  4. MathSofiya Group Title
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    where?

    • 2 years ago
  5. MathSofiya Group Title
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    i see it

    • 2 years ago
  6. SmoothMath Group Title
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    Kind of a lot of mistakes. Let me see you try again.

    • 2 years ago
  7. SmoothMath Group Title
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    \[\large T_3 = f(a) +\frac{f'(a)}{1!}*(x-a) + \frac{f''(a)}{2!}*(x-a)^2 + \frac{f'''(a)}{3!}*(x-a)^3\]

    • 2 years ago
  8. MathSofiya Group Title
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    I fixed it

    • 2 years ago
  9. SmoothMath Group Title
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    That's just by the definition of the Taylor series.

    • 2 years ago
  10. SmoothMath Group Title
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    Much better =)

    • 2 years ago
  11. SmoothMath Group Title
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    Okay, that's the first part. Now for the second part, we need to get M.

    • 2 years ago
  12. MathSofiya Group Title
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    \[f^4(x)=sinx\le M\]

    • 2 years ago
  13. MathSofiya Group Title
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    \[\left|R_3(x)\right|\le \frac{sinx}{4!} {\left|x-1\right|}^4\] shoot me

    • 2 years ago
  14. SmoothMath Group Title
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    Hold on. We need to get an actual value of M. Remember how we get that?

    • 2 years ago
  15. MathSofiya Group Title
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    it's the absolute value of f^4 (x)

    • 2 years ago
  16. MathSofiya Group Title
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    don't give up on me quite yet please..... sin(1) or I have to do something with \[x \ge 0.8\]

    • 2 years ago
  17. SmoothMath Group Title
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    Let's step back, slow down, and understand what it is that we're doing. Just some theory.

    • 2 years ago
  18. MathSofiya Group Title
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    \[f^{(n+1)}\le f(lower limit)<\]

    • 2 years ago
  19. MathSofiya Group Title
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    \[f^{(n+1)}\le f(lower limit)<M\]

    • 2 years ago
  20. SmoothMath Group Title
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    Here'es the idea. We've made a taylor polynomial to approximate the function. Now we're trying to make a statement about how accurate the approximation is. Taylor's inequality allows us to do that. Here's how. We look at the derivative one higher than our approximation, and we look at it on the interval that we're interested in. If we can pick out a number, call it M, that the derivative is less than on the whole interval, then that allows us to use that M to limit the error.

    • 2 years ago
  21. SmoothMath Group Title
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    The basic statement is: \( \large f^{n+1} \le M\) Therefore, \(R_n < \text{some thing dependent on M}\)

    • 2 years ago
  22. SmoothMath Group Title
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    The lower the M I'm able to pick, the better. Why? Because it allows me to put a lower cieling on the possible error.

    • 2 years ago
  23. SmoothMath Group Title
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    ceiling*

    • 2 years ago
  24. MathSofiya Group Title
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    yep. So we pick the lowest limit that we're allowed 0.8 in this case

    • 2 years ago
  25. SmoothMath Group Title
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    I don't think you're right.

    • 2 years ago
  26. SmoothMath Group Title
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    Let me talk about the maximum of a function for a bit.

    • 2 years ago
  27. MathSofiya Group Title
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    ok

    • 2 years ago
  28. SmoothMath Group Title
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    Okay, I'm thinking just a few examples will help you to see what it is we're doing.

    • 2 years ago
  29. SmoothMath Group Title
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    |dw:1342915184406:dw|

    • 2 years ago
  30. SmoothMath Group Title
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    Text on right says max=10

    • 2 years ago
  31. SmoothMath Group Title
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    |dw:1342915323520:dw|

    • 2 years ago
  32. SmoothMath Group Title
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    |dw:1342915451038:dw|

    • 2 years ago
  33. SmoothMath Group Title
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    Tell me that you understand. Because I can't make the words to explain this.

    • 2 years ago
  34. MathSofiya Group Title
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    I'm really sorry, John. I'm really trying to understand this. Ok here is what I understand. I see that you have drawn three unique functions, and each function was given an interval, and a maximum was chosen. The maximum that you have chosen is the M. The derivative should be less than M on the whole interval.

    • 2 years ago
  35. MathSofiya Group Title
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    That M allows us to put a cap on the error

    • 2 years ago
  36. MathSofiya Group Title
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    What we're trying to do with M is to put a cap on the error. That's why we need to find an actual value for M.

    • 2 years ago
  37. MathSofiya Group Title
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    That value M is greater than the next derivative up.

    • 2 years ago
  38. MathSofiya Group Title
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    hold on smooth

    • 2 years ago
  39. SmoothMath Group Title
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    Well, let me plot the 4th derivative for us.

    • 2 years ago
  40. MathSofiya Group Title
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    no hold on

    • 2 years ago
  41. SmoothMath Group Title
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    http://www.wolframalpha.com/input/?i=plot+sin%28x%29+on+x%3D.5+to+x%3D1.5

    • 2 years ago
  42. MathSofiya Group Title
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    x=1.2 will give me the ultimate cap. \[f^4(x)=sinx\le sin(1.2)<1.932\]

    • 2 years ago
  43. MathSofiya Group Title
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    ultimate cap for our interval

    • 2 years ago
  44. SmoothMath Group Title
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    sin(1.2) is not 1.932.

    • 2 years ago
  45. MathSofiya Group Title
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    0.932 :P

    • 2 years ago
  46. SmoothMath Group Title
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    That's impossible, since sin oscillates between 1 and -1.

    • 2 years ago
  47. SmoothMath Group Title
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    Good good =)

    • 2 years ago
  48. MathSofiya Group Title
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    typo dude...chill out

    • 2 years ago
  49. MathSofiya Group Title
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    and I didn't look at your wolfram ;P

    • 2 years ago
  50. MathSofiya Group Title
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    Victory?

    • 2 years ago
  51. SmoothMath Group Title
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    Almost there.

    • 2 years ago
  52. MathSofiya Group Title
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    The example is what confused me. They picked 7 instead of 9....but 1/7 is greater than 1/9 that's why they picked 7

    • 2 years ago
  53. MathSofiya Group Title
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    remember the interval \[7 \le x \le 9\]

    • 2 years ago
  54. SmoothMath Group Title
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    I do remember. Please realize that the maximum is not always going to be at one of the endpoints.

    • 2 years ago
  55. MathSofiya Group Title
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    I understand that now

    • 2 years ago
  56. SmoothMath Group Title
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    You really have to look at what the function is. It's going to be different every time and you have to analyze the function to figure out where the upper bound is.

    • 2 years ago
  57. MathSofiya Group Title
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    Yes sir.

    • 2 years ago
  58. SmoothMath Group Title
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    Okay good =)

    • 2 years ago
  59. MathSofiya Group Title
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    \[\left|R_3(x)\right|\le \frac{.932}{4!} {\left|1.2-1\right|}^4\]

    • 2 years ago
  60. MathSofiya Group Title
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    yes?

    • 2 years ago
  61. MathSofiya Group Title
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    \[\left|R_3(x)\right|\le 0.000054\]

    • 2 years ago
  62. SmoothMath Group Title
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    Looks good, pal.

    • 2 years ago
  63. MathSofiya Group Title
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    Thank ya! You're amazingly patient. Thank you soo much John. :)

    • 2 years ago
  64. SmoothMath Group Title
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    My pleasure! =D

    • 2 years ago
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