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MathSofiya

  • 3 years ago

Approximate f by a Taylor polynomial with degree n at the number a. \[f(x)=sinx\] a=1 n=3 \[0.8 \le x \le 1.2\]

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  1. MathSofiya
    • 3 years ago
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    I'm skimming through my book and don't seem to find a concrete answer...what the difference between MacLaurin and Taylor again?

  2. MathSofiya
    • 3 years ago
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    MacLaurin is a type of TAylor?

  3. timo86m
    • 3 years ago
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    maclourin is at 0 tho

  4. KingGeorge
    • 3 years ago
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    A Maclauris series is a Taylor series at \(x=0\)

  5. MathSofiya
    • 3 years ago
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    so I have the equation f=sinx and I'm trying to approximate that line with degree 3 and number 1?

  6. KingGeorge
    • 3 years ago
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    To approximate one with a Taylor series, use the formula \[\sum_{i=0}^n \frac{f^{(n)}(a)}{n!}(x-a)^n\]

  7. KingGeorge
    • 3 years ago
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    Excuse me, \[\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i\]

  8. MathSofiya
    • 3 years ago
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    \[\sum_{n=3}^{?} \frac{sinx}{3!} (x-1)^3\] this doesn't look right...where did I go wrong?

  9. timo86m
    • 3 years ago
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    \[\sin \left( 1 \right) +\cos \left( 1 \right) \left( x-1 \right) -1/2 \,\sin \left( 1 \right) \left( x-1 \right) ^{2} \]

  10. KingGeorge
    • 3 years ago
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    So you have \(a=1\), and \[f^{(0)}=\sin(x)\]\[f^{(1)}=\cos(x)\]\[f^{(2)}=-\sin(x)\]\[f^{(3)}=-\cos(x)\]

  11. timo86m
    • 3 years ago
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    oh wait :D are you in radians or degrees sofiya

  12. KingGeorge
    • 3 years ago
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    Use the formula, and we get \[\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3\]This can, of course, be simplified a bit.

  13. KingGeorge
    • 3 years ago
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    Be back in a couple minutes.

  14. MathSofiya
    • 3 years ago
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    k

  15. timo86m
    • 3 years ago
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    \[\sin \left( a \right) +\cos \left( a \right) \left( x-a \right) -1/2 \,\sin \left( a \right) \left( x-a \right) ^{2} \]

  16. KingGeorge
    • 3 years ago
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    Remember we're going to \(n=3\) and not \(n=2\).

  17. MathSofiya
    • 3 years ago
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    I'm kinda lost...what's my goal here? to write it in the condensed form \[\sum\] or to add or simplify the written out form?

  18. KingGeorge
    • 3 years ago
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    To add and simplify the expanded form.

  19. timo86m
    • 3 years ago
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    king george check inbox

  20. timo86m
    • 3 years ago
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    is that right what i sent you?

  21. KingGeorge
    • 3 years ago
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    One minute.

  22. MathSofiya
    • 3 years ago
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    \[\frac{\sin(1)}{0!}(x-1)^0+\frac{\cos(1)}{1!}(x-1)^1-\frac{\sin(1)}{2!}(x-1)^2-\frac{\cos(1)}{3!}(x-1)^3\] = \[sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}\] Like this or in the\[\sum\] form?

  23. timo86m
    • 3 years ago
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    sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2

  24. KingGeorge
    • 3 years ago
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    The first way @MathSofiya. You don't want sigma notation in your answer.

  25. timo86m
    • 3 years ago
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    sin(a)+cos(a)*(x-a)-(1/2)*sin(a)*(x-a)^2-(1/6)*cos(a)*(x-a)^3 with 4th approximation choose a=0 and x-(1/6)*x^3

  26. KingGeorge
    • 3 years ago
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    @timo86m We're told that \(a=1\)

  27. MathSofiya
    • 3 years ago
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    so this way would be correct I guess \[sin(1)+cos(1)(x-1)-\frac{sin(1)(x-1)^2}{2}-\frac{cos(1)(x-1)^3}{6}\] and i have to find a way to add this fact to the equation \[0.8 \le x \le 1.2\]

  28. KingGeorge
    • 3 years ago
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    I have no idea what \(0.8\leq x\leq 1.2\) is supposed to do to our solution. As far as I'm concerned, that's irrelevant since \(x\) is a variable. We've just restricted the domain.

  29. timo86m
    • 3 years ago
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    @KingGeorge Taylor is all about knowing a value at a such as 0 90 and such for sin(x) so that you get an appoximation. nobody knows what sin(1) is but we know cos(0) sin(o) and such.

  30. KingGeorge
    • 3 years ago
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    Taylor series works with all \(a\). Whether we can compute \(\sin(1)\) by hand or not is irrelevant. If we can't compute it, leave it as \(\sin(1)\).

  31. MathSofiya
    • 3 years ago
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    ooohhhh can I pick any number between 0.8 and 1.2 for x to test something?

  32. KingGeorge
    • 3 years ago
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    Sure.

  33. timo86m
    • 3 years ago
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    it does work for all a but it is not practical.

  34. timo86m
    • 3 years ago
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    i think it is just f(x)=x-(1/6)*x^3 and the answer is f(1)=5/6

  35. MathSofiya
    • 3 years ago
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    or that might be to estimate the accuracy...the next question reads: "Use Taylors Inequality to estimate the accuracy of the approximation \[f(x)\approx T_n(x)\]when x lies in the given interval"

  36. KingGeorge
    • 3 years ago
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    That's probably what it's referring to in that case. I can't say I'm familiar with Taylor's Inequality though. @timo86m It specifically asks for \(a=1\), and not 0.

  37. MathSofiya
    • 3 years ago
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    \[\left|R_n(x) \right|=\left|f(x)-T_n(x)\right|\]

  38. MathSofiya
    • 3 years ago
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    is the absolute value of the remainder

  39. timo86m
    • 3 years ago
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    a 1 could be what they wanted sin(x) evaluated at. it makes no sense to have cos1 sin1 because it has irrational number.

  40. KingGeorge
    • 3 years ago
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    That's why you don't evaluate it. We merely leave it as \(\sin(1)\) or \(\cos(1)\).

  41. timo86m
    • 3 years ago
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    it said approximate

  42. KingGeorge
    • 3 years ago
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    A Taylor series of finite length is by definition an approximation, whether we evaluate \(\sin(1)\) or not.

  43. MathSofiya
    • 3 years ago
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    if \[\left|f^{(n+1)}(x)\right|\le M\] then \[\left|R_n(x) \right|\le \frac{M}{(n+1)!}\left| x-a\right|^{n+1}\]

  44. MathSofiya
    • 3 years ago
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    There are three possible methods for estimating the size of error: 1. The method above 2. Alternating Series Estimation Theorem...if it's an alternating series 3. Use graphing device

  45. MathSofiya
    • 3 years ago
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    since this is an alternating series we can use method 2 I guess

  46. KingGeorge
    • 3 years ago
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    It says use Taylor's inequality though, so I'm inclined to think method 1 is probably what we want. (as long as that is Taylor's inequality)

  47. MathSofiya
    • 3 years ago
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    sure

  48. MathSofiya
    • 3 years ago
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    It doesn't explain what M is

  49. KingGeorge
    • 3 years ago
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    So do we want to find \(R_n(x)\)?

  50. MathSofiya
    • 3 years ago
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    yes that would be a start |sinx-T_n (x)|

  51. MathSofiya
    • 3 years ago
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    but they also said that \[f(x)\approx T_n(x)\]

  52. KingGeorge
    • 3 years ago
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    I think we should be using \(x=0.8\) to get these approximations since \(|\sin(1)-\sin(.8)|>|\sin(1)-\sin(1.2)|\).

  53. KingGeorge
    • 3 years ago
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    So if we look at \(|\sin(0.8)-T_3(0.8)|\), we get \(.248284...\)

  54. KingGeorge
    • 3 years ago
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    But I don't think that's all we need.

  55. MathSofiya
    • 3 years ago
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    Are we assuming that f(x) is approximately T_n(x) therefore its sin(x)-sin(x) and in f(1)=sin(1) and T_n(.8)=sin(.8)?

  56. KingGeorge
    • 3 years ago
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    Since \(f(x)\approx T_n(x)\) we know that \(T_n(0.8)\approx \sin(.8)\). They aren't exactly the same, but they're about the same.

  57. MathSofiya
    • 3 years ago
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    The above statement makes sense...but when do I know to plug 0.8 in or 1.2 in...I guess I'm missing the bigger picture

  58. KingGeorge
    • 3 years ago
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    I'm really not sure where to go from here. Let me see if I can get someone else to help out with this last part.

  59. MathSofiya
    • 3 years ago
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    I know that's not sin(x) or cosine x...but I'm trying to come as close to a line as possible

  60. MathSofiya
    • 3 years ago
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    ok sounds good.

  61. KingGeorge
    • 3 years ago
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    Current question I'm unable to help on is "Use Taylors Inequality to estimate the accuracy of the approximation \(T_n\) when \(0.8\leq x\leq 1.2\)?"

  62. MathSofiya
    • 3 years ago
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    Should I post it as a new question?

  63. KingGeorge
    • 3 years ago
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    Might be a good idea. This thread is getting a bit cluttered.

  64. MathSofiya
    • 3 years ago
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    sure

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