anonymous
  • anonymous
For the equation f(x)=arcsinx find the Taylor Polynomial of degree 3 of at c=1/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
sin^-1 derivatives.... thats gonna be messy i think
amistre64
  • amistre64
4 iterations of it eh...
amistre64
  • amistre64
well, 3

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More answers

amistre64
  • amistre64
y = sin^1(x) (y' = 1-x^2)^(-1/2)
anonymous
  • anonymous
f(x)=arcsin(x) f'(x)=1/Sqrt[1-x^2] f''(x)=x/(1-x^2)^(3/2)
anonymous
  • anonymous
f(x)+f'(x)(x-c)+f''(x)((x-c)^2)/2!
amistre64
  • amistre64
y'' = x(1-x^2)^(-3/2) yeah, that was easier than i expected :)
amistre64
  • amistre64
y''' = 3x^2 (1-x^2)^(-5/2) right?
anonymous
  • anonymous
so all you are doing is taking the derivative of the the equation to the 3rd prime?
anonymous
  • anonymous
\[f(x)+f'(x)\frac{(x-c)}{1!}+f\text{''}(x)\frac{(x-c)^2}{2!!}+f\text{''}(x)\frac{(x-c)^3}{3!!}\]
amistre64
  • amistre64
\[\sin^{-1}(x) + (1-x^2)^{-1/2} [x-(1/2)] + 2x(1-x^2)^{-3/2} [x-(1/2)]^2 + 3x^2 (1-x^2)^{-5/2}[x-(1/2)]^3\]
amistre64
  • amistre64
i forgot the factorials..... i loathe typing these things out lol
anonymous
  • anonymous
That's why I typed them in mathematica and copy them here
anonymous
  • anonymous
whats the difference between your two answers?
amistre64
  • amistre64
well; mine is missing factorials under them for starters lol
anonymous
  • anonymous
actually same, amistreo actually plug in for f(x),f'(x)
amistre64
  • amistre64
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!
anonymous
  • anonymous
Oh ok makes sense thanks so from here how do i determine the accuracy of this polynomial?
amistre64
  • amistre64
htat i got no idea about; i just read the material on HOW to do them the other day :)
anonymous
  • anonymous
There is actually a formula that's pain to solve, let me find it
anonymous
  • anonymous
ok
anonymous
  • anonymous
It would be the accuracy at x=\[\sqrt{2}/2\]
anonymous
  • anonymous
Oh, that's easy if you know x
anonymous
  • anonymous
sorry to make you scramble
anonymous
  • anonymous
arcsin(Sqrt[2]/2)
anonymous
  • anonymous
sin of what angle is Sqrt[2]/2
anonymous
  • anonymous
?
anonymous
  • anonymous
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
anonymous
  • anonymous
So sin(45degree or radian pi/4) is Sqrt(2)/2 which means arcsin(Sqrt(2)/2) is 45degree or radian pi/4.
anonymous
  • anonymous
To find accuracy \[\left| pi/4 - what u got up there \right|\]
anonymous
  • anonymous
oh ok gotcha
anonymous
  • anonymous
There are other kind of question where they ask you to find error bound, which is pain to solve
anonymous
  • anonymous
oh yea we haven't gotten that far yet.
anonymous
  • anonymous
oh you will and you will hate it

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