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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
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin^1 derivatives.... thats gonna be messy i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04 iterations of it eh...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = sin^1(x) (y' = 1x^2)^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=arcsin(x) f'(x)=1/Sqrt[1x^2] f''(x)=x/(1x^2)^(3/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)+f'(x)(xc)+f''(x)((xc)^2)/2!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y'' = x(1x^2)^(3/2) yeah, that was easier than i expected :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y''' = 3x^2 (1x^2)^(5/2) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so all you are doing is taking the derivative of the the equation to the 3rd prime?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(x)+f'(x)\frac{(xc)}{1!}+f\text{''}(x)\frac{(xc)^2}{2!!}+f\text{''}(x)\frac{(xc)^3}{3!!}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sin^{1}(x) + (1x^2)^{1/2} [x(1/2)] + 2x(1x^2)^{3/2} [x(1/2)]^2 + 3x^2 (1x^2)^{5/2}[x(1/2)]^3\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i forgot the factorials..... i loathe typing these things out lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's why I typed them in mathematica and copy them here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the difference between your two answers?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well; mine is missing factorials under them for starters lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually same, amistreo actually plug in for f(x),f'(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin−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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh ok makes sense thanks so from here how do i determine the accuracy of this polynomial?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0htat i got no idea about; i just read the material on HOW to do them the other day :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is actually a formula that's pain to solve, let me find it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It would be the accuracy at x=\[\sqrt{2}/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, that's easy if you know x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry to make you scramble

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sin of what angle is Sqrt[2]/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0So sin(45degree or radian pi/4) is Sqrt(2)/2 which means arcsin(Sqrt(2)/2) is 45degree or radian pi/4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find accuracy \[\left pi/4  what u got up there \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are other kind of question where they ask you to find error bound, which is pain to solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yea we haven't gotten that far yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you will and you will hate it
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