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MathSofiya

  • 3 years ago

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.....

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

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

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

  4. MathSofiya
    • 3 years ago
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    where?

  5. MathSofiya
    • 3 years ago
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    i see it

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

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

  8. MathSofiya
    • 3 years ago
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    I fixed it

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

  10. SmoothMath
    • 3 years ago
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    Much better =)

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

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

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

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

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

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

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

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

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

  20. SmoothMath
    • 3 years ago
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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.

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

  22. SmoothMath
    • 3 years ago
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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.

  23. SmoothMath
    • 3 years ago
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    ceiling*

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

  25. SmoothMath
    • 3 years ago
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    I don't think you're right.

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

  27. MathSofiya
    • 3 years ago
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    ok

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

  29. SmoothMath
    • 3 years ago
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    |dw:1342915184406:dw|

  30. SmoothMath
    • 3 years ago
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    Text on right says max=10

  31. SmoothMath
    • 3 years ago
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    |dw:1342915323520:dw|

  32. SmoothMath
    • 3 years ago
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    |dw:1342915451038:dw|

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

  34. MathSofiya
    • 3 years ago
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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.

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

  36. MathSofiya
    • 3 years ago
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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.

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

  38. MathSofiya
    • 3 years ago
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    hold on smooth

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

  40. MathSofiya
    • 3 years ago
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    no hold on

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

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

  43. MathSofiya
    • 3 years ago
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    ultimate cap for our interval

  44. SmoothMath
    • 3 years ago
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    sin(1.2) is not 1.932.

  45. MathSofiya
    • 3 years ago
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    0.932 :P

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

  47. SmoothMath
    • 3 years ago
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    Good good =)

  48. MathSofiya
    • 3 years ago
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    typo dude...chill out

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

  50. MathSofiya
    • 3 years ago
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    Victory?

  51. SmoothMath
    • 3 years ago
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    Almost there.

  52. MathSofiya
    • 3 years ago
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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

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

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

  55. MathSofiya
    • 3 years ago
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    I understand that now

  56. SmoothMath
    • 3 years ago
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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.

  57. MathSofiya
    • 3 years ago
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    Yes sir.

  58. SmoothMath
    • 3 years ago
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    Okay good =)

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

  60. MathSofiya
    • 3 years ago
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    yes?

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

  62. SmoothMath
    • 3 years ago
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    Looks good, pal.

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

  64. SmoothMath
    • 3 years ago
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    My pleasure! =D

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