## anonymous 5 years ago For the equation f(x)=arcsinx find the Taylor Polynomial of degree 3 of at c=1/2

1. amistre64

sin^-1 derivatives.... thats gonna be messy i think

2. amistre64

4 iterations of it eh...

3. amistre64

well, 3

4. amistre64

y = sin^1(x) (y' = 1-x^2)^(-1/2)

5. anonymous

f(x)=arcsin(x) f'(x)=1/Sqrt[1-x^2] f''(x)=x/(1-x^2)^(3/2)

6. anonymous

f(x)+f'(x)(x-c)+f''(x)((x-c)^2)/2!

7. amistre64

y'' = x(1-x^2)^(-3/2) yeah, that was easier than i expected :)

8. amistre64

y''' = 3x^2 (1-x^2)^(-5/2) right?

9. anonymous

so all you are doing is taking the derivative of the the equation to the 3rd prime?

10. anonymous

$f(x)+f'(x)\frac{(x-c)}{1!}+f\text{''}(x)\frac{(x-c)^2}{2!!}+f\text{''}(x)\frac{(x-c)^3}{3!!}$

11. amistre64

$\sin^{-1}(x) + (1-x^2)^{-1/2} [x-(1/2)] + 2x(1-x^2)^{-3/2} [x-(1/2)]^2 + 3x^2 (1-x^2)^{-5/2}[x-(1/2)]^3$

12. amistre64

i forgot the factorials..... i loathe typing these things out lol

13. anonymous

That's why I typed them in mathematica and copy them here

14. anonymous

15. amistre64

well; mine is missing factorials under them for starters lol

16. anonymous

actually same, amistreo actually plug in for f(x),f'(x)

17. 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!

18. anonymous

Oh ok makes sense thanks so from here how do i determine the accuracy of this polynomial?

19. amistre64

htat i got no idea about; i just read the material on HOW to do them the other day :)

20. anonymous

There is actually a formula that's pain to solve, let me find it

21. anonymous

ok

22. anonymous

It would be the accuracy at x=$\sqrt{2}/2$

23. anonymous

Oh, that's easy if you know x

24. anonymous

sorry to make you scramble

25. anonymous

arcsin(Sqrt[2]/2)

26. anonymous

sin of what angle is Sqrt[2]/2

27. anonymous

?

28. 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

29. anonymous

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

To find accuracy $\left| pi/4 - what u got up there \right|$

31. anonymous

oh ok gotcha

32. anonymous

There are other kind of question where they ask you to find error bound, which is pain to solve

33. anonymous

oh yea we haven't gotten that far yet.

34. anonymous

oh you will and you will hate it