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MathSofiya

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.....

  • one year ago
  • one year ago

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

    • one year ago
  2. SmoothMath
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    Which part is giving you trouble?

    • one year ago
  3. SmoothMath
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    It looks like you made some mistakes in constructing the Taylor Series.

    • one year ago
  4. MathSofiya
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    where?

    • one year ago
  5. MathSofiya
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    i see it

    • one year ago
  6. SmoothMath
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    Kind of a lot of mistakes. Let me see you try again.

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

    • one year ago
  8. MathSofiya
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    I fixed it

    • one year ago
  9. SmoothMath
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    That's just by the definition of the Taylor series.

    • one year ago
  10. SmoothMath
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    Much better =)

    • one year ago
  11. SmoothMath
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    Okay, that's the first part. Now for the second part, we need to get M.

    • one year ago
  12. MathSofiya
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    \[f^4(x)=sinx\le M\]

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

    • one year ago
  14. SmoothMath
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    Hold on. We need to get an actual value of M. Remember how we get that?

    • one year ago
  15. MathSofiya
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    it's the absolute value of f^4 (x)

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

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

    • one year ago
  18. MathSofiya
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    \[f^{(n+1)}\le f(lower limit)<\]

    • one year ago
  19. MathSofiya
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    \[f^{(n+1)}\le f(lower limit)<M\]

    • one year ago
  20. SmoothMath
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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.

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

    • one year ago
  22. SmoothMath
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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.

    • one year ago
  23. SmoothMath
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    ceiling*

    • one year ago
  24. MathSofiya
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    yep. So we pick the lowest limit that we're allowed 0.8 in this case

    • one year ago
  25. SmoothMath
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    I don't think you're right.

    • one year ago
  26. SmoothMath
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    Let me talk about the maximum of a function for a bit.

    • one year ago
  27. MathSofiya
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    ok

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

    • one year ago
  29. SmoothMath
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    |dw:1342915184406:dw|

    • one year ago
  30. SmoothMath
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    Text on right says max=10

    • one year ago
  31. SmoothMath
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    |dw:1342915323520:dw|

    • one year ago
  32. SmoothMath
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    |dw:1342915451038:dw|

    • one year ago
  33. SmoothMath
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    Tell me that you understand. Because I can't make the words to explain this.

    • one year ago
  34. MathSofiya
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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.

    • one year ago
  35. MathSofiya
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    That M allows us to put a cap on the error

    • one year ago
  36. MathSofiya
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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.

    • one year ago
  37. MathSofiya
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    That value M is greater than the next derivative up.

    • one year ago
  38. MathSofiya
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    hold on smooth

    • one year ago
  39. SmoothMath
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    Well, let me plot the 4th derivative for us.

    • one year ago
  40. MathSofiya
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    no hold on

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

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

    • one year ago
  43. MathSofiya
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    ultimate cap for our interval

    • one year ago
  44. SmoothMath
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    sin(1.2) is not 1.932.

    • one year ago
  45. MathSofiya
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    0.932 :P

    • one year ago
  46. SmoothMath
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    That's impossible, since sin oscillates between 1 and -1.

    • one year ago
  47. SmoothMath
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    Good good =)

    • one year ago
  48. MathSofiya
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    typo dude...chill out

    • one year ago
  49. MathSofiya
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    and I didn't look at your wolfram ;P

    • one year ago
  50. MathSofiya
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    Victory?

    • one year ago
  51. SmoothMath
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    Almost there.

    • one year ago
  52. MathSofiya
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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

    • one year ago
  53. MathSofiya
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    remember the interval \[7 \le x \le 9\]

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

    • one year ago
  55. MathSofiya
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    I understand that now

    • one year ago
  56. SmoothMath
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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.

    • one year ago
  57. MathSofiya
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    Yes sir.

    • one year ago
  58. SmoothMath
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    Okay good =)

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

    • one year ago
  60. MathSofiya
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    yes?

    • one year ago
  61. MathSofiya
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    \[\left|R_3(x)\right|\le 0.000054\]

    • one year ago
  62. SmoothMath
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    Looks good, pal.

    • one year ago
  63. MathSofiya
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    Thank ya! You're amazingly patient. Thank you soo much John. :)

    • one year ago
  64. SmoothMath
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    My pleasure! =D

    • one year ago
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