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anonymous

  • 5 years ago

For the equation f(x)=arcsinx find the Taylor Polynomial of degree 3 of at c=1/2

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  1. amistre64
    • 5 years ago
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    sin^-1 derivatives.... thats gonna be messy i think

  2. amistre64
    • 5 years ago
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    4 iterations of it eh...

  3. amistre64
    • 5 years ago
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    well, 3

  4. amistre64
    • 5 years ago
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    y = sin^1(x) (y' = 1-x^2)^(-1/2)

  5. anonymous
    • 5 years ago
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    f(x)=arcsin(x) f'(x)=1/Sqrt[1-x^2] f''(x)=x/(1-x^2)^(3/2)

  6. anonymous
    • 5 years ago
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    f(x)+f'(x)(x-c)+f''(x)((x-c)^2)/2!

  7. amistre64
    • 5 years ago
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    y'' = x(1-x^2)^(-3/2) yeah, that was easier than i expected :)

  8. amistre64
    • 5 years ago
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    y''' = 3x^2 (1-x^2)^(-5/2) right?

  9. anonymous
    • 5 years ago
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    so all you are doing is taking the derivative of the the equation to the 3rd prime?

  10. anonymous
    • 5 years ago
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    \[f(x)+f'(x)\frac{(x-c)}{1!}+f\text{''}(x)\frac{(x-c)^2}{2!!}+f\text{''}(x)\frac{(x-c)^3}{3!!}\]

  11. amistre64
    • 5 years ago
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    \[\sin^{-1}(x) + (1-x^2)^{-1/2} [x-(1/2)] + 2x(1-x^2)^{-3/2} [x-(1/2)]^2 + 3x^2 (1-x^2)^{-5/2}[x-(1/2)]^3\]

  12. amistre64
    • 5 years ago
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    i forgot the factorials..... i loathe typing these things out lol

  13. anonymous
    • 5 years ago
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    That's why I typed them in mathematica and copy them here

  14. anonymous
    • 5 years ago
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    whats the difference between your two answers?

  15. amistre64
    • 5 years ago
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    well; mine is missing factorials under them for starters lol

  16. anonymous
    • 5 years ago
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    actually same, amistreo actually plug in for f(x),f'(x)

  17. amistre64
    • 5 years ago
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    sin−1(x) +(1−x^2)^(−1/2) [x−(1/2)] +2x(1−x^2)^(−3/2) [x−(1/2)]^2 ---------------------------- 2! +3x^2 (1−x2)^(−5/2) [x−(1/2)]^3 ------------------------------- 3!

  18. anonymous
    • 5 years ago
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    Oh ok makes sense thanks so from here how do i determine the accuracy of this polynomial?

  19. amistre64
    • 5 years ago
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    htat i got no idea about; i just read the material on HOW to do them the other day :)

  20. anonymous
    • 5 years ago
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    There is actually a formula that's pain to solve, let me find it

  21. anonymous
    • 5 years ago
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    ok

  22. anonymous
    • 5 years ago
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    It would be the accuracy at x=\[\sqrt{2}/2\]

  23. anonymous
    • 5 years ago
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    Oh, that's easy if you know x

  24. anonymous
    • 5 years ago
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    sorry to make you scramble

  25. anonymous
    • 5 years ago
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    arcsin(Sqrt[2]/2)

  26. anonymous
    • 5 years ago
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    sin of what angle is Sqrt[2]/2

  27. anonymous
    • 5 years ago
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    ?

  28. anonymous
    • 5 years ago
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    i dunno its part of the same question...first it askes to find the taylor polynoial like you did than it said find the accuracy of the polynomial at x=sqrt2/2

  29. anonymous
    • 5 years ago
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    So sin(45degree or radian pi/4) is Sqrt(2)/2 which means arcsin(Sqrt(2)/2) is 45degree or radian pi/4.

  30. anonymous
    • 5 years ago
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    To find accuracy \[\left| pi/4 - what u got up there \right|\]

  31. anonymous
    • 5 years ago
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    oh ok gotcha

  32. anonymous
    • 5 years ago
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    There are other kind of question where they ask you to find error bound, which is pain to solve

  33. anonymous
    • 5 years ago
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    oh yea we haven't gotten that far yet.

  34. anonymous
    • 5 years ago
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    oh you will and you will hate it

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